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H: Identifying a connector
Can anyone identify this connector?
Female part:
The crimp parts look like this:
And the male part looks like this:
Context:
These came with a Smoothieboard v1.1. I keep messing up the crimp connections, so I want to order some spare connectors. The documentation does not seems to identify them. If there's any sort of site that has a directory of connector types, that would be good to know too.
AI: It looks like a “Molex” 4-position connector. It seems to be part connector series “2695” (if you look up ‘Molex 2695’, you get plenty of results). If you go to Molex’s page directly for that part, they even provide a nice “Mates with/use with” section detailing the corresponding headers / crimps / pre-crimped leads).
Here is a link for what I think you want:
Molex 4-Position female connector
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H: Ideal opamp closed loop problem
The inverting circuit with the T network in the feedback
is redrawn in Fig. P2.30 in a way that emphasizes the
observation that R2 and R3 in effect are in parallel (because
the ideal op amp forces a virtual ground at the inverting
input terminal). Use this observation to derive an expression
for the gain by first finding and
For the latter use the voltage-divider rule applied to R4 and
(R2 || R3)
Above is the question
I cant understand why i cant an Req=R2||R3+R4 in feedback circuit.
But using a volatge divider considering R2||R3 and R4 to calculate Vx is valid.
AI: Hints: -
why i cant an Req=R2||R3+R4 in feedback circuit.
Your formula finds the Thevenin equivalent resistance but looking into the R4 resistor from the op-amp output node. That will be of some use in your derivation.
You also need to recognize that the Thevenin equivalent voltage at the junction of R3 and R4 is a fraction of that at the op-amp's output. So, find Vx in terms of Vout and then recognize that Vx feeds via R2 creating the negative feedback.
Or regard the op-amp output as being Vx and derive the op-amp gain to Vx (R2 as feedback) then, when you have that formula, convert it to the real op-amp output using the reverse of the attenuation made by R4 and R3||R2.
Then you are good to go.
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H: Tube amplifier preamplifier
I received this little two tube preamp
http://www.gikfun.com/electronic-diy-kits-c-7/6j1-tube-preamplifier-board-20-stereo-multiple-diy-kit-p-730.html
It comes with 9 470uF caps. The symbol on the board for them is a circle, with white strips across one of the legs. Does anyone know if that indicates the positive leg of the capacitor?
Also, I don't read Chinese characters. Which set of RCA plugs are the input?
No word back from GikFun to these questions yet.
There is no schematic posted to work out the details myself.
Thanks
Updated - added a picture of the labels
Updated - added a picture showing two of capacitor pins that both connect to the RCA plugs common plane - and to the common pins of the AC input jack.
Ah ha! Found an e-bay listing that confirms Tony Stewart Sunnyskyguy's answer. Thank you!
AI: Google says
Input + Output
输入 + 输出
Stripe is -VE
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H: Atmega328 AU has a low Voltage output pin
For a project I am programming an Atmega 328 AU on a custom made PCB. After a lot of struggles I managed to get a bootloader and a program on the chip. I programmed some IO pins and they seem to work fine. The output pins that I use give: LOW = 0 Volt High = 5 Volt which is desired. However, one pin, pin 9 of the chip, gives: LOW = 0 Volt, HIGH = 1 Volt. The 1 Volt does not trigger my opto-coupler so I would like this pin also to give HIGH = 5 Volt. Is there any way to change this in software?
I use the Arduino IDE for programming.
Thank you guys.
edit:
Still not figured out. Ill add some schematics. Note: The supply is not that flat with my own power supply, but I could add a bigger capacitor. However, for the opto-coupler is does not matter and the output will correctly trigger the opto-coupler for the 5V.
AI: The package type of ATmega328-AU is Plastic Thin Quad Flat Package (TQFP). The pin configuration is different from the 28 SPDIP package types. On 28 SPDIP package, pins 9 and 10 are crystal input. In this case, you could not be used like Arduino Pin.
On Atmega328-AU is different, you can use Pin 9 like Arduino Pin 5.
Your test code can be like this:
void setup() {
pinMode(5, OUTPUT);
digitalWrite(5, HIGH);
}
void loop() {
}
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H: How to be sure that a ceramic capacitor can manage a RMS current?
I'm in the middle of a design and I need a capacitor of:
10µF
Rated at 10V
Has to handle a ripple current of 250mA RMS, at 400kHz
Most electrolytic capacitor data sheets give a maximum RMS value for the current. However, I would rather use a ceramic capacitor for this application.
I'm about to use a X7R from Kemet, package 0805, the C0805C106K8RACTU. In its datasheet I've found:
The data sheet indicates a Dissipation Factor of 10%, measured at 1kHz.
I can now estimate the ESR at 400kHz with the usual formula $$ ESR = {{DF}\over{100}} \cdot {{1}\over{2\pi f C}} = 0.004\Omega$$
Dissipated power is then $$W = ESR \cdot RMSi^2 = 0.25mW$$
To check if I'm within range, I should either:
Find in the data sheet some indication of the maximum dissipated power by the device. But this information is not present.
The data sheet does provide the maximum working temperature: 125ºC. To use this information I also need to know the thermal resistivity in ºC/W, but this information is not present.
I would go to another manufacturer, but I have not seen yet any capacitor data sheet providing this information.
How do I know what is the RMS limit for the current in the case of a ceramic capacitor?
EDIT:
I understand that my example is quite in the safe zone. However, my question is more general: Where does the safe zone end, according to the datasheet?
I've seen this similar question, but there is no definite answer. One comment propose to check the data sheet (and this exactly the sense of my question). One answer says to compare the dissipated power with a Pmax value. Again, in ceramic capacitors, I can't find this kind of information.
AI: The allowed RMS current appears to be rarely mentioned in ceramic capacitor datasheets. Have a look around for additional information provided by the manufacturer on their website. KEMET for example provides a lot of information in their design tool KSIM. They also provide the most important data in additional specsheets, e.g. C0805C106K8RACTU.
You can directly reference the max RMS current characteristic:
Other manufacturers like TDK don't directly provide the current curve, but the ESR and Ripple Temperature Rise curves. They also state in their specification, that the temperature rise at the capacitor surface should be below 20°C. Example plots:
Source:TDK C2012X7R1A106K125AC Characterization Sheet
So using the ESR graph and the Temperature Rise graph at the same frequency, you can easily calculate the maximum power dissipation:
$$P_{max} = ESR \times I_{RMS}(20°C)^2$$
And with the ESR at your frequency of operation you can then calculate the maximum RMS current.
Popular manufacturers like Murata, Samsung, AVX etc. all provide this kind of information. But the datasheet that is provided on the websites of the distributors is often more like a catalog than actual data. Make sure you check the manufacturers website for this kind of information for the specific capacitor of interest.
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H: ADC input filter design for DC input signal
My plan is to use an ADS1115 ADC to read a low impedance 0-5V output signal from a level transducer. The ADC data sheet gives guidance for ADC input filtering. This data sheet, and even other input filter design notes all assume a relatively high frequency input signal (High frequency relative to my application).
My application is a level transducer that will mostly be 0v for the majority of its life. I will only be monitoring the input signal for a high level event. If a high water level were to occur, it would happen very slowly, it would take over 10 minutes just for the level to increase an inch. Once the level reaches a high threshold; however, level reading accuracy is critical.
My design is power critical, runs on battery. I chose my sampling rate to be 64 samples per second. The entire system will wake from sleep, take a few samples, average them, and go back to sleep (sleep duration will be a few seconds). The transducer is mounted directly onto PCB via one or two inches of wire. Industrial environment, very noisy environment as wel.
Given this application, is a RC circuit really necessary? I would imagine just removing the resistor from the RC circuit and letting just capacitor filter all AC noise would be enough for my application? I would of course still use the differential capacitor as per the data sheet. Or would it be better to make a RC filter with a very low cut off? That was my original plan, but the use of precision resistors and required class 1 caps would eat into board space and allotted budget; I didn't want to go this route unless it would be necessary.
Thank you for taking the time to read!
AI: An anti aliasing filter should be always present. It would be better to sample at high sample rate, use a FIR filter to eliminate the environment noise 50/60Hz, thus the anti-alias filter can be a small RC with low TAU value. Sampling at 64sps would imply large RC filter and it is very close to 60Hz, so most probably you will pick lots of noise/garbage.
See this example:
Youtube , source PDF
You could use 1kHz sampling rate with a 300Hz cutt-off anti-alias RC filter.
$$f_{cuttoff}=\frac{1}{2\pi RC}$$
You said, your source has low impedance, so you don't need a buffer. Let we neglect the source impedance (=0) and we calculate as the entire resistance is the filter resistance R=1kOhm.
$$C=\frac{1}{2\pi Rf_{cuttoff}}=\frac{1}{2\pi\cdot 1k\Omega\cdot 300Hz}\approx 5.3\cdot 10^{-7}F\approx 0.56\mu F$$
simulate this circuit – Schematic created using CircuitLab
Above it is an anti-alias filter 300Hz cutt-off. Then you sample and filter with low pass FIR filter, for example 128 taps. You would get a precise measurement without environment noise.
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H: Does the "electricity saving box" work?
https://electrical-engineering-portal.com/the-real-truth-behind-household-power-savers
concludes
Power factor correction devices improve power quality but do not
generally improve energy efficiency (meaning they would not reduce
your energy bill). There are several reasons why their energy
efficiency claims could be exaggerated.
First, residential customers are not charged for KVA – hour usage, but
by kilowatt-hour usage. This means that any savings in energy demand
will not directly result in lowering a residential user’s utility
bill.
yet some reviewers such as on http://amazon.com/dp/B07FSQTBTQ claim that it has saved them money.
What could be the reason? Is it that some electricity meters measure kVA, or that people who want to reduce their bills adopt other measures and incorrectly attribute savings to the device?
AI: Power factor correction uses capacitors to add leading reactive power (VARs or kVARs for large loads) to counter lagging reactive power required by inductive loads like fluorescent lights or motors.
Reactive power goes back and forth between source and load. The capacitors act as a form of battery decreasing reactive power supplied by source.
It does not change the real power W or kW. You pay for energy in kWh. Energy = P t
So power factor correction will benefit the power company, but have no impact on your power bill. Actually, the powers involved are meaningless for homes.
Claims are not reality.
Large consumers require power factor correction if pf is < 0.9, because they pay for kW in kWh, but power companies must provide kVA. Best pf = 1. Companies with a poor pf must pay a surcharge for kWh used.
GreatScott! debunks this. Chinese Power Saver - Does it actually save power? He even mentions claims.
Essentially the device connects a high voltage 2μF capacitor across the mains terminals. Rest of the parts discharge capacitor when disconnected from circuit
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H: What is a unipolar inductor?
Can somebody explain what a 'Unipolar Inductor' is? I'm trying to understand the process by which the rotational energy of a black hole is converted into energy to power relativistic jets.
In the book I read (https://1lib.eu/book/2339827/854387 [pages 157, 158, 159]) it is said that the black hole in the magnetic field of the surrounding gas behaves like a spinning conductor in a magnetic field, therefore like a unipolar inductor (apparently). I already asked this on the physics site but that has for now been left unanswered, so I thought I`d ask here about the analogy the book uses.
AI: I will take a shot. What in physics is called a unipolar inductor is not any kind of device that in electronics is called an "inductor". Rather, it is is a homopolar generator.
When an electron (or any charged particle) moves in a magnetic field, a force acts on it that is perpendicular to its direction of motion, and perpendicular to lines of the magnetic field. If the electrons are bound, that is "attached" to some atom, this force will distort the shape of the electron "cloud" around the atom. However, if the electron is "free", the force will cause the electron to veer from the direction it is travelling in.
Some of the electrons in metals are free. In a "homopolar generator", a metal disk rotates with it's axis parallel to the lines of a magnetic field. The free electrons in the metal disk will tend to either move toward the center, or away from the center depending upon the orientation of the magnetic field and the direction of rotation of the disk. In either case, the accumulation of electrons in one part of the disk, and a corresponding dimunition of electrons in another part of the disk create an electric potential. In the homopolar generator, "brushes" provide an electrical contact between the disk and wires, and transfer this electric potential to a circuit.
Electrons, as well as ions, move freely in plasma. If plasma is spinning, in a magnetic field, the electrons will tend to move inward or outward, just like in the human invented homopolar generator. Positively charged particles will move in the opposite direction as the electrons. This will create an electric potential in the plasma.
How all this relates to rotating black holes, and the jets that emanate from them, I am not competent to say. But you can take away that 1) rotation, plus 2) freely moving charged particles, plus 3) magnetic field, generates 4) an electric potential.
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H: Modifying an irregular length bit field in a register in C/C++
Suppose I wanted to modify the field ADD[7:0] in the following register:
I wrote this to make things more convenient.
{
const uint8_t Character = USART_TERMINATION_CHAR; //8-bit ASCII termination character
//masks to modify register
uint32_t Zero_Mask = ~Character;
uint32_t One_Mask = Character;
//shift so mask will apply to appropriate location in "CR2" register
Zero_Mask = Zero_Mask << 24;
One_Mask = One_Mask << 24;
MODIFY_REG(USART1->CR2, Zero_Mask, One_Mask); //USART_CR_ADD[7:0], address or character match symbol
}
But if the field, in question, was an irregular number of bits, for example DIVR[6:0]:
How would I actually do this? Because in the code snippet above I was relying on the uint8_t type being the same length as the field. But there exists no 7-bit data type and if I use something like uint8_t the mask is going to be modifying bits outside of the field due to the presence of extraneous bits.
AI: You don't need to use the new value to make a mask. Instead you simply need to clear all bits in the region you care about, and then the set mask is the new value.
For example, for DIVR1, you would do:
MODIFY_REG(<reg>,0x7F<<24,(newVal & 0x7F)<<24);
To understand this further, lets split it up into what's going on. Under the assumption you are using an STM32 type processor (just a guess), the MODIFY_REG macro expands to:
WRITE_REG((REG), (((READ_REG(REG)) & (~(CLEARMASK))) | (SETMASK)))
Which is equivalent to:
uint32_t tempVal = REG; //Read the register value
tempVal &= ~CLEARMASK; //Clear all bits that are set to 1 in the mask
tempVal |= SETMASK; //Set the new bits
REG = tempVal; //Assign the value back
The clearing step basically will take any bit set to a 1 in the CLEARMASK, and set the corresponding bit in the temporary value to 0. It doesn't matter if one of those bits is already zero.
With all the bits in your register chunk clear, then the SETMASK becomes simply your value to set the chunk to. Any bits that are 1 in the mask will be set as expected, any bits that are 0 will already be zero in the temporary value as you've already cleared them.
So the first example expands to:
uint32_t temp = <reg>;
temp &= ~(0x7F << 24); //Clear bits 30:24 inclusive (0x7F is a value of seven consecutive ones)
temp |= ((newVal & 0x7F) << 24); //Sets the new value into bits 30:24.
<reg> = temp;
Note that we mask the new value to ensure that any bits beyond the seven we are interested in are not accidentally affected. You could also do:
uint32_t mask = 0x7F << 24;
MODIFY_REG(<reg>,mask,(data << 24) & mask);
Which would be equivalent.
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H: Firmware not embedded on the device?
Recently reading about Linux and its folder /lib/firmware, I read that firmware is not needed to "live" on the device, that some network devices dont have any ROM to store a firmware. How is this possible if its definition is that is software that is present on the device and it software that runs on the device? Or maybe its called firmware buts its some kind of driver?
I understand that software drivers are the ones which know the "api language" of the device and what data or instructions send to this to do something but its needed a firmware on the device because its a software "listening" to these instructions from a physical port on the device and executing some code after these instructions are received. I dont know if I got it wrong
Thanks
AI: This is done various ways. Think of it as a bootloader perhaps. Either purely in hardware or combination of hardware and software, the device would need just enough functionality to show up on whatever bus (USB, PCIe, etc)(get enumerated), but also be identifiable as some product. Then the driver will download the latest application firmware so that it does not have to be burned to flash every time an update is available and also flash is expensive.
It can be a case with usb for example that the logic in the chip can enumerate as a simple device, the firmware is downloaded into sram, then the logic or that software then re-enumerates by electrically disconnecting then reconnecting but reconnecting as per the software/firmware has chosen to use the device.
PCIe does technically have a re-enumeration thing but not the kind of thing you would use for this.
So it is not a chicken and egg problem, there is logic or firmware or a combination of the two that gets the item on the bus initially but the full featured product firmware is then downloaded.
If there is firmware on the device in order to facilitate this it may be a mask rom or a rom (or flash) on the part that the chip vendor has provided for the chip customers. Or it may be the case that the chip level product does not support anything like this but the system design for the product is to have just enough firmware to get started and then the rest downloaded live.
So if there is software/firmware involved (in getting on the bus in the first place) there is some form of non volatile storage containing code on the product, so it is not a case of there is zero rom/flash but maybe less rom/flash than if the whole of the firmware lived on the product.
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H: Can't secure supply of 32mhz Automotive grade oscillator - why is this?
This is my first automotive PCB. I designed in a high temp crystal from Digikey. Stock then went to 0 with big lead times, and then my alternate went EOL. I then spun the board with a TCXO. All three parts I had picked went to 0 stock. I then sourced from a manufacturer, and then this went EOL.
Isn't a high temp 32Mhz oscillator a common thing in automotive just like a 32Mhz oscillator is in a consumer product? It sounds like it isn't. I never had trouble in consumer. Can someone tell me what I don't know about sourcing this part and not having to re-source it every 3 months?
It is hard without a mentor, or experience in automotive, to find who knows what I need to know. So far the buyers don't know the answer, and the manufacturers are not responding (if they know).
AI: Isn't a high temp 32Mhz oscillator a common thing in automotive just
like a 32Mhz oscillator is in a consumer product?
In general there are many more parts meeting commercial specs than there are with an automotive qualification. This is not unique to oscillators specifically.
When selecting a part you should check where the product is in its life cycle. You use tools like "part intelligence" to find out this information. This will allow you to make an estimate of how long the part will be available.
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H: What is beam coupling coefficient for linear beam klystron?
What does beam coupling coefficient exactly tell? For eg. a voltage gain coefficient tells how much times the input was amplified. So what does beam coupling coefficient tell, like does it tell how much the beam is coupled with something?
AI: In a linear beam two-cavity klystron acting as a microwave amplifier, the electrons are velocity modulated by the applied input RF signal at the input of the buncher cavity. The beam coupling coefficient describes the degree to which the electrons undergo the process of velocity modulation. It is quite similar to the modulation index which determines the degree to which the carrier is modulated with respect to the message signal.
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H: Arduino resetting or malfunctioning when controlling 48V DC motors with relay module
I'm building an Arduino controlled crane machine out of old crane machine parts.
The motors in a crane machine are 48V DC and I am controlling them with an H-bridge configuration using an 8 channel relay module I got off eBay which is controlled by an Arduino Nano. I'm using an ATX power supply to power the system but I'm using a high power boost converter to get up to 48V for the motors.
The issue I'm having is when the motors are moving, 50% of the time the Arduino will either reset, or will skip a few lines of code, or sometimes the Arduino freezes and will keep going in one direction indefinitely.
It's only programmed to move to the centre of the cabinet, drop the claw until the down switch activates then raise the claw until it hits the top after that, it goes back to the corner, but it only rarely finishes it, usually it resets or keeps going in 1 direction.
I've tried 2 different power supplies, I tried powering the Arduino and relay module over USB, I tried 2 different Arduino Megas. When I try lowering the voltage on the boost converter, it doesn't stuff up as often and when I power it to 12V it hardly does it at all but 12V is way too slow for the motors. When there is no power at all for the motors, it doesn't reset at all, but obviously that isn't useful when you're trying to make a crane machine.
AI: There's a few things that could be causing your problems.
The most likely is that you're getting a lot of electrical noise on the digital input from the crane machine. That noise is coming into the Nano over the limit switch wires.
You have the motor power wires and the limit switch wires in one bundle from the Nano to the crane machine. The motors will cause a lot of electrical noise on the power wires. With the power wires parallel to the signal wires, some of that noise will make it to the signal wires.
You need to first reduce the amount of noise produced by the motors. There would normally be a capacitor across the motor terminals right at the motor. If there's not one there already, install like a 100 nanofarad capacitor rated for 250V (I think that's the next higher up from 50V) across the motor right at the motor.
At the Arduino end, you need to clamp anything that makes it that far. Put a small (few hundred ohms to 1 kiloohm) resistor in series with the limit switch inputs, and a small capacitor (1 nanofarad or so) from the digit input of the Nano to ground. While you are at it, put a Schottky diode from the digital pin to the Arduino 5V.
Another thing you could do is to use a shielded cable to run the limit switch wires through. A shielded cable has a braided conductor around the signal carrying wires. The braid is connected to ground. It prevents electrical noise from getting into your signal wires.
Try to neaten up your wiring - keep high current conductors away from low current signals.
Keep in mind that breadboards have very limited current carrying ability - don't use the breadboard as a junction for all of your ground wires.
Breadboards are also prone to bad contacts - vibration may cause sporadic loss of power or other connections to the Nano, resulting in erratic behaviour.
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H: VBUS Pin on USB Communication
I have been eyeing USB communication in microcontrollers especially the dspic33 series. Now, when I looking over the physical connector of the USB, I came to know that there are 5 pins used.
VBUS,D+,D-,GND,USB ID pins. In these 5, I got little confusion about the VBUS pin. As per the datasheet of the MCU I am using, it is not self-powered for the VBUS pin. We have to supply externally 5V to the VBUS pin.
On seeing the architecture of the USB engine, the supply given through the VBUS pin is passed to the analog comparator. This is because the VBUS pin in the MCU is to monitor the state of USB communication, stated in the datasheet. No more info.
So, what are the roles played by the VBUS pin in the USB communication? The microcontroller I am using is capable of playing OTG role.
From the above info I gathered, I need clarifications on these three points.
What is the use of VBUS pin in the microcontroller if it is used as a host? What are the operations are done by USB engine when MCU acts as host?
What is the use of VBUS pin in the microcontroller if it is used as a device? What are the operations are done by USB engine when MCU acts as device?
How the Vbus pin is used in OTG mode?
The answer need not to be MCU oriented. The relevant and general answers are also welcome.
Thanks for the help!!!
AI: A host must provide 5V power supply to device. A host will wait for a device to appear before doing any communications to device. Host always initiates any communication.
