text stringlengths 83 79.5k |
|---|
H: What does an "E" mean in a resistor value? (ie 150E)
I'm trying to build a dc motor controller circuit as described here.
But it requires 2 "150E" resistors
From looking online I believe this means 15 ohms but I'm curious as to why, and why I couldn't find anywhere explaining such.
AI: E means ohms. 150E is 150 ohm.
See link |
H: Can you help me identify a flaw in my DC motor brake design
Can you please look at the schematic below and tell me what I am doing wrong.
The p-channel MOSFET always shorting no matter what I do also checked with different components and got the same result.
Thanks,
AI: Can you please look at the schematic below and tell me what I am doing
wrong
The IRF5305 (P ch MOSFET) is upside down and the body diode will always conduct.
As for anything else, if you could take some time to explain why you might want to actively short out the motor with the P ch MOSFET, this might help. |
H: How does an MCU pin detect a voltage?
If I have an MCU pin connected in between 2 resistors (a voltage divider) how does the MCU pin actually detect the voltage? I am using an ATMega328 and reading the voltage from the divider by an ADC enabled pin.
I believe my understanding is broken but here is what I'm thinking...
the pin is floating but is set for input (has internal pull-up enabled)
the top resistor in the divider takes some voltage (voltage drop)
now, the wire that connects the middle of the voltage divider to the pin has a potential of x volts (it could potentially push by x volts)
BUT, no current will flow through the connecting wire to the pin as the pin isn't at ground (there is nothing to "pull" the current towards it)
if no current flows to the pin, how is voltage detected? If the current is 0 then according to Ohm's Law voltage would be 0 too
AI: the pin is floating but is set for input (has internal pull-up enabled)
is a paradox. Either it's floating, or it's pulled up. These things are opposites. "Pull-up" means there is a resistor that connects the pin to a fixed voltage; floating means that the pin is not connected to either ground or any non-zero voltage.
For ADC pins, pull-up doesn't make sense β you don't want to measure your supply voltage, you want to measure an external voltage!
Hence, there's no resistor voltage divider.
So, let's have a look at the datasheet, p.310:
It says there, the input resistance of the ADC is 100MOhm.
So, yes: Practically no current will flow into the ADC. That's a good thing, too, because if current flows into the the ADC, then that would change the voltage on the input, thus destroying the measurability of the signal of interest.
So, the question is: How does the actual measurement take place?
Now, that is a different question, and depends on the architecture of the integrated ADC. In case of your ATMega328, you're dealing with a successive approximation ADC. Which means, that via means of an analog circuit, the voltage at the moment you measure is "buffered", saved to a capacitor, and then, a DAC's output value is changed successively until it matches the stored voltage (very rough overview). I'd like to point you to the internet on successive approximation ADCs. |
H: Amplify sinewave to 6A
I am trying to create a 6A, 60Hz sinewave. A 50W audio amplifier with the standard 4Ξ© minimum output resistance is available, but I deduce it is only able to supply an absolute maximum of 3.5A. My idea was to use a transformer on that output and increase current that way. The signal will be applied to a measurement device, meaning minimal load only. Finding the right transformer is getting me headache. What is the right way of creating such low-voltage, high-current, virtually no-load AC signal, what would you do? Cheap & easy preferred.
AI: One method that would optimize the load would be to buy a 50VA or better toroidal power transformer with a 14-16V secondary.
You can test-power the transformer from the mains as intended and place a few turns of wire threaded through the core. Measure the voltage induced, divide by the number of turns to determine the volts per turn.
You can then wind a secondary with your desired ratio (say 4:1) using wire capable of handling 14A. Say the volts per turn is 0.25, original secondary is 15V, and you want a 4:1 ratio (3.75V), so you would wind 3.75/0.25 = 15 turns of heavy wire (which is where you connect your load). Connect the amplifier to the original 15V secondary.
The transformed impedance is divided by the square of the turns ratio, so a 250m\$\Omega\$ ohm load will appear as 4 ohms at the amplifier with a 4:1 turns ratio. You can go as low as a single turn, which will give you very high current (potentially), but you will have to keep the resistance extremely low. In my example, the turns ratio would be 60:1 and the current could be as high as 200A but the resistance would have to be kept to 1.5m\$\Omega\$ total including the winding.
Be sure to insulate the original primary winding connections because they will have approximately mains voltage on them at full input voltage.
Be aware that many audio amplifiers are utterly incapable of supplying their full rated RMS power for any length of time, so you may need to do the optimization I suggest.
Note: A small nit - we don't call this amplifying because there is no power gain. You will actually lose a bit of power in the transformer. |
H: Decoupling capacitors values in mixed-signal circuit
I know theory about decoupling capacitors (ESR, ESl, parallel capcitors etc.)
I have read Electromagnetic Compatibility Engineering, H.W. Ott.
My question is about practical design. How shall i choose appropriate capacitors to achieve the best results ?
I have to design mixed signal pcb. It is measurement system with analog front-end, ADC, DAC, FPGA. There will be a few chips which will be working with different frequencies (shall I take into consideration each freq?).
Do engineers in this type of design calculate capacitors carefully (taking into considerations for example impedance peaks in resonance frequency) or put a few different range capacitors ?
I would like to solve this problem in correct way not only for the best system accuracy but also for knowledge for the future deign :)
Could I kindly ask more experienced people for an advises and answers ?
Cheers,
e2p
AI: Normally when I do a design I use the decoupling capacitor recommended by the manufacturer datasheet. No deep analysis is usually involved.
In cases where I am doing something more custom I use a great web based tool called K-SIM.
http://ksim.kemet.com/
In its simplest use, it can calculate the impedance, and ESR of a capacitor across frequency.
Given a set of real ceramic capacitor part numbers, and the quantity of each part, it can calculate the impedance of the set of capacitors in parallel across frequency. It takes into account the non-ideal properties of the capacitor when doing the calculation.
It can also do other calculations such as ripple current, temperature rise, S-parameters, etc. |
H: Unknown connection to digital counter 4060
Hey friends! I'm supposed to build this circuit but I'm not really sure what should I do about the thing that is inside the red circle I marked.
Original site is here:
http://www.eeweb.com/blog/extreme_circuits/dicing-with-leds
I have no idea what that triangle means and where does it connect with digital counter 4060. I'd appreciate some explanation. Thank you!
AI: The triangle indicates a logic inversion on the input.
The symbol for the part uses a convention that was invented 20 or so years ago (30?), and was intended to show, on the symbol, all you needed to know about the part. Where there are multiple logical functions in the part, these are split apart and shown separately.
In the case of the 4060, it has both an RC oscillator, and a binary counter. The oscillator is shown separately inside its own little box.
Pin 12, the Master Reset, operates to both clear the counter AND to disable the oscillator. The oscillator runs when the reset is low. The convention is that enables are shown in positive logic, so an inverter is shown on that input. As the input goes to two functions, it's shown as two inputs. Helpfully, one doesn't have a pin number on it, to show it's not a real input.
Why don't we see more of this type of diagram? The convention never really caught on, but it is still used. To show diagrammatically all the complexity of some circuits resulted in a few new squiggles being invented, and then combined in difficult to parse ways. Everybody, as soon as they saw one of these symbols, would just head to the data sheet to read what it did. It was an unnecessary complication, and not up to the job. VHDL for design of FPGAs was catching on at the same time, and was far more appropriate for complicated stuff, and simple logic stayed with simple symbols (mostly!). |
H: Slow rise time on MOSFET
I'm doing this project, where i control the speed of a motor with a 0-3.3 V PWM signal from a microcontroller. In the test setup i use a function generator (Digilent Analog Discovery). I would like the PWM to run at 50 kHz, but for better measurements, I have lowered the frequency to 15 kHz, 50 % duty cycle.
My problem is, that there seems to be some sort of capacitance in my circuit, lowering the FET's rise time, when i test it.
The schematic:
"Opdrifts_PSoC-Blæser_signal" is the microcontroller. It is here i supply the - PWM in my test setup.
"R6" is where the motor will be. In the test i use a 10 kOhm resistor.
"Scope" is where i measure relative to ground.
The measurement:
Output signal
Rise time is measured to about 13 us.
Hope somebody here can help me out.
AI: The MOSFET drain has a lot of capacitance. The diode adds a bit, as does your 'scope probe. You should use a much lower resistance than 10K to simulate the motor.
Also, the 1N4007 is unsuitable for 50kHz, the reverse recovery time is of the order of 5-10usec (unspecified on datasheets). Use a Schottky diode such as 1N5819 for a low voltage motor (for higher voltages than about 30V-50V you can use an ultrafast silicon rectifier rather than a higher voltage Schottky) . |
H: K2961 Mosfet Equivalent
I've been searching for a replacement for a Toshiba K2961 MOSFET for a few days, but I can't seem to find any.
The datasheet for it is available here:
http://pdf1.alldatasheet.com/datasheet-pdf/view/213272/TOSHIBA/2SK2961_06.html
Perhaps someone can help me find an equivalent model of MOSFET to replace it?
Thanks in advanced, any suggestions are appreciated!
AI: It's in a rarely found Japanese extended TO-92-style through-hole package (probably an internal leadframe tab for higher power dissipation) and has relatively low input capacitance and relatively high Id/Idp.
Frankly, I think you'd be better searching for the exact part if you're not able to reverse engineer the circuit.
Look for a 2SK2961. The 2S is a Japanese thing - they leave it off. You can find replacement parts for a dollar or two. |
H: Where to put the Flyback Diode in a Relay Module?
I'm working with a relay module to switch a solenoid with an Arduino, Im having problems because the arduino get some noise in an LCD but I'm not sure where to put the diode, there is 3 pins, In, Vcc and GND. I think there should be a diode between In and VCC/GND but looking at the internet there is a diode between VCC and GND, shouldnt it be in the In because there is a direct connection with the Arduino?
Edit1 (sorry for my bad drawing)
I add my diagram, this the connection with the problem, and it only get noise when the valve is connected with the 12v source
AI: There is undoubtedly already a catch diode in that module, so adding one won't help one bit.
You should get a module with opto-isolation and supply the relay coil power from a separate power source with no connection to the Arduino power.
Also keep the wires to the contacts well away from the wires to the LCD module.
You should, however, connect a diode across the solenoid (assuming it is a DC type!) so that it is reverse biased.
If it is an AC type then there is not much that can be done. An RC snubber will help a bit, as will a MOV or TVS but they are not greatly effective in reducing this kind of electrical noise and things will only get worse as the contacts wear. |
H: Multi-cell li-ion charger
I have a problem with LI-ION technology.
I'm planning to use four 18650 cells in series, and I want to build a circuit for charging them. The final application is a sort of small UPS; I should keep the 18650 batteries charged and, when the main power goes off, switch to the 18650 batteries; when the power comes back the power switches back to the main adapter and the batteries can be charged again.
As far as I know, the four cells should always be balanced, but I haven't found any IC producer (TI, Analog, Maxim, Linear, Microchip) that offers a charger with cell balancing.
So, here are my questions:
Is there any IC capable of charging four li-ion cells in series keeping them balanced?
I found some cheap balanced charger on ebay (just search for BC-4S15D). Will this fit my needs (so charge and balance four 18650 batteries and keep them charged until needed)?
Thank you
EDIT: Another idea came to my mind: could I use "floating" grounds to charge each battery separately? I mean, one single battery assemble is
one 18650 battery
one 5V li-ion charger (e.g. one based on a TP4056)
one step down converter (e.g. this)
Then these assemblies are attached in series, and all the positives (only positives) of the step down converters are attached to a 24V supply.
I mean something like this
simulate this circuit β Schematic created using CircuitLab
Since the max voltage on the "last ground" is 12.6V, the 24V input has enough room to generate the 5V voltage.
Do you think this design can have some problems?
AI: Your idea of grounding the step-down regulator to the cell it is charging is quite clever, but it won't work properly because its supply current also flows through all the cells below it. This will cause the lower 3 cells to get varying amounts of uncontrolled charge current that their associated chargers cannot get rid of, which will eventually cause them to become unbalanced and possibly overcharged (which is potentially very dangerous).
To do it properly you need 4 fully isolated power supplies, or use a single 4 cell charger and wire a balancer onto the battery. Cheap 4 cell balancer boards are readily available on eBay and elsewhere. These often also have over-charge and over-discharge protection.
If you use a protection board then the charger can just be a regulator with adjustable current and voltage, like this.
A charger built with these components may be more efficient, more reliable, and possibly even cheaper than a bunch of step-down regulators and TP4056 chargers. |
H: Is the Nyquist plot a plot of Re(G(s)) v.s. Im(G(s) or Re(G(jΟ)) v.s. Im(G(jΟ))
It seems that people in control theory are divided over whether the Nyquist Plot/Diagram (that is the plot, not the contour!) is a plot of $$\text{Re}(G(s)) \text { v.s. } \text{Im}(G(s))$$ or $$\text{Re}(G(j\omega)) \text{ v.s. } \text{Im}(G(j\omega))$$
After a quick internet search, I have found that
The popular web resources that uses the former convention includes:
http://lpsa.swarthmore.edu/Nyquist/NyquistStability.html
http://www.facstaff.bucknell.edu/mastascu/econtrolhtml/freq/freq6.html
http://sandeepsingh.net/Control%20Systems.htm
People who uses the latter convention includes:
http://www.cds.caltech.edu/~murray/books/AM05/pdf/am08-complete_22Feb09.pdf pg 273
http://www-control.eng.cam.ac.uk/gv/p6/Handout6.pdf
http://www.roymech.co.uk/Related/Control/Nyquist.html
The rest does not seem to care about this distinction and does not label their plots properly.
In any case, from a purely academic perspective, which is correct? I need as many electrical engineers to weigh in as possible.
Side note: if this is any worth I think the former is correct since we are not only mapping the Nyquist contour on the imaginary axis but also over whichever semicircle happen to be on the contour, and those are not on the imaginary axis!
AI: From a practical point of view the Nyquist plot is the mapping of the imaginary axis from \$-j\infty\$ to \$j\infty\$ through the function 1 + G(s).
From an academic point of view it is the mapping of the Nyquist contour through 1+G(s).
The Nyquist contour is again the imaginary axis from \$-j\infty\$ to \$j\infty\$ and then an arc that starts at \$0+j\omega\$ and goes clockwise to \$0-j\omega\$ with \$\omega\$ approaching infinity. The idea is to get a contour that encompasses the whole right half plane. Then Cauchy's principle of argument can be used to determine the difference (!) between the number of zeros and poles in the enclosed region.
If G(s) has low-pass characteristic then points at infinity will map into 1 since G(s) approaches zero. All practical systems have low-pass characteristic so the semicircle can be omitted. |
H: Measurement with oscilloscope best practices
I own a DS1102E oscilloscope with 1x/10x switch on the probe, and I need to measure voltage output from the MEMEs driver similar to MAX14515 (I can not find the exact datasheet for MAX14574). The driver provides 0-140V biased at 70V at 5kHz wave.
Is it safe (for the chip and the oscilloscope) to measure the voltage between GND and outputs (p.5-2 says 400V max input, just want to make sure I got it right)?
Should I have 1x or 10x attenuation on my probe when measuring this voltage magnitude at the given switching frequency?
I would appreciate the explanation on under what conditions 1x/10x setting is best fitted (i.e. high frequency, low voltage, high voltage, input/output impedance/capacitance...).
AI: Yes, presuming the chip can drive the probe impedance (usually 1M ohm for 1X and 10M ohm for 10X)
You probably want to use the 10X setting. The probe should present a higher impedance in the 10X setting and therefore load the circuit less. It also divides down the voltage that's passed along to the scope so that you're less likely to saturate the input stage.
Generally you want to use 10X. The only time 1X really makes sense is when you are trying to measure a really small signal level that gets buried in noise with the 10X attenuation. Most of the scope probes that I have are fixed 10X with either 10M or 2.2M impedance. |
H: What is the UART TX interrupt for?
I know that the RX interrupt is obviously used to save polling, but why the TX one too?
AI: The main goal of the TX interrupt (really an END OF TX) is to send the content of a buffer (multiple bytes) automatically. When implemented in a proper way:
Enable the TX interrupt.
The user code starts transmission by sending only the first byte in the buffer.
At the end of TX (of the first byte), an interrupt will be generated.
In the TX ISR (Interrupt Service Routine), the code must send the next byte in the buffer and update the buffer index.
At the end of this transmission, a new interrupt occurs, and so on, until the entire content of buffer is sent "automatically".
Disable the TX interrupt.
The exact behavior depends on the microcontroller. That is a general description. |
H: How to power a DC brushless motor without a speed controller
I have a brushless motor but It doesn't power directly with a 9 volt battery, but I don't have a speed controller what else can I use to power it.
AI: Brushless motors do not have commutators so they cannot be powered directly from DC sources such as batteries. You'll have to procure a brushless motor driver from somewhere in order to use the motor. |
H: simple synchronous single winding motor
I am baffled how this motor works...It has a single axial winding that turns
a disk. The disk is not a strong magnet but somehow this motor has torque
enough to swing a floor fan. I see on the internet that this type of motor is also
used for microwave ovens to turn the tray..
The flux is flowing axiallly through the disk but the disk is not conductive so I
don't think eddy currents are producing the torque. Somehow the flux is being redirected I guess but how can that be???
From the image you can see the disk is sitting in the center of the many turn winding(covered by the white plastic). There are tabs that come up from the
bottom of the motor plate that are used to hold the winding in place. Is it possible that these act as induced stator poles???
AI: It appears that this may be a claw-pole motor. A claw-pole motor would have metal prongs surrounding the rotor that form alternating north and south poles by ever other prong entering the rotor area from opposite sides of the coil. There should be an air gap between the prongs and between each prong and the housing on the opposite side of the coil. There does not seem to be any mechanism to insure that the motor starts in the same direction every time it is turned on. It does not matter which way a microwave oven tray turns. A fan or pump can also be designed so that id doesn't matter which way it turns.
This is not a version of a shaded pole motor. It is either a permanent-magnet synchronous or reluctance synchronous motor.
Here is a picture of the rotor of an automotive alternator. The north and south poles are formed in the same way.
This diagram of a claw-pole stator is a little more difficult to understand, but is shows the same type of construction. |
H: VHDL case with different outputs
Is it a valid and synthesizeable VHDL code?
case IR(10 downto 7) is -- RD
when "0000" => R0 <= RTA;
when "0001" => R1 <= RTA;
when "0010" => R2 <= RTA;
when "0011" => R3 <= RTA;
when "0100" => R4 <= RTA;
when "0101" => R5 <= RTA;
when "0110" => R6 <= RTA;
when "0111" => R7 <= RTA;
when "1000" => R8 <= RTA;
when "1001" => R9 <= RTA;
when "1010" => R10 <= RTA;
when "1011" => R11 <= RTA;
when "1100" => R12 <= RTA;
when "1101" => R13 <= RTA;
when "1110" => R14 <= RTA;
when "1111" => R15 <= RTA;
when others => null ;
end case;
As you see in each case the output signal is different: R0, R1, R2, etc. I thought maybe I have to put all the outputs in each case; Something like this:
case IR(10 downto 7) is -- RD
when "0000" =>
R0 <= RTA;
R1 <= 0; -- The problem here is that I don't want R1 to be 0
-- but to retain its previous value
R2 <= 0;
R3 <= 0;
R4 <= 0;
R5 <= 0;
R6 <= 0;
R7 <= 0;
R8 <= 0;
R9 <= 0;
R10 <= 0;
R11 <= 0;
R12 <= 0;
R13 <= 0;
R14 <= 0;
R15 <= 0;
when "0001" =>
R0 <= 0;
R1 <= RTA;
R2 <= 0;
R3 <= 0;
R4 <= 0;
R5 <= 0;
R6 <= 0;
R7 <= 0;
R8 <= 0;
R9 <= 0;
R10 <= 0;
R11 <= 0;
R12 <= 0;
R13 <= 0;
R14 <= 0;
R15 <= 0;
.
