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H: Essential Nodes in a Circuit
My book says above that there are three essential nodes in the circuit above. I am seeing four essential nodes in total which are marked by those four black dots. Since each of those black dots connect three or more circuit elements? Can someone explain to me how there are 3 essential nodes instead of 4 essential nodes?
AI: The bottom two Nodes are effectively at the same potential. So they can be considered as a single Node.
Regards |
H: Vout = 12V, Iout= 15A, 180 W Switching regulator
I have experience of designing DC-DC converter upto 5A.
But now for one of my application, I need to design circuit with following specs:
Vin = 24 - 25 V,
Vout = 12V,
Iout = 15A,
Pout = 180W
So, Is there any switching regulator capable of such high power rating?
( Any application note will extend my knowledge for High Power Design )
Thank you in advance.
AI: Use your experience at 5Amps .Run several together if you are happy with your 5A products.You have heard words like "polyphase ,Interleave ,Ripple cancelling ,Split phase etc " They all do the same thing and you wont have an EMC suprise assuming your 5A ones are quiet.There are chips that do this if you want to use a chip .Your power components will be the same as your 5A product which eases stocking problems .I am assuming that your 5A product is SMD and automaticly assembled so the increase in parts count wont be a problem .What I am suggesting is nothing new and is often done on computer core supplies. |
H: DC Motor consuming too much current
I'm trying to use an H-bridge, which is connected to a wall power supply, to power a DC motor. The motor was originally powered by a 12V battery pack in a toy.
Now my motor is consuming too much current (2.5 A instead of 1A) (the theoretical maximum current of the power supply is 2A)
the motor even started smoking :(
Why is the motor consuming too much current? It should only take what it needs from the power supply right ?
thanks for your help, there was an overvoltage :(, I destroyed my motor.... The motor worked with 5v and I fed them with 12v
AI: So without any more information this is what I would suggest to check.
Check the voltage on the motor leads while it is in the toy as well as your current application. Just because something has a 12v battery doesn't mean that the motor runs on 12v, there could be some voltage regulator hidden away.
Make sure the motor can spin freely. If something is binding up the motor or a magnet inside has come lose it can create a very large load on the motor which will drastically increase its current consumption. It should spin without any clicking or grinding.
Some questions:
Can you see a model number or any identifying marks on the motor, you may be able to look up a datasheet to confirm the voltage it needs.
In your new system, is the motor under load or spinning freely? As I said earlier, the more stress you put on the motor the more current it will draw. |
H: Output power vs idle power of an audio amplifier
Let's say I have a AB class power amplifier that is able to give an average output power of 30W on a 6 ohm loudspeaker. Now in general, how much power should the amplifier dissipate when there is no signal applied?
I am asking this because I have an amplifier that dissipates about Vdc*Idc =30W when it is idle. The maximum amplitude (when clipping starts) of the output signal is 20V on a load of 6 ohm and with Ltspice I found that the maximum output power v(t)*i(t) is about 66W and the average power is about 30W. To me it seems like the amplifier dissipates too much when idle, doesn't it?
AI: It all depends on the power level where the amp transitions from Class A to Class B modes of operation, i.e. where the inactive transistor cuts off.
There is a conflict between efficiency (transition to Class B at very low power, at the risk of some crossover distortion) and sound quality (high idle power, where the amplifier may remain in Class A up to 5 or 10W).
Many of the "Class AB" amplifiers seen online are practically Class B and will show some crossover distortion as designed, as they are designed such that both output transistors are not "on" simultaneously. (Note that the Wikipedia description of Class B may be true for RF amplifiers but disagrees with 80 years of actual Class B audio amplifier design (p.49 of link)).
Some amplifiers allow adjusting the quiescent (idle) current via a pot - see RV24 on this schematic - after assembly and after replacing a transistor, you measure the output stage current and adjust the pot for 20mA (or other recommended figure). study the circuit for yours and see if this is possible.
TL/DR Your idle power is on the high side, but not unrealistic for a Class AB amp designed for higher quality. If it's adjustable it may simply be maladjusted, or "tweaked" in an attempt to reduce distortion. |
H: Is it necessary for "LDR R0, [R0]" when saving from r15 in arm?
From ARM® Developer Suite Assembler Guide
Saving from r15
In general, avoid saving from r15 if possible.
If you do save from r15, the value saved is the address of the current
instruction, plus an implementation-defined constant. The constant is
always the same for a particular processor.
If your assembled code might be used on different processors, you can
find out what the constant is at runtime using code like the
following:
SUB R1, PC, #4 ; R1 = address of following STR instruction
STR PC, [R0] ; Store address of STR instruction + offset,
LDR R0, [R0] ; then reload it
SUB R0, R0, R1 ; Calculate the offset as the difference
If your code is to be assembled for a particular processor, the value
of the constant is available in armasm as {PCSTOREOFFSET}.
AI: LDR reads a word from memory.
LDR R0, [R0] reads a word from memory at the address held in R0, then stores the value in R0 (which happens to be the same place the address was stored).
It's not explicitly stated, at least in the part of the documentation you quoted, but before this code snippet you should set R0 to an address that you want it to use as scratch space. |
H: Baud rate vs. Bit rate
When we initialize a serial communication, we usually have to set the baud rate (e.g. 115200)
I notice that many people assume that 115200 means 115200 bits per second. However, each baud(signal change) can encode more than 1 bit. baud 115200 means 115200 bits per second only if each baud = 1 bit.