A device can use 5V for powering itself, or if it is self-powered, simply sense when host connection is made or disconnected. A device will always wait for communication from host.
OTG device must output 5V when in host mode and not output 5V when in device mode. In device mode it can use it for sensing connection or charge batteries.
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H: Assigning binary value to decimal value using counter
I don't know muach about verilog, but i started to study it lately. So if these codes doesn't make sense at all, forgive me :D
I am writing a verilog code for Sinus lookup table.
for instance,
'd1 <= 16'b0000000000010001;
'd2 <= 16'b0000000000100011;
'd3 <= 16'b0000000000110101;
'd1 is decimal value 1 and 16 bit equivalent value of it * 2^8 in binary.
and so on...
I think I couldn't provide the relation between counter and decimal values.
Here comes my codes...
module TopModule(clk, count ,reset);
input reset;
output reg [3:0]count;
input clk;
always@(posedge clk)begin
if(reset)
count=1;
else begin
for(count=1; count=90; count=count+1) begin
'd1 <= 16'b0000000000010001;
'd2 <= 16'b0000000000100011;
'd3 <= 16'b0000000000110101;
'd4 <= 16'b0000000001000111;
'd5 <= 16'b0000000001011001;
'd6 <= 16'b0000000001101011;
'd7 <= 16'b0000000001111100;
'd8 <= 16'b0000000010001110;
'd9 <= 16'b0000000010100000;
'd10 <= 16'b0000000010110001;
'd11 <= 16'b0000000011000011;
'd12 <= 16'b0000000011010100;
'd13 <= 16'b0000000011100110;
'd14 <= 16'b0000000011110111;
'd15 <= 16'b0000000100001001;
'd16 <= 16'b0000000100011010;
'd17 <= 16'b0000000100101011;
'd18 <= 16'b0000000100111100;
'd19 <= 16'b0000000101001101;
'd20 <= 16'b0000000101011110;
'd21 <= 16'b0000000101101110;
'd22 <= 16'b0000000101111111;
'd23 <= 16'b0000000110010000;
'd24 <= 16'b0000000110100000;
'd25 <= 16'b0000000110110000;
'd26 <= 16'b0000000111000000;
'd27 <= 16'b0000000111010000;
'd28 <= 16'b0000000111100000;
'd29 <= 16'b0000000111110000;
'd30 <= 16'b0000000111111111;
'd31 <= 16'b0000001000001111;
'd32 <= 16'b0000001000011110;
'd33 <= 16'b0000001000101101;
'd34 <= 16'b0000001000111100;
'd35 <= 16'b0000001001001011;
'd36 <= 16'b0000001001011001;
'd37 <= 16'b0000001001101000;
'd38 <= 16'b0000001001110110;
'd39 <= 16'b0000001010000100;
'd40 <= 16'b0000001010010010;
'd41 <= 16'b0000001010011111;
'd42 <= 16'b0000001010101101;
'd43 <= 16'b0000001010111010;
'd44 <= 16'b0000001011000111;
'd45 <= 16'b0000001011010100;
'd46 <= 16'b0000001011100000;
'd47 <= 16'b0000001011101100;
'd48 <= 16'b0000001011111000;
'd49 <= 16'b0000001100000100;
'd50 <= 16'b0000001100010000;
'd51 <= 16'b0000001100011011;
'd52 <= 16'b0000001100100110;
'd53 <= 16'b0000001100110001;
'd54 <= 16'b0000001100111100;
'd55 <= 16'b0000001101000110;
'd56 <= 16'b0000001101010000;
'd57 <= 16'b0000001101011010;
'd58 <= 16'b0000001101100100;
'd59 <= 16'b0000001101101101;
'd60 <= 16'b0000001101110110;
'd61 <= 16'b0000001101111111;
'd62 <= 16'b0000001110001000;
'd63 <= 16'b0000001110010000;
'd64 <= 16'b0000001110011000;
'd65 <= 16'b0000001110100000;
'd66 <= 16'b0000001110100111;
'd67 <= 16'b0000001110101110;
'd68 <= 16'b0000001110110101;
'd69 <= 16'b0000001110111011;
'd70 <= 16'b0000001111000010;
'd71 <= 16'b0000001111001000;
'd72 <= 16'b0000001111001101;
'd73 <= 16'b0000001111010011;
'd74 <= 16'b0000001111011000;
'd75 <= 16'b0000001111011101;
'd76 <= 16'b0000001111100001;
'd77 <= 16'b0000001111100101;
'd78 <= 16'b0000001111101001;
'd79 <= 16'b0000001111101101;
'd80 <= 16'b0000001111110000;
'd81 <= 16'b0000001111110011;
'd82 <= 16'b0000001111110110;
'd83 <= 16'b0000001111111000;
'd84 <= 16'b0000001111111010;
'd85 <= 16'b0000001111111100;
'd86 <= 16'b0000001111111101;
'd87 <= 16'b0000001111111110;
'd88 <= 16'b0000001111111111;
'd89 <= 16'b0000001111111111;
'd90 <= 16'b0000010000000000;
end
end
endmodule
And this is my testbench
module TestBench();
reg clk,reset;
wire count;
uut uut(.clk(clk),.reset(reset), .count(count));
initial begin
clk=0;
forever #10 clk=~clk;
end
initial begin
reset=0;
#1810;
reset=1;
end
endmodule
AI: Okay i figured out the solution.
Here are top module and testbench for these code.
`timescale 1ns / 1ps
module TopModule(clk, aci ,reset,sonuc,en);
input en;
input reset;
input [7:0]aci;
output reg [15:0]sonuc;
input clk;
always@(posedge clk)begin
if(reset)
sonuc <= 'd0;
else begin
case(aci)
'd1 : sonuc <= 16'b0000001111111111;
'd2 : sonuc <= 16'b0000001111111111;
'd3 : sonuc <= 16'b0000001111111110;
'd4 : sonuc <= 16'b0000001111111101;
'd5 : sonuc <= 16'b0000001111111100;
'd6 : sonuc <= 16'b0000001111111010;
'd7 : sonuc <= 16'b0000001111111000;
'd8 : sonuc <= 16'b0000001111110110;
'd9 : sonuc <= 16'b0000001111110011;
'd10 : sonuc <= 16'b0000001111110000;
'd11 : sonuc <= 16'b0000001111101101;
'd12 : sonuc <= 16'b0000001111101001;
'd13 : sonuc <= 16'b0000001111100101;
'd14 : sonuc <= 16'b0000001111100001;
'd15 : sonuc <= 16'b0000001111011101;
'd16 : sonuc <= 16'b0000001111011000;
'd17 : sonuc <= 16'b0000001111010011;
'd18 : sonuc <= 16'b0000001111001101;
'd19 : sonuc <= 16'b0000001111001000;
'd20 : sonuc <= 16'b0000001111000010;
'd21 : sonuc <= 16'b0000001110111011;
'd22 : sonuc <= 16'b0000001110110101;
'd23 : sonuc <= 16'b0000001110101110;
'd24 : sonuc <= 16'b0000001110100111;
'd25 : sonuc <= 16'b0000001110100000;
'd26 : sonuc <= 16'b0000001110011000;
'd27 : sonuc <= 16'b0000001110010000;
'd28 : sonuc <= 16'b0000001110001000;
'd29 : sonuc <= 16'b0000001101111111;
'd30 : sonuc <= 16'b0000001101110110;
'd31 : sonuc <= 16'b0000001101101101;
'd32 : sonuc <= 16'b0000001101100100;
'd33 : sonuc <= 16'b0000001101011010;
'd34 : sonuc <= 16'b0000001101010000;
'd35 : sonuc <= 16'b0000001101000110;
'd36 : sonuc <= 16'b0000001100111100;
'd37 : sonuc <= 16'b0000001100110001;
'd38 : sonuc <= 16'b0000001100100110;
'd39 : sonuc <= 16'b0000001100011011;
'd40 : sonuc <= 16'b0000001100010000;
'd41 : sonuc <= 16'b0000001100000100;
'd42 : sonuc <= 16'b0000001011111000;
'd43 : sonuc <= 16'b0000001011101100;
'd44 : sonuc <= 16'b0000001011100000;
'd45 : sonuc <= 16'b0000001011010100;
'd46 : sonuc <= 16'b0000001011000111;
'd47 : sonuc <= 16'b0000001010111010;
'd48 : sonuc <= 16'b0000001010101101;
'd49 : sonuc <= 16'b0000001010011111;
'd50 : sonuc <= 16'b0000001010010010;
'd51 : sonuc <= 16'b0000001010000100;
'd52 : sonuc <= 16'b0000001001110110;
'd53 : sonuc <= 16'b0000001001101000;
'd54 : sonuc <= 16'b0000001001011001;
'd55 : sonuc <= 16'b0000001001001011;
'd56 : sonuc <= 16'b0000001000111100;
'd57 : sonuc <= 16'b0000001000101101;
'd58 : sonuc <= 16'b0000001000011110;
'd59 : sonuc <= 16'b0000001000001111;
'd60 : sonuc <= 16'b0000001000000000;
'd61 : sonuc <= 16'b0000000111110000;
'd62 : sonuc <= 16'b0000000111100000;
'd63 : sonuc <= 16'b0000000111010000;
'd64 : sonuc <= 16'b0000000111000000;
'd65 : sonuc <= 16'b0000000110110000;
'd66 : sonuc <= 16'b0000000110100000;
'd67 : sonuc <= 16'b0000000110010000;
'd68 : sonuc <= 16'b0000000101111111;
'd69 : sonuc <= 16'b0000000101101110;
'd70 : sonuc <= 16'b0000000101011110;
'd71 : sonuc <= 16'b0000000101001101;
'd72 : sonuc <= 16'b0000000100111100;
'd73 : sonuc <= 16'b0000000100101011;
'd74 : sonuc <= 16'b0000000100011010;
'd75 : sonuc <= 16'b0000000100001001;
'd76 : sonuc <= 16'b0000000011110111;
'd77 : sonuc <= 16'b0000000011100110;
'd78 : sonuc <= 16'b0000000011010100;
'd79 : sonuc <= 16'b0000000011000011;
'd80 : sonuc <= 16'b0000000010110001;
'd81 : sonuc <= 16'b0000000010100000;
'd82 : sonuc <= 16'b0000000010001110;
'd83 : sonuc <= 16'b0000000001111100;
'd84 : sonuc <= 16'b0000000001101011;
'd85 : sonuc <= 16'b0000000001011001;
'd86 : sonuc <= 16'b0000000001000111;
'd87 : sonuc <= 16'b0000000000110101;
'd88 : sonuc <= 16'b0000000000100011;
'd89 : sonuc <= 16'b0000000000010001;
'd90 : sonuc <= 16'b0000000000000000;
endcase
end
end
endmodule
`timescale 1ns / 1ps
//////////////////////////////////////////////////////////////////////////////////
module TestBench();
reg clk,reset;
reg [7:0]count;
reg en;
TopModule uut(.en(en),.clk(clk),.reset(reset), .aci(count),.sonuc(sonuc));
initial begin
clk=0;
reset=0;
count<=0;
#100
en <= 1;
end
always@(posedge clk)begin
if(reset)
count<=0;
else if (en)
count<=count+1;
end
always
#10 clk=~clk;
initial begin
#1810;
reset=1;
end
initial begin
#10;
count<=count+1;
end
endmodule
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H: Why does a pure resistance have an imaginary current and voltage?
While solving the below circuit:
The solution is as shown below:
I noticed the following while analyzing it:
Why does the voltage across a pure resistance and the current through it have imaginary parts?
What do these imaginary parts represent?
Similarly for a pure reactive load (inductive) what does the real part of the voltage Vx mean?
AI: Why does the voltage across a pure resistance and the current through
it have imaginary parts?
The imaginary part tells you that the current flow through the resistor is lagging the voltage applied to the series network by a certain amount. As far as the resistor is concerned, both its voltage and its current are totally in phase but, relative to the applied voltage applied across the series network of R and L, they are both lagging.
What do these imaginary parts represent?
The imaginary part indicates by how much the current in both L and R (the same current) is lagging the applied voltage to that series network.
Your math is correct apart from forgetting to add 90 ° to the inductor voltage value
|
H: Do you know the name of this item for crimping two cables?
I just need the name of this component.
AI: They are specifically known as U-joint crimps. The cables are overlapped into a U-shaped peice of metal, which is then crushed around them.
Butt crimps do a similar job, except there are two crimping points with the cables butted together and crimped into their own end.
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H: How to short certain pins on a 7x9 array
I’m planning to control an 80s electric typewriter via a Raspberry Pi. I want to emulate the keyboard matrix by shorting one of seven 5V pins to another one of nine that feed into a microcontroller. Eg, shorting pin 0 and 16 prints out a full stop; 5 and 12 print ‘w’ etc.
How can I best control this?
AI: There's the conceptually easy way, that would take as many elements as you have keys to emulate, and the trickier more efficient way.
Conceptually easy. Find out what voltage drop you can tolerate at each closure. Place a relay, or FET, or optocoupler output closure at each intersection that will give you less than this voltage drop. This is expensive in components, as not only do you need many 10s of switches, but you also have to control them from your Pi.
More efficient. The keys obviously belong to a multiplexed array. Take the row drive pins into your Pi, and drive one of the column pins at the right time, when the relevant row pin is active. Ideally you'll have an oscilloscope so you can see what's happening, and to debug any timing problems you run into. Potential timing problems are that your Pi will will not be able to respond instantly to a row drive going active. Your typewriter may tolerate this lateness, or may not. You may have to experiment with your column response timing.
A hybrid, which does not trouble the Pi for a fast response. Connect a multiplexer to the drive pins, and a demultiplexer to the response pins, and program their routing from the Pi. These will almost certainly respond fast enough. If they have sufficiently low resistance, then you could use analogue muxes like HC4051 to do both jobs.
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H: Buffer amplifier and input impedance
It is said that the input impedance of a buffer must be high enough so that it can isolate its input voltage and carry it to its output for driving purposes.
What I do not understand is why we treat all the independent sources as shorts while calculating input impedance. I know that input impedance is the impedance seen by input voltage present at buffers input.
There is, however, a huge difference between treating independent sources as shorts while calculating input impedance, and not doing so. Someone on this website said you can try finding input impedance without treating independent sources as shorts and get the same result as if you did.
I do not believe that. Can you enlighten me on the use of input impedance and the logic behind its calculation?
Input impedance found by treating sources as shorts cannot be used as an impedance replicate of a circuit connected to some other circuit which we try to solve without writing lots of node equations and therefore has no practical use. Also I looked at posts regarding it on this website and none answered my question.
AI: What I do not understand is why we treat all the independent sources as shorts while calculating input impedance.
Hopefully, this answer will make the reason clearer.
Someone on this website said you can try finding input impedance without treating independent sources as shorts and get the same result as if you did.
They are correct.
When we refer to the input impedance of a circuit, we are (almost always) referring to the small signal input impedance. That is, the input impedance gives us the ratio between how much the current will change if we make a small change in input voltage or vice versa.
$$Z_{in} = \frac{\Delta V_{in}}{\Delta I_{in}}$$
or, using the terms of calculus
$$Z_{in} = \frac{dV_{in}}{dI_{in}}$$
If we are given a circuit, such as this:
simulate this circuit – Schematic created using CircuitLab
We can calculate the relationship between the input voltage and the input current as follows.
$$I_{in} = \frac{V_{in} - V_{internal}}{R_{in}}$$
or
$$V_{in} = V_{internal} + R_{in}I_{in}$$
Differentiation gives us
$$Z_{in} = \frac{dV_{in}}{dI_{in}} = \frac{dV_{internal}}{dI_{in}} + \frac{d(R_{in}I_{in})}{dI_{in}} = \frac{dV_{internal}}{dI_{in}} + R_{in}$$
But since \$V_{internal}\$ does not depend upon \$I_{in}\$,
$$ \frac{dV_{internal}}{dI_{in}} = 0$$
So
$$Z_{in} = R_{in}$$
Which is exactly the impedance we would get if we simply treated the voltage source \$V_{internal}\$ as a short circuit.
If you understand how this works with one internal resistance and one internal voltage source, then it shouldn't be difficult to see that the same principal applies when there are a network of internal resistances and voltage sources.
The results we get "the long way" is equivalent to simply treating the internal voltage sources as shorts. Since it is also generally quicker to just do the latter, in practice, that is what we generally do. We treat internal voltage sources as shorts when calculating input impedance.
|
H: Power measurement on multiple loads
I'm designing a circuit for driving multiple loads (min. 16, up to 64) like the one shown below
simulate this circuit – Schematic created using CircuitLab
The input source is from a multi-channel DAC driven by an ATMega-like uC that can range from 0 to 5 V, and it should be able to generate a sawtooth with a minimum frequency of 1 kHz. As for the maximum frequency, this will be a trade-off between the resolution and speed of the scan.
With this design, the circuit should be able to provide up to 50 mA to the load. The load resistance can vary from 100 to 3000 Ohm.
The problem is, I need to know accurately how much power is delivered to the load and I have a tight requirement of max 100 mW for each.
For this reason, I'd like to introduce a monitor on the voltage using a multi-channel ADC and send it back to the uC. In this way, besides having a power monitor, I'd have a closed-loop feedback for better control.
However, I'm afraid that placing the monitor in this way would cause a number of issues:
(minor) Output resistance of the current source would be affected by the monitor (e.g. using a voltage divider before the ADC)
(major) Need to share the SPI bus between the DAC and the ADC, lowering the maximum frequency I can obtain.
Before considering another uC, do you have any suggestion on how you would implement such functionality?
Would you also consider the possibility to disable the monitoring for open-loop control?
AI: If you do use this type of schematic from TI it should work (on a more limited power supply): -
In the comments below the question I explained why the proposed circuit won't work (take note of op-amp type and the components in the orange boxes in my schematic). Those boxed components will certainly be needed if R3 is going to be as low as 10 Ω.
Also take note that the TI design uses an OPA2376 op-amp and this limits the power supply to no-more than 5.5 volts. You need to find a rail-to-rail op-amp that has decent spec and can operate from a 20 volt supply (as per the faulty diagram in the question).
I added a few notes and basically, if you monitor V(R1) and you have accurate resistors you can reliably predict load current: -
$$I_{OUT} = \dfrac{V(R_1)}{R_1}\times\dfrac{R_2}{R_3}$$
Use a high impedance voltage monitor circuit to measure this voltage. A 1 MΩ input impedance will degrade R1 to a 0.5% lower value for instance. I'd go for at least 10 MΩ input impedance for the monitor on R1 in my schematic.
For the output voltage monitor, keep with an input impedance of 10 MΩ or greater.
Need to share the SPI bus between the DAC and the ADC, lowering the
maximum frequency I can obtain.
That's something that cannot be answered here.
Would you also consider the possibility to disable the monitoring for
open-loop control?
That's up to you. Adding a few bells and whistles is always a good thing to do at an early stage rather than try and bodge in something later on.
It's almost imperative that you use a simulator to test this before committing to a PCB layout.
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H: FET source body connected to the same potential
I have learn that the FET looks like this:
However industrial FETs are not built like that.The source and the substrate have common potential.However if the source and substrate have common potential that means the conductive channel formed is not wide enough to connect the source with the drain there is a gap between the n type channel formed and the source.So how does it really work?
AI: The source and the substrate have common potential. However if the source and substrate have common potential that means the conductive channel formed is not wide enough to connect the source with the drain there is a gap between the n type channel formed and the source. So how does it really work?
It works precisely how you say it does not. The conductive channel formed is wide enough to connect the source with the drain. The channel might be a little asymmetrical, like this:
(image from electrical4u.com)
Provided the gate voltage is high enough there is no gap and the conductive channel is, well, conductive.
|
H: How to improve the accuracy of measured ADC voltage in a circuit connected to Vcc
The question is based on the following setup.
As depicted there is an IC to measure the real value of the 3.3V supply voltage accurately. The measured value of the 3.3V source is transferred via SPI to the microcontroller. This microcontroller is measuring the voltage divider between 3.3V and GND and has a reference voltage for its ADC.
simulate this circuit – Schematic created using CircuitLab
Problem: If the 3.3V supply voltage is changing the ADC measurement from the microcontroller is influenced.
Question: Can I compensate a changing 3.3V source with the measurement of the IC if the accuracy if this chip is within my limits.
AI: If you can get occasional more accurate measurements of the 3.3V you can correct the measurement via the voltage divider, which can be made more frequently. The details, such as filtering, would depend on how the 3.3V changes (how rapidly, and how much). It could be difficult if the changes are relatively large and spikey
Maybe you could even do a one-time self-calibration and store the factor in an EEPROM. What makes sense depends very much on the details of what you are trying to do. For example, for one application we added monitoring of all supply rail voltages and currents because the instrument was going to be very much inaccessible during operation.
I suggest making the maximum allowable correction factor modest, yet large enough to counter changes due to ADC reference tolerance, drift, temperature drift, and resistor tolerance. And consider what has to happen if the accurate measurement is late or is never received, or is clearly out of whack- system-level issues.
|
H: With Bluetooth turned off in PC settings, do keyboards and mice still emit radiation?
Do you know if a wireless mouse emits radiation in wired mode? Using the USB cable, With Bluetooth turned off, but with the switch set to "On"?
And also do you know if a wireless keyboard, also with Bluetooth turned off, emits radiation? Or just uses the radiation that already is available for the PC to be connected with wireless?
Thanks!
AI: With this kind of thing you need to define "radiation". We're going to discuss radio frequency electromagnetic interference. (RF EMI) For product testing categories, this is usually divided into
intentional emission
unintentional emission
low-frequency
Everything which works by radio will be an intentional emitter. Wifi, bluetooth, wireless keyboards, etc. These are usually in a narrow band and at a very limited power level. The most powerful one you'll encounter in a normal home is a cellphone; these used to be powerful enough to bleed into your hifi (Why does GSM cause speakers to buzz?).
Beyond that, every bit of digital electronics is an unintentional emitter .. at very low power levels. Testing for this is part of CE certification. Devices which emit enough RF to interfere with radios are not permitted.
Below about 250kHz is not considered as part of this testing. Anyone in a building wired for mains electricity will be subject to a 50Hz field, and this can be quite easily picked up.
There's no evidence that any of this is harmful. The only thing which is subject to harm testing is the amount of radio energy absorbed by the head of somebody talking on a cellphone; this is covered by the EU "SAR" limits.