.
.
when others => null ;
end case;
If I don't have all the outputs (R0 to R15) into each case the VHDL compiler gives me a warning: "Found 1-bit latch for signal . Latches may be generated from incomplete case or if statements. We do not recommend the use of latches in FPGA/CPLD designs, as they may lead to timing problems."
I need to get rid of this warning.
AI: Verilog and VHDL are hardware description languages, but they do look similar to procedural programming languages. This can sometimes lead to code that is technically valid but not synthesize-able.
In your first example code, each output not explicitly specified with a new value, implicitly retains its previous value (because that's what would happen in a procedural programming language). That is the "implicit latch" that the synthesis tools are warning about. This HDL code might be technically valid, but may not be synthesize-able (or maybe not in the way that is intended). The problem is that the synthesis tool can't do what you want without also synthesizing the actual memory to store the values, and it's not clear that is what you want. Apparently you didn't tell it to create any memory to store those values.
Your second example code explicitly determines the value of each output in each case, so the synthesis tool understands exactly what you want. In your second example, only one output gets the RTA value, and the rest of the outputs are 0. This is valid and synthesize-able, however you mention in the comments that this isn't really what you want either, you actually want to update one register while retaining the values of the other registers.
To write HDL code that updates the value of a selected register, you need to declare the actual memory elements. I'm not a VHDL guy, but in Verilog I'd maybe write something like this:
// declare R0..R15 as register storage (memory), not just signal wires
reg[7:10] R0;
reg[7:10] R1;
// etc...
reg[7:10] R15;
// could be written more efficiently as a register file instead of individual registers
// update a register selected by IR[10:7] with data RTA
always @(posedge clock)
begin
case(IR[10:7])
4'b0000:
R0 <= RTA;
break;
4'b0001:
R1 <= RTA;
break;
// etc.
4'b1111:
R15 <= RTA;
break;
endcase
//
end |
H: VHDL: Using type on ports
Is it poosible to use user-defined typed in input/output of VHDL enitities?
ENTITY test IS
PORT(
input : IN std_logic_vector (0 TO 63);
test_input: out MY_TYPE;
);
END test ;
AI: Yes, declare the type in a package, and use the package before the entity, and also in the modules that instantiates the module, so the same user defined type is available. |
H: Considerations when using internal pull-up/down resistors
I am working with STM32 micro-controllers. These micro-controllers have the option to set the input pin as pull up or pull down:
Earlier I used to give external pull up or pull down resistor wherever required but now I am wondering whether I can do away with external components and use the pull down feature of the micro-controller itself.
Will that be a good thing to do?
Is there some use case where I should take extra precaution or not use this feature at all?
AI: Yes... You can... It's a proven feature.
If you see pin structure in STM32 Reference manual, you can see that it has required components --
If this satisfies your requirements, then, you can do away with external components.
Edit Remember that at Power up, Pins will be tristated. If it does not hamper your application, it is good! That's why it's good practise to init IO section first and then go ahead with other peripherals. |
H: Pull up resistors on UART
I was going through an app-note AN2606 where I came across this connection diagram:
As per my knowledge, UART is push-pull type and Tx provides the pull-up required and hence we don't need to use any external pull ups. Am I missing something here?
AI: While the microcontroller is in reset, its I/O pins will be configured as high-impedance inputs. So the pin used for a UART serial Transmit Data (TXD) will be floating during this time. This can lead to noise causing rubbish to be transmitted by the RS232 line driver.
Some line driver ICs have internal pull-up resistors on their TTL/LVTTL-side input pins to prevent this and produce an RS232 idle state. Otherwise, a pull-up resistor can be added.
A pull-up resistor on the UART serial Receive Data (RXD) input pin is not necessary because the RS232 line receiver is always driving that pin.
Note that the reset period is usually a relatively long time and that any I/O pins used as output pins are prone to this problem and must be considered for a pull-up or pull-down resistor. |
H: How to multiply fixed point numbers of different format
Assume I have two 16 bit fixed point numbers in following format being multiplied.
Q3.13 * Q10.6
What is the correct way to multiply them, i.e do I right shift the first number to align it as Q10.6 or left shift the second number to align it as Q3.13? In both cases I will lose data.
May be the correct way is to put them into 32 bits and then shift them and then multiply them. In this case, no data is lost but result will be 64 bits after multiplication.
What Q format the result have after multiplying Q3.13 and Q10.6?
AI: This isn't particularly different from multiplying two fixed point numbers with the same format. You need to do a multiplication which preserves the most significant bits, then shift the binary point back to the desired output format.
So, do a 16x16 => 32 bit multiplication. The binary point is then at position 13+6 = 19, so you have a Q13.19 format number.
Assuming you want Q10.6 format output, you shift right by 13, optionally check for overflow, then take the lower 16 bits. |
H: Understand H-Bridge
I am trying to understand the H-bridge, I have edited the circuit and followed the advice in this thread. Is this a good H-bridge design?
simulate this circuit β Schematic created using CircuitLab
I have also added a digital short-circuit protection. How can this circuit be improved?
AI: The H bridge is all about the controlled switching of current direction. Those switches could be toggle switches, relay switches, BJTs, MOSFETs etc.
The generic H-Bridge circuit
.
The bridge has two sides (labelled A & B in the diagram) - this is a full bridge arrangement. On each side there is a high side switch and a low side switch. In many circuits these will be P MOSFETS or PNP BJTs. There are also circuits which use N MOSFETS and NPN BJTs. Its usually these high side switches that create interfacing problems due to the higher voltage required at their inputs to switch them properly.
If the circuit had only one side it would be called a half bridge.
Note that the MOTOR is shown as an inductor. This is a reasonable approximation as the H-Bridge's "load".
When an inductor (e.g. motor, solenoid, relay coil) is turned ON the current builds up and stores energy in the form of a magnetic field. In the case of a motor this magnetic field causes the rotor to turn.
The problem occurs when we try to turn the current OFF. The magnetic field collapses and induces a very high voltage with the opposite polarity to the original supply. This back e.m.f. can be several hundred volts and can easily destroy semiconductors or cause severe pitting/arcing on mechanical switches.
Diode protection:
The four diodes offer a safe, short circuit path to this back e.m.f. preventing any damage. In forward current direction they do not conduct.
H-Bridge operation:
Circuit 1 shows all the switches in the OFF (or open) position. The PWM can be open OR closed, it has no effect because there is no current path through the motor. This is the free wheeling set up.
Circuit 2 shows the bottom two switches (SWLA and SWRA) closed. This puts a short circuit across the motor. This acts as a brake because the (free) turning motor is acting as a generator and the short circuit acts a load absorbing the energy produced. This could also be achieved by using top two switches SWHA & SWHB. Note there is no current path between the supply (+V) and ground (0V)
Circuit 3 shows the motor in forward drive. SWHA and SWLB are closed so the current can flow through the motor when the PWM switch is closed. It is vital that SWHB & SWLA are open. Speed is controlled by the mark/space ratio or pulse width of the signal which opens and closes SW PWM.
Circuit 4 shows the position of the switches for reversing the motor.
The key to safely operating an H-Bridge is to ensure that the control signal (some form of logic/level switching interface) can never allow one side (SWHA,SWLA or SWHB,SWLB) or both sides (SWHA,SWLA and SWHB,SWLB) to be closed at the same time. This would place a short circuit across the supply thus releasing the magic blue smoke. |
H: UART signals connected to RJ45 connector?
I'm looking at a schematics and found something strange to me.
The schematics connects UART TX/RX signals to a RJ45 connector (leave other unused pins unconnected).
Does anyone know such application?
What's the purpose of such design?
Thanks.
David
I can't get the exactly schematics, just a block diagram like this:
AI: UART TX/RX signals to a RJ45 connector [...] Does anyone know such application?
Yes, it is common with various IT devices e.g. terminal servers, some rack-mount and blade servers etc. The specific connections (wiring) between the serial signals (often RS-232 rather than UART level), and the RJ45 pins, can vary.
What's the purpose of such design?
RJ45 connectors and suitable cabling are common, and take less space on a PCB and back panel than other "standard" serial connectors like DE-9 (some servers which have very limited space on their back panel and PCBs).
Of course using such connectors for a serial interface has the disadvantage that they can be mistaken for an Ethernet connection.
Your newly added block diagram does not show the signal labelling expected for a typical RS-232 serial connection from a "UART to RS232" block. Your signals labelled RX_DP and RX_DN suggest they are the positive and negative differential pair for the RX signal. RS-232 is not differential. Therefore the various parts of the block diagram do not match (i.e. non-differential RS-232 with (implied) differential signal pin labels on the RJ45 connector). If understanding that detail is important to you, then more information is needed. |
H: Read (High/Low) Arduino Output with an Raspberry GPIO Input Port
I want to trigger events on the raspberry (3) with an Arduino. Because of very contrary statements to this, i am kinda Lost. Arduino outputs 5V, that should be read by an Input of the Raspberry. Just slow Low/High Values, kinda like Buttons presses.
Different Possibilities I found:
Voltage Divider from 5v to 3,3v for Raspberry Pi (for example: 10k and 18k, for ~3v on the Input): That would the way that I think it is the easiest. But I stumbled over a lot of statements that say they would not recommend that, because of burned GPIOs. And the pullups of the Raspberry GPIO should be a problem. Later I found that the raspberry has software set able pull up/down resistors. It feels like lots of people don't know that they can disable them?
Voltage Level Breakout boards: That would be the easy way to go, but costs money and I don't need a bidirectional connection.
Levelshifter ICs: self explanatory.
Circtuit with a Transitor: Self explanatory, that would be my choice if Voltage Divider don't work.
So perhaps you can explain me what fact is against using a Voltage divider to connect these two boards. I just want to understand the issue, if there is one. :)
Cheers Michael
AI: As you said, a resistive divider would be the easiest. Clamp diodes are cheap to add and provide extra protection.
simulate this circuit β Schematic created using CircuitLab
If you are using different power supplies for each of the devices and want isolation between them, you can use a common optocoupler such as 4N25, PC817, etc..
simulate this circuit |
H: How does one arrive at the following transfer function for a certain circuit?
I'm trying to understand the derivation of the transfer function for the following electrical system:
Where \$k\$ is some forward gain element and \$G_{s}\$ is the transfer function of some plant. According to an instruction video, the transfer function for this whole system is $$T(s) = \frac{Y}{R} = \frac{k G_{s} }{1+kG_{s}} .$$
I am wondering how this expression for \$T(s)\$ can be derived. It probably has something to do with the control mechanism, but I am not an electrical engineer so I'm not sure how to derive this expression. I do have a background in mathematics though, so maybe you can take that into account in your answer.
AI: Just write out the control system into an equation and solve for Y/R ([Signal out] / [Signal in]
Sorry I have not figured out the math function yet.
Y = (R - Y)kGs
Y = RkGs - YkGs
Y + YkGs = RkGs
Y*(1 + kGs) = Rk*Gs
Y/R = kGs / (1 + kGs) |
H: What is the function of this cable surrounding the CRT of old televisions
I have opened old televisions, and each time, I'm wondering what is the function of this cable, roughly insulated with some tape.
Note: I know what are the risks of opening televisions.
AI: It's part of the degaussing system. "Degaussing" refers to removing any residual permanent magnetic field. This is important because electron beams are deflected by magnetic fields. Residual magnetism in parts of the unit can cause strange abberations and color fringes in the picture.
To degauss the unit, a strong AC current that gradually diminishes is run thru a large coil. What you see is part of that coil system. The AC current causes a varying magnetic field. This pulls and pushes on anything that might retain a small permanent residual field. These pulls and pushes are gradually made weaker so that in the end the residual field in anything that might retain one has converged to zero.
The degaussing "control system" was often just a positive temperature coefficient thermistor in series with the coil, and that connected directly to the AC line. When first turned on, the thermistor is cold and conducts a lot of current. This warms it up, so that it gradually conducts less current thru the coil over a number of AC line cycles.
You can often hear a short hum from these TVs right after being switched on. |
H: Question on battery basics (I'm confused and not sure how to state my question)
I'm new to the mathematics behind the electrical engineering and quite confused. I have done some searching around and haven't found anything, probably because I'm asking the wrong question.
On a 9V battery, it says '9V'. What does that mean? Does it mean it always has a voltage of 9V (Around that since the battery drains), or that up to 9V can be pulled?
I guess this also hooks into another question: If you have an LED in a circuit, What decides how much voltage it gets? Does the LED limits the amount of power going through it, or does the battery force a particular voltage and current through it, which can be modified by resistors to lower the amp and up the volt?
Sorry if I am completely asking wrong.
AI: You are over-thinking all this, and/or have some bad mental models. You also provide this dilemma: that it must either be "always has 9 V" or else "up to 9 V can be pulled." That's not the dilemma and I don't even know what you mean when you say "pulled."
A battery is based on some internal chemistry. There are some molecular ion potentials based upon some pretty basic physics which, luckily for us humans, means that a battery chemistry has the effect of providing a relatively stable voltage. You can even get a battery by sticking two dissimilar metal rods into different parts of a lemon, for example. (Not a "good" battery, though.)
Batteries that are designed to supply a voltage also have a "compliance current" specification for them. And sometimes, the specification will describe how well they work at several different currents: with low current situations lasting longer and high current situations lasting much shorter, over their useful lives.
When a battery is applied to a circuit (like an LED circuit), the battery's fresh chemistry setup tries to provide a roughly stable voltage to that circuit. But if the circuit requires too much current for the chemistry and physical design of the battery, then the voltage will droop. Sometimes, it will droop a whole lot, while still providing some current. Usually, batteries that are tied to a load that could use far more current than they can manage to supply, have their internal chemistry operating at such a pace that it impairs their lifetime of service, too. So it's important to stay within their designed specifications regarding the applied load.
A very simple way to visualize a chemical battery is to think of it as an ideal battery with a series resistor attached. In this way, if the circuit tries to draw too much current then this internal series resistor will "drop" some voltage before the circuit itself gets access to it. But this is just a very simple approximation. A real battery is much more complex and modeling them is an art of sorts.
But the basic idea is that a battery (or any practical voltage source) isn't perfect and has limitations. But designers depend upon the idea of a range of reasonable voltages when faced with a range of reasonable load currents. So, for example, I might design a circuit for a 9 V battery expecting no lower than 7.5 V and no more than 9.2 V from it over its lifetime of use. No, it won't ALWAYS provide 9 V. It will provide higher voltages early in its life and lower voltages later on as its chemistry gets used up. I have to decide how low I can accept, before my circuit stops working correctly. The lower I can manage, the longer the battery lasts. But the voltage also drops a lot more quickly as it gets near the end of its life, too. So I have to make a reasoned judgment about where to draw that line. |
H: Avalanche ruggedness
How is a high avalanche ruggedness physically achieved for a MOSFET and what are the trade-offs in other parameters to get a high avalanche rating. Also, how should one interpret the avalanche energy listed in the datasheet? Another way to state this may be, for a given design, how do you know how high a rating you will need? Thanks.
AI: This topic can go very deep! I'll try to say a bit about it, according to my understanding, to at least get you started.
Power FET manufacturers began advertising their transistors as "avalanche rugged" in the mid-80s. While there was - and still is - some confusion as to what this means, many modern power FETs have a degree of robustness in the face of avalanche conditions.
Physically, avalanche robustness is achieved by preventing the mechanisms that would otherwise degrade or destroy a FET under breakdown. The two that I am aware of are snap back and hot-carrier injection. When a FET experiences snap-back, it essentially turns itself on under breakdown and can't easily be turned back off, resulting in destruction. Hot-carrier injection is a phenomena which can degrade a FET or alter its characteristics over time, even if it doesn't go into "deep breakdown". The phenomena is, again, a fairly complex one, but basically high energy carriers are generated in the drain depletion region and - if they are close to the gate - can actually penetrate the gate and impregnate it with permanent charge.
Both of these phenomena can be avoided if the breakdown current is made to run through a safe path where it doesn't generate problematic hot carriers near the gate, and doesn't activate the parasitic snap-back BJT. An avalanche rugged FET don't snap back, so for VDS above their avalanche rating, it acts a bit like a TVS diode. This is achieved physically by altering the doping profile of the backgate or adding an auxiliary drain-body diffusion that is designed to run substantial current without thermal hot-spotting issues and without activating any parastic BJTs. As for the design trade-offs, I am not expert enough to say - my guess is that it adds die area and perhaps additional parasitic capacitance.
When a rugged FET avalanches, VDS will clamp near the breakdown voltage. Any current that runs through the FET at this time will result in significant power dissipation (ID*BVDSS). But often, power supply designers only need to break the FET down for a short time. This situation comes about, for example, when there is residual energy in the inductor of a switching mode power supply after a switching event. Designers can calculate the energy in the inductor that needs to be dissipated, and datasheet curves allow them to calculate the junction temperature rise that will result. As long as the transistor can tolerate the energy delivered to it in breakdown without thermal issues, then it can survive even periodic breakdown events.
In case it is not clear, this about surviving in the face of transient avalanche events, NOT DC avalanche conditions.
Materials from FET vendors will give you a better practical guide of how to interpret data-sheet curves and design for avalanche. I recommend the following two white papers:
http://www.st.com/content/ccc/resource/technical/document/application_note/05/13/69/ee/aa/87/49/b6/CD00100956.pdf/files/CD00100956.pdf/jcr:content/translations/en.CD00100956.pdf
http://www.vishay.com/docs/90160/an1005.pdf |
H: Opto coupler keeps on breaking
Good day,
I am building an H-bridge (picture below). My opto couplers were ILQ74. I am using 560 ohm current limiting resitors for the opto coupler diodes, with an input voltage of 3.3V (from MCU). My calculations proved that even 100 ohm would be enough. However, it already happened to me twice that one of the opto coupler diodes brakes (with two different opto couplers). I measured the resistance over it and I get a value of 3 ohm (definitely broken). Could it be just a "bad batch" or something else?
AI: I agree with the comments, you are exceeding the reverse voltage spec of the opto LEDs and this is most likely what is causing the failures you see. The LEDs in the optocouplers are rated to 3 V reverse voltage maximum, and you hit them with 3.3 V.
There appears to be another problem with your design; you have insufficient current flowing in the LEDs to ensure the output can saturate. Assuming a +V of 12 V then to saturate the optocoupler output transistors requires just less than 12 mA. The optical conversion efficiency is only 35%, so this requires a minimum of about 4 mA through the LEDs. Normally you would increase the current to ensure you met all conditions, and I'd suggest the minimum LED current should be 8 mA.
To get 8 mA through the LEDs using 3.3 V logic with an LED VF = 1.2 V you need an R value of 270 Ohms maximum.
The top pair of optocouplers can be configured as self protecting since they are only driven by the FWD and REV signals.
The bottom pair of optocouplers need protection diodes across them. The maximum source or sink current is about 16 mA so this should be ok for any microprocessor output.
I hope the schematic makes sense since I only included the LED circuit stuff.
simulate this circuit β Schematic created using CircuitLab |
H: What does the recommended values on a motor mean?
If a motor has a voltage and amperage on it, what do those values mean? (Is voltage how much voltage the battery must be rated? Is amperage how much amperage the motor will use ['pull' from the battery], or how much the battery must produce?)
Does voltage correlate to speed or torque?
Does amperage correlate to speed or torque?
I'm new to electrical engineering, so please don't use complicated terms; assume I don't know. Thank you.