If each baud is 5 bits, then having baud 115200 means a bit rate of 115200 * 5 bits per seconds.
If what I said above is true, what is the baud to bit ratio for UART and USB3.0 assuming the baud rate is set to 115200?
It cannot be 1 to 1 because 115200 bits per seconds is 14.0625 KBps (kilo byte per second), which is too slow.
AI: A minor point: by definition, a Baud (named after Emile Baudot) is one symbol per second. So your statement,
baud 115200 means 115200 bits per second only if each baud = 1 bit.
If each baud is 5 bits, then having baud 115200 means a bit rate of
115200 * 5 bits per seconds.
is correct in spirit, but you should replace the word "baud" with "symbol". If you have an alphabet of 32 symbols, then each symbol encodes 5 bits, and a signalling rate of 115200 symbols per second, which is 115200 Baud (by definition), will give you 115200 * 5 = 576000 bits per second.
The commonly used phrase "Baud rate" is grammatically incorrect in essentially the same way that "ATM machine" is incorrect. The definition of Baud has the concept of rate built into it, so saying "Baud rate" is grammatically the same as saying "symbol rate rate" (which might make sense if you're talking about a changing symbol rate, but that's not how most people use it). Likewise, "ATM" has "machine" already incorporated into it, so "ATM machine" is equivalent to saying "automated teller machine machine" (which, again, could make sense if you were talking about a machine that produces ATMs, but that's usually not what people mean).
Grammar aside, most low-to-moderate speed short-range signaling over wire uses binary signalling, so the symbol rate equals the bit rate. Long range signalling over wire (e.g., cable modems) and via radio is more likely to use multi-bit symbols, and hence the bit rate will be higher than the symbol rate. |
H: What's the meaning of "x" in RxD and TxD of UART?
RxD means Receive Data.
TxD means Transmit Data.
Where the "x" come from?
AI: Comes from the 'good old days' of radio when Morse code ruled. Abbreviations abounded (c.f. text messages!), and TX was transmitter or transmit; RX was receiver or receive \$\cdot\cdot\cdot-\cdot-\$ |
H: How can I send keyboard input to USB port or Serial Port?
Now, I am using Windows 8 and have made Arduino buttons to operate as an keyboard's 4 arrow keys(up,down, right and left). From this moment, I am going to send this keyboard input event out to the USB port or 9-pin serial port(RS232) installed on the backside of a desktop.
I guess I may need to use windows header and serial communication but thoroughly have no idea, what program or language(C is preferred.) do I have to use. Also even a hardware connection.
Could I get a insight for my goal?
AI: I basically see two options here:
Use your Arduino to become a virtual keyboard (a HID device just like your regular keyboard). This will look to your PC like you have an additional keyboard connected - probably this helps for a start.
Use the serial port on your Arduino to send out the data, then read it on your PC using some tool (this is basically your idea). It's pretty easy to do this using C#, it already provides a class called "SerialPort" which basically does the reading for you once you've set it up correctly. From here on, it depends what you want to do with the data you send. An easy example can be found here. |
H: Hand soldering QFP component
I recently came across this video on YouTube, in which the guy is soldering a .5 mm QFP component.
Hand soldering a .5 mm QFP Chip
As I haven't done a lot of soldering in my life (and never used flux), I have a few questions.
The guy in the video uses only flux without solder to do all of the soldering. How is this possible? What kind of flux is he using? As far as I know, flux is only used as a cleaning chemical during the soldering process. Correct me if I'm wrong.
He puts flux over the PCB traces and the space between them. If the flux is conductive, how is it possible that there are no short circuits created?
AI: First a warning to readers: That video is an advertisement for a product.
The board you are looking at comes pre-tinned. Applying solder flux removes dirt and oxidation, and makes it easy to use the tin already on the PCB. The flux itself is not conductive. |
H: Does an op amp needs an Input Bias Current to operate?
I'm confused about the source of Input Bias Current, is it generated by op amp, where the current "leak" from the op amp. or it's a must to operate the op amp? for example if I have in a data sheet a 1pA input bias current, should I provide a 1pA to the inputs of op amp? I know that the bipolar op amp needs a current to operate, but what about FET and CMOS op amp?
In short question: Should I build circuit to "provide" this current, or it's just given by the op amp itself?
Also, what is the difference between input bias current and the leakage current?
Note: if the type of op amp make a difference, I'm interested in
LTC6078/LTC6079
AI: Input bias can flow both into or out of an op-amp's inputs, and in certain op-amp topologies may do either. The consequence is that you need to design your circuit accounting for the current, i.e. if your source impedance is 100kΩ and you've got 1pA of input bias current, a non-inverting buffer will have a voltage error of (100kΩ * 1pA) = 100nV.
As you can see, input bias current has a negligible effect for low impedance sources, and can have a very significant effect for large impedance sources. This is also why you cannot capacitively couple op-amp inputs and expect them to work; the inputs must sink or source this tiny current for the input topology to work.
Edit: To more directly answer your question, input bias current is not something you need to "provide" explicitly, rather, consider your circuit as sourcing (or sinking) this current. In the process, an error term will be generated. Another way to think of this is that the op-amp has a finite input impedance, and therefore your circuit needs to account for that. If your source has a very high impedance and the op-amp input impedance is too low (which is the same as the op-amp having a relatively large input bias current), then the voltage divider formed by the source impedance and the op-amp input impedance will be further away from 1, which results in a voltage error. |
H: My H-bridge can't control my 3v motor
I'm trying to control my 3v motor from my 5v arduino.