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H: How to protect the circuit from Ethernet ESD, overvoltage, etc
Creating a carrier board for a RPI Compute Module4, I am not sure how to protect the circuitry from possible ESD /over-voltage coming through Ethernet connector.
In normal circumstances, I would apply some zener diode and in-serie resistors, but I don't think this will work with a 1Gbps Ethernet (response time, impedance, etc).
I also looked for optocouplers IC, but all what I seen works only up to few MHz.
Looking at the Development board, it has a couple of IC protecting the input, unfortunately, no label for this 2.5mm IC:
Is there any "standard" IC to protect the board from Ethernet? I am not necessarily looking for a specific part number, but rather a IC type name or family.
AI: For fast transients and ESD protection, zener will not be a proper solution.
As you mentioned, how fast signal is a problem also.
Transient suppression diodes (TVS) and ESD diodes could be used as such protections.
In such fast data lines, low capacitance diodes must be selected.
Many producers has their own manuals for specific application areas like gigabit ethernet, video interfaces etc.
https://www.littelfuse.com/technical-resources_old/application-designs/circuits/ethernet-esd-protection.aspx
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H: Overvoltage protection develops heat, is ok?
considering my old question: Here, i've made the following circuit using equivalent components on SMD, here is the schematic:
And these are the parts that i've picked from LCSC:
Fuse: Link here
Zener diode: Link here
Resistor: Link here
Transistor: Link here
The circuit is receiving 5.1V from the +5V netflag (using a regulated DC power supply) and giving out 4.93V from the netflag +5V-OUT (the circuit on the +5V-OUT draw a really low current: 250mA), everything works good, but the transistor itself (on the backside) develops a lot of heat (65/68°C at 21°C room temperature).
Is that normal?
I need to be worried?
AI: The zenner diode should be choosen the bigger voltage value. It is not ideal device and some current runs then voltage is close to nominal. It cause drop voltage on resistor. Even it is around 0.3-0.5V transistor in active mode.
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H: What are the different types of digital signals
In a digital logic class I was taught that a digital signal is one whose y-component, voltage, is discretized but whose time domain is continuous. In a signal processing class, we worked with digital signals that may have 8 bits of discrete voltage values, but were also discretized along the time domain as they were samples of speech. Are both of these signals true digital signals?
AI: If you're talking about a signal in the context of digital signal processing (DSP) you can think of it like a WAV audio file: the time axis is discrete, and the amplitude (value) axis is also discrete. Note this does not mean the discrete steps are uniform. For example if the signal has floating point values instead of integer or fixpoint, then the values are discrete, there is a finite number of possible values, but the interval between each value and the next possible value is not always the same. If your signal is a sequence of N-bit LPCM integer samples, then the interval between each sample and the next is the same.
DSP signals are data, in other words abstract constructs (ie, numbers) handled by computers. They may be transmitted using any encoding you can think of over radio, wires, or optical fibers, or pigeon post, but what they really are is a sequence of abstract numbers. And they exist independent of the physical storage and encoding: for example a WAV file still contains the same data no matter if it is stored on a hard disk, loaded in RAM, etc. And it still contains the same data if you use lossless compression like FLAC to store it.
In a DSP signal the value axis is abstract (it may represent the number of potatoes produced by year, or whatever) but the time axis is also abstract. I'm not just talking about varying sampling rates, rather that in DSP time does not exist in the same way than in the real world. If the processing unit is fast enough, it can process the signal faster than real time. For example you can encode a MP3 file in much less time than it would take to play it. DSP time meets real time only when the digital signal is converted from/to real world analog signals. But once the signal is data, time is just a label that says "this sample was recorded at timestamp X".
If you're talking about a signal in the context of hardware digital logic then your signal will most likely be a single bit, 0 or 1, encoded as voltage or current, carried either by a single wire or a differential pair. In this case the digital signal is a meaning we give to a physical analog signal that exists in the real world. The analog signal is continuous in time and amplitude, but we decide that we'll consider it a "1" if it above a certain threshold, a "0" below another threshold, and a "whatever/X" if it is in-between. This is a useful simplification that allows us to design logic circuits by hiding the analog complexity. These signals are always time-continuous, since they really are voltages, not numbers. But some logic circuits, like flops, will only look at the input on the edge of a clock cycle (and a bit before and after), which means we can pretend that we have a time-discrete system. Unless there is a timing violation, metastability, or other events that break down the assumptions and make everything analog again.
If you're talking about a signal in the context of digital communication, then it will most likely be a sequence of symbols. That can be anything, voltage, frequency, phase, etc, that encodes some information like one or several bits. For example, this is a telegraph symbol that encodes a letter:
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H: How to drive a relay from a MOSFET/IRF?
simulate this circuit – Schematic created using CircuitLab
Hello,
I would drive a machine to pilot its ON/OFF status.
So in idea it should be OK but is it?
What requires attention or this is an error in concept?
Regards
AI: That (IRF9530) is a P channel mosfet. But you have an N channel mosfet symbol.
The circuit is correct for an N FET, but not for a P FET.
When you select an N FET to be driven by a digital circuit, you need to make sure that the Vgs-th is low enough that the 3.3V or 5V circuit can drive it. There are a class of FETs that are designed for digital control. I would look into one of those.
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H: Binary division restoring method
Hi i need help in understanding the binary division restoring method
From -2, we stored the value to 5 by adding the divisor which is 7
After adding 7 to get 5, why does it automatically jump to 10?
So from 5-7 = -2, restore -2+7=5 => 10-7=3
Where does this 10 from, and can anyone explain why this form of divison works?
Source:
https://userpages.umbc.edu/~squire/cs411_l10.html
AI: It's similar to the normal long division algorithm in decimal, with one modification to combine a check for less-than / greater-than with the subsequent subtraction. You can just do the subtraction regardless, and if the value is negative, then you "messed up" and shouldn't have subtracted it. Just add it back to undo the mistake, then continue on to the next digit.
For example, in normal decimal long division, if you were dividing 120 by 8, the first thing you would check is whether 8 > 1. Since it is, you keep the value '8' as the first digit of your partial value and don't do any subtraction. Next you would drop down the 2 and check if 8 > 12. It's not, so determine how many times it goes into 12, which is 1. (In binary, this answer is always 1). And so on...
Using the same example with the "restoring method", you first do 1 - 8. The answer is -7, so the result will have a 0 in that bit/digit, and you add back +8 to get back to a partial value of 1. Then drop down the 2, do 12 - 8, and get a value of +4. Since it's positive, that digit of the result is a 1.
It's not how I would normally think of doing it, but I suppose it works. The example for decimal won't always work perfectly since it's missing the test for quotient digits 2 through 9.
Also, you asked where the 10 came from. It's just the decimal value of 01010, the first 5 bits of the dividend.
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H: What type of oscilloscope can be used to troubleshoot 100BASE-T1 automotive ethernet?
I would like to implement the 100Mbps single pair automotive ethernet specified in 802.3bw, otherwise known as 100BASE-T1. I was told that in order to troubleshoot and test on this standard, the scope must be capable of 1 Ghz. Is this accurate? Why would this be necessary as opposed to an oscilloscope that is well above 100MHz such as one that is 200MHz, but still 1 GSa/s ?
AI: Sounds like you want a protocol tester more than you want an oscilloscope. But oh well, that's not the question here.
I was told that in order to troubleshoot and test on this standard, the scope must be capable of 1 GHz
No. Even wikipedia will tell you that it only requires CAT3 cabling, and that doesn't guarantee a bandwidth even close to that.
Ah, with your comment:
Only looking for network traffic analysis at this point
yeah, well, then an oscilloscope in itself is no use at all. You'll need a device that speaks that protocol. For some higher-end, there might be add-ons that implement such a protocol decoder. But that again has nothing to do with the scope's 1 GHz bandwidth or not – it's something that extends the signal processing capabilities of the scope, and "converts" it into a protocol analyzer.
Since this is an ethernet standard, in principle, a network card connecting your device A to a PC, which sniffs and forwards the packets in both directions, and another network card for PC <-> device B would sound like a saner investment than an oscilloscope.
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H: Why is my loudspeaker not working? From Make: Electronics
I am a complete beginner in the world of electronics and I am following the amazing book by Charles Platt, Make: Electronics (second edition). I am stuck at Experiment 11, when I need to connect an oscillator to a loudspeaker.
The picture shows the circuit I am trying to make. It's an oscillator. It does work with a LED, but not with the loudspeaker. I do hear a very faint click when the capacitor discharge, but no sound. The loudspeaker is a 8ohm as indicated in the book.
This is a picture.
I know it's not super clear, but consider that the top part works as it should, it's the part with the speaker that possibly has something wrong.
To check that everything was working I also added a LED and made a video (and a switch to control it), here: https://youtu.be/UkEZzxmkfes
It shows that the oscillator works ok, but the loudspeaker doesn't.
Am I connecting the loudspeaker the wrong way? Is it possible that the loudspeaker is broken?
More generally, since I am a complete noob, how would I go debugging this / giving people the proper information in order to help them helping me?
Thanks!
AI: The frequency of oscillation is determined by the 470 kΩ resistors and 3.3 μF capacitors and will be roughly \$ \frac 1 {RC} = \frac 1 {470k \times 3.3\mu} = 0.64 \ \text{Hz}\$. i.e., less than one pulse per second. The circuit has been optimised for the LED rather than the speaker.
Modify the circuit so that the frequency is 100 to 1000 times higher and you should be delighted. Note that if you put the LED back in you won't be able to see it blink (due to your persistence of vision) although if you can wave the breadboard back and forth you may be able to see a "dashed" line of light.
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H: Will Kirchoff's voltage law work in a vacuum
Kirchoff's voltage rule tells us that the sum of voltages around a closed loop in a circuit will equal zero.
If a battery sits on a table, using KVL and treating air as matter with high resistance makes a KVL loop with a voltage gain of the battery's voltage which gets lost to the air.
If this battery floats in a near-perfect vacuum like space that has little matter and no interference will KVL still be applicable to create a "mesh"/loop around the battery?
AI: Kirchoff's voltage rule tells us that the sum of voltages around a closed loop in a circuit will equal zero.
Of course, that applies everywhere, even with a lack of conductors. That's a direct result of electrical potential being a potential field.
(this assumes we're in the DC case.)
A potential field is a field, i.e.
an amplitude (here: voltage) defined on a space (e.g. a 2D plane or 3D space), where no matter which path you take, the integral over all amplitudes on the points you travel is always 0 when that path is closed.
The direct consequence is that, yes, even in absence of a current, the voltages add up to zero – just: that has little effect, since there's no charges that these voltages move.
KVL makes a simplification: with a lot of precision, we really only care about the currents that flow through conductors when there are conductors.
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H: Examples of using a circular polarizer filter for LED display
I was looking for ways to increase the contrast of a LED-based seven-segment display like this against indoor ambient light, and found this comment on EEVblog suggesting using circular polarizing filters. The basic idea is that
when circularly polarised light gets reflected (off the surface of the display), the reflection is polarised in the opposite direction, and so gets attenuated.
Are there visual examples of circular polarizers being used on LED displays to attenuate ambient light? I'd like to see how they look before purchasing them, but haven't found anything so far.
Bonus question: is this technique commonly used? If not, what is usually done to achieve the same effect? e.g. in clock radios or other consumer equipment.
AI: Polarizing filters are only useful for rejecting specular light reflected from a smooth surface. Indoor light tends to be mostly diffuse, and most LED displays are given a matt or semi-matt finish to reduce specular reflections.
I don't know of any LED display filters that use circular polarizing filters. Most use colored plastic (acrylic). Some are etched to make reflections more diffuse, or coated to absorb light rather than reflect it.
A colored filter works by attenuating light outside the bandwidth of the LED. Since ambient light has to go through the filter in both directions the attenuation is doubled, while the LED light is not significantly reduced. However in-band reflected light is not significantly attenuated by the filter either, which reduces contrast.
To improve contrast a gray filter may be applied, which reduces the LED output but reduces in-band ambient light more because it goes through the filter twice. With modern high-efficiency LEDs the reduced LED output can easily be compensated for by increasing LED current. A single filter can do both jobs simply by being 'deeper', because the stronger filtering also significantly increases in-band attenuation.
Another technique that can help with red LED displays is to use a 'purple' filter that lets some blue light through. This works by using the eye's ability to distinguish color to tell the difference between lit (red) and unlit (purple) areas.
Here is an example of a purple filter compared to a red LED (from Agilent App note 1015):-
Note how the red filter band is deliberately arranged to be on the edge of the LED output, which produces some 'gray' attenuation as well as having a sharp cutoff at shorter wavelengths. At longer wavelengths the eye already has less response so a sharp cutoff on the right is not needed.
In bright sunlight getting enough LED light to overcome diffuse reflection can be a problem, so a smooth filter surface often works better - unless sunlight reflects directly off it. Depending how the display is mounted, it may be possible to prevent direct reflection at normal viewing angles by tilting the filter, or the display may be recessed into the device or have a hood over it to stop direct light hitting it.
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H: How slowly can you clock an AT25DF641A SPI Flash?
I am have started to try to use a SPI Flash. To better understand how it works, I decided to try to bit bang it with the GPIO on a Raspberry Pi. Everything seems correct, but I do not seem to have any output from the SPI Flash. The clock signal I am sending it is approximately 1 hertz. As I troubleshoot, I thought I should ask the question:
Is there a limit to how slowly you can interface with an SPI flash?
I am currently using the AT25DF641A-SH-T (1265-1180-1-ND on Digi-Key).
AI: The way to check this is in the timing diagram which is in most datasheets with communications specifications. There are no maximum times for \$T_{CLKH}\$ or \$T_{CLKL}\$ which means the SCK can be as long as you want. There are also no maximum restrictions on CS times either so that can be held low or high for as long as you want.
Source: https://www.mouser.com/datasheet/2/590/doc8693-1385727.pdf
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H: PC ATX12VO (12V only) standard - Why does everybody say it has higher efficiency?
To my understandings, the new 12-volt only standard moves the 5V and 3.3V supplies off from the ATX power supply "box" to the motherboard.
According to PC World (which is cited by Wikipedia), Gamers Nexus, and Linus Tech Tips, the new standard "improves efficiency" by not generating the 5V and 3.3V rails.
Why?
Modern PC parts still need those rails, for example, SSDs usually use 1.5V 3.3V for main operation.
We use 5V in USB devices to charge up cellphones and power personal beverage coolers.
So the need is not going away (any time soon) just because Intel has published a new standard.
Buck converters are cheap. Buck converter controller chips are tiny.
Why not throwing a few more components into the ATX power supply "box" and down-convert from the high-demand 12V rails?
The new standard calls for the conversion to happen on the motherboard. Isn't it just the same?
Linear regulators commonly found on cheap adapters like M.2 to SATA generates 3.3V rails by consuming all the power from the rest of 8.7V as a space heater.
Aren't these more problematic and should be eliminated first?
I understand a part of their claims is that traditionally all rails are directly converted from the main line (110V or 220V).
But you can easily design a multi-stage converter using the main just once.
To summarize: We still need the 5V and 3.3V rails. We do it either in the power supply "box" or on the motherboard. We can use buck converters instead of linear regulators anywhere anytime. Why does moving the 5V and 3.3V rails to a different place -- still doing the same thing -- improve efficiency?
AI: Modern PC parts still need those rails, for example, SSDs usually use 5V for main operation.
No, NVMe SSDs as attached by M.2 get a 1.5 V and a 3.3 V rail,
We use 5V in USB devices to charge up cellphones and power personal beverage coolers. So the need is not going away (any time soon) just because Intel has published a new standard.
USB is a very good example of something that will not change, indeed.
Why does moving the 5V and 3.3V rails to a different place ...
Why not throwing a few more components into the ATX power supply "box" and down-convert from the high-demand 12V rails?
Because: you can't drive anything that draws any significant current at 5V if that comes from your ATX supply some 40cm of cable away – voltage drop in the cable. You can't drive any high-speed circuitry with a constant voltage through cabling, either – parasitic inductivity will mean you simply can't react to the quickly changing current draw.
So, you always need to convert close to your large loads. That's why there's arrays of rather beefy DC/DC converters around CPU sockets!
With the converters close to the load, you can build many converters that are optimized for the load profile of their individual load – and hence, more efficient on average – than one necessarily massively overdimensioned converter, which nearly never runs close to maximum efficiency.
Linear regulators commonly found on cheap adapters like M.2 to SATA generates 3.3V rails by consuming all the power from the rest of 8.7V as a space heater. Aren't these more problematic and should be eliminated first?
Yes, but these things are aftermarket, and I've not seen anyone use one of these in a modern system. They are effectively never a problem. "My car users sometimes throw an anchor out their window, isn't that more problematic than another liter per 100 km off for a car?" comes to mind....
But you can easily design a multi-stage converter using the main just once.
You seem to be quite the expert on converter design, then... I'd see quite a few problem building that to a high degree of efficiency.
Also, don't underestimate the idle current savings you get when you get rid of a bunch of rails that your target system probably doesn't even need.
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H: Feed-through termination
I have a very basic question about rf-electronics: I tried to understand how feed-through terminators work, but I could not find a wiring diagram online.
Intuitively it would make sense to me if such a feed-through terminator is equivalent to a t-piece with a termination at one end of it, e.g. for a 50 Ohm feed-through terminator I would expect to have a 50 Ohm termination at one output of the t-piece and the output-signal at the other end of the t-piece.
Is this a legitimate way of understanding this?
Thanks a lot for your help! :)
AI: This is a feed through terminator
courtesy of EEVblog
simulate this circuit – Schematic created using CircuitLab
This is the schematics, you can use N parallel resistors of value N x 50 Ohm.
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H: 555 Timer won't run at calculated frequency
I have a 555 timer connected to an LED that I am trying to strobe for 450ms on, 72ms off to replicate an existing strobe light I have. The issue I am having is it is strobing at around 243ms on, 42ms off. According to my calculations, in order to get that timing I would need Ra=37.2k, rb=6.2k, with C=9.3uF. My actual values are Ra=56K, Rb=10K and C=9.3uF, however. According to the formulas found online for Th and Tl, with those values, I should be getting 460ms on and 69ms off, which is close enough for me. I am reading the pulse durations with a light sensor connected to an arduino running on a separate power supply. I've tried replacing the 555 timer with a different one (I bought 5) and it doesn't change anything so I'm really not sure what is happening to make it run the way it is. Is there something wrong with my circuit or am I missing something here?
Edit: the capacitor is a 106 ceramic, which should be 10uF but I measure it at 9.3uF
AI: Firstly, the ceramic capacitor will likely be a LOT less than 10uF with bias on it, as the voltage across it increases it will drop in value. It would not be unusual to see a small capacitor lose 50% or 70% of its nominal value.
Secondly, you're drawing a great deal of current from the power supply (LED with no series resistor) and there is no bypass capacitor on the 555 pin 5 and negligible capacitance across the supply (only 10nF). NE555 (bipolar) even on their own draw big surges when they switch.
Try bypassing the supply with 10uF ceramic, bypass pin 5 with 10nF or 100nF, and add a second regulator and second set of batteries just for the LED.
Or use a microcontroller, even with the internal RC clock you'll get infinitely more predictable performance.
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H: p-channel and n-channel MOSFET small signal models
What is the difference between the p-channel and n-channel MOSFET small signal models? Is there a difference?
AI: The only difference in a given model is that signs for voltages and currents are changed as appropriate.
There are differences in the actual devices that stem from physics: holes are not as mobile as electrons. That is reflected by the performance that you can achieve in a given FET, and in the numbers that you plug into a given model.
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H: Square wave, triangle wave etc. not defined in International Electrotechnical Vocabulary (IEV.) Are they not standard terminology?
Why are there no definitions for typical terms like square wave, triangle wave etc. in the International Electrotechnical Vocabulary (IEV)?
Are these widely used descriptions not defined by international standards?
And if so:
Why?
Is there a more correct terminology one should use instead?
I have also searched through German standards and did not find any mention of terms like "Rechteckspannung" (square voltage) and the like.
AI: The IEV is intended as a resource for those who prepare standards for the use of electrical and electronic equipment. It is not intended as a comprehensive vocabulary of electrical engineering. It does include many terms describing pulsed signals of which a square wave is a special case. One must conclude that, up to this time, there has been no need to define such terms in order to generate their technical standards.
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H: What equation should I use to calculate inductance of wire coil?
I've been working on a project recently which requires me to build an inductor, however in my research I found many different equations (with a range of variables used and giving very different results to the same coil specifications) and I was wondering what equation I should use. I have attached some links below to sites with different equations on them.
https://sciencing.com/calculate-inductance-coil-6026538.html
https://www.allaboutcircuits.com/tools/coil-inductance-calculator/
https://en.m.wikipedia.org/wiki/Inductance
AI: The proper formula depends upon a) whether your coil has a core and b) whether your coil is long or short, fat or thin, etc.
The most efficient air-core coil in terms of copper is a Brooks coil. It has a square cross section, and an outside diameter twice its inside diameter. You can see an image of the coil here.
The formula for the inductance of a Brooks coil is:
$$L \approx 0.025491rN^2$$
where N is the number of turns,
r is the inside radius in cms,
and L is the inductance in uH
There are also formulae for single layer long coils, multi-layer short coils and flat "pancake" coils found at this page
This image from AllAboutCircuits, shows various coil cross sections, and their corresponding (approximate) inductance formulae.
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H: MOSFET Selection criteria considering conduction losses, switching losses and working frequency
I am trying to choose a suitable MOSFET for my full-bridge resonant converter for wireless power transfer. The circuit topology is something like the following. Specifications: rated power \$-~500 W\$, DC supply \$-~50 V\$, switching frequency \$-~200~kHz\$
I need some help to determine the best-suited MOSFET for high-efficiency considering conduction and switching losses.
Here is my approach so far:
I have been following MOSFETS Selection guide by Infineon
Decide the range of MOSFETS based on \$V_{DS}\$ and \$I_{D}\$ (selection \$V_{DS}=100V\$ and \$I_{D}@T_A=25^0C>20A\$)
Select the package type to have sufficiently low lead inductance at switching frequency.
Question 1 - What are the suitable package types at \$200~kHz\$ switching frequency? For example, Is \${\rm D^2PAK7pin}\$ good enough, or should I go with something like TO-Leadless/QFN? What is the rule of thumb for package selection at different frequencies? see below picture.