AI: It really depends on the motor, but usually it is pretty self-describing:
Lets take a look at a spec sheet for a cheap motor: https://www.arduino.cc/documents/datasheets/T010160_DCmotor6_9V.pdf
Rated Voltage 6V DC -- the rest of parameters in this spec sheet (RPM, current, etc..) apply when the motor is powered from 6V. You could power it from a different voltage but then the values in this spec sheet would not match.
Operating voltage 1.5-6.5V DC -- allowed voltage range; below 1.5V, the motor will not produce any torque; above 6.5V, the motor may overheat and likely fail
No load current β€280mA -- if motor is not loaded, it will draw 0.28A (at 6V). So the motor wastes 6 * 0.28 = 1.7 watts on internal losses, friction, etc..
starting current β€5A -- also known as stall current, this is the maximum current a motor will draw. This usually happens when motor cannot rotate, say because its shaft is blocked. Your motor driver must support this much current.
Other questions:
Does voltage correlate to speed or torque? -- In ideal motor, voltage linearly correlates with no-load speed. So half the voltage is half the speed. In the real motors this is not quite that simple because motor has internal resistance and friction, so if a voltage is too low motor will refuse to turn. Also, since motor when stalled acts like a resistor, hight voltage meaning higher stall current and higher stall torque.
Does amperage correlate to speed or torque? -- In ideal motor, amperage correlates with torque. In the real motors, the winding has non-zero resistance, so some amperage is wasted heating the motor up.
Is voltage how much voltage the battery must be rated? - the voltage refers to the voltage on the motor. This will be rated voltage for alkaline batteries, slightly more for lithium or lead-acid cells, slightly less if you use motor controller, etc... For the common case of alkaline cells connected directly to the motor via a mechanical switch, the battery rated voltage is a good approximation.
Is amperage how much amperage the motor will use ['pull' from the battery], or how much the battery must produce? Both. Motor will try to pull that much current. If the battery can produce it, great. If not, the motor will produce as much torque/speed as battery allows, and will stall if the battery simply does not have enough current.
One notable exception is combination of high quality hobby RC vehicle motor and lithium-ion batteries -- if you stall that and have no protection, either motor or batteries will catch actual fire and possibly explode. So don't do that. (in the modern RC cars, the ESC - electronic speed controller -- is the most fragile part, and thus it will protect both motor and battery. A high quality ESC will limit the current / shutdown on overheat; a cheaper ESC will just fail permanently and then you will have to pay $150 to buy a new one :( ) |
H: Arduino power and networking over cat6
I plan to run about 4-5 arduino pro micro in my backyard to handle controlling of solenoid valves for garden watering based on measurements of soil (so we keep it well watered).
My question regards the supply voltage, I wish to use cat 6 over maybe a 20m run in total to supply both power and RS485 networking over the twisted pairs.
Based on previous answers regarding cat6 I know the amperage should be fine assuming all arduino even run full without sleep around 45mA that's some 250mA.
The solenoid valve says 8w at 12v which is 0.6A, assuming the current is required to keep it open (not closed - is this correct?) which I'd hope so otherwise loss of power would cause it to open.
I guess I'd need to run 12v anyway, should I run 12v and 5v down the pairs? If I run just 12v I need a linear voltage regulator, which get quite hot does that mean they are inefficient? If I run 5v as well, how well does 5v perform over that distance due to voltage drop on cat6 pairs?
AI: In general, higher voltage means lower loss, so you do want to run highest economical voltage (but not more than 48V, this will be dangerous). I also recommending running a single voltage, but using two wires -- for example, orange/or-white for (+) and blue/blue-white for (-). This just makes wiring easier, and significantly decreases the chances of microcontrollers spontaneously rebooting due to interference or ground loops.
First, lets see if you can use 12V. 20 ft of 23 AWG cable will have the resistance 0.4 ohm (assuming 20 ohm/1000 ft, and two wires in parallel for gnd and Vcc). At 0.6A, you will lose 0.24V, which solenoids are not going to notice. If you run 5 solenoid valves at once using a single cable, they will draw 3A total, and a cable will will drop 1.2 volt, still enough to energize the solenoids. So yes, looks like we can run 12V only.
Note that if you needed more power, you would have to go to 24V or 48V, and have 12V regulator in each box.
Now to get 5V, you can use linear or switching regulator. The advantages of linear regulator is that it is simple and already present on some arduinos ( not on pro micro). At 45mA, it will waste 0.3W, which is not a lot and easily dissipated. However, given that switching 5V regulators are less than $2 each on ebay, I highly recommend you just buy a dozen of them and use them everywhere. |
H: PWM controlled Input Voltage/Current regulated Converter (for solar panel)
(newbie question)
Briefly: In a converter (let's say DC-DC), I want to dictate a voltage to the output of solar panel, on the fly.
As I look around online, for example in some IEEE papers, Buck or Boost converters are used to regulate the second stage: the charging current and voltage. I studied them a bit. But I don't understand: Does changing their duty cycle, regulate their input voltage too? If not, what kind of IC regulates input, based on given duty cycle? Please note that I don't want an on-chip MPPT solution. But to regulate the input of my converter i.e. the output of the solar panel. A microprocessor will specify the voltage of solar panel, through PWM duty cycles.
Going through the products of T.I. or Linear I did not find such an Item.
Solar panel's typical power output would be in the range of 5w...
In this project, it is not important what the charging voltage (second stage) is, actually.
I'll gladly edit, clarify or rephrase the question or title, if you see a problem.
AI: Solar panel output voltage and current can not be controlled or dictated. The output is determined by the amount of sunlight received and the characteristics of the connected load. The solar panel output is the voltage and current where the load line intersects the solar panel V vs I curve as shown below. The load is assumed to be a fixed resistor. The buck or boost converter will draw a certain current with a certain input voltage. Depending on the load connected to the converter, in may be able to adjust itself to track the maximum power point. However, it will be limited by the ability of the load to safely and usefully absorb the available power. |
H: What is this trapezoidal logic gate?
What is this thing called? how does it work?
I've tried googling so much, even pasted the image in the google search box haha
Its used here on the MSP430:
AI: It's a multiplexer. When the input on the side has a specific value, the input corresponding to that value is propagated to the output. It's not a single gate per se but rather is composed of several gates that implement the function. |
H: Is 2's complement the only way that singed binary numbers are represented?
I have only seen mention 2's complement practically everytime signed number representation is mentioned. However, I have found that there are other ways to represent signed numbers also, these are 1's complement, excess-k and base -2.
Except 1's complement I do not really understand how the other two work. However, what I want to know is, are these other representations ever used at all? If so, where and why? I know that the strength of 2's complement is that addition and subtraction become the same. However, I am not sure about the strengths of the other mentioned methods.
AI: We all float on again...right back into the land of sign-magnitude numbers!
Twos' complement is almost universally used for integer arithmetic for the exact reasons you stated -- addition and subtraction work the same way, carries behave sensibly, and signed overflow is easy to detect as well.
However, floating point math almost universally uses sign-magnitude representation for the overall number, primarily to ease rounding considerations and remove asymmetries in significand range. The only machines that don't use sign-magnitude representation for floating point are of the prehistoric type (even things like S/360s, VAXen, and Crays that predate IEEE 754 use sign-magnitude FP). However, the ubiquity of sign-magnitude IEEE 754 FP has had an impact on integer arithmetic as well, thanks to Brendan Eich's little project called JavaScript.
You see, JavaScript's only arithmetic type is the IEEE 754 double precision float; thankfully, IEEE 754 floats can represent integers exactly up to the significand width. The one problem with this, though, is that using IEEE 754 floats for integer math means you are stuck with sign-magnitude integer math! That's right, negative integers work differently in JavaScript than in every other modern language. Think about this the next time you hear someone pitching Node.js... |
H: Is there a way to use constructive interference to create a specific sound in a specific location?
With 8 speakers, one in each corner, can a combination of waves be emitted that are silent from the sources but form a coherent emission somewhere between?
I mean that even if you walked around the room the sound of something, say a ball bouncing, would stay in the position that it sounds like its bouncing in relative to the walls and not because your position is being tracked and each speaker adjusts its volume accordingly, silencing as you approached and vice versa, but because the pressure changes in each speaker are beyond the 20 to 20k Hz range we hear until a combination of interference creates a wave that is formed at a controllable location and actually emits from there. An opposite to dead spots.
We have Phased arrays to steer sound without moving the array if we're not using the parametric array effect to 'beam' the sound but maybe some combination of these. Acoustic phase conjugations which can create a localized vibration or sound wall that could perhaps be enough of a non-linear medium to scatter the ultrasonic propagation? There is Intermodulation, primarily considered a distortion but all combined this could be possible?
AI: I don't know if I should answer my own question but no one has yet put all these together...
We have Phased arrays to steer sound without moving the array if we're not using the parametric array effect to 'beam' the sound but maybe some careful combination of these. Acoustic phase conjugations which can create a localized vibration or sound wall that could perhaps make the air enough of a non-linear medium to scatter two converging ultrasonic wave's propagation at a point in mid air via Intermodulation distortion.
All combined even with lots of maybes, this seems quite possible to me. |
H: What is the origin of the capacitance formula?
Everyone knows capacitance is defined as C = Q/V, but why? Sure I understand the utility of it and its use, but can someone provide a motivation or historical context for this definition? I'm unable to find one on the internet. What I am is asking is if someone can tell me who came up with this definition? When did they do so? What problem were they confronted with? etc...
AI: When the township of Leyden, glass jars were discovered in the 15th century to hold a charge between metal plates, the term "Charge Accumulators" became popular with friction generations creating the charges. Then when Coulomb and later Faraday found materials with higher superficial charge density, due to a higher dielectric constant they became known as "Condensers". Now the d constant is so diverse for many reasons, we just call them "CAPACITORs". I would like to talk about the historical significance of Q=CV with some personal experience.
200 Years ago there were many brilliant Mathematicians Laplace, Poisson, Green, Gauss, Ohm, Heaviside that described many characteristics of Voltage and Current using the force or motion on particles yet none of them adopted Faraday's * intuition* ( see graph at end) and the 18th century mathematicians above insisted it was a surface Force at a distance between conductors.
I won't. but, Maxwell proved they were both correct. You can read his books.
I recall in the early 70's at university I was taught nothing about dielectric properties of materials, until I read a Mil-Std Hdbk on capacitor properties after I graduated and got my first job in an Aerospace R&D dept. It was then, that I discovered that Carl Jung "aha" experience on the difference between mica, tantalum , teflon , alum. electrolytic, polyester, polyurethane, polystryrene, and all the other capacitor materials.
Did you know why Tesla's Mica insulation made the best hi9gh voltage? It was the best because he knew that he needed it to be > 99.99% pure.
That purity may have improved by now in terms of parts per milion or even billion (ppb) and that "purity" is the most critical property in all dielectrics including semiconductors for defect size in 450mm wafers and in batteries for lowest ESR, highest Ah capacity and self-leakage as well the tetrahedral tolerances of atom bond angles in Lithium Phosphate (LiPo) battery molecules and passivization chemicals that prevent self discharge.
- That impurity is also what leads to premature power transformer explosions.
Faraday wrote many books ( you can find and read) on the lab experiments properties of materials for Q=CV where he saw the forces in between particles while the Math guys all saw voltage as a force or motion from a distance between conductors. Both were correct as Maxwell discovered and he put it all together. "Holtz" also experimented in ionization and showed quantity of electricity in each discharge is independent of the strength
of the current and of the material of the electrodes, and that it depends on the nature and density of the gas, and on the distance and form of the
electrodes. He verified Faraday's discovery that the breakdown of a medium at a conductor surface was a little less when it was negative than when it is
positive. This meant electrons were -ve and more easily escaped orbit in conductors. In essence Maxwell proved all insulators are dielectrics and everything else is a conductor and explained in math how all the prior discoveries before and during this century including Newton's Laws, Faraday's Laws, Gauss's Laws, Lenz's Law, Ohm's Law were unified in what we now call Maxwell's Laws.
Yet Faraday still had some "paradoxes" that he could not solve like linked flux does not change yet an emf is induced and meanwhile Lenz accurately describe the EMF being generated from motion. It was Maxwell who explained how they cancel each other out in a motor.
Not only is it important to realize that Ic=C*dV/dt but it is actually \$Ic=CdV/dt + VdC/dt\$. Faraday's experiments showed this even though he did not do the math. Those who know about some ceramic properties may understand this significance.
I recall verifying myself in the early 80's when I was implementing ESD awareness at a large Burrough's plant in Winnipeg for large Disk Drives (HDD) and Computer systems as TE Mgr with all the best test gear. I could measure my body voltage with a little LED box and metal touch pad. First I would ground myself measure 0V then slowly raise one foot and touch measured 200V, returned my foot to the insulated ground and got near 0V again. Why? because I was changing my body shape and thus V=Q/C and while Q was constant, I changed by body voltage by changing C. ( Keep this in mind the next time you touch an LED rated for -5V, which is more robust than a 1pF FET rated for 25V.
Richard Feynman commented:
"From a long view of the history of the worldβseen from, say, ten
thousand years from nowβthere can be little doubt that the most
significant event of the 19th century will be judged as Maxwell's
discovery of the laws of electromagnetism. The American Civil War will
pale into provincial insignificance in comparison with this important
scientific event of the same decade."
This mathematical expression of Faraday's wisdom is in the appendix of Maxwell's book above when he was a young Professor of experimental Physics at Oxford.
The above shows Faraday's vision that differed from all the Mathematicians.
Note that C is defined by the space between the parallel plates and of course the dielectric constant. While the tuba effect is the stray capacitance that affects everything around. The intuition was in Faraday's books from his experiments where the mathematicians disagreed. yet both agreed Q=CV . Often today Engineers think of Caps as fixed components but they aren't and piezo ceramics are perfect example.
Thus remember that \$ \frac{dQ}{dt}= I_C(t) = C\frac {dV}{dt}+V\frac{dC}{dt}\$ is just as important than Q=CV
This may give some insights for you on the practical use of Q=CV Electronics For Dummies |
H: What determines how much power the motor 'draws'?
I'm new to electrical engineering, and probably asking the wrong question. Not sure how else to phrase it.
For the question, I'm going to use an example layout of simply a 9v battery attached to a toy DC motor.
From my understanding of my previous question, the voltage coming from the battery through the motor will be around 9v. The current will be from 0amps to however much the battery can supply without frying. What decides how much current goes through the motor?
AI: Some non-math beginner concepts:
All motors are generators. Suppose your permanent-magnet-type DC motor generates 9V when the shaft is spun at, say, 100Hz or 6000RPM.
This means that, when the motor is connected to a 9V battery, it will speed up to 6000RPM, then remain at that speed. When it's up to 6000RPM, it draws zero current (ideally.) That's assuming that bearing- and brush-friction is insignificant, and the rotor is in vacuum (so, no air is heated, with no wind blowing out of the holes in the case.) At max speed the current falls to zero because the motor(generator) is supplying 9V, and the battery is also supplying 9V of the same polarity. The two opposing voltages subtract to zero. The rotor keeps spinning at a constant RPM. And, other than friction effects, it doesn't need any energy from the power supply. For a real motor, the entire supply current is caused by unwanted friction (and also of course by any mechanical-load wattage being drawn from objects connected to the spinning shaft.)
In other words, it's not easy to calculate the current a real motor will draw from its power supply. The current depends on internal friction of bearings and brushes and wind/fan effects, as well as depending on the amount of work (wattage) the motor shaft is performing.
With DC motors, the RPM of the shaft is proportional to the drive voltage and the generated voltage. If a motor runs at 6000RPM when connected to 9V, it will run at 1/3 the speed if powered at 1/3 the volts (so, 2000RPM for a 3V power supply.) There really isn't any such thing as a 9V motor or a 24V motor, etc. Those are just the operating voltages for a minimum bearing-wear and maximum brushes-lifetime. Ideally, any DC motor can be run at much lower voltage and RPM than its spec-sheet voltage. And, any DC motor can be connected as a generator, and used as an RPM-sensor: the output voltage is proportional to rotor speed.
The rotor of any motor has mass, and can't speed up instantly. So, whenever you connect your DC motor to a 9V battery, it won't draw infinite current, and won't instantly jump to 6000RPM speed. It will however accelerate, and momentarily draw a large current while doing so.
All motors are flywheels. When first connected to a DC power supply, the motor speed increases, and kinetic energy is stored in its rotor. When hooked up and running with constant RPM, we can disconnect the supply, and the RPM will stay the same, and not instantly stop. (The rotor is an energy-storage device.) Then, depending on friction and work-wattage, the motor RPM will smoothly decrease, just like any flywheel which is performing work and losing its kinetic energy.
If we want to rapidly slow down a DC motor, we can disconnect it from its power supply, then connect it to a resistor. The motor acts as a generator (as it always does,) and drives a current in the resistor. The motor slows down, and the "braking resistor" gets hot.
In all, DC motors resemble capacitors, where the "charge" is the kinetic energy of the rotor-flywheel. DC motors draw a big current when first connected to a DC voltage. Then, the current decreases to nearly zero. Disconnect a spinning DC motor, and the power-supply voltage still remains on its terminals. What if we short those terminals? POW! A huge current appears briefly, and the rotor jerks to a sudden halt. We've discharged the stored mechanical energy, and the conductors end up at higher temperature. |
H: Designing a dental furnace control board
I have a few question about designing a heater control board . Firstly I took a power electronics lecture in the university eon ago but now I try to remember it anew. :)) Regarding my subject, I have an old dental furnace (ivoclar p80) which runs AC 230V and makes heat up to 1200 Celcius degree and it is totally 1800Watts. and I want to design my own control board for this oldy one. Actually I have searched some different boards in term of electronics component but I'm confused. Therefore may I learn which important points of should I be noted to design. For example which silicon device I should use such as power mosfet,src,IGBT so on. one that shall I use optocoupler to drive the switching device or a filter to prevent signal noise and such different questions :)) also I need to control the temp val sensitively. If you could help me it would be appreciated! Thanks in advance!
AI: Buy an off-the-shelf PID self-tuning temperature controller. Find out what the sensor used in the oven is first. You can use the controller to switch a mechanical contactor, a mercury displacement relay or an SSR. Make sure the controller matches the output device as well. Output can be a relatively long 'PWM' (a few seconds if a zero-voltage switching SSR or about 20 seconds if a mechanical contactor). If you use an SSR you will have about 8-10W to get rid of so you may need a big heatsink or a smaller heatsink and a fan.
The controllers are not particularly expensive, a few hundred dollars for a really good one, and crappy ones from China are almost free (but tend to be only type K, still probably better than something you would build on the first try).
Some have optional communications via RS485, for example. If you need to get it to talk to something, buy one with communications and you can hook it to a PC or an Arudino or a RPi etc. Some have programmable ramp and soak.
The sensor is undoubtedly a thermocouple, probably type K, N, S or R. The latter two are platinum alloy types, the first are base metal alloy types and will be much thicker wire at the sensing junction. You must maintain thermocouple wire (or proper extension leadwire), in the correct polarity, from the sensor right back to the controller terminal block. The Pt alloy sensors are very low output per degree so a better controller is indicated to maintain accuracy.
Keep the electronics cool and it will last a very long time.
If you really, really want to spend the time to develop a controller, which is not remotely a cost-effective use of time unless you intend to sell these in fair quantities, first you must figure out how to measure the sensor, including signal conditioning, cold-junction compensation and linearization, with acceptable noise level (more challenging for Platinum types) and update rate (10x per second is more than enough, less is probably acceptable. You can buy chips to do some or all of this, with varying degrees of performance and limitations.