For this i use :
5 volt power source from pc for the arduino
L9110 H-bridge module
2 AA Battery as power source for the L9110 H-bridge
Here is the setup :
however if I touch the h-bridge pin with the 5v source from the arduino (red cable on the image), nothing happens....
Any idea ? if I power the motor directly from the battery, the motor works
Maybe the problem is that I have 2 different power source
AI: You need to complete a circuit in order for the current to flow. Creating a common ground between the power sources (connecting the grounds together) should fix the problem. |
H: Erasing Flash with X-rays to remove security bits
Scenario: A chip is read/write protected by programming a security bit in its onboard flash memory. Normally this would mean it cannot be re-written. Would a dose of X-rays be capable of erasing the flash to the extent the part could be reprogrammed? So, a number of questions.
a) How much X-ray flux is required?
b) Are such X-ray sources easily available?
c) Would the X-ray flux cause permanent damage to the chip?
d) Has this been done by anyone?
AI: Floating gate flash is susceptible to X-ray, but you risk damaging the onboard charge pump if you subject the device to sufficient energy to guarantee erasure of all bits (the only way to guarantee that you have erased the security bits).
It is possible to erase memory cells at certain energy levels that will not significantly damage the cells, but the exact level is not really known and varies with process and unfortunately that energy level is what fries the onboard programming charge pump.
Floating gate device memory cells are particularly sensitive to energy levels in the < 9keV range; see this and this application notes from Spansion (originally from AMD).
It is possible to make X-ray sources in this range, but probably not cheaply, and this risks damaging the memory cells such that they are not recoverable.
I have had ordinary post-solder X-ray (tungsten source, very little energy below 40keV) flip bits from programmed to de-programmed, but it is somewhat rare.
That same source, when turned up to around 140keV, fried the charge pump and some of the flash memory cells in a power sequencer (flash based programme).
The short answer is that yes, flash can be deprogrammed by X-ray, but with no guarantee of the operation of the device if you successsfully get the device de-programmed and just what energy level for a given part is very unlikely to be known.
The erase event is quite simple (according to research available in IEEEXplore):
The floating gate is programmed by tunneling electrons across the tunnel oxide between the control and floating gates.
X-ray is simply high energy photons that produce electron-hole pairs on the floating gate - as electrons have a higher energy level than holes, more electrons than holes migrate, resulting in a net loss of electrons on the floating gate when the event ends and recombination takes place; whether this is sufficient to de-programme the cell is device dependent. |
H: LM317 Dropout Voltage
The LM317 Data Sheet doesn't appear to specify a dropout voltage in the table of electrical characteristics. 1.2V - 1.3V sounds about right for the dropout voltage; is "reference voltage" an alternate name for it?
AI: is "reference voltage" an alternate name for it?
No. The reference voltage is a different operating parameter.
Table 7.3 in the datasheet you point to gives the actual value on the second line, calling it "Input-to-output differential voltage", giving it a minimum value of 3.0V. That is to say, you should always give an LM317 at least 3V more than you've set for its output voltage.
Now, while that is a safe value for all allowable LM317 operating conditions, the actual picture is far more nuanced:
That chart comes from National Semiconductor's version of the LM317 data sheet. That's a good reference in this case because they are the originators of the LM317 design, and they published their design's schematic, so we may safely assume that regulators from other manufacturers presuming to use "317" in their part number operate in essentially the same way. (And if not, they shouldn't be using that part number!)
You may look at that chart and say, "Aha! My regulator's load is only 500 mA, so the actual dropout in my case is about 1.7V." And that is true, as far as it goes. Just beware that by designing your system to run with exactly 1.7V across the regulator, you are throwing away all design margin. If someone takes your regulator outside in winter, it will be in a dropout condition, because as the chart shows, dropout rises at lower temperatures. Not quite to 3.0V, but well above 2.0V at any rate. |
H: Desoldering a TO-220 MOSFET from a plated through-hole board with SMD complications
I'm planning on replacing a TO-220F package MOSFET on a through-hole circuit board that also has SMD components on it. In this case, an SMD capacitor is connected to the source gate of the MOSFET by what looks to me like a small solder bridge or "trace" on the underside of the board, as pictured below.
I'll need to desolder the MOSFET and solder-in the new one. My question is, will or can the solder "trace" be preserved during the desoldering process, or will I have to somehow recreate it with the new component in place, if that is even possible?
The board is TCO'99 if that is relevant (solder already present may or may not be lead-free? higher melting point and working temperature?).
I will have copper solder wick, rosin flux paste, silver-bearing solder and a twenty-dollar soldering iron with an ESD ground-clip and a fine/chisel tips to work with.
Any other tips within the scope of this question are appreciated. Primarily, I am focused on making sure that existing components are not damaged during the process and that the new one is installed so that proper functioning is restored.
The problem with the MOSFET itself is that it is overheating very rapidly, though the circuit seems unaffected. A thermal sensor is placed directly between it and another which shuts the circuit down over a certain temperature.
I located this particular component by touching it, and it does get very hot in addition to exuding an acrid burning smell after a few minutes.
My working assumption is that the component suffered critical overheating which separated some layers within it, and its thermal performance is now shot.