Next, I look at the specifications such as \$R_{DS-on}\$, \$Q_{g}\$, \$C_{oss}\$, \$FOM=R_{DS-on}\times Q_{g}\$, etc. For example, the following is the comparison of six different \${\rm Infeneon-OptiMOS^{TM}5}\$ MOSFETS (This list can be very long!).
This is where I am lost.
Question 2 - Can I just chose only based on the above \$FOM\$? What are the other criteria for setting up the boundaries? For example, What will be the maximum allowed \$C_{oss}\$ for these specifications (i.e., at 200 kHz)? I can only find generalized claims about these relationships in the literature.
PS. I have been trying to simulate my circuit with LTSpice, but the simulation was not successful with the available spice models - It takes a very long time to simulate and return an error. I am working on getting the simulations done, anyway I am interested in a faster method for such comparison. Therefore, I want to rule-out the answers "just simulate it and compare the losses"
AI: Since you have a resonant converter, you should focus more on the Rdson. The output capacitance also plays a role, during the dead time the output capacitance needs to be discharged. Be aware that the equivalent capacitance is the transformer, output caps of the FETs? as well as the reflected capacitance of the bridge rectifier
So, basically during the dead time you can consider the current through the inductance as constant.
This is the current that needs to discharge the equivalent capacitances.
If the output capacitances are large and dominant, then ic=c*dv/dt tells us that for a given C either the current needs to be large(large magnetizing current) or the dead time needs to be large.
So yes, you want to have both a low Rdson and Output capacitance for resonant topologies, but in my experience the Rdson is far important.
Btw: are you sure about the secondary series cap?
This is only required for wireless power transfer. The presence of that cap makes the control complex.
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H: Boost converter extremely low efficiency
I'm trying to use a boost converter from Pololu that steps up my 1.25 input voltage from a AAA battery to 5V in order to power an Arduino. I have found that my Arduino needs .03A to be powered.
I understand that if the boost converter were 100% efficient and when using (V*I)in = (V*I)out, I would need .12A to be drawn from my AAA battery into the boost converter. However, when measuring the current drawn directly from the battery using a multimeter, I find that 1.05A is leaving my battery and entering the boost converter, while .03A is leaving the boost converter and entering the Arduino. This implies that my boost converter is about 11% efficient. The voltage seems to be stepped up fine from 1.25V to 5V.
This efficiency is concerning since the Polulu component has an input current limit of 1.2A, and I also need to drive a small motor. I am wiring the positive and negative terminals of my AAA battery to "Vin" and "GRD" on the boost converter respectively, and then wiring the "Vout" and "GRD" pins to the "5V" and "GRD" pins respectively on my Arduino. I am not using the "SHDN" pin.
So, am I wiring the boost converter incorrectly, or is this efficiency typical?
AI: Take a look at the curves they published, What you are getting sounds somewhat close. I know you are losing voltage in the current measurement, that is the way they work. The max with one volt in is 150mA at about 60% efficiency. I would assume you are below 1 volt which takes you completely off the curve. Consider putting two batteries in series, your efficiency will go up a lot and it might even work.
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H: Can you please help identify this circular part with 2 black wires and 2 red wires?
I found this part in an old auction lot I bought. Could you please help identify what this part is?
AI: Possibly a toroidal common-mode choke in a plastic case.
Especially if it's relatively dense (ferrite core) and if red-to-red and black-to-black measure short and red to black measures open.
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H: How do people estimate node voltage and branch current by observing a circuit?
The question might sound silly, but I hope experts can guide me on that particular issue. I am struggling for the last few months. I am a beginner in analog circuit design. When I follow a tutorial, I found that by observing a big transistor (MOS)-level circuit, they can say about node voltages, branch current, input and output resistance, and sometimes gain. My question is, without drawing a small-signal model of those circuits, how do people do that? Is this because they already solved many circuits by hand (I mean not using any simulator), and that is why a concept was built? Even when I try to draw a small-signal model for a big circuit I get lost due to the circuit's complexity. What is a good way to approach this kind of circuit?
I added the figure as an example circuit!
Thanks for your time!
Figure reference: https://www.hindawi.com/journals/apec/2014/274795/
AI: A lot of it is simply seeing/solving circuits and subcircuits often, and being able to pull out those structures and consider their behavior in isolation. Even complex circuits tend to often decompose to a number of simpler elements, allowing a lot of behavior to be qualitatively described quickly, saving mathematical analysis for unfamiliar or unusual sub-structures.
These understandings of common subcircuits, of course, come from mathematically solving those subcircuits for behavior, typical operating point, input/output impedances, etc.
As an example for this circuit, one path of analysis that comes to mind is to:
Assume all MOSFETs in saturation.
Notice that Mk (likely strongly inverted) is giving a constant bias voltage to M3/M4 gates thus making M3/M4 matched cascodes for M1/M2.
Attempt to pattern-match M1/M2 to a current mirror, noting that the gates don't connect to the drain so I might not be able to immediately make that simplification.
Hand-wavily consider M1/M3 as if it were a single transistor with very high output impedance because a cascode structure behaves that way under the assumed conditions. Same for M2/M4.
Now the two sides match a current mirror structure, with high output impedance thanks to the cascode.
At each step, I redraw the circuit (mentally or otherwise) and continue making simplifications/equivalences that seem to simplify it further.
Knowing the circuit's rough structure and behavior allows me to decide what stimuli make sense for it--it can now be simulated to supplement and verify the analysis. If I get unexpected results, I revise my analysis, simulate again, and so on until my understanding and expected/actual results converge.
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H: MOC3041 TRIAC Dimming Circuit
Details: I try to build the single TRIAC circuit for ON/OFF, Dimmer(600W AC Light), and AC fan Speed Control. I trying this the first time.
Problem: I using MOC3041 to control the TRIAC ON/OFF it seems to work fine but while I Tried Dimming I found using MOC3041 cant able dim the lamp due to Zero crossing in the MOC3041. Now I try to change MOC3041 to MOC3052, Please clarify to me what are the things I need to take care of before proceeding.
What will happen to my switching if I change MOC3041 to MOC3052?
Note: snubber Circuit value - Rs-50E/1W, Cs-0.01uF/400V
Thank You.
AI: This is the main difference :
At the desired phase angle (zero cross + delay) you turn the triac ON. at the next zero cross you turn the triac OFF.
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H: Ungrounded laptop power brick
I'm travelling at the moment and the three-prong plug on my laptop power brick is going through a two-prong adapter and therefore losing the connection to ground.
The laptop stopped charging and I took it to a local repair place. The guy put it down to the power brick and sold me a replacement. This looks like a genuine HP product but could be a copy. The sticker says "connect only to a grounded outlet".
I have had several electric shocks when handling the plug. Does this mean that the power brick is faulty, or is it normal given that it is not grounded?
AI: Yes, devices with grounded plugs need ground or their mains EMI filter makes the ground to have half of mains voltage appear on e.g laptop metal parts. That is why you get shocks or may damage equipment when connecting e.g. grounded devices to your laptop while it is connected to your ground-bypassed charger.
The warning reads on almost any equipment with a grounded plug, including genuine HP chargers, and if it does not read on the device, it reads in the manuals which should be read and understood.
The point is, a device with grounded plug must not be used in ungrounded socket. I also understand this is also not always possible, and two-prong chargers that do not need ground have their own issues.
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H: Impedance in LTspice and current plotting
I wanted to try a test simulation on LTSpice to understand few things. I have a RLC parallel circuit and I wanted to see how the impedance varies according to each one of the components (very easy but I needed to run the simulation for visualization purposes).
To my surprise, when I plot I(V1) and I(L1)+I(R1)+I(C1) the peak of the currents at the resonance frequency are not the same, and I cannot seem to figure out why.
Any ideas?
]2
Thanks in advance :)
AI: If you use I(V1), you get the magnitude of the current.
So if you plot I(L1)+I(R1)+I(C1) you plot the sum the magnitudes.
Using the magnitude ignores the phase differences between the currents, the imaginary part of the current is removed and converted into the magnitude value.
Only if you would do a "proper" sum of I(L1)+I(R1)+I(C1) that takes into account the imaginary part of the currents and plot the magnitude of that sum, would the result be the same as I(V1).
When you plot I(V1) the summation is done properly (taking the imaginary part, phase etc) into account. So then at the resonance frequency, the inductor and capacitor's impedances cancel each other out and "from the outside" (looking at the impedance of the RLC tank) you only see the impedance of the resistor.
Tip: you're now actually plotting the admittance 1/Z. If you want the impedance plot, print 1/I(V1) or do as I usually do: replace the AC voltage source V1 with an AC current source with AC = 1. Then the voltage at "input" is the same as the actual impedance as V = I * Z = 1 * Z = Z (I = 1 because of the current source with AC = 1).
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H: What kind of FPGA board for sha-256 algo
I am newbie in VHDL and FPGA, i was searching for an answer to my question but didn't find any good answer.
In my projet i have a .txt file containing a hexadecimal data, i want to send this data to an FPGA, 256-hash it, and then send the hashed data back to my laptop in another .txt file.
To do this task what kind of FPGA do i need to buy?
PS: I want a speed and not expensive one, if it's possible
Help is appreciated
AI: I would expect this to be possible with the "icestick" FPGA cards. However, SHA-256 is likely to be memory-bandwidth limited on most PCs, so in almost all cases it would be faster to do it on the CPU than transmit it anywhere and back again.
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H: Why does a fuse voltage rating only signify the dielectric strength?
From what I understand, a typical fuse wire will melt once it's current rating is breached by some factor.
However, a fuse also has a voltage rating, which signifies its dielectric strength. From what I've read, this voltage rating means that, "beyond this particular voltage, the fuse may conduct electricity even if the current being drawn is much much higher".
Does this mean that 1A @ 5V and 1A @ 100V builds the same amount of heat in a fuse wire? It shouldn't theoretically since the power rating P=VI is way different, right? To put it in another way, can I use a 1A @ 220V fuse in an environment of 1A @ 5V?
AI: To put it in another way, can I use a 1A @ 220V fuse in an environment
of 1A @ 5V?
Yes you can. It's the 1 amp rating that determines at what level of current the fuse initially melts.
However, if you used a 100 volt rated fuse on a 250 VAC application, then it may blow but continue to arc and conduct unsafely.
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H: Impedance matching L Network
I‘m trying to understand exactly what an L-section Network does. I understand it tries to eliminate the imaginary part of the impedance and match the real part. However I do not understand what excatly both components do and why the inductor(in my example, more specifically in a Highpass Downward L-match) has to be a shunt inductor and the capacitor has to be in series? Thank you for any help.
AI: I‘m trying to understand exactly what an L-section Network does.
What you are describing in your question is a loss-less, high-pass, L-pad impedance matching circuit as per this example: -
The L-pad is used to match a lower impedance on the left with a higher impedance on the right. Either end can be source of course. In other words; you can match a higher source impedance to a lower load impedance or vice versa.
The formulas for L and C are dependant on the operating frequency of interest (\$\omega\$), the input impedance (\$R_{IN}\$) and the output impedance (\$R_L\$): -
$$C = \dfrac{1}{\omega\cdot R_{IN}}\cdot\sqrt{\dfrac{1}{\frac{R_L}{R_{IN}}+1}}$$
$$L = R_{IN}\cdot R_L\cdot C$$
Here you can find a calculator that saves you crunching the numbers by hand: -
That website also provides proofs for the formulas.
why the inductor(in my example, more specifically in a Highpass
Downward L-match) has to be a shunt inductor and the capacitor has to
be in series?
Well, you can make a low-pass version like this: -
Both work the same; at the frequency of interest, you can provide loss-less impedance matching as opposed to a wideband lossy impedance matcher like this: -
I understand it tries to eliminate the imaginary part of the impedance
and match the real part.
It provides loss-less impedance matching i.e. it makes the input impedance looks resistive and the output impedance looks resistive at the desired operating frequency. Either side of the mid-operating frequency, it's not quote perfect, but, it's good enough for most signals just like an antenna is not quite perfect either side of it's target mid-point frequency.
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H: Transistor maximum base breakdown voltage
In a npn transistor with 5 volts VCE and current limited to 10ma by a resistor, it needs to drive its base with a DC voltage ranging from 3 to 500 volts DC, using a base resistor of 10 K ohms the voltage in the base never exceeds 900mv, is my drive correct? In this case, how do you set the maximum current you can receive at the base?
AI: If you want the device ON with an input voltage between 3 and 500V, you are going to need more circuitry, because at the high voltage, the base current will be too much.
I suggest putting a comparator at your input so you have a constant voltage to turn on the transistor.
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H: Why is the closed gain loop -3dB from the open loop gain rather than some other value?
I am reading the book Op Amps for Everyone and in the chapter of feedback loop theory, I have this quote (see picture) where it says in reality the closed loop gain is down by 3dB at the point X as shown below.
How does he know it is -3dB and not some other value that is proportional to the gain B?
page number 85, link to the book:Op Amps for Everyone
AI: The text says it's a single pole amplifier circuit and that always means that at the cut-off frequency, due to the transfer function gain equation, the output signal is down to exactly half the power of the input signal. Half the power in decibels is \$10\log_{10}(0.5) =\$ -3.0103 dB.
That's what single pole filters do at their cut-off frequency.
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H: Resistor that can be placed like a jumper
I want to make a relay output to be configurable by the client as NO/NC or guarded by resistors (10k opened, 1k closed).
A solution I have been thinking about uses resistors that can be placed like a jumper on a pin header, like this:
A quick search for existence of this jumper type yielded nothing. Does anyone know if this is an actually available component? Any other (albeit simple for untrained end users) solutions are welcomed.
AI: You can use 3-way headers instead of 2-ways for the jumpers. That way, you can shift the jumper by 1 position to the Resistive Protection position.
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H: New STM32L433 pins pull high??? Damaging opamps
I have an STM32L433, fresh from RS components. I soldered it to my board. PA6 is connected to the output of an opamp - INA181A1.
There are another 2 INA opamps on the board, connected to other pins.
I have now had 2 INA181s die (actually I am not sure they died, since I unsoldered the first and replaced it assunming it was broken... the second I unsoldered just the output leg and may have damaged it mechanically...). The board is not even programmed yet. There is actually not yet even a programming header...
The net between the opamp and the MCU is pulled high. If I hold the MCU in reset state (connect the debug header nRES to ground) the opamp-MCU net floats. As soon as I release the reset pin, it pulls high.
The other 2 channels to not do this! The other 2 channels behave exactly as expected!
There is no board error. There is >500kOhms between the net and ground/3v3 when not powered - this pull IS coming through the MCU. I observe there is ~50mA extra drawn from my power supply between when the MCU is held in reset and when reset is released.
Are my new chips pre programmed???am I missing some line in the datasheet/reference that says "PA6 will pull high when new"?
The boards were supplied without MCU and opamps and gate drivers. I have populated them myself. I have built 2 boards with F303CB chips, which work fine, using internal opamps, and 1 with L433CC. The pinout appears to be the same in terms of power, boot0 etc.
The schematics can be seen in https://github.com/davidmolony/MESC_FOC_ESC/tree/Rearrange_Vin - this is an open source EBike controller I work on in the evenings. I have been riding around with the F303CB version, pushing ~80phase amps and getting up to ~40km/h, so the board DOES work.
Edit:
Schematic
Offending opamp is op2, U9
AI: If the MCU is unprogrammed, it will automatically go to factory bootloader.
Bootloader uses PA6 as SPI1_MISO, which will be push-pull output.
It might go high-impedance if PA4 used as SPI1_NSS is pulled high, so that is something you might try, but not much else.
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H: Why do some microcontrollers have numerous oscillators (and what are their functions)?
Currently I am reading through the Arm based ATSAM4L series datasheet, and in the sections BSCIF and SCIF I have encountered at least 8 different oscillators/clocks (see image below). I understand that an oscillator is necessary for a clock signal for the MCU, but why are so many different sources necessary for this MCU family and MCU's in general?
EDIT: And what are the functions of these specific oscillators?
AI: 32kHz ultra low power oscillator: Used for RTCs (Real-time-clock). These can run in the background and enable time-keeping, even when the controller core is sleeping. This enables to have a running clock without the high power consumption of the controller core. The 32kHz oscillators are pretty precise and typically have an error of less than a minute per month.
32kHz RC oscillator: Similar to the above, but way less precise. RC oscillators can easily be integrated into the µC itself and don't require external parts, but have not very good tolerances. The low clock speed still enables some background stuff going on, while most of the controller is sleeping to safe power.
1MHz RC oscillator: A lot faster than the 32kHz oscillator (obviously) and enables to do some faster processing. Used for normal operation in non-sleep mode and is integrated in the controller. So for this RC oscillator no external parts are needed (again). Clock is not very precise and can vary by several percent (although often it is possible to calibrate these).
Crystal oscillator up to 30MHz: An external 30MHz crystal can speed up operation by a factor of 30 in comparison to the internal 1MHz oscillator and can be pretty precise, but needs the additional external parts. The higher clock also increases power consumption.
PLL: Enables the controller to run of a precise crystal frequency and derive an even higher clock frequency, if that is needed.
It is up to the developer to choose which clock source is the best compromise of cost, power consumption, board area, precision and performance (Which is not necessarily a static decision: The controller might run a few seconds on the high clock speed to do some acquisitions, calucations, data transfers, etc. After that it might go to deep sleep mode and only the RTC is running in the background, until some event (either by the RTC or some external trigger) wake the controler up to repeat the cycle.
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H: Is it possible to know what is the dielectric voltage between earth and conductor of those cables?
I am looking to know what is the maximum voltage (dielectric voltage) that it can be applied between the conductors of the cable and the earth (not into the cable). Usually I look for a term which is called " Voltage Test " but in the datasheet below I am not able to find the informations :
https://fr.farnell.com/amphenol-ltw/hmc-05bffm-sl8a01/nmea-micro-c-5pin-f-conn-f-pin/dp/2912028
https://media.digikey.com/pdf/Data%20Sheets/Phoenix%20Contact%20PDFs/1669770.pdf
Thank you very much and have a nice day.
AI: Dielectric breakdown of plastic is > 5kV/mm but surface contam. may reduce that to 50V/mm depending...on extreme humidity and dust, normally 500V/mm
Impulse ratings <1us may be 10 x higher
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H: PWM Motor Driver output voltage is enormously load - dependent
I've built this PWM motor driver (for a motor which requires 5A at maximum 24V and which I want to drive sometimes with 10V and other times at 24V). Here you find the datasheet of the mosfets.
PS: there is also a missing 10k resistance between gates and GND
The device works and supplies correctly my motor. If I use a 50% of duty cycle, I get an average voltage of 12V across my motor.
But, If I measure the output voltage without load, or with a low - current load (such as a 200mA fan), the device works (it provides an average voltage which is determined by the PWM duty cycle) but not with the right proportion: without load (the only load is my volt - meter) I get 12V (instead of 6V) with 25% of duty cycle and 20V (instead of 12V) with 50% of duty cycle. With fan the result is slightly different but always in contrast with the right proportion.
My possible explanation: the output voltage is, in theory, a square wave. A Volt - Meter, a fan and my motor filter it in different ways, so the result is different. But, if this explanation is true,
I'm surprised of such big differences between the behaviour with these different loads.
I don't know how should I take care of this fact and design my circuit to work properly with a specific load (for my motor, it works perfectly, but maybe it is lucky....I've never thought at this aspect).
I think that different loads will change the amount of ripple, and not the average value. But, in my measures, the average voltage is different.
Can you help me to explain this phenomenon in a more detailed way?
AI: With motor switching, you should think in terms of current instead of voltage. Consider what happens when you are running an inductive load (your motor) under load. When you turn on the FETs, the output terminal is pulled to ground and current flows through your motor. When you turn the FETs off, the motor continues to conduct current in the same direction, so the current flows through the diode, turning it on so the voltage across the motor quickly changes from 24 volts to one diode drop, then to zero volts if the switch is left off long enough for the motor to stop conducting.
Now consider when you are running with no load. The FETs turn on and pull the output hard to ground. But when they turn off, but there is nothing to pull their drains back up to 24 volts except the diode leakage and whatever else you have in parallel to the motor. If you were to look at voltage waveforms with a light load, you would see not a square wave but a ramp on the "off" half-cycle. If you were to add pull-up resistors, you would then see the average that you were looking for.
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H: Amplifier circuit for Bently Nevada velomitor 330525
I'm in need of measuring vibrations and I have access to these Bently Nevada velomitors.
I'm not sure how to design a ampliefier circuit for the signal.
The datasheet is says that it is a piezo electric velomitor, but what is throwing me off is that it has power requirements of DC voltage -22 to -30 and bias current of 2,5 to 6 mA. The output bias is -12 +-3 VDC.
The output is 4mV / mm/s and I would like to amplify this by 100, so it would be 400 mV / mm/s. I cannot figure out how to build the circuit because it apparently needs negative volts and it is only a two wire velomitor.
Could someone point me in the right directions? I have dabbled with basic electronics but nothing really with negative volts and piezos.
Datasheet
AI: The user guide referenced in the document you linked explains it fairly clearly.
Traditional velocity sensors consist of either a moving wire coil surrounding a fixed magnet or a fixed wire coil surrounding a moving magnet. The Velomitor® Sensor is more accurate than traditional velocity sensors. Because the Velomitor® Sensor contains no moving parts, it is also more durable and less sensitive to transverse motion than traditional seismic transducers. Its piezoelectric sensing element and solid-state circuitry let the Velomitor® Sensor withstand years of continuous use.
The Velomitor® Sensor is a two-wire device that requires an external power supply. The power supply must provide a DC voltage of 22 to 30 Volts and a current of 10 mA. A constant current diode must be used to limit the current to the sensor to 2.5 to 6 mA. Figure 1-1 shows a simple block diagram of the Velomitor® Sensor system. [Emphasis mine.]
Table 3.1 gives the required diode part number:
3 mA (Motorola P/N 1N5309) current diode Bently Nevada P/N 00643485.
Wiring seems to be as simple as this.
simulate this circuit – Schematic created using CircuitLab
Figure 2. General gist of your amplifier.
The datasheet mentions a DC bias on the output so you're going to need to remove that before amplification. C1 does that. You'll need to pick a value that will allow the lowest frequencies of interest through without too much attenuation.
R3 and R4 bias the op-amp to half supply so that your signal can swing symmetrically about that point. You don't say what this is feeding into so I have no idea if this will work in your application.
I had never heard the term before. Thanks for the education.