Then you must implement a PID algorithm for control. There is plenty of information out on the web on PID- if you care about overshoot and good control it may be non-trivial.
After those two, you can proceed as if you had purchased a simple controller. Otherwise, implement timing and ramping for a more complex programmable controller. |
H: What is leakage current in an LED?
Is 'carrier leakage from the active region' of an LED the same as 'reverse leakage current'? If they're different what are their respective physical origins?
AI: The term leakage is applied to reverse biased diodes because ideally the current would be zero. Saturation means more input doesn't cause a change. So the leakage current is also referred to as saturation current because it is fairly independent of the amount of reverse bias. It is not, however, independent of carriers generated by other sources of energy. If the diode is heated or light is shined on it, more carriers will be freed up to contribute increasing the reverse saturation current. The limit of leakage current has to do with the limited number of minority carries near the junction. |
H: What is the standard way to represent fixed point numbers in VHDL?
Is there a native type in VHDL language similar to std_logic_vector that allows one to create a signed or unsigned fixed point number for given length of fractional and whole parts? If so, can it be synthesized?
How does one see fixed point quantities in wave window in test bench? Does one need to create a variable of type real and then use it to represent the fixed point quantity after manual conversion?
Are there functions in VHDL 2008 for conversion to and from fixed point and VHDL real type?
AI: If You'd like to implement fixed point arithmetics in synthesizable VHDL you have two ways:
Do It Yourself. It's rather hard method, but it may depend on quantity and complexity of equations You want to implement. Generally every "variable" should have two parts (integral and fractional) made of std_logic_vectors. Of course You have to implement basic (addition, subtraction etc.) mathematical operations.
Just download and use package fixed_pkg, it implements signed and unsigned fixed point types and all basic math - more here.
But first of all - do You really need fixed (or floating) point types? Most math could be done on std_logic_vector, just like we do in microcontrollers using int types.
EDIT:
Unfortunately, link to the libraries inside pdf guide from above link is dead. Someone delete the files. But it shouldn't be a problem - currently all ieee_proposed libraries are moved to standard VHDL2008, so just change the used standard in the properties of Your project. |
H: Configuration of AC Voltage Source for Sawtooth Wave
I have a little problem with circuit configuration as you can see below.
Every Voltage Source has configuration - 120V/10 kHz
And 1,3,5,7,9 - 0 Deg, 2,4,6,8,10 - 180 Deg
I need to change voltage on every source in such way that I will see Sawtooth Wave.
I tried several different ways, for example, I had changed all votage like 120,240,360,480,600,720,840,960,1080,1200
But it did not help me.
Can somebody give to me some advice or hint?
Thanks
UPD1:
So as I understand the frequencies of sawtooth wave is depends of harmonic's number.
I had changed frequencies: 10,20,30,40,50,60,70,80,90,100kHz
and I got something strange on oscilloscope:
Looks like I made new mistake
UPD2:
Finally everything looks almost perfect. I divide voltage by number harmonic
Thanks everybody for the help!!!!
AI: All your voltage sources have the same frequency (and I'm presuming that they are sine waves. If you want to create a triangle wave then you need voltage sources of different frequencies that are harmonically related: -
If you are considering using a real op-amp I wouldn't use 120V as your basic signal amplitude. |
H: I need help with designing my first PCB on Proteus?
I've placed the component of a simple power supply circuit on PCB layout screen using auto-placer and auto-router. I would like to know many things about dimensions:
Is there a standard space between each couple of components?
If No, What should be the space between two components that are next to each other?
Do .BMP images keep the dimensions as it is?
I mean If I exported my design using the following method:
Output > Export graphics > Export bitmap.
Then, I printed the exported .BMP image.
Will it have the same dimensions as if I printed it directly from Proteus using ( output > print layout)?
I feel like the lines that connect the components are very thin, Is that OK?
Can I make them thicker?
What is the easiest way to measure the dimensions? and How can I now the final dimensions (Width and length) of my BCP?
AI: My first unrequested advise is to read this excellent PCB Design & Layout Tutorial
My second unrequested advise is to keep away from auto-placing and auto-routing until you are more experienced. Those are wonderful tools, but they take a lot of decisions in your name, and those decisions have consequences you're probably unaware of. Designing a PCB is a process, and you should read about it.
Now I'll try to answer your questions as best as I can:
Is there a standard space between each couple of components? When you design a PCB you usually follow a grid, so components are regularly spaced in multiples of the grid step. As the tutorial says "100 thou is a standard placement grid for very basic through hole work, with 50 thou being a standard for general tracking work, like running tracks between throughhole pads. For even finer work you may use a 25"
BMP images have no scale information. Scale information specifies what is the dimension of one pixel. In consequence BMP don't keep dimensions. When you export your design for a manufacturer you should use Gerber format, also available in Proteus. Alternatively, use Output->Print to print your design.
Each manufacturer has a minimum layer thickness limit imposed by the technology he is using. Today manufacturers can do wonders, but you shouldn't be challenging them. If you plan to do your own PCB, then don't go under 25th. You can make lines thicker by right-clicking and change their style.
BMP don't have dimensions. Don't use them. Use the Output->Print action in Proteus. Print your design at 100% scale on paper, and then measure it with a ruler. |
H: Connect low-RPM motor to computer PSU
My ongoing (beginner) case-modding efforts have led me to desiring a spinning globe within my computer case. I would like to use a low-RPM motor (something between 4 and 20 rpm) for that and connect it to the PSU to power it.
I have an IT background, but absolutely no clue about electrical engineering and the utmost respect for all of the magic happening there. Therefore, I've searched a lot about the topic, but failed to find definite info, thus I turned here.
I haven't yet bought a motor because I wasn't sure which one would be appropriate, but I think I'd need a motor like this one:
My PSU is an EVGA SuperNova 80+ Gold G2 550W and has plenty of connection options.
My guess was that I could connect that motor to a 12V line, like the ones used for powering fans or components. I'm highly uncomfortable with trying though, so I'd really like to have an expert opinion...
if it is safe at all (I'd like my computer to live long and
prosper) ?
which motor would be best (like the one depicted above?) ?
which internal slot cable to use (12V processor, molex,
fan, ...) ?
how to connect which cable to which slot ?
Can anybody please help me?
AI: You should use a separate power supply for fiddling around on the bench. Tapping power off of a working computer is adding a lot of unnecessary risk. You can probably harvest a working power supply for free from an old computer. Check the net for how to get it to power up, you may have to put a load on the 5V bus and to short a couple wires to fake the soft power switch. Add a fuse to the 12V connection or you could conceivably burn yourself from a short. Try a 5A inline fuse. You can find easily this as an automotive part. Photos from here and here:
The motor you show, although specs would be nice, I can assure you will operate fine from a PC power supply. It probably draws less than 1A even when stalled.
You can use the Molex-style connectors that go to some fans and old-style IDE (non-SATA) hard disk drives etc.
The motor goes between the yellow wire and a black wire. If you reverse the polarity the motor will run in the opposite direction.
Mostly the motor you choose will depend on the torque and RPM you need. The one you show is 14RPM so it takes a bit more than 4 seconds to rotate through 360Β°.
It has a built-in gearbox so it will be difficult or impossible (depending on the type of internal gearing) to turn the output shaft manually without damage.
The torque rating is usually specified in N-m or oz-in and you will have to evaluate whether it's sufficient if it is called upon to drive with much torque.
This kind of motor is called a brushed DC motor and simply turns when voltage is applied. It draws more current when you start it and more current as you load the output shaft.
To get it to turn slower you can lower the voltage with something like PWM or a voltage regulator, however the maximum torque (stall torque) will drop commensurately. More sophisticated speed control methods measure the motor speed (with an encoder or tachometer) or the motor current and raise the voltage to maintain the speed more constant as the torque increases. Some schemes also use back-EMF as a sort of tachometer. |
H: strange results from transmitter radiation calculation
This table was taken from:http://www.emfwise.com/distance.php
At that site a formula:
Power Density = P/4Ξ r2
was used in order to calculate the power density(mW/m2)
according to the distance from the transmitter.
On that table the transmit power of the transmitter
is 30mW and you can see how the power density decrease
in according to the distance from the transmitter.
When i used the formula(Power Density = P/4Ξ r2)
to calculate the power density in according to the distance that
shown in the table,all the results are ok(as in the table).
For example:
30/4Ξ (0.3*0.3) =~26.5mW/m2
30/4Ξ (5*5) =~0.100mW/m2(~100uW/m2).etc.
But when i used a distance that less than 30cm,
i got a power density results that higher than the transmitted
power of the transmitter itself and this result doesn't make sense.
For example:
30/4Ξ (0.2*0.2) =~60mW/m2.
Why do i get that strange results?
and how can i calculate power density at distance less than
30cm,that won't give me strange results?
AI: Firstly, if the area of the surface of the sphere is less then 1m^2 it is entirely valid for the power density to be greater then the transmitted power.
A one watt laser beam hitting an area 1cm^2 has a power density of 10kw/m^2, expand that bean to cover a square meter and it has a power density of 1w/m^2, expand to 1000m^2 and you are down to 1mw/m^2 In each case the total power is 1W, its just as the area falls the power density rises.
Secondly, P/(4Pi R^2) only applies in the far field, which is to say once the radiation source can be approximated as a point (Or equivalently that the wavefront approximates a spherical rather then cylindrical wave).
You don't say what frequency this is, I am guessing at 2.4GHz where the wavelength is ~12cm, which assuming say 10 wavelengths for getting into the far field means that anything closer then 1.2M is going to be wrong because the emitter is not a point, the calculations become horribly more complicated for this case, it gets even worse when antennas have gain....
Also I thought the RF exposure limits were specified in terms of electric field strength, not power, at least in the HF bands, maybe it is different in the microwave bands? |
H: What wavelength will achieve higher speed in optical fiber?
I've got quite easy question (and my attempt to solve it). So there is an optical fiber with Dispersion index: D = -100ps/nmβ’km, length: L = 10km. The question is: which wavelength will achieve faster speed and the end of the optical fibre: Ξ»_1 = 1000nm or Ξ»_2 = 1001nm?
In my opinion if the speed of the wave is dependent on the refractive index which will be the same in the same fibre then the speed of both wavelengths will have the same speed. Is it true? Does dispersion have any impact here?
AI: First, when you talk about the "speed" of a signal in optical fiber, that's ambiguous. You should be clear about whether you're interested in the latency (the time it takes a signal to travel from one end of the fiber to the other) or the bit rate. In this case, it seems most likely you're interested in the latency, or propagation delay.
In my opinion if the speed of the wave is dependent on the refractive index which will be the same in the same fibre then the speed of both wavelengths will have the same speed. Is it true?
No. This is not true. The index of refraction of a material varies (at least slightly) depending on the wavelength of the light being considered.
In addition, in a dielectric waveguide like optical fiber, as the wavelength changes a different proportion of the signal power travels in the core and in the cladding, leading to (at least small) changes in the effective index of the fiber.
In fact, dispersion can be either negative or positive (also called anomolous and normal dispersion), depending on the wavelength and the design of the fiber. We can also engineer the dispersion properties of the fiber in some cases to optimize the fiber for different applications.
But all of that is irrelevant to answering the question, because the total effect is summarized in the dispersion parameter.
When you specify the dispersion as you did, D = -100ps/nmβ’km, you're saying we already know the effect of all those variations, and that effect is that the propagation delay through 1 km of fiber changes by -100 ps for every nanometer of change in the wavelength of the signal light.
So you don't need to worry about the physical mechanism. You just need to apply the definition of the dispersion parameter to decide whether a longer or shorter wavelength travels faster through this fiber. |
H: What makes PC SDRAM so much more expensive than the same capacity in a chip?
I'm picking a SDRAM IC for a custom embedded board and was surprised by the low prices. Was expecting orders of magnitude similar to PC ram of the same capacity.
For example H5TC8G63AMR-PBA chip costs almost 4 times less than this Corsair Vengeance 8GB (1x8GB) DDR3 1600 MHz (PC3 12800).
Digging into the datasheet [page 4] for the RAM on chip it states that it its configuration is 512M x 16 and the speed isn't clear from the name.
Is the speed the main reason for the difference?
Are there other factors (like a need for a PCB) that multiply the price?
AI: To put the comments into the form of an answer so that the question can get closed...
The SDRAM part linked to is a 512Mx16 device, 512M addresses, 16 bits wide giving a total of 8Gbits.
The DIMM linked has a capacity of 8GBytes, 8 times the capacity of the individual memory chip.
The DIMM also lists its speed as PC1600 with timings of 10-10-10. For the memory chip the -PBA at the end indicates that it is PC1600 with timings of 11-11-11 (the PB) in a x16 configuration (the A).
So while the DIMM is 4 times the price you are getting 8 times the memory that can run 10% faster. |
H: How did I survive this shock?
When I was about 10 years old playing with lamps, I accidentally picked up a wire completing a circuit for a lamp plugged into the wall (120 V AC U.S. standard). I'm not sure if it was the hot or neutral wire, but I had completed the wire from hand to hand. I had a shock and dropped the wire in slightly less than a second.
According to what I see online, which says milliamps can kill me, how did I survive for this long, with no burns or negative health effects? I remember immediately running up to my mother and saying I just got shocked by a house socket with a big grin on my face! (she was not happy!)
AI: You got lucky.
Electrocution isn't an exact science. There are multiple things which make a difference:
The current which flows through a person depends on the resistance. That in turn depends on whether the skin is wet or dry, the area of skin in contact with the wire, and a lot of other biology things.
The resistance of the things completing the circuit - whether you're touching a metal radiator, or standing on a rubber doormat for example.
The path of the current through the body matters too, it is currents through the heart that are dangerous, so getting a shock leg-to-leg is less risky than left arm to right leg, for example.
Release time. If you drop the wire quickly, there is less risk of damage. The automatic response to let go of the painful thing might happen quicker or slower depending on all sorts of biological things.
Health. Some people are just more susceptible to shocks than others. This might depend on build, body fat percentage, or just a pre-existing heart condition.
And of course voltage, source resistance and frequency of the source you're touching also matter. All in all, US mains is in the range where it's not guaranteed safe, nor guaranteed to kill, so it comes down to the factors above, and a healthy dose of luck.
Don't try it again. |
H: Can I use a multimeter to measure voltage even if amps of circuit is too high
I need to measure circuits which I expect to be 12V 19A and 5V 44A
I've got a basic multi-meter which can handle 10A/ 240V
Can I use this multi-meter to measure only the voltage on the circuits? As the amps are too high; if I try to measure amps; my multi-meter will blow, but would just measuring volts be safe for the multi-meter?
Thanks
AI: DMMs in voltage mode typically have an input impedance of 10Mohm. When put across a 12V source this means that only 1.2Β΅A will flow. This is well within what the DMM can handle. |
H: Can you short two output pins of a decoder (74LS154N)
Let's say for two inputs, the output values will be the same. So instead of using an AND gate, I was wondering if i can just short them. Please check the attached logisim screenshot to get a better idea of what I'm trying to do.
The 7-seg display will show "1" when the input is 0101 or 0111. Can I just short the two pins in the real decoder, like I've done in logisim? Or would it mess up the thing?
AI: It would "mess things up".
Data sheet says that the outputs are " High fan-out, totem-pole, low impedance outputs". This means that outputs actively pull to logic high, or actively pull to logic low. In your attempt to short one output to the other, one will pull high while the other will pull low. They will fight each other, draw excessive current, possibly overheat, and yield logical output that is undefined. Not what you want - don't take this shortcut - decode each state with logic. |
H: Noise in op amp design or EMI?
Sorry for the long post, but I've been trying to work this out in my head.
The circuit Iβm working on is based on EEVBlog's dummy load. The circuit works by using an op amp to set the voltage at a current shunt resistor and burning rest of the voltage across a mosfet.
The voltage is set at the op amp by filtering a PWM signal from an Atmega328P. This is my first mixed analog-digital board, so it seemed like a useful design to start with; not too surprisingly Iβm having issues with it. In short, I have a lot of noise in the lines connected/near the op amp.
Screenshots of the relevant schematic and pcb layout areas are in this imgur link. (I don't have enough reputation to post them all here.) When I built it on a breadboard all of the signals looked good. The oscope screenshots of the breadboard post-filter voltage (net N$2) and the feedback voltage (net N$1) show mean voltages within 3 mV of each other, so everything seemed in order.
Noisy feedback voltage (net N$1) on pcb with load present: -
When I assembled and tested the PCB, the post-filter and feedback voltages have extreme amounts of noise (see the feedback pcb with load image). The noise increases when the post-filter voltage increases as well. When the load voltage is removed (but still with the post-filter voltage set) the noise disappears; see the post-filter pcb no load image.
My first thought was that the PWM signal (net ISET) was causing the problem, so I turned off the PWM and applied a constant voltage to the input of the op amp (net N$2). This still resulted in the same noise problems. I then checked the load voltage on the pcb and found it was noisy as well. When I tested the load voltage on a breadboard with a power resistor, it's not noisy with a Vpp of only ~30 mV (it's a genuine Apple iPhone wall wart). Also keep in mind there were no issues with the breadboard implementation, so the load source shouldn't be the source of the problems.
My next thought is to cut the the feedback voltage trace (net N$2) and use a jumper to route it away from everything, but that would require destruction of the pcb. So before I do any permanent alterations, does anyone have any suggestions or thoughts as to why the pcb version doesn't work when the breadboard does? Does it have to do with the trace routing? Thank you for any help!
Here is some relevant information:
Op amp: Microchip MCP604 (DIP for breadboard, SOIC for pcb)
Mosfet: NXP PSMN022-30PL
PWM: 15.6 kHz, 0-5V
Load voltage: Apple iPhone wall wart, 5V, 1A
EDIT 1
I implemented some changes Andy suggested. The feedback voltage originally had a pk-pk of 680mV with a mean of 130mV! After decreasing the gate resistor to 50 ohm, the Vpp was 650mV. Reducing if more had no effect. I then removed the feedback voltage follower buffer. This further reduced the Vpp to 450mV. I have an updated schematic in the imgur link that I will try next.
EDIT 2
The oscillations were fixed by adding the inline resistor and capacitor between the op amp output and inverting terminal. I also increased the gate resistor to 300 ohms per the op amp data sheet.
AI: It's oscillating under load conditions because the 100 ohm resistor feeding the gate forms a low pass filter with the gate-source capacitance (well about 50% of it) and adds another few tens of degrees of phase shift and, at some high-ish frequency the phase margin is no longer a margin and it oscillates.
Two things make it worse: -
The 100 ohm resistor (as mentioned above)
The buffer op-amp from the sense resistors in the feedback loop.
You could try getting rid of that buffer op-amp and replacing it with a series resistor from sense resistors to the inverting input of the main-op. Then add 10nF feedback capacitance between inverting input and output of said filter.
Or, you could just bypass that unneeded op-amp and see what happens AND reduce the 100 ohm resistor. |
H: How does a capacitor reduce voltage drop in a DC motor start?
I know that a capacitor stores charge:
C=Q/V
but what i don't understand is how this would reduce the voltage drop caused by high current draw.
My theory is that the capacitor would need to be in parallel with a conductor (minimal resistance) and then connected to the motor.
simulate this circuit β Schematic created using CircuitLab
(Diode is motor)
By doing this, current will only be drawn from the capacitor when the power source can't supply enough.
If this is true why? and if it this isn't true how is capacitor supposed to be set up then? what would happen if the capacitor is in series?
AI: What you show is just a diode in parallel with a voltage source. There is no capacitor here since it's shorted. Removing the capacitor would change nothing.
It's not clear what you are really asking, but some types of motors have a "startup capacitor" in them. These types of motors run on AC, and don't have any torque when the rotation speed is 0. The capacitor unbalances the motor to cause some torque at 0 speed.