Below is the schematic for the circuit in question:
I actually took a picture of the wrong portion of the board (Q517 instead of Q507), but the components are the same.
EDIT AFTER THE FACT: Well, I ended up desoldering everything instead of snipping and soldering the new part onto the legs on the top side of the board as answered, mostly because it was too tight to work between the components. Good thing the part was already dead, because I had to really work the pads and even ripped up a trace getting all the pins out using my cheap iron. I connected the leg with the missing pad to a solder bridge built on a snippet of the copper-wire wick I was using, and, luckily, it all worked out. Lesson is to get a soldering station if you can afford it.
AI: If possible, I would recommend removing the damaged through-hole transistor by cutting its leads above the board surface, then simply trim the leads of the new one to where you can install it by simply placing them on the existing solder pools & melting the existing solder (possibly with adding a very slight amount).
This may not be the "prettiest" way to effect the repair, but since you seem to not be very experienced with soldering on PCBs with SMD devices (and don't necessarily have great SMD soldering equipment), it's probably very much the safest. |
H: If I flick the light switch on and off will it damage the light?
Ive heard that when you flick the light switch on and off repeatedly it damages the bulb. I know this is certainly not true for LED bulbs, but Im wondering if and how it damages other kinds of bulbs. I know that light bulbs run off of ac current so they are turning on and off 120 times a second so I find it hard to believe that this would damage them.
AI: Incandescent filament lamps resistance changes dramatically - up to ten times - as they warm up. The result is that there is a large inrush current initially but as the element heats the resistance increases and the current decreases to its nominal value. This explains why bulbs generally popped on switch-on rather than at a random moment when burning steadily. (The thermal inertia of the element is high enough that it doesn't cool down between AC mains half-cycles.)
During rapid switching the filament would not cool completely before the next switch-on so it would not be stressed so much. The longer the 'off' periods the harder it is for the lamp.
My domestic experience was that lamps on dimmer switches lasted 'forever' relative to the other lamps in the house. This was due to the rotary dimmer being turned up over, say, half a second and gradually increasing the current while the element warmed up. There was no sudden inrush current.
Lamptech have an article on the subject with some graphs and response times. They also contradict me regarding longer off times. (Ah well.) |
H: Bipolar Square Wave Generator
The design requirement is to generate a +/-12V square wave from a 0-3.3V square wave at 1kHz.
Here is what I came up with:
PILOT_CTL is the 0-3.3V square wave and the end of R6 should be the +/-12V output.
I built the circuit, plugged it in and Q1 promptly blew up. I believe this is because I tied the drains of the NFET and PFET together. Looking at the DMG6601LVT datasheet I can see that the Turn-Off Delay Time for the NFET is 31.2ns while the Turn-On Delay Time for the PFET is only 1.7ns. I think this means I am shorting +12V to -12V for ~30ns through the FET which is enough to make it go up in smoke.
Any suggestions on how to improve the design to avoid this problem?
Followup
I implemented the suggestion of adding resistors to each drain individually like so:
Q1 no longer blows up and a reasonable square wave is produced:
Looking closer however the rise time is nearly 40us
As was mentioned previously, this is due to the gate charge having to dissipate through a 10k resistor. Obviously I could use smaller pullup/pulldown resistors but this will only improve by one order of magnitude before < 1k resistors start burning too much power. Any suggestions? Is a new circuit topology required?
AI: Your squarewave is at low current due to R6 being 1K .If you place some 180 ohm resistors in the drains of Q1 and make R6 820R you still have an output impedence of 1K .Q1 will now stop blowing up so you can start to refine your circuit . |
H: How does an ohmmeter measure different resistances?
An ohmmeter has set of range settings for the resistances to be measured.
When we switch to any range, what behavior of the test reference is being adjusted?
In other words; I'm wondering if we change lets say from a 200 to 1k ohm range, what exactly are we changing? Are we adjusting the voltage applied to the resistance under test or are we adjusting the current flowing through the resistance under test?
(If we want to measure a diode's bulk resistance I may need to know what is applied to the ends of the diode not to damage it)
AI: Generally the meter uses a different reference resistor for each range.
These could be built with (say) 100 ohm + 900 + 9k + 90k +900k + 9M in series. Starting from the 100 ohm, each tap point gives a total of 100, 1k, 10k etc. The ADC converter in a meter actually measures the ratio of two values -- usually an unknown input voltage and known reference voltage. Either of these can be varied.
So, if you connect the above R string in series with the unknown resistor and pass any current through it, the ratio of the voltage across the unknown R and the known reference R will equal the resistor ratios. The ADC just displays this (scaled) ratio.
So, most meters apply about 1-3 V across that whole string and display the result.
simulate this circuit – Schematic created using CircuitLab
Here, depending on which switch is closed, the meter will be on a different range. Some current will flow through the whole string; the voltage across the upper portion will generate the reference for the meter; the voltage on the lower part is measured by the meter w.r.t. the reference and displayed as resistance. |
H: What does Fujitsu call an "RFID LSI"?
Fujitsu offers ferroelectric RAM modules, including in "RFID LSI" format. I have no idea what those are, and the engrish documentation doesn't help at all.
AI: RFID = Radio Frequency IDentification
LSI = Large-Scale Integration (chip)
In this context LSI probably just means that it's at least a moderately complex chip, rather than being a precise category of complexity. This is fairly typical Japanese-English terminology. |
H: Diode preventing coil from latching?