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H: How to safely use a transistor to control fans
I'm working on a project where I control two 12V 80mm fans with an ESP32. I've taken inspiration from various blog posts about fan control on the Raspberry Pi with a transistor (e.g. https://fizzy.cc/raspberry-pi-fan/).
As depicted in this blog post, the transistor is used to control GND connectivity of the fan, to switch it on/off.
The resistor used to "protect" the transistor varies a lot depending on the source (from none, to 680 Ohm, to 1kOhm as used here).
I'll be using the power from a USB power supply (5V out) and a step up module (to ~7V) and control the GND connectivity via the ESP32. Now my problem is, that I find no explanation on how to calculate the appropriate resistor.
Question: How can I calculate the value for the resistor, to safely operate the transistor (BC547B, Bipolar Transistor, NPN, 45V, 100mA, TO-92, 3-pin)?
edit: I would also have a PN2222ABU transistor available.
AI: The first step is to determine the expected current that the fan will draw. Let's call it 200 mA (just a wild guess on my part, but you should be able to find or measure it).
Next find the minimum beta of the transistor (I'll assume 50, just for example). In order to pull 200 mA with a beta of 50, you'll need 200/50 = 4 mA of base current.
In order to get 4 mA of base current, divide the voltage drop from control to Vbe (if output of the control is 3.3 V and the Vbe is 0.6 V, then the drop is 3.3-0.6 = 2.7 V) by the base current. 2.7 V / 4 mA = 675 ohms.
Hope this helps and welcome to the site.
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H: A JFET based light sensor
I am working on a LDR based JFET project which would turn on automatically when it is sufficiently dark.
Accordingly, I have designed this circuit. Is this practical? Also when the LDR value reaches 0 ohms the gate-source junction will be fully negative, but since the junction is reverse biased will it consume any power from the left side battery? I am not interested about the resistor values here, I just want to know whether the topology is correct.
AI: The battery polarity is drawn differently than it's said in the question text (noticed already by commentators). If we assume the text is what's wanted it will not work. LDR resistance variations do not affect Vgs because there's no gate current ==>> light variations have no effect to the fet conductivity, the current of the led stays the same in dark and in light.
You need a voltage divider to generate voltage variations with LDR. If you use a bipolar transistor instead of fet you can control the base current with light on LDR.
You can set easily predictable voltage treshold and hysteresis if you use a comparator IC to sense what voltage a voltage divider with LDR outputs. Hysteresis is needed to avoid ON-OFF oscillations near the treshold.
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H: Why voltage multiplier decreases current?
As far as understand, a voltage multiple increases the voltage but decreases current.
Can someone please explain why current decreases?
AI: A voltage multiplier has a lot of capacitors. Capacitors have impedance. The capacitors are needed to add up the voltage, but the also limit the current that the multiplier can pass.
Conservation of energy plays a role. Since the voltage multiplier produces a continuous output, the power out can't be more than the power in. That means you have to get less current when the voltage goes up. That's a hard limit you can't get around.
You can get around the impedance problem to some extent. You can use larger value capacitors, or higher frequency drive, or both. The result is lower impedance and more current output.
The impedance of a Cockcroft-Walton multiplier as you showed in the question can be calculated from the following equation:
$$Z = \frac {4n^3 + 3n^2 - n }{2 \pi fC} $$
Where Z is in ohms, f is in hertz, C in farads, and n is the number of stages.
As you can see, the impedance goes up drastically as the number of stages increases.
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H: Flash or EEPROM
EEPROM vs SD Flash
I am working on a Arduino program (a game, a slot machine) that needs to store 16 bitmaps (the slot machine symbols) that are 32 * 16 bytes each (total 8192 bytes). The bitmaps are static/immutable so does not need to be changed. The only requirement is that they need to be accessed fast. Thus performance prevails capacity.
Currently, I am working with an SD Card Module for Arduino by Vellman (SPI). I noticed that, opening a file, and especially seeking a byte in the filestream causes performance penalties.
So my question is: would an EEPROM (or 2, regarding the amount of data) be a better, or at least a faster, storage solution than a SD card given that:
The data is only written once.
The data is read a lot of times.
I need to get the bytes in a nonsequential order.
Current config: ATmega32U4 (Pro Micro) at 16MHz with 32 kB
program + libs needs about 30 kb.
Background: App is driving 4 * ST7735 TFT displays (SPI), Clock (SPI) and 6 buttons (one pin) and currently, and SD card reader (SPI).
AI: Yes, do consider anything else than an SD card, because an SD card is totally overkill for storing 8192 bytes.
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H: Treble bleed frequency cutoff
I built an attenuator for my guitar amplifier. I took the circuit from here but left out the 4 ohm toggle since my amplifiers all have 8 ohm outputs. Below is a simplified schematic of the circuit (with the L-pad set to 20 dB attenuation):
The 4.7u capacitor is supposed to be a treble bleed. My intuitive understanding is that the capacitor allows the high frequencies to bypass the L-pad while the low frequencies are still attenuated. However, I don't notice any difference with the switch open or closed while playing.
I tried to measure the frequency response of the whole attenuator with a signal generator and a scope. The amplitude of the output signal did not change when toggling the capacitor for any frequency up to 20kHz. However I realise that the signal generator and scope have very different output and input impedance compared to the amplifier and speaker, so I don't really trust these results.
My next idea was to calculate the expected cutoff of the treble bleed, but I am not sure how exactly the high-pass filter equation applies here. My understanding is that the speaker is part of the filter, so
$$
R = \frac{1}{\frac{1}{0.89}+\frac{1}{8}}
$$
$$
f_c = \frac{1}{2\pi\times4.7\text{u}\times R} = 42281
$$
42 kHz is of course too high for audio, so now I am not sure if my analysis is correct.
What is the correct way to analytically determine the treble bleed cutoff frequency?
How can I reliably verify this using the signal generator and scope?
AI: The amp is 0 Ohms and the speaker may be neglected.
But the ESR of the capacitor must be <0.1 Ohm which rules 99% of electrolytics, so a metal filter cap from a microwave oven or store must be used.
In theory, the cap is not a LPF but a 4kHz HPF
The -20dB pad rises to -17 dB at 4kHz
Simply use 7.2 with C to compute breakpoint.
But to make a LPF is harder with a pad as R becomes 0.89 and needs a cap 8x bigger as a shunt.
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H: Voltage Divider returns incorrect voltage reading
I have created a simple voltage divider circuit as shown below:
Using the simulator, I can see that the voltage measures in at 0.838935V, which is expected.
However, when constructing this circuit in real life, I seem to be only reading ~0.513V on my digital multimeter. This project was initially supposed to read the battery voltage on an ADC, but this seems to be the root cause on why it is constantly returning a lower voltage.
Any ideas on why possibly this is happening?
AI: The answer is simple. Your real voltmeter is not ideal thus, it will have an internal resistance around \$1M\Omega\$. A better voltmeter will have internal resistance around \$10M\Omega\$
So your circuit will look like this:
simulate this circuit – Schematic created using CircuitLab
And the output voltage will be :
$$V_O = 7.88V \times \frac{560k\Omega||1M\Omega}{4.7M\Omega +{560k\Omega||1M\Omega} } \approx 0.56V$$
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H: Op-Amp Current Consumption
I’m using the LM7321MAX operational amplifier to supply a 12V digital signal from a 5V source. For the op-amp supply voltage: is it better to use something close/equal to my output value (like 12V) or further away (like 22V)? I know the internals of an op amp are mostly BJTs so am wondering if I’ll draw the same supply current in either case, with the 20V variant therefore using more power. Thinking the 20V->12V will happen across those BJTs
Thanks for your help!
AI: There are a number of curves showing the supply current for different supply voltages and for different temperatures and common-mode voltages, so you can predict the typical supply current. It increases with increasing supply voltage, but fairly gently.
As far your your load current, you'll need enough swing to satisfy your actual requirements, and again the datasheet provides a great number of curves. Once you have enough voltage to (worst-case) supply your required output swing, then additional voltage simply results in more power dissipation in the chip, since the additional output current is drawn directly from the power supply terminals (and possibly a bit more).
The total current drawn is the sum of the two, so it's Is + k\$\cdot\$Iout(Vsupply) where k may be a bit higher than 1, and this the power dissipation in the chip is (Is \$\cdot\$ Vsupply) + k\$\cdot\$Iout(Vsupply-Vout).
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H: Can I charge my van's car battery directly from its lithium ion "house battery"?
I have a 1991 GMC van that I've converted into a poor man's RV. I've got a Yeti 1000 lithium ion house battery, and a 100 watt Renogy solar panel on the roof. When the battery has died in the past, I have hooked the ~20v solar panel output directly to the car battery and successfully charged it (checking voltage on the car battery regularly to make sure I don't overcharge). Can I do the same thing using the 12V output from the house battery? I'm mostly concerned with damage to the house battery, which costs over $1000.
My exact question is this: If my 12v lead acid starter battery goes dead, can I connect the 12v output (I measure 12.4v) of the Yeti 1000 directly to my starter battery to charge it enough to start the car, as long as I am careful to monitor it to prevent overcharging? And is there any way this might damage my yeti 1000?
AI: New question.
Maybe, the 10A 12V limit will be exceeded with a dead battery so a 10A 12V lamps in series might be possible to use as a current limiter to get 50% charge @ 12V.
“ The Goal Zero Yeti 1000 Lithium can be charged in 9 - 18 hours by plugging into your vehicle's 12V outlet using the Goal Zero Yeti Lithium 12V Car Charging Cable. NOTE: Do not attempt to charge your Yeti Lithium from a 12V source using any other cable. Doing so may cause damage to the unit”
it must be able to measure battery current independent of the load to limit the exposure time during CV soak overcharge.
keep a record of voltage in each cell monthly. When one drops, it is a sign of end of life. The battery is only as good as the weakest cell. Each cell must very well matched. If the entire string is ok but 1 cell is overcharged or under, this stage of aging will deteriorate rapidly under the balancer does its job.
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H: In this circuit what is the function of this resistance in parallel with a capacitor
I understand that the capacitor produces a resistance proportional to the 50 Hz of the mains input current, in this case 220 volts
But why is that resistance installed in parallel, what function does the resistance have?
Thanks
AI: It is most likely a high value resistor to discharge the capacitor when the circuit is disconnected from the mains AC supply. This is purely for safety and not functionally required for the circuit.
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H: How is it that the induced EMF in an alternator is in quadrature with the main flux?
Looking at the simplified diagram of an alternator, the flux "linked" with the conductor is maximum at 90, 270 degrees, is also when induced EMF is maximum. So isn't the main flux in phase with the induced EMF, or are my assumptions flawed?
It's important to note that in an alternator, motional EMF is created which depends on the spatial angular displacement between an uniform magnetic field and a conductor perpendicularly placed. A time varying EMF is induced only because the angular displacement is a function of time, not the magnitude of the magnetic field itself.
AI: At school I was taught that the voltage induced was proportional to the rate at which magnetic field lines were being cut by the windings. In reality the field lines don’t literally exist, but it’s a useful tool at an intuitive level. The EMF is in proportion to the rate of change in magnetic field that the coil experiences. Does that help?
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H: Watts P.M.P.O. and audio equipment power
I am a beginner in electrical engineering, and to begin learning I decided to look into audio amplifiers.
I have seen that the power of audio amplifiers is often described in the units of watts, watts RMS, and watts PMPO.
So far, the only one of those units that I think I understand is watts.
For instance, I just recently made my own audio amplifier circuit. If I think: "What is the power of my amplifier supposed to be?", then my thought process goes like this:
The transistors I used are the TIP120, which have a maximum rated
power of 65W. Then, I decide that 60W is a safer maximum which would be
reached if my amplifier runs 19V through one of my 6Ω speakers.
Therefore, the power of my amplifier is 2x60W.
Then, I look at the datasheets of audio amplifier ICs, like the TPA3116 that says: "50W into a 4Ω load at 21V", which makes me conclude that in that setting it'll output 14V maximum into the speaker, which I think makes sense.
But then, I look at the datasheet of the TDA7294, and they talk about "RMS Power" and "RMS Music Power", units which I couldn't find clear information about.
Then, I look at home audio systems, like a Panasonic Stereo System I have which has "9500 Watts P.M.P.O." written on the front.
9500W? That sounds stupidly high to me, but then I look again and it has 300W written on the back of it.
If I'm correct, maybe a 1kW home stereo system could make sense, but a 2kW or greater one would not exist.
So...
Are my conclusions about my audio amplifier circuit and the TPA3116 correct?
Are my conclusions about audio equipment power correct?
Are WattsPMPO, WattsRMS and RMSmusicpower, real useful units?
AI: Audio has tons of "specifications" which really are marketing bullshit.
To specify the output power of an amp is more subtle than it looks.
Max output voltage: gives you an idea of the peak output power.
Max output voltage into 4/8 ohms resistive load: same as above, but it will be a bit lower since it takes into account the output current capability of the output stage into a resistive load.
"RMS power"
Note "RMS volts" has a meaning, which is Root of Mean of Square of voltage, ie \$ \sqrt{ \int v^2 dt } \$.
RMS volts (or amps) have a specific, well defined physical meaning: no matter the signal shape, if you know the RMS value you can calculate power into a resistive load by a simple \$ V^2/R \$. So you can run a heater on 230VDC or 230VAC RMS, you'll get exactly the same power. Note the RMS value of a DC voltage is the DC voltage.
But "RMS power" is misleading. It does not mean \$ \sqrt{ \int p^2 dt } \$ (with p as power). Calculating the RMS value of power would give a number, but it wouldn't mean anything useful. What "RMS power" usually means in audio jargon is "Average apparent power delivered to load of specified impedance over a specific time without overheating." But there is no agreed upon standard about load impedance, duration of test, etc.
Still, "RMS power" is useful. Say an amp has a peak output power of 100W into 8 ohms resistive load, this means its maximum "RMS" power would be 50W. But if the spec sheet says 25W instead, then you know it will deliver a burst of peak power, but not for very long. Maybe the power supply caps or the transformer or the heat sinks were "cost-optimized", and it can output 50W "RMS" for 100ms but not for a minute.
Because transformers, heat sinks, big caps, and chunky power transistors cost money, and money is expensive, you get "music power" which basically means "hey if you want 100W we will deliver that for 1 second but the power supply will give up soon and if you insist the amp will overheat but we can print a nice number on the brochure".
PMPO watts
This is basically marketing bullshit. Take everything that looks like a watt inside the device, add it all up, round up an order of magnitude or two, and you got a number. An intern from marketing will add a zero or two. Basically, any mention of "PMPO" on the box means it contains garbage and the specs are so ridiculous they have to wing it in order to sell it. Engineering-wise, it is about as useful as saying a Tesla can reach Mach 2 if you shoot it into space with a rocket and watch it do atmospheric reentry.
"What is the power of my amplifier supposed to be?"
In a resistive load, that's simple. Maximum current comes at max output voltage, so when you know how much voltage the output transistors drop at max current then you get the peak output voltage and you get output power from that.
With a real loudspeaker, that's a rather complicated question because the impedance curve is wiggly: a "8 ohms" speaker will go from 6 to 16-20 ohms over the frequency range. So for a known output voltage, current and power will depend on speaker impedance and thus on frequency. Besides, these are reactive loads, so current is out of phase with voltage. Resistive loads are easy because as output voltage and current increase, voltage drop over the power transistor decreases. Thus maximum dissipation in the output transistors is easy to calculate, and max current occurs at minimum voltage drop across the transistor.
But when current is out of phase with voltage, output transistor instantaneous dissipation can be much higher. For example if current lags voltage by 90° then a transistor can have the full supply voltage and max output current at the same time, which is much worse. This means Safe Operating Area is an important thing to consider (look at your transistor datasheets).
If I'm correct, maybe a 1kW home stereo system could make sense
No, it does not...
If you want to design an amp, the first thing to do is to stick a 0.1R resistor in series with your speaker, and measure current and voltage with a scope while playing music at your favorite levels. Then calculate average and peak power from the scope measurements. Most likely, if your speakers are reasonably efficient, you'll measure about 1W average and 10-50W peaks. Then you know it isn't necessary to have a kilowatt amp.
Edit: answering your comment...
Curves on the left show collector current labeled Ic(Q2), voltage Vce(Q2), and instantaneous power Pd(Q2) in the upper power transistor with resistive load and inductive load. I set the inductance to a high enough value to add a substantial phase lag.
Ic(Q2) shows two different current curves with inductive load and without. Note the current peak is lower in this example with the inductor, due to extra impedance.
Pd(Q2) shows instantaneous power, for resistive load Ic peak occurs at minimum Vce but for inductive load it does not, and peak power dissipated in the transistor with inductive load is much higher because it is Ic*Vce and the Ic peak no longer occurs at minimum Vce.
Plotting Ic and Vce on top of the datasheet SOA graph shows we're on the edge of what is allowed:
Pd(Q2)
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H: What is the purpose of RC components to ground in this class AB amplifier?
I'm talking about R6/C4. It seems like it would be filtering of some kind as the C4 will have it's voltage at the highest for DC or low frequency input, but I don't really understand the purpose.
AI: What is the purpose of RC components to ground in this class AB
amplifier?
Q1 is a common-emitter stage that supplies all the voltage gain in the circuit. Normally, for a common-emitter stage, its voltage gain would be R7/R6 but, with the capacitor added across R6, the voltage gain (and distortion) will be increased due to the reactance of the capacitor being small in the mid-band of frequencies.
Note for Kyle B: it doesn't result in negative feedback when you add the capacitor - it results in a localized reduction of negative feedback around the Q1 stage. Neither does it improve linearity - it actually reduces linearity and creates more signal distortion.
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H: Is there such a thing as a 'Constant Power Source'?
Academic question: Is there such a thing as an electrical constant power source, which delivers energy at a fixed rate?
Specifically ... there are voltage sources and current sources, which maintain (an approximation to) a reliable fixed voltage/current respectively. A current source will deliver electric charge at a fixed rate, charging a capacitor's voltage linearly; a voltage source will deliver magnetic flux at a fixed rate, 'charging' an inductor's current linearly.
I'm trying to figure out if there is a supply that could deliver energy at a fixed rate - a power source - so its voltage changes to whatever is necessary to deliver a current such that \$VI\$ is a fixed amount.
(I've tried Googling this, but the search is swamped by power supply ads, domestic power advice etc).
Guessing at its properties...
it would treat voltage and current symmetrically! So a fixed power source would charge up a \$1\mu F\$ at the same rate as a \$1\mu H\$ inductor.
it would charge a capacitor's voltage at a rate of \$ V \propto \sqrt{T}\$, because the energy is increasing linearly in time and \$E = ½ CV^2\$.
same with an inductor's current.
any resistor placed across it would dissipate the same heat regardless of the resistance - by definition. Doubling the resistance would increase the voltage by a factor of \$\sqrt{2}\$ and decrease current by the same.
its impedance is ... undefined! Open circuit voltage is infinite (like a current source); closed circuit current is ... also infinite (like a voltage source). Hmm.
For extra points: any other interesting properties? Are they actually a thing? What would they look like?
AI: Academic question: Is there such a thing as an electrical constant
power source, which delivers energy at a fixed rate?
Just consider a boost converter operating at a fixed duty cycle from a fixed DC supply with discontinuous inductor current (DCM) : -
Each time the MOSFET turns on, current ramps up in the inductor and, after a certain length of time, the energy acquired by the inductor is a specific and known value.
When the MOSFET deactivates, that inductor energy is released into the load. If this is done at a constant rate, then a fixed energy value is transferred to the load so-many times per second. That's the same as delivering a fixed output power.
It's probably less easy to see but, a flyback converter does this with a slightly higher level of "perfection" compared to the boost converter above. The boost converter will always deliver a certain residual energy to the load even when the MOSFET is never activated. However, the flyback's output doesn't piggy-back on the input voltage hence, it is a "more ideal" form of power or energy delivery system: -
In the picture above, the flyback converter is also shown to be equivalent to the inverting buck-boost converter. They both deliver a constant power to the output and, that constant power is dependent on duty cycle and input voltage.
The voltage transfer function of an ideal DCM flyback converter with turns ratio 1:1 is this: -
$$\dfrac{V_{OUT}}{V_{IN}} = D\cdot\sqrt{\dfrac{R_L}{2\cdot L_P\cdot F_{SW}}}$$
And, if you rearrange it you find this: -
$$\dfrac{V_{OUT}^2}{R_L} = \dfrac{V_{IN}^2\cdot D^2}{2\cdot L\cdot F_{SW}} = \text{power out}$$
Hence, the power out is proportional to duty cycle squared and input voltage squared. Of course, if \$R_L\$ rises too high, the output voltage of a practical design reaches a value where the output capacitor breaks down but, within practical limits, the flyback converter operated at a fixed duty cycle and fixed input voltage, is a constant power output device.
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H: Why doesn't the fan work when the LED is connected in series with it?
In this picture, I have connected a fan and an LED to a battery. The LED works when I switch the circuit ON but the fan doesn't. If I connect the fan directly, things work.
Fan not working
Fan works
What could be the reason? Is battery power the issue? To be honest, I don't think that an LED takes that much power that the fan doesn't move even a bit.
AI: Figure 1. Extract from 288 Amazing Science Station user manual.
A little care has to be taken when using LEDs to limit the current to a safe value. Connecting, for example, a 2 V green LED across a 3 V battery may destroy it. Limiting the current through the LED is usually done by adding a series resistor to the LED. I suspect that the 288 Amazing Science Station has current limiting resistors "secretly" added in series with each of the LEDs on the station to prevent accidental destruction. For a 3 V supply this might be a 100 Ω resistor which will drop 1 V at 10 mA - plenty of current to make the LED glow brightly.
When you connect the LED in series with the motor the maximum current that can flow is 10 mA (or whatever the LED current is set at) and the resistance of the motor will reduce this a little. There just isn't enough current to run the motor.
Note that in Experiment 3 the fan and the LED are connected in parallel with the battery rather than in series as you have done. I can't make out what the additional component is but they seem to have added additional resistance to the LED for this circuit.
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H: MIC29712 in constant current mode
[EDITED] I ALSO ADDED THE LAMP TO MY DESIGN WHICH I FORGOT
I am trying to use a MIC29712 in constant current mode to draw 4A for my lamp. The power supply is 15V and can give up to 8A.
Here is the MIC29712 datasheet.
What I think I can do is this:
Since the ADJ Reference is to GND, I will tie the Vout to ADJ, so the Vout will be 1.240V (Vref), hence the V1 will be 1.240.
simulate this circuit – Schematic created using CircuitLab
Does this look correct? I ordered the parts and waiting for them to arrive.