This capacitor would reduce efficiency at normal operating speed, so there is usually a way to switch it out of the circuit. A common means is a centrifugal switch.
Added
If you really mean a capacitor in parallel with the power supply a DC motor is connected to, then that's just a capacitor holding up a supply. There is nothing special about a motor being connected to that supply.
A large capacitor across a supply provides extra charge to the load when the supply voltage drops. This helps the supply look more beefy to the load than it really is, at least in the short term. In effect, the supply/capacitor combination is capable of larger short term current than just the supply alone.
A motor draws a surge of current at startup, so a capacitor can help. However, this motor initial current surge is "long", so in most cases a unrealistically large capacitor would be needed to make a significant difference.
A small capacitor across a motor can help to reduce emissions. The capacitor keeps the voltage more steady, and keeps the high frequency noise current circulating close to the motor. The time over which such a capacitor can make a meaningful difference in holding up the voltage is so small that this only does anything useful at frequencies that can radiate. |
H: Intel 8052 (and friends) minimum set up
I recently got a batch of old microcontrollers (8052 and Z8671) and I just want to get a minimum working set up for them.
Since these microcontrollers already have most everything baked into them (BASIC, RAM, ROM, Serial Ports), it seems like it'd be a simple matter of plugging them in to an FTDI chip and going about myself.
Looking at the Intel 8052 datasheet, it seems like all I need is a power source, crystal oscillator, and a serial port. Since the FTDI FT232RL is a very common and readily accessible chip, I'm using that for both power and serial I/O.
Here is what I'm using as a schematic.
With all that in place, and trying to use both minicom and picocom, I get no response from the chip at all.
What is it that I'm missing? I've tried using a similar set up for the Z8671 and haven't gotten a response as well.
For reference, here is the pinout of the 8052
AI: Since these microcontrollers already have most everything baked into them (BASIC, RAM, ROM, Serial Ports)
What makes you think that these microcontrollers have BASIC built in?
There was one somewhat rare version of the 8052 that had a BASIC interpreter built in. If you have one of these, it will be clearly marked as a "P8052AH-BASIC", like this one:
These parts were neither standard nor common. The ROM in most other 8052s will contain an application-specific program that's unlikely to be useful to you. |
H: Is 55ns SRAM fast enough to drive low resolution VGA?
I can't seem to wrap my head around the math.
Assuming I want a 640x480x256 VGA signal, and I had an FPGA or fast micro-controller to drive it, would 55ns be able to handle it?
Meaning each byte of the 8-bit memory would translate to one pixel. Also assuming my R-2R ladder was setup to something like RRRGGBBB.
I have a very specific reason to use DIP SRAM. I can find speeds down to about 15ns but small sizes. The 512k I want to buy tops out at 55ns.
If someone could give me examples of how the bandwidth is calculated, it might help me decide to use the 512k or maybe do something different with multiple 15ns chips.
Thanks!
AI: 640x480x60Hz uses a 25 MHz pixel clock so you'd need 1 byte every 40 ns (for 73 % of the time).
You also want to be able to write display data to your RAM so let's assume we want two RAM byte accesses in that 40 ns: one to read and display a pixel, one to R/W display data.
So you can either buy faster RAM or wider RAM. A 32-bit wide RAM using 55 ns chips could serve up four bytes in one 60 ns read and they'd last 160 ns.
This illustrates the arithmetic, you can explore the permutations further. If you're using an FPGA, you can do some clever prefetching stuff to burst-read multiple dwords and get more R/W slots in between them. |
H: Explanation of Conditions of Equivalent Electric Circuits
Two electric linear circuits are equivalent when they have the same voltage-current relation at the terminals of interest under all circumstances.
In order to check if two circuits are equivalent across two terminals you have to
Turn off the current sources and find the equivalent open-circuit resistance across the terminals in both circuits
Short-circuit the terminals and find the short-circuit current flowing there
What I can't find is why those two procedures ensure that the circuits are equivalent under all circumstances?
Why we only check for open-circuit resistance and shot-circuit current?
If you could link material or sites or books (even better) explaining the reason I would be glad.
AI: This is only true for linear circuits. If you graph the current versus voltage in the VI plane you get a line. For the open circuit, you get a point on the V axis. For the short you get a point on the I axis. Since all circuits that draw a line through the same points are the same line, they are equivalent. |
H: Question about how data sent from the tmp102 is processed
I wired up a TMP102 sensor(on a breakout board) from sparkfun. The wiring made sense, standard i2c wiring. Here is the TMP102's datasheet. I used this guide as a guide for the program that I wrote. The program that is running on the particle photon(the microcontroller board that I am using), has a confusing line of code. The line of code that I am referring to does this: The bits of the MSB that are received by the microcontroller(Most significant byte, not bit) are shifted left, 8 times and then Or'ed with the LSB(least significant byte). The bits of the result are then shifted right 4 times. Lastly, the whole thing is multiplied by 0.0625. This line of code sums up what I just said, and can be seen in the guide linked above:
int temp = ((( MSB << 8) | LSB) >> 4) * 0.0625;
I got everything to work(it measures temperature), but I don't know why it works. My question is: why is the above line of code required? Why do these operations need to be completed?. And as a side question, if the bits in a byte are shifted left 8 times, wouldn't that result in a value of 0000 0000? If my question is a bit vague, please tell me.
AI: The ( MSB << 8) | LSB) is understood: you want to concatinate the two bytes together. Apparently, in LSB only the [7:4] are valid data. So, you >>4 to get rid of the [3:0] bits. Now, the 0.0625 multiplication is not clear, but it is the same thing as dividing by 16, or performing another >>4 bit shift. The end result in the temp is the MSB only.
MSB: 1010 1010
LSB: 0001 XXXX
MSB<<8 1010 1010 0000 0000
LSB 0001 XXXX
when ORed
1010 1010 0001 XXXX
when shifted >>4
0000 1010 1010 0001
when multiplied *0.0625
0000 0000 1010 1010=MSB |
H: Switching between +9V and -9V (using 5V logic)
I got some linear actuators today and found out that when I give them +9V they extend and -9V makes them contract. This is easy enough to test by hand, simply by wiring the actuator up to the battery backwards for the negative voltage, but when I hook it all up I'm going to be driving all of them from my Atmega1284, which uses 5V logic.
In anticipation of this, I got a bunch of relays that run on 5V logic as well so I can control whether or not there's +9V running through with my 5V microcontroller. However, I have no way to programmatically provide -9V.
My question here is how can I switch between giving +9V and -9V when all I can do from my microcontroller is either push out 5V or 0V? Will I need more than just my relays? Is there some spiffy EE trick to make this work? Thanks!
AI: You can use two SPDT relays to select +9 or -9 or 0V.
Both energized = off, Both de-energized = off, and if one is energized and the other is not it will drive the actuator in one direction or the other.
simulate this circuit β Schematic created using CircuitLab
To suppress electrical noise, add a bridge rectifier (W04G, for example) across the actuator (the AC input terminals) and connect the + terminal to +9 and the - terminal to 0V |
H: Multiple power supplies, why link negative
I'm new to electrical engineering, and I still have many stupid questions. Here's one of them.
This is what I'm reading. https://itp.nyu.edu/physcomp/labs/motors-and-transistors/using-a-transistor-to-control-high-current-loads-with-an-arduino
In it, it says to link the external power supply's ground (negative) to the Arduino's ground (negative). Why do we do this? Also, could you link the positives instead?
The ground of the motor power supply should connect to the ground of the microcontroller, on the breadboard.
Also, why do we have to do this:
Next, add a diode in parallel with the collector and emitter of the transistor, pointing away from ground. The diode to protects the transistor from back voltage generated when the motor shuts off, or if the motor is turned in the reverse direction.
Thank you.
AI: In the application you linked, the transistor is part of two circuits, so we need a common reference between the circuits to enable the transistor to operate correctly.
simulate this circuit β Schematic created using CircuitLab
If there was no connection between points A and B, the microcontroller could not control the transistor bewcause there would be no path for the base current from the microcontroller to get back to the microcontroller's power supply.
If the connection between points B and C was removed, the relay would not operate, regardless of the state of the transistor, because there would be no path for the relay coil current to make it back to the 12 volt battery.
Having the common ground connection between the two power supplies allows the circuit to operate as required.
It would be possible to have a scenario where the positive terminals of the supplies have to be connected instead of the negative - it all depends on where you need a common reference point.
...edit after revised question...
Diode D1 is a flyback or spike suppression diode. When the transistor turns off, the magnetic field in the motor or relay (or other inductive device) will colllapse rapidly, and would create a high voltage spike on the transistor's collector. The diode will conduct the current that the collapsing field generates, preventing the voltage spike from forming. |
H: What is the purpose of a flyback diode?
I'm new to electrical engineering, and I still have many stupid questions. Here's one of them.
This is what I'm reading. https://itp.nyu.edu/physcomp/labs/motors-and-transistors/using-a-transistor-to-control-high-current-loads-with-an-arduino
Next, add a diode in parallel with the collector and emitter of the transistor, pointing away from ground. The diode to protects the transistor from back voltage generated when the motor shuts off, or if the motor is turned in the reverse direction.
I thought before that polarity could be reversed in transistors, but now I'm confused. Can it not be? Why would it hurt the transistor, and how would it? Would it instantly break it? How does doing this fix it?
Thank you.
-edit to defend not being duplicate-
The proposed duplicate does not answer my question because it does not give how the transistor is affected. From searching online, I couldn't find this either; I'm probably searching the wrong thing. It also does not answer whether it is instant or how it fixes it.
AI: In normal use, a small current on the base of the transistor allows a much larger current to flow between the emitter and collector.
In this case, however, we have a large voltage applied across the emitter and collector, with (presumably) little or no current flowing through the base, so the transistor is trying to stop current from flowing between the emitter and base.
In this case, it's probably useful to think of the transistor as a pair of diodes, with (depending on whether it's an NPN or PNP transistor) either their anodes or cathodes connected to each other.
One of these diodes is going to conduct current from the emitter/collector voltage. The other is going to try to block it. One question, then, is what's the breakdown voltage of that "diode" (that junction of the transistor). If the voltage being applied exceeds that value, you're liable to toast the transistor.
Now let's consider that added diode, and what it does.
We connect the diode so the transistor's power supply is trying to push current in the direction that the diode won't conduct. Therefore, when the transistor is operating normally the diode basically has no effect at all1 (unless we choose the wrong diode, such as one with a breakdown voltage lower than our power supply voltage).
For a voltage in the opposite direction, however, the diode looks/acts pretty much like a dead short. This means all the current from that reverse voltage flows through the diode. Since it acts like (nearly) a short circuit, no voltage can be developed across the transistor, preventing any damage to the transistor.
1. "basically" meaning that, for example, it will add a little extra capacitance. If we were dealing with a high enough frequency, we might need to figure that into calculations about how the circuit works--but for a typical motor driver, we're dealing with low enough frequencies we don't normally care. |
H: Precise line regulation using linear voltage regulators
I am trying to build a linear 24/12V power supply. One of the requirement for it is for the line regulation to be very precise (an error of 0.0005 V is considered acceptable). I wanted to use a linear regulator Ic, but I haven't found any that fit my needs.
What can I use to build such a supply?
Thanks!
AI: From what you're saying, it sounds like you want to do this from ac wall power? Please clarify. Although you say you want to use a linear voltage regulator, I would caution against that. It may be better to combine a transformer, a full bridge rectifier, and a switching power buck (or boost) power supply that meets your quality requirements.
Linear voltage regulators will have ripple, and because many things don't use them anymore, they aren't subject to the kind of innovation that switching supplies get. Linear power supplies remove power as heat. This can be calculated by the equation (Vin - Vout) * I = Heat (in Watts). It is because of this wasted power that most higher voltage things don't use linear supplies. Of course, arduino boards use them because the draw very low current, and it much less space consuming.
If your goal is less supplying power and more a precise voltage, consider a precision voltage reference. These can be as cheap as $5 if you're willing to buy from china.
In conclusion:
Linear power supplies step down voltage with considerable ripple based on many factors, like supply voltage, and also produce heat
This heat makes them a poor choice for situations requiring high voltage, long battery life, or both.
A switch mode power supply may be good if you can find one with that precision if you still want efficiency
Also, consider a precision voltage reference if you need very high precision. |
H: RC anti-aliasing filter for a pulse train
Many data-acquisiton boards are recommended to use anti-aliasing filters between the analog inputs and the ADC.
Here what I understand from anti-aliasing:
If one has a DC like signal and the freq. of interests of the signal is not more than 10Hz one can use an RC low pass filter with a 20Hz cutoff freq.
But what if the signal is a pulse?
I want to make an RC aliasing filter for pulse frequencies changing between minimum 3Hz up to 500Hz.
My sampling freq is 4kHz.
In this case, one might use an RC filter with a cutoff freq of 1kHz but this could effect the rising and falling edge shapes significantly. I still want the pulse shape as sharp good as possible.
Is there a rule of thumb for an RC antialasing filter if the input is a pulse in daq systems?
AI: Do you know about the Fourier transformation ? If not, look here.
The Fourier transformation relates signals described in the time domain (for example: a pulse) to their frequency spectrum, i.e. the frequencies they contain.
If your pulse signal has frequencies components above a certain frequency which you filter out (with your anti-aliasing filter) then you change the signal. Indeed steep edges in the signal result in high frequency components.
An ideal pulse or squarewave signal with rise and fall times of almost zero will have frequency components up to infinitely high frequencies !
You will have to figure out what you need, what are the rise and fall times of your pulse. If these are for example 1 ms (0.001 second) then the bandwith of this signal is in the order of 1 kHz (1 / 1ms) so a 4 kHz filter will not affect this signal much.
If the rise and fall times of your pulse are 100 ns you need a 10 MHz bandwidth. |
H: Intuitively, why does gain margin and phase margin infer instability in feedback system?
On MATLAB https://www.mathworks.com/help/control/ref/margin.html, gain margin and phase margin are additional gain in the system and additional delay in the phase of the system such that the system will be unstable.
Can someone explain with an example as to:
why increasing the "gain" of your system will cause instability,
why delaying a signal will cause instability? This doesn't seem to be very intuitive at all. Because if I delay my sine by 2Ο I get back my sine again.
How does gain and delay translate into physical components inside of a feedback system (say a circuit). Intuitively, a gain is an op amp, what about a delay?
Also, what does instability here refer to? Are we talking about the circuit going into oscillation or blow up behavior?
AI: why increasing the "gain" of your system will cause instability
If you have a servo control mechanism and you set a demand, the servo should rotate (or move) to the position demanded and all is good. However, if you have too much gain, the servo rapidly moves off in the direction needed but overshoots the target due to the momentum built up in the rapid acceleration. This may eventually settle down or it may just continue to oscillate (unstable).
why delaying a signal will cause instability? This doesn't seem to be
very intuitive at all. Because if I delay my sine by 2Ο I get back my
sine again.
You've basically described the reason why sustained oscillations might occur - basically you don't want a delay because a delay is likely to cause problems as you have described i.e. sustained oscillations.
How does gain and delay translate into physical components inside of a
feedback system (say a circuit). Intuitively, a gain is an op amp,
what about a delay?
Gain might be from an op-amp or, you might have a totally digital control with ADCs and DACs. An op-amp might be used as an integrator rather than strictly speaking a gain stage AND, if you apply gain, integration and differentiation (maybe three op-amp circuits) you get a PID controller: -
Pretty picture taken from here (a public domain image).
If you study the above you will see it moves through three phases.
Purely applying gain until there is potentially too much overshoot but there is still a basic control inaccuracy
Then applying an integration term to improve the basic control accuracy but it risks creating too much overshoot
Applying a differential term to the above to restore sensible operation by dramatically reducing overshoot.
Also, what does instability here refer to? Are we talking about the
circuit going into oscillation or blow up behavior?
I think I've covered this. |
H: Questions on acquiring signals in Peak Detect mode in an oscilloscope
I just learned that one must set the Acquire feature of a scope to "Peak Detect" to see the possible glitches of pulses.
But in the following document it also mentions that:
"This mode is effective when at the horizontal scale setting of 5 ms/division or slower." Here is the source:
http://www.tek.com/manual-topic/acquire
How reliable is this information?
Do one really needs to set the horizontal scale to 5 ms/division or more to make use of the peak detect mode? Why is that?
AI: Since there is a no continuous sampling above Peak Mode, glitches may be missed with , capture, processing time, display and repeat missing events
Peak Detect. Use Peak Detect acquisition mode to detect glitches as narrow as 10 ns and to limit the possibility of aliasing. This mode is effective when at the horizontal scale setting of 5 ms/division or slower
.
Note: If you set the horizontal scale (seconds/division) setting to 2.5 ms/div or faster, the acquisition mode changes to Sample because the sample rate is fast enough that Peak Detect is not necessary. The oscilloscope does not display a message to tell you that the mode was changed to Sample
However , this scope can miss glitches <10ns unless they are continuous rep. rate sync'd to fundamental or harmonic of sweep speed with sampling rate up to max rate.
Although, if there are no triggers, a glitch can be reliably captured on manual trigger mode at max sample rate up to 1/2 of sample time (best case)
Displayed below is an 8ns glitch at -10.8V with 1ns rise time, that would not be captured in slow "Peak mode" on your DSO. (not slow poke)
Note the interpolation by straight lines on this old LeCroy.
Whereas your DSO would produce nice textbook waveforms on single shot capture with 50 Ohm termination only. |
H: Powering a MCU from a battery without a regulator
I have seen some development boards (for example. BL652 dev kit) for low power chips have battery power connected directly to the MCU without a regulator.
For the example case, the battery used is a 3V CR2032. The datasheet for the MCU defines the following parameters:
datasheet page 16.
Absolute Maximum Ratings Min Max
Voltage at VDD_nRF pin -0.3 3.9
datasheet page 17.
Recommended Operating Parameters Min Typ Max
VDD_nRF 1.8 3.3 3.6
I'm interpreting this as "If your battery voltage drops to a value between 0-1.7 it isn't defined what will happen".
Why this worries me is because I've seen regulators having the Power Good pins and have found no explicit statements in the datasheet that the MCU from the example won't be damaged by the undervoltage.
How can I decide if a regulator is needed between a battery and a load, to guarantee there are no damages when the battery voltage starts dropping?
AI: If your battery voltage drops to a value between 0-1.7 it isn't defined what will happen
This is often true, but it won't, for sure, destroy anything. Because, if it was destructive, the min Vdd in "Absolute Maximum Ratings" would have been given as a positive value (which I have never seen in any datasheet, and I hope I'll never see that in my life - it wouldn't make sense).
So at this point, you are guaranteed the MCU won't be destroyed with undervoltage. However, it could still behave erratically (potentially damaging other external circuitry).
Now, in this kind of MCU, there is often a feature called "brown-out detection", or, sometimes, "undervoltage lockout". This is a feature that monitors the supply voltage and guarantees that the chip is held in reset state when the voltage is under a given level (sometimes programmable).
Good news: There is such a feature on the specific chip you're using. See chapter 5.1 in the datasheet you linked.
Therefore, you don't need to have a regulator with "power good" detection or an additional supply monitor circuit in your specific case.
Note that, if the MCU didn't have the brown-out detection included, there are tiny chips that just offer this feature (often combined with a timed power-on reset generator) without being voltage regulators. |
H: Monostable out of NAND gates
For the diagram below, I can see that it functions as a monostable (I have built and tested it), but I can't really see why it works, I mean, I can see that there is an RC network, which a monostable has, but I don't see how it links up with the logic to make a monostable.