So I've come across a case where an input from a little 5v I/O system will trigger the coil of a reed switch but not it's direct equivalent that has a built in flyback diode and changing the resistors that are in series with the input doesn't seem to work. Far as I can tell it's a CV supply though. Thoughts on getting it to work?
AI: Well, the obvious suggestion is to make sure that the polarity is correct.
With the diode in there you have to pay attention to the polarity, otherwise bad things will happen. If you're substituting for an equivalent part 'socket' without a diode, you have a 50/50 chance of it working. |
H: Why do I have to use an additional resistor with a photoresistor?
I'm totally new to electronics and I wonder why we need to put a resistor in series with a photoresistor to measure the variation of light? I mean, photoresistor is already a resistor, why do we have to decrease the voltage in the circuit with an additional resistor? Thanks in advance for your answers.
AI: EDIT: Added example for calculating voltages in a voltage divider
Because if you want to measure the resistance of something, you need to apply voltage to it.
And if you apply voltage, you need to somehow measure that voltage, and by simply measuring between the photoresistor's terminal which is on the \$+5\;V\;(V_{cc})\$ and the terminal which is on \$GND\$, you get exactly \$+5\;V\$, there is no changing voltage, no matter how small or how large the resistance of the photoresistor is.
simulate this circuit – Schematic created using CircuitLab
You measure 5V in the schematic above.
You solve the problem by using a voltage divider:
simulate this circuit
Now you can measure the voltage drop on the resistor, and from that value you can guess the amount of light the photoresistor recieves.
Example:
In the second diagram you can see that the voltage is applied across a \$50\;\Omega\$ and a \$100\;\Omega\$ resistance. Because Ohm's law says that \$U=R\cdot I\$ and the current must be equal in a series circuit, the same amount of current flows through \$R_1\$ and \$R_2\$.
In a series circuit, current stays the same, but voltage is shared between the circuits.
We can write down the following equation:
\$U_{R_1}\$ = \$R_1\cdot I\$
You could ask how can we calculate the voltage if we don't know the current.
Well, we don't know the current, but we can calculate it using Ohm's law.
We write down the original Ohm's law equation differently:
\$U=R\cdot I\;\Rightarrow\;I=\frac UR\$
Because in this case the total resistance is \$R_1+R_2\$ (or \$150\;\Omega\$ in our example), the equation for the current will be \$I=\frac{U}{R_1+R_2}\$.
We can use this equation to substitute the single \$I\$ variable in the above mentioned equation.
So the equation for each of the resistors will be:
\$U_{R_1}\$ = \$R_1\cdot\frac{U}{R_1+R_2}\$
\$U_{R_2}\$ = \$R_2\cdot\frac{U}{R_1+R_2}\$.
If we have \$50\;\Omega\$ on \$R_1\$ and \$100\;\Omega\$ on \$R_2\$, then the voltages on them will be
\$U_{R_1}\$ = \$R_1\cdot\frac{U}{R_1+R_2}=50\;\Omega\cdot\frac{5\;V}{50\;\Omega+100\;\Omega}=50\;\Omega\cdot\frac{5\;V}{150\;\Omega}=50\;\Omega\cdot0,0\dot3\;A=1,\dot6\;V\$
\$U_{R_2}\$ = \$R_2\cdot\frac{U}{R_1+R_2}=100\;\Omega\cdot\frac{5\;V}{50\;\Omega+100\;\Omega}=100\;\Omega\cdot\frac{5\;V}{150\;\Omega}=100\;\Omega\cdot0,0\dot3\;A=3,\dot3\;V\$.
If \$R_2\$ will change (for example less illumination) and its resistance rises to \$150\;\Omega\$, the voltages will be
\$U_{R_1}\$ = \$R_1\cdot\frac{U}{R_1+R_2}=50\;\Omega\cdot\frac{5\;V}{50\;\Omega+150\;\Omega}=50\;\Omega\cdot\frac{5\;V}{200\;\Omega}=50\;\Omega\cdot0,025\;A=1,25\;V\$.
\$U_{R_2}\$ = \$R_2\cdot\frac{U}{R_1+R_2}=150\;\Omega\cdot\frac{5\;V}{50\;\Omega+150\;\Omega}=150\;\Omega\cdot\frac{5\;V}{200\;\Omega}=150\;\Omega\cdot0,025\;A=3,75\;V\$.
The more the resistance of the photoresistor rises, the more voltage will drop across it.
If we give the photoresistor more illumination and its resistance falls to \$75\;\Omega\$, then the voltages will be
\$U_{R_1}\$ = \$R_1\cdot\frac{U}{R_1+R_2}=50\;\Omega\cdot\frac{5\;V}{50\;\Omega+75\;\Omega}=50\;\Omega\cdot\frac{5\;V}{125\;\Omega}=50\;\Omega\cdot0,04\;A=2\;V\$
\$U_{R_2}\$ = \$R_2\cdot\frac{U}{R_1+R_2}=75\;\Omega\cdot\frac{5\;V}{50\;\Omega+75\;\Omega}=75\;\Omega\cdot\frac{5\;V}{125\;\Omega}=75\;\Omega\cdot0,04\;A=3\;V\$.
The lesser the resistance of the photoresistor gets, the less voltage will drop across it (and more voltage will drop across the other resistor).