I am asking becouse I could not find any constant current schematic for this regulator. I am mainly worried about the ADJ pin. Is it really referenced to GND (I think this is the case,) or is it referenced to Vout?
AI: Part 1 - before OP edited the question
This answer was given before the OP changed his mind and added a lamp in series with the input supply to the regulator.
I ordered the parts and waiting for them to arrive.
Did you order a sizable heatsink for the excessive power dissipation of ~45 watts?
With 1.24 volts at the output, the device will drop 12 volts minus 1.24 volts between input and output. That's 10.76 volts. Given that you are using a load resistor that takes 4 amps, that current also flows through the regulator hence, the power dissipated in the regulator is: -
$$P_D = 10.76 \text{ volts} \times 4\text{ amps} = 43.04 \text{ watts}$$
There will also be about 100 mA flowing into the ground pin and that means another 1.2 watts of power dissipation.
Given that the device has a thermal impedance to case of 2 °C/watt and a maximum junction temperature of 125 °C, you need to be looking at a pretty substantial heatsink. At a local ambient temperature of (say) 25 °C and an "infinite heatsink", 45 watts will raise the internal junction to about 115 °C. Hence you might have to find a heatsink that has really good thermal characteristics to avoid exceeding the 125 °C junction temperature.
I estimate that the thermal resistance of your heatsink needs to be less than 0.22 °C per watt from the case mount point to an external ambient temperature of 25 °C maximum. Then you have the problem of keeping the local ambient at no more than 25 °C. Given that the heat dissipated will likely cause a localized rise in ambient, I think you will need some forced air cooling to make this work without meltdown.
Part 2 - after the OP edited the question
Now that the 2.5 Ω lamp has been added, it's clear that there is another problem; 2.5 Ω and 4 amps requires 10 volts leaving 2 volts at the regulator input. Given that the enable input needs to be guaranteed to be above 2.4 volts to activate the device, there is that to fix.
Personally speaking, and having looked through the data sheet, I don't think this device will operate at such a low input voltage (irrespective of the enable input problem) when trying to deliver 4 amps to the output. The data sheet isn't all that clear. But, what can be gleaned are these things: -
In purple is the implications that the device won't really work that well when the input supply is below 3.8 volts.
In red is a table that shows that the device manufacturer aren't considering this device capable of delivering a regulated output lower than 2.85 volts.
The two appear to be related; in most areas of the data sheet, specifications are given with the assumption that the input voltage is at least 1 volt higher than the output voltage so, if the minimum output voltage is 2.85 volts then an input voltage of 3.85 volts is needed. I am even surer now that expecting anything sensible with the input supply at only 2 volts, is wishful thinking and naive.
Part 3 - might still have a heat problem
But, there's also the problem of running the lamp at say 2 amps (should the objective be to control the lamp's brightness). With 2 amps through the lamp, the voltage seen at the input to the regulator will be at least 7 volts (and may be a volt more due to the non-linear filament resistance seen in incandescent lamps).
Now, the power dissipation in the regulator will be about 7 volts x 2 amps = 14 watts. This power and an infinite heatsink will be fine but, not using a heatsink means that the device's thermal resistance of 2 °C / watt will raise the junction temperature to 53 °C (at an ambient of 25 °C) and that will surely cause the local ambient to rise until eventually the junction hits 125 °C. In other words, you are still probably going to need a heatsink or small fan to keep localized ambient to reasonable values.
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H: What is the meaning of inferred absolute zero temperature? Why does the resistance temperature graph it is obtained from behave that way?
Studying basic direct current circuits, I've come across the term inferred absolute zero temperature on computation of resistances that change due to temperature. Based on my readings, I understand it to be a predicted value as the word "inferred" means. It predicts the value of the absolute zero temperature based on using a segment of the graph that seems to have a linear slope through extending that line to where the temperature is zero.
Although that is my understanding, I decided to ask it in StackExchange because I've searched for an explicit definition of this term and no source has ever defined it in such a way. You have to go through a lesson of resistances changing by temperature, find the terminology and read the paragraph to get it. I mean, for such a common lesson I can't believe that I couldn't find any source that just defines the terminology in a friendly manner or even have a Wikipedia article that defines it as there it gets a little more clear and precise.
Can anyone confirm this concept for me and define it more explicitly? By explicity I mean that say that the term means this, and not focus on some other topic where I have to read and it is only implied and that the terminology is just put there.
I got the image from this website.
https://enjoy-electrical.blogspot.com/2014/09/inferred-absolute-zero-temperature.html
Is it also right to say that, resistance and temperature in actual testing do not have a clearly defined relationship? Then their relationship also looks like a curve where a segment in the middle seems to have a linear behavior, which is where the concept of getting the inferred absolute zero temperature came from? Why does the resistance temperature curve behave like that? It's just that for me it kind of feels convenient that the middle segment has a linear behavior.
AI: I think you are making much ado about nothing. As stated in the website, they are approximating the behavior of the device with a straight line to make it easier to convert resistance into temperature. As a result, there is an error at the extremes of temperature because the actual resistance vs temperature departs from the straight line. They show that the approximation has an error when carried out to absolute zero which is expected since it is only an approximation. They use the word "infer" because they are using an approximation beyond its useful range. This is often done with a variety of sensors which behave linearly over only a portion of their range and there is nothing mysterious about it. As to why not all sensors are not linear is a subject for physics, not engineering.
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H: Replace LED chip with reversed polarity pin-out
I want to upgrade the LED chips of a projector.
Red: PT-120-R-C11 --> PT‐120‐RAX‐L15
Green: PT-120-G-C11 --> PT‐120‐G‐L11
Blue: PT-120-B-C11 --> PT‐120‐B‐L11
But the red LED chip (PT‐120‐RAX‐L15) has reversed polarity pin-out.
Swapping black and red cables should solve this problem, am I right?
Do I also have to swap these cables?
Another picture:
© cine4home
AI: YES, reverse the connector.
I believe it serves as a reminder to adjust the color icc profile for the projector for color correction. (opinion on reason)
The obsolete RED part is a 630 nm deeper red but only 2000 lumen.
The new RED is 620 nm more amber red but is 3000 lumen. Blue and green meanwhile are 5000 lumens and the low efficacy of red is common.
This datasheet is short on engineering reasons but has the usually marketing BS about mercury-free bulbs ( yet fails to mention the gallium arsenide content.)
So my opinion is perhaps what I might have done to alert field service people to the issue but hide the obvious reasons.
There is no damage to the reverse connection, but it won't work. and if replaced with the wrong colour, it won't work well either. So do two wrongs make it right?
The normal rated current in the forward mode is 27 Amps.
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H: Finite input impedance of voltmeters
I have a question regarding the impedance of voltmeters. I learned that a voltmeter can be modeled as follows, where \$R1\$ is the input impedance:
simulate this circuit – Schematic created using CircuitLab
I noticed that many multimeters have a \$10\,\mathrm{M}\Omega\$ input impedance on the volt range (some even more) and wondered where the (often well defined) input impedance exactly comes from. Is it coming from the ADC?
AI: Is it coming from the ADC?
The ADC never directly connects to the probe input wires directly. It will use a buffer amplifier as intermediary and, there will also be a variable attenuator to select the different voltage ranges. The variable attenuator in conjunction with the buffer amplifier will produce the 10 MΩ input resistance seen at the probe terminals.
If it helps, here is the input selection stage of an old analogue meter (the AVO): -
Stolen acquired from this site
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H: Power consumption in voltage multipliers
Cockcroft–Walton generators and Marx generators are voltage multipliers. These circuits contain various diodes, capacitors and resistors that charge the capacitors in parallel and then put them in series to multiply the voltage. How much do these circuits consume?
I saw videos on youtube that create 180kV sparks using a 9V battery.
EDITED: here's schematic Cockcroft–Walton generator:
Marx generator:
AI: What you have to realize is that the output voltage can be very high, but the output current is usually very low.
High voltage doesn't mean that you get a lot of power out of them.
They don't "consume" much power at all. A typical Cockcroft-Walton multiplier may put out only a few milliamperes at several thousand volts.
Say you have a Cockcroft-Walton multiplier that puts out 3000VDC at 10 milliamperes. You will have to supply (3000V* 0.01A = 30W) at the low voltage side.
You wouldn't normally need that much current, though.
You want the high voltage to drive a Marx generator.
A Marx generator slowly charges its capacitors through the resistors. When the voltage gets high enough, the capacitors discharge through the spark gaps. All of the energy is released at once, then the process starts again.
More current means a faster recharge of the capacitors in the Marx generator - less current just means that you get individual discharges (or discharges at longer intervals) rather than a constant "buzz" of Marx discharges.
Since you saw an example of a 9 volt battery driving the whole thing, it can't be really consuming much power - 9 volt batteries don't "do" high power. If you push one really hard, you might get a couple of hundred milliampers out of it - at nine volts, maybe a couple of watts for a little while.
Your examples don't match what you would have seen in the videos, though.
The Marx generator shown would need about 45000 volts DC input to get to 180000 volts output.
The Cockcroft-Walton multiplier shown has a DC output far too low to drive the shown Marx generator to 180000 volts.
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H: Variable is not changing when one button is pressed. (String)
I couldn't decide whether this belongs in C++ or Arduino. But I guess it belongs here since it's Arduino programming.
Anyways, I'm having a hard time solving this issue. I have 4 buttons, 2 for up/down, 1 for select, and 1 for paging (Switch pages). But the down button is having issues. Here the code:
#include <LiquidCrystal.h>
LiquidCrystal lcd(7, 8, 9, 10, 11, 12);
int buttonApin = 6;
int buttonBpin = 5;
int buttonCpin = 4;
int buttonDpin = 3;
int page = 1;
String option1 = "Option A";
String option2 = "Option B";
String option3 = "Option C";
String option4 = "Option D";
String current = "";
String previous = "";
String arrow = "> ";
String selected = arrow + current;
void layoutMenu() {
lcd.clear();
String current = option1;
String selected = arrow + current;
String previous = option2;
lcd.print(selected);
lcd.setCursor(0, 1);
lcd.print(previous);
while (page == 1) {
if (digitalRead(buttonApin) == LOW) {
String current = option1;
String selected = arrow + current;
String previous = option2;
lcd.clear();
lcd.setCursor(0, 0);
lcd.print(selected);
lcd.setCursor(0, 1);
lcd.print(previous);
}
if (digitalRead(buttonBpin) == LOW) {
String current = option2;
String selected = arrow + current;
String previous = option1;
lcd.clear();
lcd.setCursor(0, 1);
lcd.print(selected);
lcd.setCursor(0, 0);
lcd.print(previous);
}
if (digitalRead(buttonCpin) == LOW) {
Serial.print(current); // it was always OptionA when printed to serial.
if (current == option1) {
lcd.clear();
lcd.setCursor(0, 0);
lcd.print("Selected A");
}
if (current == option2) {
lcd.clear();
lcd.setCursor(0, 0);
lcd.print("Selected B.");
}
}
}
}
void setup() {
lcd.begin(16, 2);
lcd.print(" Yanny9000 ");
Serial.begin(9600);
pinMode(buttonApin, INPUT_PULLUP);
pinMode(buttonBpin, INPUT_PULLUP);
pinMode(buttonCpin, INPUT_PULLUP);
pinMode(buttonDpin, INPUT_PULLUP);
delay(5000);
lcd.clear();
String current = option1;
String selected = arrow + current;
String previous = option2;
lcd.print(selected);
lcd.setCursor(0, 1);
lcd.print(previous);
layoutMenu();
}
void loop() {
// Redesigned the menu. No need for loop().
}
Here's the code I'm having problems with:
if (digitalRead(buttonApin) == LOW) {
String current = option1;
String selected = arrow + current;
String previous = option2;
lcd.clear();
lcd.setCursor(0, 0);
lcd.print(selected);
lcd.setCursor(0, 1);
lcd.print(previous);
}
if (digitalRead(buttonBpin) == LOW) {
String current = option2;
String selected = arrow + current;
String previous = option1;
lcd.clear();
lcd.setCursor(0, 1);
lcd.print(selected);
lcd.setCursor(0, 0);
lcd.print(previous);
}
A is up, B is down. When the up button is pressed, it selects an option named "Option A" and changes the vars: Current (Which contains the current option selected), Selected (Displays option selected), and Previous (Previous Option).
When B is pressed, it selected an option named "Option B" and changes Selected and Previous.
Issue: Current does not change when B is pressed.
Current is used by C button (Select) to select options. For some odd reason it does not change. I been trying to solve this problem for ages. Can anyone help me? I'm using an mega 2560.
AI: That's not Arduino but basic C++ question.
You are creating a new local string variable with the same name than your global variable, and changing the local variable. The global variable will not change.
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H: No voltage induced in secondary coil of transformer with 5V 1.5 kHz voltage from Wien bridge oscillator in primary coil
I have built a Wien bridge oscillator which outputs 5V at 1.5 kHz. When I hook up this output to the primary coil of a transformer I can't measure any voltage in the secondary coil.
Schematic:
simulate this circuit – Schematic created using CircuitLab
I thought that the problem might be a lack of current and therefore power in the primary coil. (How would I change that?)
How can I get a voltage induced in the secondary coil with this oscillator (or possibly another oscillator design?)
AI: The problem with your circuit is that the oscillator is not starting reliably when your load is your transformer. When the oscillator does not start, the output voltage is in the uV range.
One design goal of a Wien bridge oscillator is that it start reliably. Often there is another design goal which is that the amplifying element should not saturate. These competing design goals are often met by adding circuitry that stabilizes the amplitude of the oscillations.
When the oscillations are weak, the amplitude stabilization circuitry provides high positive feedback to increase the amplitude. When the oscillations are at the desired level, the positive feedback is reduced so that the overall closed loop gain is 1. If the oscillations are above the desired level, the positive feedback is reduced further, so that the oscillations dampen.
Components that are used for amplitude stabilization include small incandescent light bulbs, thermistors, back to back diodes, and JFETs.
Incandescent light bulbs provide amplitude stabilization through the varying resistance they have with temperature. If oscillation is weak, the filament is cold, and the resistance is low. If the oscillation is strong, the current heats the filament, which in turn increases its resistance.
One possible solution to your oscillator not starting when loaded with the transformer is to increase the value of R2 (or alternatively decrease the value of R1). You may double it, or even more. The down side is that this will surely cause the op amp to saturate if oscillations start. This will cause the waveform to deviate considerably from a pure sine wave. Saturation may also alter the frequency of oscillation somewhat. You may or may not care about these effects.
Another possible solution is to increase the number of turns on your transformer primary. This may not work without also increasing the value of R2 (or decreasing that of R1). When a transformer secondary only has a volt-meter across it, the transformer acts essentially like an inductor. The inductance seen at the primary is too low, which causes the op-amp to be heavily loaded, and this decreases the feedback. That in turn causes the oscillator not to start. Increasing the number of turns on your transformer will increase the primary inductance, and that is probably a good thing even if it is insufficient to start the oscillator.
Another solution is to use back to back diodes to achieve amplitude stabilization. I have provided a circuit here:
simulate this circuit – Schematic created using CircuitLab
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H: How to select the correct CM choke?
Say I want to reduce noise on the data line for WS2812B led strips.
Is it a viable option to use CM chokes to reduce this noise?
According to the docs, WS2812B data speed is 800Kbs. Which I guess translates to 800Khz on the data lines.
I read cm chokes can reduce noise. But I am not sure which ones I should use.
Should I get one which has a resonant frequency of 800Khz? Or will that then have the exact opposite effect of what I am trying to achieve?
AI: There are several things to consider here.
First of all, the applicable frequency. It's not depending on the data rate as you think, but rather it's depending on the signal rise and fall times. Usually this frequency is at least 10 times higher than the base data rate dependent clock frequency.
I'm not sure if CM chokes have a specified resonant frequency. They are intentionally lossy components like ferrite beads. You need to select a part that has as high an impedance as possible within the frequency region of operation.
As mentioned in the comments, beware that the data signal is indeed differential or the signal integrity will suffer severely.
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H: Trying to understand push-pull output mode and open-drain output mode in a MCU's GPIO pin
I have a MCU (STM32H7xx) whose GPIOs, when configured as output, can be configured either as open-drain or push-pull. Before seeing this, I though that open-drain meant hihg-Z, however I don't see how that works just reading the STMicro documents. Let me explain:
This is the GPIO schematics:
And this is the explanation I'm reading (attached as image to not lose the formatting):
Now, with regards to the push-pull output mode, my understanding is that the pull up/down resistors are disabled, the mosfets are used to output VDD or VSS (which is ground). I'm happy with that.
What I don't understand is what is the goal of open-drain mode, let me explain: the doc. says that you need to enable the pull-up resistor to work in this mode. In that case, I understand that when you enable the NMOS, VDD is grounded through the pull-up resistor, therefore the output pin is driving a 0. However, if the NMOS is disabled, it will be driving VDD through the pull-up resistor.
Questions: Is the above correct? If so, what is the point of this mode? It seems to achieve exactly the same behaviour as the push-pull one. How do you get a high-Z output?
Sources:
stm32h7 43/53 reference manual - see 11.3.10
STM32 GPIO configuration - see 3.3.2
AI: First misconception : Push-pull mode does not mean the weak pull-up and pull-down resistors are disabled. On some MCUs they can be, but this MCU has separate control for resistors and output mode. The resistors can be enabled even in push-pull mode, but it just makes no sense because the strong transistors can push or pull much stronger and will override the weak resistors.
The push-pull mode means the output is either strongly pulled low to VSS, or strongly pulled high to VDD.
The open drain mode simply stops driving the high FET, so either the pin is strongly pulled low to VSS, or left floating high-Z. Depending on internal pull-up resistors or external ones, that is then responsible to pull the data high. If you want high-Z, use open-drain mode so it won't push high, or set the pin as input and it won't push high or pull low.
Open drain mode is compatible with other items that are open-drain. In the case of I2C bus for example, the IO pin must be in open-drain mode, and what is even more handy is the fact that while it is a 3.3V MCU, the open-drain mode allows (some pins, not all) to be pulled up to 5V, to for example allow the 3.3V MCU to talk directly on a 5V I2C bus with other 5V I2C devices.
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H: Over voltage and reverse polarity protection
I was trying to make an over voltage and reverse polarity protection circuit and create a PCB that I can easily use in some of my projects.
My level of electronics it's not so good, and that's why I'm here.
I thought one circuit, but as long as I know it's not so good:
simulate this circuit – Schematic created using CircuitLab
There's a bunch of circuits arround the internet that involves transistors, but I don't know which one I would pick (or even an IC, as it was discussed previously here, but I couldn't understand which configuration should I use).
Here's my requirements:
It should output around 5V (if the input it's too low it can output less voltage, but if it's greater it should be limited). I want to use it with 74LS ICs, so the input voltage should be 5V, but it can go from 4.75V to 5.25V. The current used it's, at most, 1.5A.
The input voltage can go from +40V to -40V.
I'm using a workbench power supply, so no need to over current protection (it's already integrated).
It should be as simple as possible (~5 components). I don't want to solder >25 components.
I don't care if something blow up. As long as the external circuit stays safe, it's OK for me.
Is there an IC or circuit that suits my needs?
Please go nice on me, this isn't my specialty. Thanks!
AI: A diode reverse voltage protection in a 5 volt circuit is not ideal. Even if you use a good Schottky you are still going to loose ~ 0.5V and the reverse current in some schottky's may still damage your circuit.
Overvoltage protection with a Zener is also not ideal. You need some margin so you must use a 5,6V zener, if not it will conduct "nominally". A voltage at 5.6V may exceed your ratings. (And you need a fuse - if not the zener will overheat and blow up)
What you want is an efuse. As your voltage is similar to USB there are many types, so I am sure you can find one that fits your need. One example is NIS6150. Your biggest challenge might actually be to find one that is solderable by hand. Most of these components are tiny.
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H: Circuit design - How to fix power supply overheating and reduced performance?
Goal
I'm trying to make a IOT radiator valve so I can manage the heating of every room in my house automatically with a system I'm building.
Context
I've pulled apart a cheap "smart" radiator valve to salvage its 3.0V (25 Ohm) motorised worm-screw component, that adjusts the flow of water through the actual TRV valve that is on the radiator (to regulate its heat).
I'm using a 3.3V battery power supply (0.5 Amps) to power the ESP-01S control board I'm using along with the wormscrew motor.
I've allocated two GPIO pins on the ESP-01S board (represented as A and B below) for controlling the direction of the motor, since it is not a servo. I've coded the board so that the pins will never be both +ve at the same time (to prevent the following control circuit from short circuiting).
Above is an image of the circuit diagram I've developed to handle the polarity switching that the motor requires, in order for the flow of water through the TRV valve to be increased/decreased on demand. When the wormscrew is fully retracted, the valve is closed, but when it is fully extended, the valve is fully open.
When A is "on" (+ve), B will be "off", resulting in A's NPN transistor (Aₜ) to be on, while B's NPN transistor (Bₜ) stays off. This allows current to flow from A to Ground through the motor making it spin in one direction and vice-versa when the polarity of A and B are swapped.
The transistors are 'S8050'.
The problem
The motor spins fine when connected straight to +ve pin and ground. However in my circuit, it runs much slower and the power supply for the board and motor gets so hot that it almost blows out and it's scalding to touch. Any ideas to why this is happening and how I can stop it would be much appreciated!
I've played around putting some resistors on the base pins of the transistors to reduce the current going through them, but that just makes the motor stop all together. Is the resistance of the circuit too high and so it is pulling too much power, or is the opposite true?
I am a complete noob when it comes to electronics, but I don't want that to stop me making this project for my Dad, so any advice would be greatly received! If I have missed any crucial information, please feel free to ask and I will try my best to supply it.
AI: Ohm’s Law tells you it draws 3/0.25=12 A to start and transistor is far too small in power rating to handle cool operation. The current gain drops towards 10% of hFE as a switch so your base current is under-driven and Rce is 1.5 Ohms
Try <50mOhm Nch 1V=Vth FET instead which is necessary.
Your battery won’t last long either.
If Base =0 Ohms you get slightly lower Vce(sat) and slightly more RPM. It will also slow down with a battery charge. With a CMOS driver ~ 50 Ohm @3V that's (3-0.7=2.3) >> 2.3V/50= 46mA then you can drive motor best with what you have and run cooler than what you had. This is OK for the short time you expect.