Any help would be appreciated
AI: A better drawing of the same circuit can be found here.
Where there is also an explanation.
In your circuit one of the inputs of U2 is connected to Vcc but that makes no difference to the behavior of the NAND gate. Make the truth table for a NAND and you'll notice that in both cases U2 acts as an inverter.
When the button is pressed, the output of U1 will become high, it was low so Ct was not charged (it discharged via Rt). Since Ct is not charged V1 will also be pulled high and a current though Rt starts to charge Ct.
As V1 becomes high, U2's output becomes low (U2 behaves as an inverter).
This low signal gets back to U1 and keeps the output of U1 high. So Ct can charge. As it charges the voltage V1 drops untill U2 flips over and it's output becomes high. If you have released the button as well, this will cause the output of U1 te become low again as it was before the button was pressed. |
H: How to calculate current consumption of this sensor?
I am calculating the current consumption of an IR ranging detection sensor. Below is the related part I captured from the sensor's specification. The working mechanism of the sensor is that it will send out an IR beam, and then measure the elapse time when it receive back the beam.
From the table below, I think it means the sensor will take 33ms to take one measurement. So what does the "Active Ranging average consumption" actually means? How to calculate the current consumption if I take measurement every 0.5s, 1s and 2s?
AI: The active ranging current is the current while measuring, the timing ranged inter measurement value is the current when not ranging. To get the total current multiply the two currents by the proportion of time you will be in each of the two states.
Assuming it takes 33ms for a range then your average power consumption in mA is:
0.033*n*19 + (1-0.033*n)*0.016
where n is the number of ranges per second.
So for 1Hz (n=1) the average current will be 0.64mA. For 10Hz (n=10) it would be 6.28mA.
With a nominal voltage of 2.8V this gives a power draw of 17.6mW Vs the 20mW number given in the table. However the 20mW is a maximum current consumption. The operating currents don't state a maximum but the standby currents do, they indicate that the maximum can be 40-50% over the typical values. Applying the same margin to the transmit currents would give 24mW power draw maximum. This is too high but it is reasonable that it's an over estimate, standby current is largely due to leakage currents which are more prone to process variability than operating currents. If we were to assume that the max operating current was only 10% over the typical then we get:
Max current = 0.033*n*19*1.1 + (1-0.033*n)*0.016*1.4
Which for 10 Hz ends up working out as 19.4mW.
So with a little bit of hand waving and reasonable assumptions it all seems to just about add up. A maximum current of just over 10% of typical seems like a reasonable tolerance to me. |
H: How much silicon is required for the most common ICs?
Where can I find information about the die surface for various ICs? I'm thinking about opamps in the first place.
The surface area is of course dependent on the features of the ICs (for opamps rail to rail, speed, noise) and on the max current they can handle.
I think the process technology affects digital electronics, but not much analog circuitry.
So, where can I find the surface area for various common ICs and especially opamps?
This is out of curiosity, I don't have a need for any project.
AI: Parts that are available as bare die, this information should be in the datasheet.
Some older parts included this information even if they weren't generally available as bare die, since it might be useful for estimating reliability of the part.
For example, here's the AD621 (actually an in-amp, not an op-amp, but it was the first PN I thought of):
So, that one is about 9.1 mm2. (I dunno how they screwed up the inch measurements so badly). |
H: Mask some part of devices in schematic
I have a question about working with Altium Designer Software. I put some devices in schematic, but I do not want all of them importing to PCB. I mean I want to mask some part of schematic in such way that software does not import those parts into PCB. Would you please tell me how to do it?
Thanks in advance
AI: Right-click on the component and select properties. Set the characteristic you want in the drop-down box on the dialog:
Mechanical parts will not show up on the PCB but will be included in the BOM. |
H: Changing the frequency to an induction motor
I am currently having an issue with an induction motor driving a conveyor belt too fast.
It the connected to a Frequency inverter which is connected to a plc.
The inverter is sat to output 50 hz. But i need it to run in 4 hz or less.
Would it do any harm running it so low?
Anything i should be aware of before i change it?
This is the ratings on the Motor :
AI: A current design inverter designed to control the speed of an induction motor will be able to operate the motor at rated torque at nearly zero speed. The problem will be keeping the motor from overheating without the motion of the rotor moving air though and over the motor. You can operate the conveyor very briefly at the speed you want, but not continuously.
If the motor makes a lot of noise, the inverter may not be adjusted properly, but some additional noise is normal. You can check the current, but many ammeters will not work properly with the inverter waveform and frequency. The inverter may have a motor current display. The inverter manual should help with getting it running properly. |
H: Differential gain of amplifier with current mirror and NO LOAD
Please consider the differential amplifier (Q1&Q2) with current mirror active load (Q3&Q4) as shown below:
simulate this circuit β Schematic created using CircuitLab
I know many questions have been asked already regarding this circuit on EE SE, but I didn't find an answer to my question.
My question is: what will the output voltage be when subjected to differential input WITHOUT ANY LOAD AT THE OUTPUT? Framed differently, what is the differential gain with no load?
In my understanding, the currents in the collectors of the 2 right transistors will not agree and thus the output impedances at the collectors will come into play. However, I do not know how to calculate this output impedances and how they will affect the output voltage.
Side note: If you think that asking about the output voltage with no load is meaningless than consider a load whose input impedance is comparable or larger to the output impedance looking into the collectors.
AI: Assuming that you are talking about the gain at low frequencies ("DC gain") a derivation of the gain will result in an expression like
$$
Av = g_m (r_{o2} || r_{o3})
$$
where \$g_m\$ is the transconductance of Q2 or Q3 and the \$r_{o2}\$ and \$r_{o3}\$ is the output resistance of these transistors.
The gm of the transistor is easily found to be
$$
g_m = \frac{I_C}{V_T} \qquad V_T = \frac{kT}{q}
$$
The output resistance is a property of the transistor and usually described by the Early voltage \$V_A\$. Using the Early voltage the output resistance is
$$
r_o = \frac{V_A + V_{CE}}{I_C} \approx \frac{V_A}{I_C}
$$
since usually \$V_A \gg V_{CE}\$.
Assuming that both transistors have the same ro the gain is
$$
Av = g_m (r_{o2} || r_{o3}) = \frac{I_C}{V_T} \frac{V_A}{2 I_C} = \frac{1}{2}\frac{V_A}{V_T}
$$
With an Early voltage of 100V and a kT/q of 26mV we get a gain of about 2000. |
H: Matte Film On LCD Displays
What is the technical name of the matte material/film that is put on LCD Displays?
Also what would be the method of applying such film on a blank lcd screen (if possible)?
EDIT: (Not my picture, but it illustrates what I'm talking about, left is film and right is no film)
EDIT 2:
Just in case you were wondering, I'm planning on using this for a few portable tablet-like monitors that i want to also use outside and this seems good for reducing glare and improving overall look of the monitor. The screens i current want to use this on are 16.5 inch and 5.5 1080p lcd displays
Would i be able to achieve that? What are The Downsides?
AI: It is called diffuser. It homogenizes the light from the back or the edges of the display to create a homogenous apparel of the background illumination.
Alternatively it is possible, that you mean the anti glare or matte foil, which reduces specular reflection on displays.
According to your edit2, I'm quite sure, that you mean anti glare foil, which is a good choice for outdoor applications. You should also consider a rather bright background illumination, as it needs to concur with the huge amount of daylight which is reaching the eye of the user. |
H: Arduino: Did I damage analog joystick clicker by using analog instead of digital input?
I am using a thumb joystick for Arduino UNO, trying to work with it.
There are x and y values that I get with analog input A0 and A1, and one z value for the click that I should have gotten from the digital input D7, but unfortunately I wired it to another anolog input (A3).
This is the wiring from the specification
So, now that I fixed my mistake, I get the x from A1 and y from A2, but z is always zero. Is there a chance that the joystick part is physically damaged? Or worse, my Arduino UNO? How can I check?
int sensorPin = 5;
int value = 0;
void setup() {
pinMode(7, OUTPUT);
Serial.begin(9600);
}
void loop() {
value = analogRead(0);
Serial.print("X:");
Serial.print(value, DEC);
value = analogRead(1);
Serial.print(" | Y:");
Serial.print(value, DEC);
value = digitalRead(7);
Serial.print(" | Z: ");
Serial.println(value, DEC);
delay(100);
}
Chinese specific specification is here. I wish it was also available in English! :-/
AI: Did I damage analog joystick clicker by using analog instead of digital input?
No. Neither mode will draw damaging amounts of current through the device.
But it does look like you've set pin 7 to be an output, and are trying to read digital values from that pin. Is that what you intended? :)
Make it an input instead and generally give your code a close examination line by line.
Further info:
Setting a pin to be an input will configure your microcontroller to use "High-Z" mode on that particular pin. This means that the pin becomes a high impedance input ("Z" is the standard letter for impedance). This means that almost no current is able to be either drawn from, or sunk into, that pin.
While in High-Z state (also known as "tri-stated", or simply as "input") this pin is able to do either digital input readings (logic levels... 0's and 1's) or analog readings (an 8-bit analog input is able to give you a value between 0x00 and 0xff), which you can scale mathematically to a voltage between 0V and VCC. |
H: 2n2222 NPN low side switch not working with 31V
I'm attempting my first slightly more advanced circuit and failing. I have a led string which I'm powering with a 31V 0.21A power adapter that came with it. Now I'm trying to switch it on and off with a 2n2222 NPN transistor which I'm controlling with a Raspberry Pi 3.3V GPIO. Initially, I had an optocoupler in there also to keep my Raspberry safe but took it away to confirm it isn't the problem.
I have connected the emitter to ground, collector to the LEDs and the GPIO to base via two 220 resistors, achieving 440 ohms.
Now, putting the GPIO to high or low doesn't affect the lights (at least noticeably). Instead, the lights are steadily burning with a faint glow. Removing the GPIO from base pin turns off the lights completely. I'm having a hard time understanding what's going on.
Some questions that might solve this:
I based my resistor calculations on this tutorial assuming 100 hFE. I'm not sure about the hFE value. The 2n2222 datasheet doesn't really tell what it is with 31V and 0.21A. Any help here?
The 31V is quite high (weird choice by the light manufacturer in my opinion) but the transistor can handle up to 40Vce and 1A. Am I misunderstanding something here?
One option is that I somehow got high and low the wrong way. Does the behavior sound like it could be caused by an inversed circuit?
Solution
Thanks to @WillDean I realized I was using two grounds which was messing up a whole lot of things. I didn't realize this is an issue before the obvious was stated.
AI: You need to connect 0V of your RaPi to the emitter connection on your power supply because you cannot drive dc current into the base without a return path for that current.
But, before you do that make sure that the 31 V dc supply can connect to 0V on the RaPi safely i.e. is the 31 V supply isolated from ground?
It may be OK if it isn't but tread carefully because I don't want you to destroy your RaPi. |
H: listed ac motor horsepower doesn't match voltage and amperage
In my search for a bandsaw, I've come across a model that has an AC motor whose plate displays 3/4 Horsepower, but also 115/230 volts, 11.2/5.6 amps. You can see that the math suggests (115 x 11.2 = 1288, 1288 / 746 = 1.7) a horsepower rating which is double what is stated. For the money, I'd much rather assume the higher number. How can the two very different figures be reconciled?
AI: You need to consider both efficiency and power factor. Output power = V X A X eff. X pf. 0.44 is not unreasonable for eff. X pf, but it would probably be higher for a modern motor. Service factor is allowable sustained overload. You can operate a motor at the service factor without any immediate bad effects, but it will be running hotter and a lot of operation at the service factor will result in a shorter than expected service life. |
H: How does a bathroom scale automatically power on when I step on it?
When I step on a bathroom scale how does the circuit turn on? I've disassembled these types of weight scales looking for obvious hardware switches, but none exist. I suspect the ADC is configured to generate an interrupt and wake the micro controller from a low power mode. If this is the case, how would this spec be translated to a common micro controller such as a MSP430F67 with its built-in 24-bit ADCs?
AI: The scale is never really "off". Instead, it just turns off the display and other peripheral circuits such as the ADC and goes into a low-power mode internally. Every now and then (most likely based on a timer interrupt), it powers up the ADC briefly and checks the sensor to see if the weight has changed, and if so, it turns the rest of the peripherals, including the display, back on.
In low-power "sleep" modes, many microprocessors (and especially the MSP430 series) have a level of current consumption that compares favorably with the self-discharge rate of the battery. |
H: What does low-impedance mean in the CMOS inverter circuit?
For the CMOS inverter the text states "once the transients have settled, a path always exists between VDD and the output realizing a high output (βoneβ), or, alternatively, between VSS and output for a low output (βzeroβ). This is equivalent to stating that the output node is always a low-impedance node in steady state."
What does the last statement about low impedance mean?
AI: Mosfets can be used as switches. In a cmos buffer one of the switches is to Vdd and the other is to Vss. A switch has two states; open or high impedance; or closed or low impedance. So when output of the cmos buffer is high, the switch to Vdd is closed. We can tell that this is low impedance by adding a load and seeing the effect. At no load the output will be Vdd. If we add load up to the Ioh specification, the output will stay close to Vdd indicating a low impedance path to Vdd. Similar for low to Vss. |
H: Shorted 12V wire to GND in ATX power supply. Did I fry it?
I'm a total noob, please forgive me.
I have several components that require 12V DC, and I decided to use an old computer power supply to give me the juice.
I snipped and stripped a matching pair of yellow and black wires...
After marveling at the results of hotwiring the electronic lock, I decided to test out the EL wire I bought ages ago.
However, I could not resist touching the wires together to see what a 12V spark looks like and now the power supply's fan stopped.
I tested the outlet, and it's fine. Did I just fry my power supply? Or perhaps I blew a fuse in the thing?
Why did my PS stop working and how might I fix it?
AI: Posting comments as an answer, since apparently it helped OP solve his problem.
Some PSUs have output short protection, some don't. Some would have survived, others wouldn't.
There's really no way to tell just from the info in the question. I've accidentally shorted some PSUs before and most of them have survived. It really will depend on the model. If it doesn't turn back on again then obviously something is damaged inside, possibly a fuse, but it could also be one of the output devices, in which case it'll be much harder to repair.
When I say "survived" I mean that most of the times when PSUs with output protection are shorted, they turn off, you power cycle it (remove its power input/disconnect the it from the outlet), and they will turn back on again if the fault condition was removed. I looked up this unit on google and apparently its a very cheap PSU, some buyers even received it with some parts of the wiring missing apparently.
I'd tell you to check the fuse but I think the energy in the PSUs capacitors would be enough to blow a component inside of it. |
H: Arduino Nano: How to decide voltage divider values that are safe for the input of the ADC
I am building a simple voltmeter circuit that needs to measure up to about 35 volts. Looking around the web, I am finding all sorts of suggestions for voltage divider values, like 1 Mohm / 10 Kohm, and the one I am currently experimenting with, which is 10K/2K. I am having issues with allowing enough time between readings for the Arduino Nano to settle down, of course, but that is not the question. Using 10K/2K it maxes out at a reading of 1020 on A6 with an input of 27 volts.
I do not understand why one posting says to use resisters in the Mohm range and the other (that I am using) is a total of 12k.
Question: Is there a rule of thumb that should be used to decide what resistances should be used for the divider for measuring this level of voltage?
And how important might it be to keep the resistance between Vcc and GND higher than what I am using?
--
EDIT: EE asked me if this was the same as another suggested Q/A - and this is only different because it is specific to the Arduino's ADC input impedance. The other answer only talked about the ratios.
The answers to this are great for my particular application.
--FURTHER EDIT: Based on the collective berating from the powers that be, I would mention that this started in an Arduino forum where I was concerned about the particular ADC, then it got moved here. Calling this a duplicate is incorrect. It is very specific to the Nano application. I wanted to make sure my small values wouldn't cause problems by being between Vcc and GND. Plus, I did not know the proper input impedance for the Nano's ADC. It all got answered here, nicely.
Thus I believe it stands as being unique here. I can find no duplicates of this particular question. Being marked this way with no explanation is frustrating. If there really are duplicates then I would like to see their answers. But finding none, this one adds to the community and I believe it should not be marked as a duplicate here.
And the good thing is that one respondent brought up smoothing capacitors. So now I am using one, and it helps stabilize the readings. So this has been a positive experience.
The problem now is that Arduino people won't necessarily know to look here for this answer when they try to use their Arduinos this way. So there will likely be duplicate questions there which just got answered here.
AI: There are two things to consider:
The ratio of values
The input impedance of the ADC
Point 1 has hard and fast formulae. No rule of thumb here.
Point 2 is a little more "thumby". You want R2 (the lower resistor) to be low enough that the input impedance of the ADC doesn't skew the readings much, yet large enough that you dont get excessive current draw through the divider. As a rule of thumb I work to a 10% rule. R2 should be no more than 10% of the input impedance. That means for an ADC with 1M input impedance you are looking at an absolute maximum of 100K for R2. Anything more and you start to see bad drift. Better is to go lower, but you don't want to go too low. 10K is a good value. Only 1% of the input impedance yet high enough to not waste current.
TO get more specific refer to the datasgeet.
The ADC on the ATMega328P is:
... optimized for analog signals with an output impedance of approximately 10 k⦠or less. If such a source is used, the sampling time will be negligible.
The output impedance of a voltage divider is basically the two resistors in parallel \$(\frac{R1 \times R2}{R1 + R2})\$, but if R2 is considerably smaller than R1 then it will be approximately R2, so about 10KΞ© is the maximum the ADC likes to have as R2, though it can be higher if R1 allows the impedance to drop to 10KΞ©.
For instance, if you have your example of 1MΞ© / 10KΞ©, the impedance is \$\frac{10,000,000 \times 10,000}{10,000,000 + 10,000} = 9990\Omega\$ which is about right for the ADC. |
H: Are built-in ADC converters in MCUs as reliable and accurate as their standalone counterparts?
This question is inspired by this comment by KyranF:
"built-in 24-bit ADCs" I wouldn't trust that as far as I could throw it.
How does a bathroom scale automatically power on when I step on it?
As the title plainly states, is there any foundation for this comment and are built-in ADCs any less accurate than a dedicated chip, and if so why? What parameters on an ADC should be looked at for accuracy and precision?
AI: Are built-in ADC converters in MCUs as reliable and accurate as their standalone counterparts?
Why shouldn't they? You know, every device you buy has their specs, and they will strive to follow them, because if they don't, they risk their company's good name or worse.
β¦ comment by ${some Person on the Internet}
I ${ranting about something} as far as I ${totally unverifiable problem}
Well. Yes. There's someone on the internet voicing a negative opinion without backing it with facts. That's not really new[citation needed] β but anyway, I think it's safe to simply ignore this.
Look at whatever you want to digitize. Make sure you properly filter it. How high is your noise after that? You don't really have to have an ADC that can do better than the dynamic range between that average noise amplitude (\$\sigma\$ for things that follow central Gaussian distributions) and maximum signal amplitude. Figure out that dynamic range, and then figure out the Effective Number of Bits you need (ENOB). Often, your requirements are even lower β the bathroom scale example is very good: To be honest, a bathroom scale more accurate than 200g is senseless β you simply can't tell the difference whether a person drank a glass of water or gained weight. Let total range be 1 kg β 200 kg, then you need but a resolution of \$\frac 1{1000}\$ of the maximum value β in other words, a 10 bit converter will happily do. People still tend to buy the devices that say "\$\pm 10\text{ g}\$", so you'd need another factor of 20 to make marketing happy β that's about 4.5 bits. So, with an ADC that has 14 bits, you'd be better than any customer needs, and marketing wouldn't even have to lieΒΉ.