As you can see, we moved from \$3,\dot3\;V\$ to \$3,75\;V\$ when the photoresistor's resistance rised then the voltage dropped to \$3\;V\$ when the resistance fell. |
H: Using optocoupler with MOSFET for dimming a LED
I have a blue LED which I want to dim using an Arduino, an optocoupler and an N-MOSFET.
For achieving this I assembled the circuit on a breadboard as shown in the pictures, using a YwRobot breadboard power supply to provide power to the optocoupler's transistor, and the MOSFET. I used the Fading example sketch from the Arduino IDE.
When I try to dim the LED in this basic configuration: \$(PWM_{out} → Resistor → LED → Ground)\$
The LED slowly lights up, reaches its maximum, then slowly dims, then the cycle repeats as it should in this case.
But in the circuit it behaves weirdly.
It doesn't dim, it's on full brightness, but when the cycle reaches again zero voltage, it's turned off. No sign of any change in the brightness.
I tried modifying the delay of the cycle and I used a multimeter to measure the LED's voltage. As soon as the cycle moved from turned off state \$(0V)\$ to a slightly higher voltage (I used 1 unit increments in analogWrite and 4 second delay) the LED was on full brightness \$(2,8V)\$.
I don't know what's going on. Maybe I don't need one of the resistors on the optocoupler's BJT side? I tried taking the \$R_3\$ resistor out but then the LED was always on, didn't even bother to turn off so I put it back.
Questions:
What should I change in the circuit to make the LED dim like in the
basic configuration?
Do I need current limiting resistors for the BJT both on the emitter
and the collector side \$(R_2, R_4)\$ in this case? I know that in
the case of FETS, the \$I_D=I_S\$, so one resistor \$(R_6)\$ is
enough.
Here are the schematic and the Fritzing project:
Datasheets:
4N35M optocoupler datasheet
IRLZ34N n-mosfet datasheet
AI: power mosfets have significant gate capacitance and your gate drive is significantly unbalanced.500 ohms to charge and 600K to discharge that's about 100:1 no wonder the PWM osn't showing results.
replace R3 with 1K replace R2 with a wire, that will give a stronger disharge drive and should work better. |
H: Regarding CPU speed required for USB applications
I am working on a project for developing a USB keyboard, and I am new to USB protocol and I did initial study regarding basics of data transmissions through USB.
I am now checking for USB supported microcontrollers required for my application and any development kits associated with it.
While checking some, I came through usb keyboard with Teensy which uses a microcontroller AT90USB162-16AU which has a CPU speed of max 16MHz.
But for USB isn't the required clock speed of 48MHz?
What am I missing here?
AI: The USB in the AT90USB162-16AU has its own clock circuitry. The 48 MHz USB clock is generated with an on-chip Phase Locked Loop (PLL). The PLL in this case always multiplies its input by 6. Thus the PLL clock register must be programmed by software to generate an 8MHz clock on the PLL input as shown below:
So if an 8 MHz crystal is used, the PLL clock prescaler is set to divide by 1. If a 16 MHz crystal is used, the PLL clock prescaler is set to divide by 2. In either case, the input to the PLL is 8 MHz, and the output is 48 MHz.
One should not use the internal 8 MHz RC clock when doing USB as it will not be accurate enough. |
H: Something wrong with a 30 vdc variable power supply design
Recently I have designed a variable 30 vdc output power supply
When I tried it on the breadboard it worked fine and gave me a maximum voltage around 29.8 vdc
When I tried to design the PCB ( its my first PCB ever )
And it has a lot of mistakes I know
But I don't know why but the maximum output voltage is just 20 vdc
And when I increase the potentiometer above that voltage
The multimeter reads nothing (open loop I guess)
So what could be the problem ?
Is it one of the capacitor ? Is it damaged because of the heat applied by the soldering iron ?
Can I fix it ?
AI: Check C4 is inserted with the correct polarity.
Check voltage on LM317 input is more than 30 volts. |
H: What is the use of the 0.1Ohm Sense Resistor in Battery Charger
In the below circuit, can someone please point out the use of the 100m Ohm resistor?
Can i replace it with a 0Ohm Resistor?
AI: The purpose of the 0.1-Ohm resistor is to provide a measurable voltage that corresponds to the current being supplied to the battery. Note that pin 1 is labeled SENSE. The fast charge current is determined by the value of that resistor, so replacing it with a 0-Ohm resistor will probably cause some very bad things to happen to your battery, the charging circuit, or both. |
H: PLC rung definition and evaluation
I was reading a book about PLCs (Programmable Logic Controllers).
I found this paragraph: "A ladder diagram is read from left to right and from top to bottom... The top rung is read from left to right. Then the second rung down is read from left to right and so on....
Each rung must start with an input or inputs and must end with at least one output.".
In another source i found that a rung is a single line...
I am not sure if i understand well...
A rung can consist of multiple lines, right? For example is this a rung?
A B X
--| |----| |--+-------------( )-
|
C |
--| |---------+
Let's say both A and B are 0 and C is 1.
In this case, how it is evaluated?
From left to right and from top to bottom:
It can not be:
1. Evaluate (A AND B)
2. Set output X
3. Evaluate C - can't go back?
I think it should be:
1. Evaluate ( A AND B ) OR C
2. Set output X
AI: You are correct in your execution sequence and your statement that a rung can be several lines.