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H: Biasing an inverting Op-Amp using a voltage regulator
When biasing an inverting op-amp input using a voltage regulator (say, L7805), at the non-inverting pin, can we use the same regulator to power other ICs or should we only use it on the op-amp?
Also, in the case of an inverting amplifier, does the op-amp needs an input resistor at the non-inverting pin or can directly connect the regulator output and the op-amp?
Thanks in advance.
AI: I would say that this is not great practice, but is probably ok for non-critical applications. There will be applications where its not ok to do that, but it depends in what configuration you are using your opamp, and how clean the signal needs to be.
If you are using it in any sort of negative feedback situation (like an inverting amplifier), noise on the non-inverting input will lead to noise on the output. Connecting the non-inverting input to power other devices will likely introduce such noise. If you are doing this, you should filter that input, like this:
simulate this circuit – Schematic created using CircuitLab
The values are a bit arbitrary, and you might not need two capacitors, but I might do it just to be safe. If you have a specific opamp in mind, you should check the input current leakage specification, and you may be able to increase R1 to add more filtering.
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H: Does my IR LED need a ground?
I have a circuit that connects an IR led to my Arduino. The positive leg goes to +5v and the negative goes (via a resistor) to a digital pin. However, I have since seen some pictures of IR circuits that also connect the LED to ground. Do I need to do that, and if so, how do I do it? (So far, I've built 2 of these circuits and nothing has happened in the month since.)
AI: You can have the GPIO drive low to light the LED OR drive high to light the LED, depending on how you connect the LED.
Some I/O technologies can drive a low with more current, so connecting the LED to VCC (+5V) was preferred. This is generally not the case for modern technologies, the low and high current are usually about the same. Some older designers still have habits from the old days and tend to connect LEDs to VCC.
If you need a transistor driver, a low side switch is often easier to make. For a low side switch, the LED is connected to VCC.
simulate this circuit – Schematic created using CircuitLab
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H: How does my 802.11 a/b/g/n wireless card determine the modulation to use?
I was thinking yesterday about how the 802.11 a/b/g/n standards all use different modulation schemes, but my wireless card manages to detect frames from networks that may use any of those modulations. So, when my card detects a signal containing a frame from my neighbor's 802.11b network, how does it know to demodulate it with, for example, DSSS and not OFDM? Does it just attempt all modulations? Does it use some kind of signal sensing?
AI: DSSS vs OFDM: these are different standards (a/g/n vs b, which is obsolete). This situation where you'd need to detect the differences never arises: a beacon from an access point is either of these standards, and that defines what the whole network does. Since the channels for these different modes don't even match, there can't be interoperability. A card that supports both types of networks simply has to listen for both types.
Does it use some kind of signal sensing?
Err, everything a receiver does is signal sensing...
Still, within a modern WLAN, the stations have freedom to use different modulations within OFDM: BPSK, QPSK or QAMs. Atop of that, there's a variety of different forward error correction rates that might be used.
That's decided based on the channel quality estimates that both stations communicate.
The modulation used in a frame is noted in the header symbols at the beginning of the frame, so that the receiver always knows how to demodulate the payload.
The header itself uses a fixed modulation (BPSK).
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H: How to calculate fuse i2t rating when current is known?
How do I convert the fuse current rating to the i2t value? For example, in this circuit simulator, when adding a fuse, the only values that can be specified are the i2t value and resistance. The fuse I am attempting to simulate says 32mA.
AI: You simply get these parameters from the fuse data sheet or compute from the curves.
Here I gave you some fuses to blow and sliders to increase the surge load.
Now you enter 0.000047 for i2t and compute R from 3200mV/32mA = 100 then add your load and watch the fuse get hot. Time sample around 500us for real-time .
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H: Bench Power Supply for ESC and KV on motor
I have a bench power supply it can provide ~10 amps and ~30 volts. Working on an autonomous car project I have an ESC to manage a brush-less motor.
Two questions I cannot seem to wrap my head around:
The ESC has a 35A requirements. With the bench power supply I have I can only go to 10A. If I am powering the ESC with this what are the risks? Over heating the supply? I set the volts to 13v based on the 35A specs.
The motor stated 3600kv or 36v (if I understand correctly). Do I need to be concerned with this? The ESC manages that so as long as I manage the 35A the motor is fine. Why do I need to know this metric?
AI: The ESC has a 35A requirements. With the bench power supply I have I can only go to 10A. If I am powering the ESC with this what are the risks? Over heating the supply?
The 35A specification means that the ESC can safely draw 35A from the supply. The amount it actually draws will depend on the motor you drive with it, and the load you put on that motor.
I set the volts to 13v based on the 35A specs.
That's somewhat nonsensical. Current depends on the load and the voltage, and you haven't tied those together yet. You can have 35A at any voltage if you get to choose the load.
The motor stated 3600kv or 36v (if I understand correctly). Do I need to be concerned with this? The ESC manages that so as long as I manage the 35A the motor is fine. Why do I need to know this metric?
If you're deriving the 36V from the Kv = 3600 -- no. Model airplane motors are rated in with a voltage to speed constant called Kv, which is in units of RPM/volt. For a motor that's driven with \$v\$ volts p-p at the coils, it will turn \$v \cdot K_v\$ RPM. That's pretty close to the speed it'll turn with \$v\$ applied to the battery terminals of the ESC and the ESC at full throttle.
Given that Kv rating, the motor is probably designed to run off of a single Li-Ion cell. That's consistent with a propeller speed of a bit more than 10,000 RPM, which is typical. 12V with that Kv would be over 40,000 RPM -- I wouldn't attempt that without a scatter shield, or at least without making sure that there wasn't anything in the path of the debris when the motor explodes from centrifugal effects.
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H: Intuitive way of "visualizing" how the energy is stored in an inductor?
I've been trying to more or less understand intuitively how energy is stored in an inductor, but I don't seem to get anywhere decent.
In a capacitor I understand, I believe: an external battery pushes electrons and holes (going with the electron/hole theory, even though it's only electrons) to opposite sides and they remain on the 2 plates of the capacitor, forced to be there by the battery. Where we disconnect the battery and let the capacitor discharge through a resistor, there is already a difference in potential in the 2 terminals of the capacitor, created by many +s in one side and many -s in the other side. So as soon as there is a path for the current to go, it will go at full speed and start decreasing, since the difference in potential is also decreasing as the current goes (the number of +s in one side is starting to equal the number of -s in the other side), until Nature goes back to the equilibrium state of same potential in both terminals and we are in current of 0 A.
I'd imagine it as something like this:
But how can I "visualize" this in an inductor? Maybe I don't have the necessary concepts clean enough in my head, so I can't see it clearly as maybe I'm supposed to?
Any help is appreciated!
EDIT: Just to say, the idea here is to understand intuitively, so I don't need to memorize the formulas to know what's gonna happen in some part of a circuit.
AI: I'm going to start by discussing capacitors then move to inductors. I'll avoid complex equations (to save myself as well as this discussion.) I may even discuss duals between capacitance and inductance and how that relates also to physical ideas such as mass, force, velocity, and momentum. (That will treat them in isolation with the purpose of providing some intuition, only.) But when I get some time for that. I think it is important to start with the area where we each may feel more comfortable.
The following will be more qualitative and will discuss ideas that aren't found in textbooks. I'm doing that because it may help you visualize, despite the risks of taking a non-standard approach.
Capacitors
For capacitors, you probably know already that the plate area (assuming both plates have the same area), the distance separating the plates, and the physical matter (inserted to replace a vacuum or air) used in between the plates relate directly to its capacitance value. These ideas of plate area, plate separation, and inserted medium (if any) aren't hard to grasp. The formula is kind of easy, too:
$$C=\epsilon_{_0}\frac{A}{d}$$
It's important that you think of the above as implying a vacuum in between the plates. It's possible, of course, to stick bits of matter in between them, too. And who knows? Something different might happen. It turns out that for certain types of matter, the above equation does appear to require an added factor, the relative permittivity. This is just a fudge-factor that must either be 1 or larger than 1 and tells you something about the inserted material's ability to respond to a charge difference between the plates.
The new equation with its fancy new fudge-factor is:
$$C=\epsilon_{_r}\epsilon_{_0}\frac{A}{d}$$
And it works great, if you know \$\epsilon_{_r}\$. If you fill in the entire volume in between the plates (no gaps, etc) then you can just use the material's \$\epsilon_{_r}\$. Of course, if you do something screwy like filling oddball portions of the volume then matters aren't so simple. (But people mostly don't do things like that. They try and keep it simple and cheap to make.)
What does the fudge factor do? What does it mean? What does it relate to, physically?
Well, the charge is separated by a gap that keeps them from crossing over and neutralizing themselves. They want to, but can't. This sets up a field in-between the plates, with the field lines perpendicular to the plate surface and passing straight through from one plate to the other. There's also a field elsewhere too, surrounding the capacitor in the air around it and those field lines point all over the place, depending on where you are (and/or the time you are there for changing situations), as rotating vectors as you move through that space. But the round-trip path integral requires that the lines in between the plates point directly from one plate to the other and that is the important point that matters to a dielectric medium that may be inserted into that volume.
When something is inserted there, the matter within the material may be able to form dipoles. Molecules are a little more complicated athan a dipole. I have to admit that using the idea of an idealized dipole is a bit of a simplification. But the simplification is pretty close and good enough for most uses.
Now, before I go further, let's go back to the case where there's only a vacuum or air in between the plates. When a capacitor is charged, there is a certain amount of usable energy stored somewhere. Where? Well, in the vacuum in between the plates. I want you to firmly fix that in your mind. It's not stored in the plates, where there is no electric field at all (if there were, electrons would be moving around and we know that does not happen.) Instead, it can be reasonably visualized as being stored in the vacuum in between the plates.
Now let's return to the dipole idea, where we've inserted some kind of magic material that can form dipoles when exposed to the electric field lines between the plates. Some molecules or groups of molecules in a dielectric can rotate and align themselves with the electric field lines. It's a response to the field. In rotating into alignment, these dipoles can be thought of as forming a kind of "short-circuit" or "shortcut" that "bridge-over" (if you like) some of the vacuum. Vacuum, where the energy is stored, also has resistance let's say. And the molecules form a short circuit that bypasses this vacuum resistance and reduces the effective gap distance.
Now, according to the equations above, if the distance between the plates is less then the capacitance value is increased, other things being the same. So if you accept this way of thinking about it, it can explain why a dielectric increases capacitance.
Some materials won't polarize and form dipoles, though. So those are not considered to be dielectrics. And nothing I know of forms dipoles oppositely directed against the electric field line direction, so unless someone knows something I'm missing we cannot reduce the capacitance of a vacuum-gap capacitor by inserting an anti-dielectric.
From this, it's not so much about the little charges themselves but about fields. The fields that exist when there is a charge difference between the plates. Also, it's useful to imagine here that energy cannot be stored in matter but instead only in vacuum. And that if you fill that vacuum up with something that can form dipoles, then there is less vacuum within which energy can be stored -- or, in effect, a smaller effective gap.
So there is a gap you can measure with a tape measure. That's one number. But there's also a gap you cannot measure directly, which is the effective gap within which energy is stored. The value of \$\epsilon_{_r}\$ is just a quantity telling you by how much the measurable gap is shortened in order to get the remaining true vacuum gap, which is the only gap a capacitor cares about.
Note that this isn't strictly true, either. It's a simplification. Matter that rotates into alignment can rotate back out of alignment, too. And it takes some energy to achieve rotation and releases some when the field lines are diminishing and the molecules can "unwind" a bit. For capacitors, this effect can result in some heating as molecules moving about is pretty much the definition of heat energy. And if you knock them around by oscillating the field then some of that will wind up as extra vibrational energy in the capacitor. Also, there's a question about how much "memory" these molecules might have -- perhaps they don't completely go back to where they were before the capacitor was charged, once discharged. They may "remember" something. But with most capacitors this effect is pretty minor and so capacitors usually are not considered to have hysteresis.
(This hysteresis detail is not true for core materials used in inductors where hysteresis is an important detail you often cannot ignore.)
There are more effects (minor for the most part), such as electric field fringing at the edges of a capacitor's physical design. But they aren't usually important.
Inductors
For inductors, you probably know already that the coil of wire and the physical matter (again inserted to replace a vacuum or air) that the wire coil surrounds relate directly to its inductance value. Here, things are a little bit more complicated to grasp. The equivalent to plate separation, but for an inductor, is the magnetic path length.
The problem with inductors is that there aren't any monopole magnetic charges. So you cannot just stick magnetic charges on some surface. So, here, you cannot have these magnetic field lines between two plates.
With the capacitor's case, I mentioned that there is a path around the outside of the capacitor, too, but that for all intents we can just focus on the field lines between the plates and ignore the rest. But with an inductor, we do have to focus on the entire magnetic path and not just a small segment of it.
This means that the simplest inductor is a toroid. That's because the entire magnetic path is a simple circle -- the average circumference around the toroid. (That \$2\pi\,r\$ thing.)
(Okay, let me take that back. The simplest inductor is a solenoid that extends to infinity. But since infinitely long inductors are hard to come by and short ones add some complexity I don't want to deal with here, it's back to that toroid.)
The toroid formula here is pretty simple, too:
$$L=\mu_{_0}N^2\frac{A}{d}$$
Well, \$d=2\pi\,r\$ so I really should have written this:
$$L=\mu_{_0}N^2\frac{A}{2\pi\,r}$$
That looks worse, but conceptually it is not. I've replaced \$d\$ with the circumference which is \$2\pi\,r\$, but that's not a conceptual problem of any kind. \$A\$ is just the cross-sectional area of the toroid. Then there is the added \$N^2\$ factor. You can probably imagine why the number of loops would have an effect. However, the reason the value is squared is a little beyond the scope of where I want to go. So let's just call that a slight added complexity. But it's not that much.
Once again, though, keep in mind that the energy of an inductor is not stored in the wire. It's stored in the vacuum volume within which the magnetic field resides. Here, that's the interior of the toroid.
So, what happens when a material with permeability is inserted into the vacuum volume? Well, in certain cases things "line up" again, just like those dipoles before. In this case, it's more complicated. So the term domain is used to represent collections of small bits of matter that can align themselves with a magnetic field. Electrons orbiting individual atoms also can line up (and they do to some degree.) But suffice it that the basic idea remains. Some magnetic domains line up with the magnetic field lines and form "shortcuts" through the vacuum along the lines of the magnetic field that causes them to align. This shortens the effective magnetic path length and therefore increases the inductance.
Modify the above equation to:
$$L=\mu_{_r}\mu_{_0}N^2\frac{A}{2\pi\,r}$$
Here, \$\mu_{_r}\$ is just a number that tells you by how much the vacuum path length is shortened. That can be by a factor of 1000 or more.
So, once again there is a magnetic path you can measure with a tape measure. That's one number. But there's also a magnetic path length you cannot measure directly, which is the effective length within which energy is stored. (The total vacuum volume that stores the energy will be this shortened length times the toroid's cross-section area.)
With materials we can use to increase the inductance, with domains that align, there is hysteresis. If you start with such a material with domains aligned in random directions, then apply a magnetic field that forces them into some alignment, and then remove that magnetic field, they will partially return to their earlier state (or something close enough for now) but they will not entirely return to their earlier randomized condition. Some "memory" of the prior aligning field will still hang around.
If you reverse the field, you'll first have to overcome that memory and then re-align them oppositely. Removing the field will leave some memory of that field, too, which was opposite. This process continues over and over when AC is applied to an inductor.
Just as with any material, too, rotating the domains does leave some vibrational energy in the material -- heat.
A changing magnetic field induces a non-Coulomb electric field. That means currents can flow, under AC conditions. Since some materials we use (iron) are conductive, these currents do in fact flow and will also generate heat. The first order non-Coulomb effect is called an Eddy current. This isn't something you have to worry about in a capacitor because, by definition, a dielectric is an insulator and currents just don't flow well in those. But for inductors, and at higher frequencies, it's a problem. Ferrites help with this because conductive particles are mixed with non-conductive bits that help block these currents.
Finally, there are also practical limits on the strength of the magnetic field (the number of Teslas) that can be supported. Different materials will place differing limits here, as well. Once this is exhausted (i.e., all of the magnetic domains have done all of the "lining up" they can do and there's nothing left over, anymore), then the inductor starts behaving as though there was no core material, at all. More like an air-core inductor. The process is often gradual so that the effect isn't sudden. Some oscillating circuits actually depend upon this "feature" in order to work.
Simplification
Before I go on, I want to return to the capacitor and dig one level below my qualitative hand-sweeping above. The reason is that I want to point out a new concept that you should burn into your every thought about the world around you -- the idea of emergent phenomena.
Sometimes, theories describing well one level of experience and that are well verified, experimentally, have no meaning whatsoever at a deeper level. Instead, these ideas are emergent -- usually as a result of large number population statistics.
The concepts of temperature and entropy from statistical thermodynamics are well established and extremely important. They just work. But the ideas of temperature and entropy have no meaning whatsoever at the atomic level. They just don't exist. There is energy. And energy is an important concept both at the atomic level and in thermodynamics. So there is some common ground. It's just that individual particles don't have the idea of temperature. They may have velocity, momentum, energy, etc. But not temperature or entropy. Those two arise out of the statistics of quadrillions (and more) particles that interact with something else that also includes similar large numbers of particles.
For example, a cup of water and a thermometer. These both have huge, inconceivable numbers of individual particles in them. And when you insert a thermometer into a cup of water to measure its temperature, you are asking about "what happens with the quintillions of particles in the thermometer when immersed into septillions of water particles that will all be randomly bouncing around and affecting each other?" That has meaning because of the large numbers of possible starting states and the myriad results that are possible, almost all of which will produce a "reading" on the thermometer that you expect to see within error bounds. There are a few outlier initial states that would yield different results. But the likelihood of that happening is so low that the entire universe's lifetime would have expired before it occurs. So it is valid to talk about temperature and entropy as if they exist, at this emergent level. (Because outlier behaviors are extremely rare.)
(Note: The 0th law of thermodynamics is worth a nod at this point, too.)
Just remember this general rule: no matter what level you think you understand something, there's a deeper level at which you don't and out of which what you think you know emerges!
The capacitor is just such an example. I hand-waved above about dielectric dipoles and so on. But deeper down, it is of course at lot more interesting and complex. And further down yet, even still more exciting and complex. Let me give you a taste of just the next level down, drawn mostly from ideas expressed in Chabay & Sherwood's excellent book for learning physics called "Matter & Interactions."
Here's my rendition of a combo of what they write about:
I've tried to show a cloud of electrons in the two metal plates, but where the left side of both plates has just a few extra electrons near the surface (and just a few less electrons near the surfaces on the right side.) In between, I'm showing the polarized dielectric dipoles. And around the outside around the bottom I'm showing the e-field direction vectors as you move around the outside of the capacitor along some curve.
(Keep in mind that out of the very, very large number of conduction band electrons present on both plates, there are slightly fewer of them on the left plate than on the right plate.)
I've also added a green and an orange dot at two interesting locations within the dielectric. The green dot is right in the middle of one of those dipoles. From its perspective the vector points as indicated above. The orange dot is between two dipoles and from its perspective the vector points the other way, also as indicated above. As you move around within the dielectric, there are some very complex e-field directions, as you can imagine. I've only illustrated two simpler ones.
The question is, what's the net direction of the e-field within the dielectric? It seems very hard to compute that. But if you look at the direction vectors and imagine summing them as you move yourself along a closed path circling around the outside and then through the middle of the capacitor to finish up where you started, we know this sum must be zero. (It's impossible for it to be non-zero.)
But you can see that two vectors on the outside (furthest left and furthest right) both point in the same direction. So the sum around the exterior and ignoring the dielectric for a moment, must be non-zero and the net pointing to the right. Therefore, the net direction of the dielectric itself must be pointing to the left by just exactly the right amount needed to cancel the exterior path sum.
I told you that capacitors are interesting. Note that my earlier hand-waving didn't get close to this level of detail. And there's still deeper levels, yet, too.
Momentum, Mass, Energy, Etc.
You may have heard that the energy on a capacitor is \$\frac12 C V^2\$ and that for an inductor it is \$\frac12LI^2\$. You may also know that the kinetic energy of a particle is \$\frac12 mv^2\$. It seems interesting that there is some similarities, just the same.
$$\begin{align*}
\begin{array}[t]{r}
{\text{momentum:}}\vphantom{q=C\:V}\\\\
{\text{momentum:}}\vphantom{\text{d}\,q=C\:\text{d}\,V}\\\\
{\text{force:}}\vphantom{\frac{\text{d}q}{\text{d}t}=C\:\frac{\text{d}\,V}{\text{d}t}}\\\\
{\text{mass:}}\vphantom{C=\frac{\text{d}q}{\text{d}V}}\\\\
{\text{velocity:}}\vphantom{q=C\:V}\\\\
{\text{acceleration:}}\vphantom{q=C\:V}\\\\
{\text{energy:}}\vphantom{\frac12 C \,V^2}
\end{array}
&&
\overbrace{
\begin{array}[t]{c}
q=C\:V\\\\
\text{d}q=C\:\text{d}V\\\\
I=\frac{\text{d}q}{\text{d}t}=C\:\frac{\text{d}\,V}{\text{d}t}\\\\
C=\frac{\text{d}q}{\text{d}V}\\\\
V\\\\
\frac{\text{d}V}{\text{d}t}\\\\
\frac12 C \,V^2
\end{array}
}^{\text{capacitor}}
&&
\overbrace{
\begin{array}[t]{c}
p=m\:v\vphantom{q=C\:V}\\\\
\text{d}p=m\:\text{d}v\vphantom{q=C\:V}\\\\
F=\frac{\text{d}p}{\text{d}t}=m\:\frac{\text{d}\,v}{\text{d}t}=m\:a\\\\
m=\frac{\text{d}p}{\text{d}v}\vphantom{C=\frac{\text{d}q}{\text{d}V}}\\\\
v\\\\
a=\frac{\text{d}v}{\text{d}t}\\\\
\frac12 m\,v^2\vphantom{\frac12 C \,V^2}
\end{array}
}^{\text{particle}}
&&
\overbrace{
\begin{array}[t]{c}
\phi=L\:I\vphantom{q=C\:V}\\\\
\text{d}\phi=L\:\text{d}I\\\\
V=\frac{\text{d}\phi}{\text{d}t}=L\:\frac{\text{d}\,I}{\text{d}t}\\\\
L=\frac{\text{d}\phi}{\text{d}I}\vphantom{C=\frac{\text{d}q}{\text{d}V}}\\\\
I\\\\
\frac{\text{d}\,I}{\text{d}t}\\\\
\frac12 L\,I^2\vphantom{\frac12 C \,V^2}
\end{array}
}^{\text{inductor}}
\end{align*}$$
Also, look up Lagrangian mechanics and the principle of least action. (I started with momentum for a reason.)