Note your signal's bandwidth, and derive a minimum sampling rate from that (typically \$f_\text{sample} > 2\cdot f_{bandwidth}\$).
Your ADC's datasheet, be it a built-in one, or be it a separate one, will clearly state what it can do. It will state an ENOB or noise power, and it will clearly state the raw number of bits you get (which is > ENOB due to the noise in an ADC).
Don't let trolls on the internet troll you. The market has a clear indication for you: There's so many engineers that wanted to have ADCs integrated in their microcontroller, that basically every microcontroller at least has a variant with a built-in ADC. That might mean having those is actually useful.
Sure, they're not designed to give you 14 effective bits at 200MS/s, or 20 bits with a bias current of 10pA at 1MS/s β but that's not your use case, is it?
ΒΉ they usually don't give a damn about that, though, for hard-to-verify senseless features in consumer products. Buy a consumer stereo and read the claimed sound output power. Use at full volume until batteries are empty. Patent that technology of glorious energy creation as perpetuum mobile.
Marketing is basically the opposite of that random person you quoted: They go out and make positive remarks unbased on any factual thing, but on a random aspect of the system they don't really understand. |
H: Oscilloscope saving data unexpectedly
I'm using a Tektronix TDS 1002B oscilloscope.
I was investigating some glitch issues on pulses/square-waves and decided to save the time-series data to a USB drive in CSV file format to plot in MATLAB.
Input to the scope is a square-wave or pulse.
When I set the scope's Acquire feature to 'Peak Detect' setting, and observe the signal on the scope's screen live I only see over-shootings at the rising edges but the pulse seems it has no glitches.
But then I save this signal data(using Save feature) to a USB drive in CSV file format to plot in MATLAB.
In the following example, the input to the scope is a 50Hz 4Vpp square-wave.
When I set the Acquire to 'Peak Detect' and saved the data in CSV format, here is what I plot in MATLAB:
And here in the zoomed view I see glitch on each falling edge:
On the other hand, when I set the Acquire to 'Sample' and saved the data in CSV format, here is what I plot in MATLAB:
As you see the square-wave is clean and free of glitches:
I tested this measurement with different inputs I mean different function generators and pulse outputting devices at various frequencies.
But I always see same type of glitch always on the falling edges when the Acquire is set to 'Peak Detect' and saved as in CSV format.
I made the same test with a chinese scope Siglent SDS1102CML and when I saved the data in Peak Detect mode I don't see any glitches.
What would the reason be for those glitches after each falling edge. Is that an oscilloccope error?
Edit:
Following PlasmaHH's advice I checked the measuremtns with more horizontal settings. When I decrease the division enough, at some point the saved data for falling edges are clean, glitches disappear.
When the horizontal is set to 50us here is the data saved:
And here is the zoomed view of a glitchy falling edge:
When the horizontal is set to 5us there is no glitch, here is the data saved:
When the horizontal is set to 250ns there is no glitch, here is the data saved:
Edit 2
Horizontal is set to 5ms per division. The input is 50Hz square wave.
Here are the results:
Each falling edge has glitch. And here is the zoomed view:
And here below is the data from CSV file where the first glitch occurs:
t = [-0.01006
-0.01004
-0.01002
-0.01
-0.00998
-0.00996
-0.00994
-0.00992
-0.0099
-0.00988
-0.00986
-0.00984
];
V= [2.76
2.76
2.8
-2.72
2.76
-2.76
-2.68
-2.76
-2.68
-2.76
-2.68
-2.76
]
plot(t,V)
AI: It seems that this scope saves the min and max values per sample when in peak detect mode, as long as the sample rate is not at the maximum setting (one of the main features of peak detect mode, you get the extremes during the sampling period).
Ordering it this way in the csv file will make it seem to jump back and forth, but if you sort it again the other way round, then you will notice that for your falling edge it will look like a falling edge over a few samples, but then again it will look like these glitches on the rising edge.
To properly plot it, I think you want to take every other value and plot one as min, the other as max. |
H: Identifying this input feedback topology
I'm confused between whether this is voltage or current mixing. I'm thinking it is current mixing since the input current is subtracted from the source current. But why cant it be voltage since Rb is connected in series with the input loop ?
AI: OK, I know what you mean - in spite of an uncomplete circuit diagram. As you can see, the feedback path and the input path meet in a node which is common to all 3 resistors. In this node, two currents are superimposed - hence, we have "current feedback".
From general feedback theory we know that the input impedance is reduced in case of current feedback (in contrast to voltage feedback). As can be seen, this is the case for your circuit (Miller effect). This is another indication for current feedback.
At the same time, the resistor RE provides voltage feedback (the emitter voltage VE determines the controlling voltage VB-VE=VBE).
However, the question remains, if the whole scheme (current and voltage feedback within one stage) makes much sense. But that`s another question. |
H: Architectures for High Throughput Data acquisition with embedded systems
Goal: to acquire a high-throughput ADC-generated data flow (1 Msample/s @ 16bits) into a System-on-Chip processor. Purpose: real-time data visualization and some (minor) processing in real-time
What are the pros and cons of the various architectures to achieve the goal.
Architecture 1
A Dual Port RAM, taking data from an FPGA which glues the SPI output of the ADCs and the port 1 of the RAM. The SoC IC is connected to this RAM on port 2 using an external memory bus and sees the sampled data in its own memory space. The processor is 100% free from managing the ADC. The sample data is visible in a double buffering scheme from Soc end with interrupts informing the SoC when the buffers are full.
Architecture 2
As in Architecture 1, making the dual port RAM visible on a PCIe bus as an endpoint device (as opposed to using the External Memory Bus which we have seen in several SoCs).
Architecture 3
Use an Soc with embedded SPI controllers and DMA. Program the DMA to be triggered by the End-of-Conversion signal of the ADCs and to move data from the SPI FIFO to memory.
Architecture 4
Use a dual or quad-core processor and devote one core to control the SPI interface with the ADC and poll a GPIO port to detect end-of-conversion.In fact, this solutions implements by software DMA-like functionality.
Architecture 5
Your solution?
AI: First of all, 1 MS/s at 16 bit is just 2MB/s β that's really not too much for USB2 to carry. There's no need for dual port RAM, if we're talking about devices that would lend themselves to visualization or has PCIe like your Arch2 suggests, in my opinion.
The fact that you're doing visualization implies you don't care about latency β what's half a millisecond to the human eye? So, you're pretty free with respect to choice of sample transport.
So:
Arch 1
Lots of components, including an FPGA that does nothing but write a lowly 1 million samples per second to a RAM interface. I'd say, if you go that way, use a feasibly fast bus, and that would include simple SPI or QSPI, and a bit of RAM with the FPGA to implement a ring buffer. No need for dual-port RAM β you'd need to communicate information like "ok, there's new samples available for you" or "no, nothing to fetch right now", anyways.
Arch 2
PCIe sounds like a huge overhead here. Again, the rate we're talking about is 2MB/s.
Arch 3
If your ADC, and your SoC allow you to do that, start with that! Certainly sounds like the easiest, lowest-component-count solution. Often, this doesn't work for electrical reasons. SPI is absolutely a normal interface for an embedded system to have, so I'd assume that it'd be rather easy to find a controller that has it.
Problem remains that you'd still need someone to e.g. generate your sample clock etc.
Arch 4
well, yeah, as you say, a less great version of 3.
Arch 5
1MS/s isn't really high-throughput. In fact, I remember writing firmware for a now defunct ARM cortex-M0 project that ran the internal ADC at 500kS/s and pushed the data through USB2 to a PC. With a slightly more capable MCU, you should be able to do the same. That way, you'd have cheap-as-hell device dedicated to handling ADC data and stuffing it in USB packets, and you'd just have to write a couple lines of Python or C to run on your embedded device to ask the microcontroller for USB bulk packets full of data. Bonus: you can clock down your main CPU whenever you want to, and it will have no effect on the sampling.
Arch 6
Kinda easy. You can all do minimal visualization, sampling at several megasamples per second (complex) and a bit of analysis on ARM cortex-M4, with the help of a bit of glue-FPGA (without own RAM, iirc). This is proven by the open design of the HackRF one. I think it might be worth for you to look into this. From my perspective, it sounds like you'd basically just want to throw out all the RF stuff in that, and use it as is. You'd even get drivers and firmware for free!
HackRF hardware components diagrams from the project wiki
Above diagram is simplified, as mentioned, there's a small "glue" FPGA between the ADC/DAC hybrid and the LPC Cortex-M4, as the schematic will tell you. |
H: Can I use an AC motor to generate electricity which is mounted to a bicycle by a belt or I need it to be a DC motor?
I am trying to build a generator connected to a heater (250 Watts = 0.335HP). The energy goes to the motor from a bicycle.
So the question is: should I use an AC motor or a DC one?
AI: Anything other than a permanent-magnet motor is going to be difficult to use as a generator. There are permanent-magnet AC motors, but they are not very common. If you are trying to generate power from the effort of a single person, you are looking for at least an order of magnitude too much power. It will take a pretty strong athlete to generate even 250 watts over a significant period of time.
If you manually turn the shaft of a permanent magnet motor, you can usually feel a certain amount of force that tends to hold the shaft at specific positions. If you connect an analog voltmeter between two terminals, you should see the needle move a little bit when you turn the shaft.
Electric motors used in vehicles are usually permanent-magnet commutator-type DC motors. The largest motors of that type would probably be used for electrically-driven radiator fans. The engine cranking motor is larger, but that is likely to be a series-connected wound-field DC motor. That will not work as a generator unless you can re-connect the field for parallel connection. Some wheelchair motors may be permanent-magnet motors. |
H: Switching between two power sources
I am designing a circuit, which can use either 5V from USB, or some higher voltage (7-12V) from a battery. If both power sources are connected at the same time, I want some kind of electrical switch, that would disconnect the battery and power the circuit just from the 5V USB. But the circuit still has to work, when only one of the sources is present. I tried to design a switch using two mosfets, but I was not able to figure out a working circuit.
Can such a circuit be build by only using two mosfets? Also, do not worry about the voltage regulation, that is taken care of.
AI: There are dedicated chips that you can get which will do this but doing it with discrete parts would look something like this:
simulate this circuit β Schematic created using CircuitLab
All part numbers are the defaults rather than recommended parts.
Without the USB disconnected R1 ensures M3 is off. R5 turns M2 off and R2 turns M1 on. The end result is that the output is connected to the battery.
With USB powered M3 turns on. This pulls the gate of M2 down turning it on which in turn pulls the gate of M1 high and turns it off. Power then flows through D1 (which should be a schottky diode) and to the output.
D1 also protects the USB from over voltage while the voltage on C1 drops to USB levels. Note, if the battery is under the USB voltage then the body diode of M1 will feed power into the battery. This is outside of your stated operating range but if it is a possibility add a diode to prevent it.
C1 should be sufficiently large to prevent the output voltage dipping too far during switch over.
And now I wait for everyone else to point out the problems with this circuit (or point out how to do it with half the parts) since I'm sure I've overlooked something... |
H: Do I need buffer amps and zeners for SPI pins in a tiny IC chip?
I am currently working on a project using the MMA9555L pedometer chip (from Freescale Semiconductor Inc.) and establishing communications through an SPI bus going into an Arduino Uno. Below is a picture of the circuit I intend to use initially which is from the data sheet of the chip.
Before I power it up I am not sure about one thing. Arduino pins outputs 5V for digital signals however all the IO and SPI pins in this chip works at 1.8V. Will I need Zener diodes or voltage regulators as well as an Opamp in buffer amp configuration (to make sure current is limited) for each of the data pins? For reference, I will be doing the initial testing on a breadboard so no fancy (small form factor) components initially.
Sorry if my question sounds silly, I am only a mechanical engineer and do not have much experience with IC circuitry.
AI: Opamps/zeners: no that is too complicated, what you need are levelshifters.
That can be in the form of a chip, some resistors (for 5 V to 1.8 V) or a transistor and a resistor (1.8 V to 5 V). You can also buy small ready-made levelshifter boards on eBay. An example.
or this one from Adafruit2 |
H: Pulse width parameter in VI Characteristic curve of Diode
I was referring to the datasheet of IN4001 from Vishay
In Fig.4 (Typical Instantaneous Forward Characteristics), pulsewidth=300usec 1% duty cycle is given. I would like to know what this parameter is? In datasheet, there is no mention of pulse width it looks.
AI: The pulse width and duty cycle is an explanation of how they measure the instantaneous forward voltage parameter. This is how you would set up the test equipment to verify that the diodes meet specifications.
You can't measure the voltage without putting current through the diode. You can't put current through the diode without changing the die temperature, so the forward voltage for longer pulse widths and larger duty cycles would appear to be lower or higher because the die heats during the pulse or over a number of pulses.
For low currents the forward voltage decreases with temperature as you might expect for a diode, but for extremely high currents the Vf temperature coefficent actually goes through a null at something like 10-20A and becomes positive because the non-ideal resistive term dominates, and it has a positive tempco.
The Tj, although specified as 25Β°C cannot remain constant. |
H: GL850G without over-current protection
i am using the GL850G in gang mode for a circuit without over current protection to extend my available USB ports.
on power up the crystal starts oscillating for around 40ms then stops.
then i using a external USB hub with a GL850G it is working correctly.
my guess would be what it has something to do with the overcurrent protection?
PWREN1# is always high, i also tried to pull-up/pull-down OVCUR1# but PWREN1# is not changing state. any ideas?
AI: The crystal stops because the host is not ready to output any USB activity yet. This is the mandatory SUSPEND mode for all USB devices: when upstream port has no activity for more than 3ms, any USB device must enter SUSPEND mode, which usually requires clocks to stop. That's why XTAL stops. The initial state of upstream port on this hub should be DP0=pullup high. You will likely see XTAL activity again if you just short the DP0 to ground (which would simulate the USB_RESET bus state). The hub IC should wake up and start XTAL oscillations.
This crystal issue has nothing to do with ganged or other power modes. PWRENx port control outputs are HIGH because they are inactive, and will go LOW (to enable downstream power in case of downstream high-side power switches) after the hub will be enumerated and all ports will be enabled by host. |
H: Over voltage and reverse voltage protection of multiplexed ADC with voltage follower
I am designing a protection circuit for the ADC section of a design I am working on.
There will be 8 analog signals being read by a single pin of the MCU via an 8 channel multiplexer. The analog signals each go through a voltage divider such that the current to ground is about 1 mA before getting fed into the multiplexer. The multiplexer output feeds into an op amp configured as a voltage follower (buffer with gain 1) so that the leakage current on the ADC input clamping diodes doesn't affect the ADC reading.
The ADC is on-board my ARM MCU, which can tolerate voltages of -0.3 to 3.3 V.
I would like to add some sort of negative voltage and over voltage protection circuit before the multiplexer to protect both it and the op amp. I think the best candidate would be to use a set of clamping diodes on each signal going into the multiplexer. This would entail 16 individual diodes in total (2 for each ADC channel). Then each analog signal path would look like this:
simulate this circuit β Schematic created using CircuitLab
Ideally, this would allow for a wide range of AC or DC analog signal to be fed into the voltage divider (at left). My design spec would be for it to tolerate -230 to 600 V AC or DC analog signal.
Instead of soldering 16 discrete Schottky diodes onto my PCB, I was hoping to simply use a TVS diode array to do the job. I was thinking of using the PUSB2X4Y in this application. Since it has 4 clamping channels, I could simply use two of these to clamp the voltages on each of the 8 total analog channels in question.
The data sheet doesn't specify how much continuous current each diode can withstand, so I'm not sure if the part will work or not. Every TVS diode array that I have looked at on Digikey only lists the pulse current that each diode can withstand... not constant. Anyone know what ballpark figure to plan for on the constant current front? Is the common practice to not use TVS diode arrays and opt for a multitude of discrete Schottkys to get the job done?
The other thing that I'm not sure about is how to account for leakage current. Will the leakage current through D1 "cancel out" the leakage current through D2 in normal operation or will it end up affecting the voltage being fed into the multiplexer? Same question with D4 and D5, though I'm not worried about those leakage currents since I have the op-amp buffering that input.
AI: Better to use silicon diodes for D1 and D2 rather than schottkies, on leakage grounds. I know their Vf is larger, on paper probably exceeding the maximum mux input voltage, but their leakage will be orders of magnitude lower. Some silicon diodes are advertised as low leakage. However, there's little point in striving for a leakage lower than your mux will provide. Note that both diode and mux leakage currents tend to increase exponentially with temperature, sometimes a horrible looking data sheet figure for worst case at temperature will be OK if you will only be using your system at ambient.
Use a resistor between the diode clamp and the mux input to limit the current into the muxes protection diodes, once D1 or D2 are clamping.
Not all multiplexers are equal, some have robust input protection as they're designed for this type of use. Some specify their input protection didoes can tolerate a high current. Do a wide search and read the data sheets carefully.
Do not assume leakage currents will cancel. Leakage is an uncontrolled parameter.
Don't forget to use an adequate voltage rating for R3, your common or garden resistor is usually only good for 200v. Use several in series, or one rated for a much higher voltage, they're not too expensive. Spikes of 1500v are common on mains.
Something like PUSB2X4Y has pulse specifications in the amps. If it can take a microseconds pulse of 4.5A at typically 3.8v across it, then you can assume it will take 10s of mA all day, every day, without embarrassment.
In case you haven't spotted Matthew's comment below, here is what I think he was suggesting, in the left-hand diagram, back to back diodes across a buffer op-amp.
simulate this circuit β Schematic created using CircuitLab
Although the op-amp output could be thought of as more current-capable than the inputs, it still has substrate diodes and a maximum current specification, so is also in need of protection itself. Some amplifiers even have inputs specified to well outside the rail for input protection, but only 0.3v overvoltage and weedy current spec on the output pin.
Going on from that basic idea, my take on the principle is shown on the right. The diode string D3-6 provides voltage clamping to 'a little outside' the rails, R2 protects the diodes, R4 protects the amplifier input, R3 protects the amplifier output and bootstraps the voltage across diodes D4 and D5 so that their leakage to the input is minimal. With such bootstrapping, the diodes could be almost anything, even big rufty-tufty bomb-proof rectifiers.
The two diodes in series suggests that care is needed in layout if protection is to extend to fast pulses. Consider the SOT-23 BAV99, two series diodes in one package, to implement the pair D3,4 and the pair D5,6. They are specified continuous >100mA, typical 10mS pulse 800mA, which for any reasonable R2 sounds adequate. BAT754S is an alternative in schottky. Similar currents, but much lower clamping voltage.
You do not actually need an op-amp per channel if your multiplexer leakage is low enough. The circuit at the bottom shows the single buffer following the mux driving all of the input protection diodes. Note that the multiplexer leakage appears at the amplifier input, whereas using a per-channel buffer eliminates mux leakage.
The 'on' channel is receiving the correct bootstrap voltage. The 'off' channels will probably be getting the wrong voltage, and the 'inner' protection diodes may well conduct. This is not a measurement problem, as the channel we want is correct. It may, or may not, be a problem to what is driving those inputs, to have our nominally high impedance inputs yanked off to a different voltage. If we assume it's a very feeble current source (we are concerned about leakage, so we know it's not a low impedance source) with a large capacitance to ground, it may take a long time after selecting that input before the voltage has returned to its correct value.
Actual leakage measurements for diodes at 15C.
diode -2/-5v leakage slope resistance over +/- 10mV
----- -------------- ------------------------------
1N4148 4nA 30Mohm
BAT42 35nA 1Mohm
BAS116 <10pA (30v) >>20Gohm
The BAS116 conduction continued as 40pA 300mV, 45nA 450mV, 16uA 640mV. The BAS116 typ/max spec at 25C is 3pA/5nA, and 3n/80n at 150C.