Ladder vs text
In the old days many of the PLCs programmers could switch between ladder and text or 'op-codes' (I can't remember the correct terminology). Mitsubishi still support this. Allen-Bradley does in that you can edit a rung of ladder with a text editor and I've used this function to create ladder logic in Excel (incrementing inputs and outputs by formula) and pasting the results back into the ladder editor. The text version gives a better clue as to the execution. Your code, for example, becomes:
A B X
--| |----| |--+-------------( )-
|
C |
--| |---------+
OP VAL ACC Comment
ST A 0 0 ! Store A in ACCumulator
AND B 0 0 ! Logical AND of B with ACCumulator
OR C 1 1 ! Logical OR of C with ACCumulator
OUT X 1 ! Copy ACCumulator to X
Tricky branches require BST (branch start) and BND (branch end) or equivalent which allow intermediate results to be pushed onto a stack.
A quick web search threw up this site which isn't brilliant but may be of some use.
NOT contacts
For anyone else new to this, 'not' contacts are represented by a '/'. See X1 and NOT X2 in the example below. (Many PLCs use 'X' for input and 'Y' for output.)
X1 X2 Y0
--| |----|/|--+-------------( )-
|
X3 |
--| |---------+
The OP is correct in a sense that it might give more of a visual clue to the order of execution if the ladder was drawn as follows:
X1 X2
--| |----|/|--+
|
X3 | Y0
--| |---------+-------------( )-
The brainwave of ladder logic, however, was that it presented logic in the form of an electrical schematic that electricians were familiar with. An electrician would 'know' that current could flow 'up' the or branch to the coil. One of the common problems in electrician training was to get across the concept that, while the ladder listing looks as though everything operates simultaneously, in fact it is being executed sequentially.
Taking advantage of sequential scan
The home-made one-shot example below shows an example of using sequential logic which might not work with real relay logic.
! C0 fires for one scan when X1 turns on
! provided the one-shot memory is off.
X1 C1 C0
--| |----|/|----------------( )-
! C1 'remembers' that X1 was on.
X1 C1
--| |-----------------------( )-
! Y0 fires for one program scan.
C0 Y0
--| |-----------------------( )-
The logic should be fairly clear. Trying to do this with relays would create a race condition (although there would be other solutions in that case).
Program execution order
Typically PLCs task execution order runs as follows:
Read the inputs.
Execute the logic.
Write the outputs.
Handle other tasks, etc.
Each 'scan' on newer processors takes < 1 ms which has come down a lot from the early days where 20 - 50 ms would have been common for larger programs.
Online status monitoring
For those new to the subject: the ladder diagram is also very useful for online debugging. The status of the contacts or 'power flow' is indicated by changing colour or wire representation.
X1 X2 Y0
--| |--==|/|==+=============( )-
|
X3 |
==| |=========+
In the example above we can see that we have an open contact in the first branch (indicating that X1 is off) whereas the lower branch has a complete circuit through X3 to Y0 which turns on. This gives an easy to understand visualisation of the program logic. |
H: M-Bus (Meter-Bus) Arduino / RPI
Is there some one that have manage to connect to a M-Bus (Meter-Bus)? With a Arduino or RPI? Maybe a M-Bus to serial is the way to go?
AI: If you want to connect as a master:
For a quick start: go with the RPI and have a look at the openmuc jMBUS lib:
https://www.openmuc.org/index.php?id=48
I used it in combination with a ready to use (commercial) M-BUS-master-module offering a RS232 interface:
http://www.solvimus.de/en/smart-metering/m-bus-oem-master/
pro: only needs one 24V DC supply, con: expensive (~30 €).
There is a simple MBUS-master circuit shown as part of rscadas libmbus: https://github.com/rscada/libmbus/blob/master/hardware/MBus_USB.pdf
In http://openenergymonitor.org/emon/node/1944 user Trystan Lea showed how they used the interface successfully. He also referenced a (required) step-up DC/DC converter (XL6009). Sorry - I was not able to get a direct link to the relevant post - look for '29/05/2015 - 15:40'.
To implement a slave:
Use the TSS721A (TI - http://www.ti.com/lit/ds/symlink/tss721a.pdf) mentioned by @Rev1.0 or the NCN5150 (ON Semiconductor - http://www.onsemi.com/pub_link/Collateral/NCN5150-D.PDF). |
H: Simulation of TDR using LTSpice
I'm trying to simulate a TDR based on the figure below, and one of the scenarios is a short circuit. The simulation should last for 140ns.
The parameters are
\$Z_A = 50 \Omega \$
\$T_A = 5ns\$
\$Z_B = 75 \Omega \$
\$T_B = 10ns\$
Rise time of source signal \$\tau_r = 0.1ns\$
I took this website as reference to come up with the schematic below:
As nothing about the voltage source was given other than the rise time, I didn't really know what to put as its parameters, so I took most of the parameters given in the reference website.
I then probed the wire between R2 and TL-A to get the graph.
First of all, is my simulation even correct? This is the first time I'm using LTSpice, and I'm completely new to this electronics topic.
Second, as I've only seen TDR with one transmission line, I'm not sure how to interpret the graph.
a) I thought this circuit is similar to 3 resistors in series - R2 (\$50\Omega\$), ZA (\$50\Omega\$) and ZB (\$75\Omega\$). So shouldn't the incident voltage at the point between R2 and TL-A be \$\frac{50}{50+50+75} * 2V = 0.571V\$?
b) I understand that the time taken for the signal to come back to the probe point is twice the transmission delay. Isn't this the case be it 1 or 2 transmission lines?