I'm sure you know that the capacitor's charge is conserved. You can kind of think about the capacitor's charge as being related to potential energy in a physical system and think about the inductor's Webers as being related to the kinetic energy in a physical system. (You could choose the other way, if you like.)
Regardless, I like to think of the Webers (or volt-seconds) in an inductor as the magnetic dual of the capacitor's electric charge. It's a little weird at first, because charge is countable (in our minds) but volt-seconds seems to depend upon time (and it does) and isn't countable in the same way. But for all intents and purposes, \$L I\$ is magnetic charge and \$C V\$ is electric charge. These things are conserved just as momentum in a physical system is also conserved.
Similarly, the current into a capacitor is force. When you apply that force to a capacitor, it accelerates, changing the voltage. The voltage across an inductor is a force. When you apply a voltage across an inductor, it also accelerates, changing the current.
It's also useful (especially when you consider an LC tank) to think of the inductor's energy as kinetic and the capacitor's energy as potential. The LC tank is a simple system that converts one to another and then back again. This is perhaps very similar to the idea of a comet in a very highly elliptical orbit. At apoapsis, almost all of the kinetic energy has been converted to potential energy. Then as the comet accelerates back towards the sun, this potential energy is converted into kinetic energy. At periapsis, then, almost all of the potential energy has been converted to kinetic energy. And the comet then continues, over and over. An LC tank is kind of like that, except that the polarities flip such that there are four states instead of two if you include polarity. If just energy, then two states just like the comet.
A really good paper to read is Introduction to Quantum Electromagnetic Circuits, Vool and Devoret where the authors discuss a recipe of sorts for finding Lagrangians without having to guess about it. Lots of classical physics there, so don't worry too much about that word in the paper's title that suggests otherwise.
Somewhere in that paper, they choose differently than I did above, saying that the inductor's energy represents potential, and not kinetic, energy. But like I said, you get to choose. So I prefer to think of the Coulombs on a capacitor as position coordinates and the Webers of an inductor as momentum coordinates, which swaps those roles. (But the authors are a lot smarter than I am. So perhaps you should listen to them, instead of me.)
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H: 9v oscillator circuit that produces a frequency of 500 Hz
I am a gr 10 student that really has no idea what I am doing, however, I have made a circuit schematic and just need to know if it is right I guess.
AI: Aaron, Have you ever heard about a circuit simulator? It is a good way to learn a lot about circuits before you ever have to build one out of physical components.
I recommend you search for a free software package called LT-SPICE. This cool program lets you draw a circuit and then run a simulation on the circuit and look at the subsequent voltage and/or current waveforms produced by the circuit.
LT-Spice can be used with a model of the 555 timer chip so it should be a great learning experience for you to get familiar with a circuit simulator and the experiment with your specific circuit.
I decided to use LT-Spice to simulate a circuit similar to yours but made a number of changes to get it to oscillate at very close to 500Hz.
Changes Made:
Eliminated the L-C circuit on the output because it is not clear what that is intended to produce.
Purged the trim-pot on the pin 5 Control Voltage pin. You can apply frequency trimming later.
Adjusted RA, RB and C1 to standard component values to get the oscillation frequency to be very close to 500Hz.
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H: Good design for reading a DIP Switch once
I want to read an 8bit DIP switch at MCU startup, after that it will be disregarded. I have enough pins on the MCU to read this directly. My question is; is it worth while to stop the power for the switch after it is read? Power is not a huge consideration but as it will be on 24x7, running on large batteries I at least don't want wasteful usage. What would be the best practice? I could sacrifice one more pin on the MCU to do this.
Originally I was thinking of a multiplexor, but the price difference for an MCU with more pins is so small it would be easier to go that way. Thanks!
AI: Everything here assumes the switches are connected between microcontroller pins and ground, so that a closed switch reads zero.
Option 1:
Drive all your pullup resistors from a microprocessor pin. This'll take an extra pin, but when you're done with it you can pull it low.
Option 2:
If you have a modern microcontroller, it has every kind of pull-up, pull-down, etc. Usually the pins will be inputs on start-up (check the data sheet). Assuming the processor allows it, set the pins to inputs with pull-ups. Then wait a bit, then read your switches, then set the pins to pull-downs or as outputs and set them to 0.
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H: How does a switch's series voltage affect discharge time in a relaxation circuit?
I made this relaxation oscillatory circuit in LTspice using a voltage controlled switch (sw) that switches off at \$ 100V \$ and back on at \$ 50V \$. To the left, the sw has \$ V_{ser}= 0V \$ series voltage and it takes the capacitor \$ C_1 \$ approximately \$ \approx 0.7ms \$ to hit the drop-out voltage at \$ 50V \$. The math confirms the discharge time I measured in the LTspice trace:
\$ V_{d} = V_{c}* e^{-t \over R_{th}C } \$
\$ t = -\ln {V_d \over V_c} R_{th}C_1 = -\ln{50V \over 100V} * 10^6 * 10^{-9} = 0.693ms \approx 0.7ms\$
where \$ V_d \$ is the drop-in voltage and \$ V_c \$ the cut-in voltage. \$ R_{th} \$ is the Thevenin resistance the capacitor sees - that is, \$ R_2+(R_1||R_{on}) \approx R_2 \$.
Now if I add a series voltage (\$ V_{ser} = 20V \$) to the switch sw (schematics on the right), I spotted no difference in the charge-up time but a significant increase in the discharge time (\$ \approx 1ms \$. I don't understand either case.
If the capacitor starts discharging at \$ 100V \$ and there's this series voltage of \$ 20V \$, the capacitor is effectively discharging a \$ 100V - 20V = 80V \$ of initial charges. The capacitor should then hit the drop-out voltage quicker than that without adding the series voltage in the switch.
\$ t = -\ln {V_d \over V_c} R_{th}C_1 = -\ln{50V \over 80V} * 10^6 * 10^{-9} = 0.47ms \$
But the simulation gives \$ \approx 1ms \$
How does the switch's series voltage affect the relaxation circuit - specifically, why does it increase discharge time, whereas the charge-up time seems not to be affected?
AI: Due to the fact that in your circuit \$V_∞ = 20V\$ instead of \$0V\$.
You need to use this general formula for the capacitor charging/discharging phase:
$$V_C(t) = V∞ + (V_{start} - V∞) \times \left(e^{\frac{-t}{RC}}\right)$$
Where:
\$V_{start}\$ initial capacitor voltage.
\$V∞\$ steady-state final voltage.
Because now you have been using the simplified version that assumes \$V_∞ = 0V\$.
And if we solve it for the time we get this:
$$T = RC \times \ln \frac{V_{start} - V∞}{V_C - V∞} = 1\text{ms} \times \ln\frac{100V -20V}{50V - 20V} \approx 1\text{ms} \times 0.98 \approx 0.98\text{ms}$$
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H: Replacing two ideal with two linear voltage sources
I am currently learning how ideal sources may be replaced with linear sources to make the application of other circuit theory possible.
Unfortunately, I don't understand how the following conversion is possible.
This is the original circuit.
simulate this circuit – Schematic created using CircuitLab
And my prof. converts it into the following, unfortunately very quickly and with little commentary:
simulate this circuit
with:
$$ V_1^* = V \frac{R_3}{R_1+R_3} $$
$$ R_1^* = \frac{R_1R_3}{R_1+R_3} $$
$$ V_2^* = V \frac{R_4}{R_2+R_4} $$
$$ R_2^* = \frac{R_2R_4}{R_2+R_$} $$
Ultimately, the question in this exercise is what voltage-drop occurs over \$R_5\$, but my question relates more to the conversion above, and less to the accomplishment of said goal. I only mention it, since it may allow some simplifications which would not be possible, would we seek information about other specific nodes or currents, for example.
Let's only consider \$V_1*\$ and \$R_1*\$ since the conversion for the other voltage-source is equivalent.
The voltage itself of course stems from a voltage-division between \$R_1\$ and \$R_3\$. Yet I don't quite see how this comes about, since, going from the cathode and through \$R_3\$, our remaining voltage should at first glance be smaller, the larger \$R_3\$ is. A second glance gives me even more trouble, since I don't see a way for current to travel from the upper cathode to the upper anode while going through the rest of the circuit (since it would ultimately have to be at node A again before going through \$R_1\$ and into the anode.
When calculating the inner resistance, the resistors \$R_1\$ and \$R_3\$ are clearly considered to be parallel. But, considering one of the voltage-sources, I see no way that current could travel from cathode to anode without going through both resistors.
I assume that in both cases my problem is that I consider only one of the sources at a time, while I should consider them both at once. They supply the same voltage, and are oriented so that no potential-difference occurs between their cathodes, which means that there is no potential-difference between their anodes as well. In this exercise they actually stem from a single source, which got split into these two to make a simplification of the circuit possible.
So, why can we make the above replacements, and why are the relations the way they are? Am I right that I should consider both sources at the same time? Some explanation and/or pointing to relevant theory would be highly welcome.
AI: For general notes on the thevenin-theorem, you may consult Christianidis Vasilleios answer.
In the case of my exercise, and the solution provided by my professor, the procedure was related, but a little different, and I want to explain it now:
Thevenin's theorem states that we can replace any bipolar circuit with only linear elements by an equivalent circuit, containing only one voltage source and one inner resistance.
Let's name the node between the two voltage-sources C, and cut out the upper voltage-source, together with resistors \$R_1\$ and \$R_3\$. This is the cut-out circuit:
simulate this circuit – Schematic created using CircuitLab
This is clearly a bipole, and only contains linear elements, so we may form an equivalent circuit, containing only one voltage source and one resistor.
For the resistor, let's name it \$R_1^*\$, short the voltage source. This will lead to:
$$ R_1^* = R_1||R_3 = \frac{R_1R_3}{R_1+R_3} $$
For the voltage, note that whatever load \$R_L\$ we may put between A and C, the voltage-drop across it will equal the voltage-drop across \$R_3\$. (Since \$R_L\$ is in parallel to \$R_3\$.
$$ R_L = R_3 $$
There will also be a voltage-drop across \$R_1\$. Here the voltage-divider rule comes into play:
$$ \frac{U_3}{U} = \frac{R_3}{R} $$
or:
$$ \frac{U_L}{U} = \frac{R_3}{R_1+R_3} $$
This leads to:
$$ U_L = U \frac{R_3}{R_1+R_3} $$
which will be the equivalent voltage for our cut-out circuit.
Plugging this new circuit back into the original one will give:
simulate this circuit
The same may be done with the lower voltage-source together with the respective resistors \$R_2\$, \$R_4\$ leading to the result of the question.
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H: Help wiring relay to LED driver - L/N on driver to L/O on relay?
I have a Shelly Dimmer 2 wifi-controlled relay, and I'm a bit confused as to wiring it to my LED module. The front of the relay looks like this:
This is the LED lamp I am trying to connect. At one end is a normal 2-prong 240V mains plug (i.e. AU plug, no earth), it goes to a LED driver, then to the LED module itself:
Closeup of the LED driver
There is a "L" and a "N' on the left input side.
I'm just trying to understand the correct way to wire it up to the relay.
Should I connect "L" on the driver to "L" on the relay? And then "N" on the driver to "O" on the relay?
Or is there another way to do this?
I don't need a physical switch in my case, I just need to use wifi to control the relay to control the LED.
(Shelly also talks about using a https://shop.shelly.cloud/bypass-wifi-smart-home-automation if you don't have a neutral, and your load is < 10W. I assume the LED draws very little at idle, so that my be needed here).
AI: From the manual you linked to:
When using the device without neutral, Shelly Dimmer 2
requires at least 10W of power consumption, in order to operate. If the connected light has a smaller power consumption, then Shelly Bypass is required for the operation of the Device.
So as (or if) you think your LED array is going to draw less than 10W, you should wire it with a neutral if you can otherwise you'll need to add their bypass device. The option to wire without a neutral is really for retrofitting this device to a house where neutrals often aren't run to the switches because they don't need to be with conventional wiring.
In addition, wiring it WITH a neutral will allow you to operate the wi-fi functions without any load connected, or if the load should fail at some future point. Without a neutral, the dimmer always requires some current flowing through the load in order for its internal electronics to run.
Wire it in accordance with the manual, with L of your driver to O and the neutrals connected together:
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H: Controlling AVR reset via voltage divider
I have an AVR atmega32u4 which is being run on 3.3v. I program it using a avr usbisp, which runs at 5v.. The problem i'm having is driving the reset pin.. I currently have this schematic where PROG_RESET is the 5V/GND reset pin on the programmer, and AVR_RESET goes to the AVR IC's reset pin...
What I am seeing is: programming works brilliantly, but the chip never boots, as AVR_RESET is constantly held at ~100mV - low enough for the IC to stay in reset. Even when only connecting power. Briding the 10K resistor doesn't seem to boot it either.. What values of resistors should I change to, to make the reset divider functional?
AI: Using a resistor in this way will not work. The 10k resistor and 310 resistor form a potential divider which means that when PROG_RESET is floating, there will be \$V=\frac{3.3\times310}{10k+310}=0.1V\$ at the reset pin.
Instead you should use a level shifting circuit which either drives the RESET line high when the programmer is not connected, or has a high-impedance output.
If you want a simple modification to your existing circuit, you could replace your two low-value resistors with an N-channel MOSFET, using the following circuit:
The MOSFET could also be replaced by a diode with Anode to AVR_RESET, and Cathode to PROG_RESET. In this case, it should be a schottky type diode with a low forward voltage drop (maybe < 0.5V) to ensure the AVR reliably detects a logic 0.
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H: How to repurpose an old PCB?
I am brand new to electronics and would like to know if there's any way to reprogram the microcontroller or reuse a used PCB. I extracted this PCB from my old Bluetooth speakers. Here's the image:
Also, I don't recognize which part is used as the Bluetooth or FM module or antenna on the PCB. My question might be really basic or silly but I don't know much and am trying to learn, so please help.
AI: I am brand new to electronics and would like to know if there's any way to reprogram the microcontroller or reuse a used PCB.
Not universally. Most devices that contain a microcontroller will have programmed the microcontroller to disable any further programming. Or, the microcontroller was 1-time-programmable to begin with.
I don't even think there's a programmable microcontroller on your board, at all! There's simply application-specific controllers that integrate Bluetooth, SD card reader, and audio logic, and do everything a company without any own knowledge of these rather hard fields needs to produce a cheap-as-possible Bluetooth speaker.
Look at that PCB: It's really quite cheaply made, and doesn't need many components to do what is a rather complicated task!
Reusing such a (not really great-looking) PCB also doesn't pay: Never quite what you need, usually. If you want to get started into microcontroller programming:
All microcontroller companies make "evaluation boards", to help people get started using their microcontrollers.
And the ones from ST for their STM32, called Nucleo, are cheap, so start with one of these that contains the peripherals you want.
Also, I don't recognize which part is used as the Bluetooth or FM module or antenna on the PCB.
The zig-zag line on the bottom right of your picture is a very crude 2.4 GHz antenna, so that's your bluetooth. The closest IC to that is the Bluetooth IC.
FM antennas (you mean FM in the broadcast band ca 88–108 MHz) are far too large to fit on the PCB, so whatever does the FM uses the audio or power supply cabling as antenna. That works well because the audio frequencies are much, much smaller than the radio frequencies.
My question might be really basic or silly but I don't know much and am trying to learn, so please help.
Not silly, just reflecting you've not got much experience. That's normal!
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H: Do AC current sources exist and why or why not? For both actual and theoretical
I got confused when I encountered this problem on Chegg.
https://www.chegg.com/homework-help/questions-and-answers/5-find-ac-current-source-iin-polar-form-b-find-vt--c-choose-one-source-current-leads-sourc-q19121964
Question:
5.
a) Find the AC current source, Iin in polar form.
b) Find VT .
c) Choose one:
i) The source current leads the source voltage.
ii) The source current lags the source voltage.
My question isn't about solving the problem. Do AC current sources exist? I've tried searching and they don't pop up except for the Chegg problem I saw. I also recently took a Circuits class and three basic active components discussed to us are the DC voltage source, AC voltage source and DC current source. I don't get if they exist at all, and if so why is information about it not as common?
EDIT: People might answer only about the actual AC current source. I am also talking about theoretical. Like can I do the same thing in Chegg where you put a wave symbol and an arrow and call it a theoretical AC current source? How would you then use it in computations? Do you just attach a phasor to it and use current in the x-axis instead of voltage for vector diagrams?
It's because I'm curious on both. Yes I want to know whether they exist in real life but I also want to know their "existence" on our theoretical computations. Are they there but scarcely used? Or do they not exist at all on common electrical circuit related courses?
AI: Ideal AC and DC voltage and current sources are useful for analysis (theoretical) purposes.
It could be said that an AC generator that supplies power to a grid operates as an AC current source. There are also AC motor controllers that act as current sources. However it may be more accurate to say that they are DC current sources that control motors through inverters. That is a DC current source drives an inverter that supplies current to a motor. Grid connected solar and wind power inverters are similar.
Re edit:
Like can I do the same thing in Chegg where you put a wave symbol and an arrow and call it a theoretical AC current source? How would you then use it in computations? Do you just attach a phasor to it and use current in the x-axis instead of voltage for vector diagrams?
Yes
Re Comment:
Here is an example from Scott, Linear Circuits, 1960. It clearly shows an AC current source, but the symbol doesn't include as sin wave symbol. I also found textbook problems that simply identified the source i(t). Some of those were sine wave sources and others defined some other periodic waveform.
Also note that a current transformer act as a current source.
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H: JFET series switch
The picture shows a series JFET switch. Is r_d itself the load or does the load have to be attached at the port v_out? If the second one is true then what is the purpose of r_d? Can't we replace r_d with the load we want?
AI: my question is whether r_d itself is the load or the load has to be
attached at the port v_out?
The only sensible option (without further context) is that \$R_D\$ is an attached load. It certainly cannot be part of the JFET because it makes no connection to ground. The Vout circle symbol is a little confusing but, I believe, it's just showing the point where you measure Vout.
\$R_{DS}\$ is typically the drain-source on-resistance of the JFET.
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H: Amplifying the output voltage from a coil (10Hz-20kHz) using an AD620 module
Disclaimer: I’m computer scientist.
I would need some help to get an AD620 amplifier module to work.
Some context is available here in a previous question.
Here is a physical view of the amplifier:
The documentation was unfortunately in Chinese, but I translated it. It is available here.
I managed to amplify the output of my generator (TieDie Handyscope H3-5) and adjust gain, but not with the exact same setup depicted in the image. I needed to connect the ground of the input terminal to the ground of my power source.
I didn’t manage to amplify the signal I need to amplify.
I want to amplify the varying voltage at the terminals of a coil that I didn’t manage to see with my oscilloscope (TieDie Handyscope H3-5). I was expecting that I would see at least the ambient 50Hz noise.
AI: I didn’t manage to amplify the signal I need to amplify.
If you are talking about amplifying the coil signal (previous post) then I see where you possible went wrong. I've downloaded the schematic and added what you need to add: -
Basically, as I understand, you connected your coil directly across both inputs. If you do this then you need to add bias resistors from one or both inputs to ground to make the input bias currents (small but still relevant) pass the ground and correctly apply DC levels to your inputs correctly. You might get away with 1 MΩ resistors too but don't go lower than 100 kΩ in case it upsets your experiment.
There was also a small error in the schematic that I fixed although I expect the module you bought to be OK.
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H: Why am I getting 239 VAC between GND and line when the device is turned off?
I noticed a weird thing that I don't understand.
I'm working on an electronics project which has inside of it a 230 VAC (I'm in Europe) to 12 VDC transformer.
For the convenience, the project is plugged to the Eaton ePDU G3, which is a sort of a very smart network-controlled power strip which, among others, allows to power each socket on and off. The PDU is connected in turn to the UPS, that is, a unit with a battery inside which keeps powering the appliances in a case of an outage. Finally, the UPS itself is plugged into a surge suppressor. Everything is grounded properly.
When the device is powered on, everything works as expected. When I measure the difference between the two wires, this is what I get:
GND ↔ Line: 239 V
Neutral ↔ Line: 239 V
GND ↔ Neutral: 1 V
However, when I turn the device off through the PDU, I get:
GND ↔ Line: 239 V
Neutral ↔ Line: grows from 4 V to 36 V for something like thirty seconds, then drops to zero, grows again, etc.
GND ↔ Neutral: decreases from 236 to 215 V for something like thirty seconds, then goes back to 236 V and starts decreasing again.
The measurements are taken with an inexpensive multimeter, which may explain why adding the values from the last two rows doesn't always equal 239 V. I don't have an oscillograph, so I can't see the exact pattern.
I imagined that the mechanical relay which turns the appliance on and off in the PDU is acting on the Line, breaking the connection when the socket is turned off. In practice, the relay acts on the Neutral instead. This creates an electric shock hazard: it looks like the appliance is powered down, but there is still a 239 V differential between some of the wires inside.
What could be the reason for a PDU to do it this way? Am I missing something?
AI: The power strip probably uses a TRIAC or other solid state switch device to turn the AC on and off at the outlets. As with all electronic components, there is leakage current through the device when it is "off". Not enough to hurt someone, but enough to get the attention of a meter with a megohm or multi-megohm input impedance. Attach a small load, such as an incandescent night-light bulb, across the meter input and the reading should drop to near zero.
OK, I wrote that before I caught the significance of the second part of your post. It still applies to why you get a voltage reading when things should be off, but not to why the main disconnect is in the Neutral rather than the Line. For that I have no answer except poor manufacturing oversight - it could be a wiring error. The next step is that the power strip is designed that way, and that is bad. Someone around here who knows more about European wiring practices might be able to add more.
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H: Where is the antenna in this remote control board?
This is a PCB of a (radio?) remote control.
AI: The antenna is the rectangular loop trace at the top edge of the PCB. Transistor Q1 is the oscillator. L1 is probably used to provide DC power to Q1, and C1 and C2 are part of the matching and resonating network.
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