That means, at that temperature, and making the assumptions of reverse leakage varying by a factor of 2 up and down, and 3mV voltage follower offset, you could assume the following leakage
diode no bootstrap bootstrapped
----- ------------ ------------
1N4148 6nA 1pA
BAT54 50nA 3nA
BAS116 <10pA <<10pA
I made those measurements with a Β£8 meter with 10M input resistance and a 200mV range, so 10pA per LSB, not difficult (obviously can't tell the difference between 0 and 10pA!). I suggest you do the same with your chosen diodes and at higher temperatures. |
H: Advice on 4th-Order Bandpass Filter
I am trying to design a 4th-Order bandpass filter with the following specifications; centre frequency 30 MHz, bandwidth 5 MHz and of course minimising loss.
I present my current design below which I determined using the ELSIE simulation design program. The problem is I am not very confident in electronics and I am not convinced about my circuit diagram.
simulate this circuit β Schematic created using CircuitLab
If anyone has any advice or can spot any mistakes I would appreciate any thoughts!
Cheers!
Quick Edit: I have attached the output schematic of my ELSIE simulation in case it is useful to anyone:
AI: You may be constrained by Network Analyzer impedance of 50 ohms. But your filter need not be so constrained. End-matching networks allow you to optimize these goals:
(1) Choose filter components whose unloaded Q is high (at 30 Mhz).
(2) Choose filter components whose nominal value has high accuracy.
(3) Choose filter component values whose reactance dominate circuit parasitic reactances.
(4) Minimize passband attenuation.
One approach (perhaps Elsie allows this) chooses source and load resistance that gives optimized component values. Usually, both source and load are equal (and for your filter type, larger than 50 ohms). Then end-matching networks are added to the design to accommodate 50 ohm source & load Z. This does have the disadvantage of adding (at minimum) two extra components. Familiarity with available commercial components is of paramount importance in choosing an optimum filter impedance. And estimating circuit parasitic reactances is important too.
You will likely find that a filter that can meet design tolerances will need to be tune-able - because fixed, standard component values that Elsie spits out are not available, and because their manufactured accuracy is too loose. If production volume is high enough, then you might move to a monolithic surface-acoustic-wave (SAW) filter and avoid tuning. This solution is chosen by TV manufacturers.
Here is an example 3rd-order approach to illustrate:
Filter design chooses an arbitrary inductor value of 265nH, unloaded Q=100. This (in my opinion) is an "optimum" component value, that also yields "optimum" capacitor values. Centre frequency=30Mhz., BW=4Mhz. Source and load resistance turns out to be 405 ohms:
simulate this circuit β Schematic created using CircuitLab
Hopefully, Elsie allows you to specify 405 ohm source/load Z. Choose source/load Z that yields realizable, optimized component values spewed by Elsie. You'll soon see that higher Z yields large inductors & small capacitors, while lower Z yields small inductors & large capacitors. This filter is then adapted to your desired source & load impedance of the network analyser (50 ohms) with end-matching networks:
simulate this circuit
Note that with all top-coupling capacitors, the attenuation outside the passband is not symmetric...attenuation is poor above the passband. Choosing end-matching to be inductive can improve this situation. You can transpose a filter too:
simulate this circuit
After 50 years, "Handbook of filter synthesis" (Anatol Zverev) is the reference for lumped-component filters (in your frequency range), but its not for mouse-jockeys. |
H: How much does the physical size and shape of a RF antenna affect its range/penetration?
I'm curious about the physical propterties of the tiny antennas inside many embedded devices (so ~2.4ghz if frequency matters).
Made this graphic to reduce typage:
Red Circle = target
Grey bar = concrete obstruction
Black bar = just dividers for the different senarios
1: regular antenna
2: 2x as long antenna
3: 3x as long, curved antenna
What would the difference be in these cases?
Does any of these changes impact the overall signal strength?
If so, roughly by how many times relative to the first one?
AI: How much does the physical size and shape of a RF antenna affect its range/penetration?
Completely. Utterly. An antenna is fully defined by its shape and size and material.
A minor modification in design can pretty drastically change an antenna.
What would the difference be in these cases?
Sorry, the answer is:
Antennas aren't that simple.
An antenna has to be designed exactly for the frequency and bandwidth, impedance and direction that it's going to be used. So, it's not like you can just take every design and make it twice or three times as long and expect it to just continue working.
As a practical example:
A very common antenna is the quarter-Wavelength monopole. If you double its length, you get a current maximum at the boundaries (where there used to be a minimum) when the feed point has a maximum β which is certainly not what you'd want when building an antenna with the same wave impedance, and still having a main beam normal to the dipole's extension.
So, no, we can't tell you how these other antenna are going to behave. You'll have to specify exactly how they'll look like, and then, you'd normally either manually analyze them, or throw a simulation at them.
Does any of these changes impact the overall signal strength?
Yes. But:
Signal Strength and Antenna shape and size aren't directly related β using a different shape will yield a different gain pattern, i.e. the antenna will radiate in different directions with a different relative power, but it's also not like it's monotonous that a simple "enlarged" antenna will yield to higher gain in the normal direction of the antenna β on the opposite, at some lengths of a linear antenna, you'd actually get zero emission in the direction that you seem to think is always the main direction.
And: by changing the size of your antenna, you also change things like the impedance, and that means you'll reduce the efficiency of the antenna!
If so, roughly by how many times relative to the first one?
Something between 0 times as much and 100,000 times as much. Impossible to say.
So: I'm afraid there's no shortcut to learning antenna and antenna system basics if you want to design antennas. Sorry.
Now, you have a complicating thing: the concrete obstacle.
Again, sadly, radio propagation isn't that simple, either. An emitter that hits the wall with very nice, plane wave beam in the center will probably be absorbed the most, whereas something that "hits" the edges of your wall will probably make use of diffraction and refraction β but that depends on the relation of wall shape, wave length, wall material (are there steel armaments) and relative positioning of wall, antenna and target (measured in wavelengths). |
H: Ripple at Resonance?
This parallel LC circuit, energized with a current source, is supposed to work as a low pass/high pass filter depending on the branch I take the output from. PSPICE simulation shows exactly that, EXCEPT there's a spike at resonance(157kHz) where the gain explodes. I did some googling and apparently this is what they call a ripple, am I correct? If so, why is there a ripple?
AI: That's not a filter, that's a resonant circuit.
As you've drawn it with ideal components, you would expect the impedance to become infinite at some frequency, 157k sounds right for those values.
With real components with loss, the impedance never becomes infinite, although it can become unexpectedly large.
A filter would be driven by a source with some resistance. That would 'tame' the resonance and give you more 'fliter-like' results. Look for filter design programs, there are a few free ones kicking about on the web. Simulate the networks they give, not forgetting the source and load resistance of course! |
H: How to determine gain from a bode plot
Given this bode plot from an OP-Amp Amplifier:
What is the method for determining the voltage gain?
AI: The method of determining gain is to look at the Y axis for the frequency of interest. A Bode plot is log(gain) as a function of log(frequency). Basically, it tells you gain directly. |
H: Shift register - counter
So i have this shift register
With the initial values of ABCD (1101)
And the question is, how will this change after 4 clock pulses.
As I understand it, this shift register goes shifting the values from left to right. My first guess was basically taking the the lsb and putting it in first place and repeating the procedure 4 times (4 clock pulses).
I would then get ABCD (1101)
This is not the correct answer though.
AI: For the correct answer consider the value shifted into the left most bit position to be the Exclusive OR of the current states of A and D. So, for the first shift since A and D are both 1's the Exclusive OR would be zero and zero would be shifted into the left most bit.
(A list of logic symbols which includes what "=1" means can be found here.)
That is to say, starting with A, B, C & D being 1101, the next state would be 0110. And the state after that would be 0011. Armed with this information it should be easy find the 3rd and 4th states. |
H: Connected 12V-1.0A transformer instead of 12V-1.5A
I'm using an Inspire 2.1 2400 audio set -
https://www.cnet.com/products/creative-inspire-2-1-2400-speaker-system-for-pc-2-1-channel-21-watt-total-black-series/specs/
The set comes with a 12V-1.5A transformer, and I just connected a same-voltage-lower-ampers (12V-1.0A) transformer into it, and everything works pretty much the same..
Does it make any internal damage? What's the actual effect of lower ampers?
Edit:
Choosing power supply, how to get the voltage and current ratings?
doesn't really answer my question .. I asked what happens if I plug in a lower amperage transformer.
AI: Most audio systems in this class use class AB linear power amplifiers. At very low volume this type of amplifier operates in class A, drawing a small constant current. At higher volume it changes to class B, where it only draws enough current to deliver the required volume to the speaker.
The Creative Inspire 2.1 2400 uses a TDA8410J, which integrates 3 class AB power amps into a single IC. The set comes with a 12V 1.5A power adapter because the manufacturer has determined it needs that much power at maximum audio output. At lower volume it needs less power, so a smaller adapter could be used so long as you keep the volume down.
If you try to run the amp at higher power than your power supply can handle a number of things may happen:-
The power supply could overheat and burn out, or shut down due to excessive current draw.
Its output voltage may drop, distorting the peaks of the audio waveform.
There may be increased mains frequency 'ripple' or hum which becomes audible at high volume. |
H: How to control the output voltage for the different types of the DC generator?
Generally DC generators are classified according to the ways of excitation of their fields. There are three methods of excitation.
Field coils excited by permanent magnets β Permanent magnet DC generators.
Field coils excited by some external source β Separately excited DC generators.
Field coils excited by the generator itself β Self excited DC generators.
How to control the output voltage for the different types of the dc generator and how to reverse the voltage polarity?
AI: The voltage of a permanent-magnet generator must be controlled by controlling the generator speed. The voltage sent to loads could be controlled by sending the voltage first to a voltage regulator or converter. The voltage of a separately-excited DC generator can be controlled by controlling generator speed, by using a variable field supply, or by a combination of both. A self-excited dc generator can be controlled by generator speed or by a rheostat or electronic circuit that reduces the field current. A combination could also be used. |
H: What does it mean to establish the state of the internal node in CMOS circuit?
This is the NAND circuit and the 'int' is the internal node of the circuit.
It is seen that A = B= 0β 1 gives the worst delay in the below table. What does it mean to establish the state of the internal node as stated in the text below.
It is not clear to me.
AI: The switching speed of logic gates is determined by the driving strength of the transistors and the capacitances that are present on each node.
Only nodes where the voltage has to change are relevant.
If input A is switched from 1 to 0, the node "int" is initially at 0 and as M2 is switched off it remains at 0 since M1 stays on. The transistor M3 only has to charge the parasitics of node F.
If input B is switched from 1 to 0, M1 is switched off and M2 stays on. The node "int" is initially at 0 but has to be charged to about VDD-Vth. At the same time node F has to be charged as well so the transistor M4 has to charge two nodes, which makes this transition slower: "The state of the internal node has to be established". |
H: Need some advice- Should I use the CORDIC FFT function or write code for an FFT function myself?
I'm currently working on designing a filter in VHDL for a project based on an FPGA. The filter would require FFT and I've heard about the CORDIC FFT function but never used it so I'm not sure how accurate it will be. So should I write VHDL code for an FFT function or use the CORDIC function? Also, does using the CORDIC FFT function require any permissions to be taken in case any work gets published?
AI: Use the vendor specific generated block that is available to you. Then read the documentation for the block. For example here is the documentation for the latest Xilinx FFT block. It will likely allow you to customize some parameters like input data size and block size.
Making your own FFT block could turn into a pretty advanced semester long undergraduate project. |
H: Tesla Car "maximum torque at 0 RPM" - is this correct?
Tesla Model S Wiki
I've been watching youtube videos on this car, and everyone states that the crazy accelleration is due to maximum torque at 0 rpm. Doing further research, this car uses an AC induction motor, not a DC motor.
From my old lecture slides, I remember that the torque curve of an induction motor is not this, but can be shifted (by varying the voltage/frequency, I can't remember).
Is the "maximum torque at 0 rpm" misinformation going around?
AI: With frequency control, there is not just one torque curve, but an infinite number of curves, one for every operating frequency. The voltage needs to be proportional to frequency. If the voltage is carefully regulated using a mathematical model of the motor with motor operating voltage, current and power factor information, the torque curve can be made to have the same shape at any speed. The required current to produce a given torque at zero speed, will be close to the current required to produce the same torque at rated speed. The motor is never operated at high slip, the operating point is always to the right of the pullout torque point.
When starting, the applied frequency is enough above zero so that enough slip is created to produce the maximum torque that the motor can safely produce. |
H: 4 To 1 MUX - Figuring out the inputs
If i have a 2to1 mux together with logic gates like this:
I can simply create a karnaugh map and test the different values of XYZ and get a function.
From this i get the function f = x'z + yz + x'y
However in the next task, I am supposed to look at this 4to1 MUX
And decide what inputs to use to achieve the same function f = x'z + yz + x'y
The four different alternatives I get to choose from on each input are:(0,1,Z,Z')
In the task I am also told to create a new karnaugh map for this 4to1 MUX.
I just don't get how I am supposed to think about the inputs to get the same function, or maybe I have misunderstood something.
AI: Start by looking at the equation for a 2-in MUX:
$$
\begin{align*}
M(A,B,S) &= A\cdot \bar{S} + B\cdot S
\end{align*}
$$
From this, you can derive some useful results (among others not listed here):
$$
\begin{align*}
1.~~~M(0,x,y) &= x\cdot y \\
2.~~~M(x,0,y) &= x\cdot\overline{y} \\
3.~~~M(x,y,0) &= x \\
4.~~~M(1,x,y) &= x + \overline{y} \\
5.~~~M(x,1,y) &= x + y \\
6.~~~M(x,y,1) &= y \\
7.~~~M(x,y,x) &= x\cdot y \\
8.~~~M(x,y,y) &= x + y \\
\end{align*}
$$
Now, your equation is:
$$f = \overline{x}z + yz + \overline{x}y = \overline{x}\left(y+z\right) + xyz $$
And you already have your required mux template laid out as:
simulate this circuit β Schematic created using CircuitLab
You could use a number of different approaches. But let's just use our imagination only and work from that.
The last mux, where \$x\$ is the selection input, selects between the following two options:
$$\begin{array}{rl}
\overline{x}&z+y \\
x & yz
\end{array}$$
Well, if you look up at the table of "useful results" above, you can easily see how to implement these.
For \$\overline{x}\$, you see from the table that either #5 or #8 would do the job here. Depending on what you feel is appropriate for your answer, pick one or the other.
For \$x\$, you can see that either #1 or #7 will do the job. Again, depending on what you feel is appropriate for your answer, pick one or the other.
For me? To reduce fanout, I'll pick the ones that have constants. So #1 and #5. Here's the result:
simulate this circuit
That's it, really. Sometimes, it helps you to just lay out a useful table for the mux function. |
H: Troubleshooting why Powerbank Doesn't Charge Cell Phone
I have powerbank that provides 5v (presumably regulated) power, as typical of most usb interface powerbanks. I have been trying out my powerbank with different phones. It works on all except one. I have tried it with various USB cables, making sure I didn't overlook anything foolish. Then, I looked at the AC adapter of the phone that it didn't seem to be compatible with. The adapter read: 5.4v DC output. The device's official website or manual doesn't tell me what the input voltage should be. So I have the following questions
Is the fact that the device's AC adapter creates a DC 5.4v output enough to assume that, this is the correct voltage for my cell phone?
If 5.4v is the input voltage, then when I attempt to power it with a 5v power bank, am I hurting it? Thus far, it just seems to be non-responsive when I try this.
It seems another explanation is polarity mismatch. I'm aware that laptops have a polarity correction feature, which could explain why I can charge this phone via my laptop. Is there any other way to tell if polarity is an issue, given that I don't have access to detailed product information (unless the official website is updated, or something)
So that's the gist of it, I'm not sure why my powerbank isn't charging my phone. And I'm having difficulty troubleshooting the underlying problem by myself. Any tips big or small will be appreciated.
AI: A device (phone) gets charged at optimal (high) rate only when it recognizes "charging port signature" on charger side. For Type-A port, there are several DIFFERENT port signatures that a charger port can provide:
USB Battery Charging v1.2 Dedicated Charging Port, where D+ and D- are floating but tied together;
USB Battery Charging v1.2 Charging Downstream Port, when a special low-voltage handshake over D+ and D- lines happens;
Apple charger signature, when a combination of DC levels is present on D+ and D- lines with high-impedance resistor dividers (75k/50k);
Sony charger signature, when D+ and D- have 5k/10k dividers to VBUS;
Qualcomm "Quick Charger" signature, which uses an intelligent IC on charger side to understand a special sequence of DC levels driven by device on D+ and D-; can provide 9V and 12V on VBUS;
Samsung charger signature (details unknown, to me);
USB Power Delivery protocol V1.2 - uses complex packet-based protocol over VBUS wire using FSK modulation method at 23.2MHz carrier frequency. The port might have additional contacts to detect plug insertion.
...
For Type-C connector, there is a newer Power Delivery protocol v2.1, which solely uses the CC wire (Communication Channel) to negotiate voltages and available currents that can be delivered via VBUS.
A charging port usually provides only one signature, and most of mobile devices have an intelligence to recognize it, usually by sequentially probing different signatures. When a phone recognizes something it understands, it will take the charge. [However, there were attempts to make a charging port that tries several signatures sequentially, as some Microchip/SMSC hubs of the past.]
We don't know what kind of signature is used in this particular power bank, so all bets are off. Now, consider that your old phone can be designed for some other proprietary signature, so it should not come at any surprise that some phones will not charge well (or at all) from this particular power bank. |
H: ATX PSU +12V outputs connected in parallel
I need +12V and around 10A for my project. Each wire on the ATX supply I bought is stranded 20 AWG and is rated for less than 10A. What would happen if I connected multiple +12V wires in parallel to divide the current among the thin wires? The PSU says that it can supply 16A out of the +12V output.
AI: Your PSU seems to have a single-rail 12V output. If you will keep the wires (yellow and black) to all your Molex 4-pin connectors of the same length (or if you plan to re-work the 12V connection to some other connector or directly to a board), you can reasonably assume that the current will be equally distributed, more or less, and you will have all 16A at the end with no problem.
If your PSU has multi-rail topology (has 12V1, 12V2, etc. labelled outputs), you might want to put small serial (equalizing) resistors (0.05 Ohms or less) between ganged outputs from these separate 12V1 and 12V2 etc. outputs, to accommodate possible voltage differences in the rail's outputs. |
H: Does "Common Mode Noise" only affect single-ended signals?
In the above illustration the common mode noise is hitting to both + and - wires.
Here is my understanding of the "common mode noise":
In a Single-ended system: the - wire is GND and + wire is carrying the actual signal, so the noise will only affect the + wire but not the - wire since it is GND. Noise then will appear at the output (difference of the + and - wires).
In a Differential-ended system(differential signalling):* the - wire and the + wire both are carrying signals which are mirrors of each other, so the same noise will simultaneously be added to the + wire and the - wire. Noise will not appear at the output.
If my understanding is correct; does that mean if one talks about "common mode noise", he is talking about a single-ended system?
AI: If my understanding is correct; does that mean if one talks about
"common mode noise", he is talking about a single-ended system?
A differential receiver has to "manage" the common mode noise and although a perfect receiver will certainly achieve this, the real world receivers don't. For instance (just an example), an op-amp may be specified as having 100 dB common mode rejection but the fine detail in the data sheet tells you that this might only be at 100 Hz and at (say) 10 kHz, this has degraded to 60 dB and maybe 40 dB at 100 kHz etc..
So no, common mode noise is a big issue for both types of system. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.