At the probe point, I thought it takes 2(5ns + 10ns) = 30ns for the reflected signal to first reach the probe point, meaning the voltage level at that point is constant until 30ns? Why is the voltage level changing at 10ns already, and for it to actually increase?
c) Why are the next voltage level changes at time intervals of 20ns? Why not at multiples of 30ns?
AI: We can sort out the times here.
At the probe point, the incident signal will be at 0.5(on) (as it is) until the signal has traversed TLA (it is a 1:1 voltage divider during transit), then had a discontinuity at TLB and returned from that point.
Note that you must use the effective impedance depending on where the signal is; at < 10nsec (for your probe point), TLB does not exist.
As the traversal time (round trip) in TLA is 10ns, that is the time you will see the discontinuity reflected from TLB at the probe point. When the signal is returning from an impedance higher than the source, it returns in-phase. The reflection co-efficient at the TLB entrance is (75-50)/(75+50) = 0.2, and we do indeed see the reflection increase by this amount.
At this time, the signal has already moved through TLB by 5 ns, and therefore it will take another 5nsec (outbound through TLB) + 10 ns (return through TLB) + 5 nsec (return through TLA) = 20 nsec for the next discontinuity to show up at the probe point for a total time of 30 nsec.
Just why you see another discontinuity into something larger than the specified load impedance (reflection co-efficient of -0.8, approxiamtely 5 ohms) is interesting and may have something to do with the specifics of each model. |
H: I2C with 1.8V levels without level shifters?
I have problems detecting an MPU-6050 I2C device on the Intel Edison Mini breakout board (it only supports 1.8V levels). According to the thread here, it might be possible to get the sensor working with 1.8V levels:
"The MPU-6050 also supports different supply (VDD) and I/O interface
(VLOGIC) voltages. If your board allows you to make separate
connections to VDD (on pin 13) and VLOGIC (on pin 8), leave VDD on
3.3V and connect VLOGIC to 1.8V. Connect the pull-up resistors on SDA and SCL to 1.8V. The MPU-6050 will then speak I2C directly at 1.8V,
eliminating the need for a discrete level shifter."
Anyone can suggest how this setup might work (i.e. make a drawing) ?
Thanks!
AI: The I²C bus pullups and VLOCGIC on the MPU chip must be connected to the 1.8 Volt rail:
simulate this circuit – Schematic created using CircuitLab
The MPU-6050 still needs 3.3 Volts on its VDD pin. Note that this will only work if your MPU-6050 module has not connected VLOGIC to VDD in any way on the PCB. |
H: What is the physical explanation of electric breakdown?
I would like to know what is the physics behind the electric breakdown in gases so they become electrically conductive. So far, all the information that I have found describes the phenomenon in a macroscopic scale. I am interested in the microscopic explanation of the phenomenon, e.g. What happens to the molecules.
AI: In a storng E field, the free electrons in the air or in the gas cloud begin to accelerate until they hit the molecule. If the kinetic energy of accelerated electron is such to expel the other electron in the molecule then you have now two electrons and one ion, positively charged.
If this gas cloud is relatively low density, with lot of free space between molecules, the electrons can gain higher speeds prior hiting the molecule thus more probability to ionize, you get a chain reaction - breakdown. As the ions have the most mass, they are very steady, they can't accelerate or move as electrons, so the most flow consists of electrons which are very movable. |
H: Updating outputs in PLCs
For all PLC manufacturers, is Mass I/O copying the choice for updating outputs?
I found that there is another possible method: the continuous I/O updating.
Can anyone tell me what is the reason for this choice (Mass I/O copying)?
Are there any advantages over the continuous I/O updating method?
AI: I had edited my response to your earlier question to cover this lightly. Here's some more detail for a basic PLC.
Task execution order
Typically PLCs task execution order runs as follows:
Read the inputs.
Execute the logic.
Write the outputs.
Handle other tasks, etc.
The 'Read the inputs' and 'Write the outputs' stages are the 'mass I/O copying' stages. The reason the logic is executed this way is that it presents a consistent state for the inputs through the program and prevents output strobing during the program scan. It also means that updating of I/O can be handled as a background task.
Problem with direct input read
UP BUTTON MOTOR UP
X1 Y0
--| |---------+-------------( )-
UP BUTTON MOTOR DOWN
X1 Y1
--|/|---------+-------------( )-
In the above example if X1 were to turn off between the first read and the second read both MOTOR UP and DOWN outputs would be energised simultaneously. Normal practice is to read all the inputs into an internal register and the program looks at that rather than directly at the inputs. This gives consistent predictable behaviour of the code and avoids race conditions.
Problem with direct output write
UP BUTTON MOTOR UP
X1 Y0
--| |---------+-------------(L)-
UP LIMIT SW MOTOR UP
X2 Y0
--| |---------+-------------(U)-
This example might not be the best code but PLC coding allows it and it can prove to be a useful tool when used with skips / jumps. The first rung turns on MOTOR UP using a latching 'relay coil'. The second rung unlatches the coil when the UP LIMIT SW turns on. It should be clear that if X1, UP BUTTON, switch is held on that immediate output writes would cause Y0 to blink briefly on every scan. Writing to the internal output memory and copying the whole lot to the outputs when program execution is complete solves this problem as the state at the end of the logic scan is what gets written to the outputs.
Immediate read
Most PLCs allow an immediate read and write which can be useful to shorten the cycle time of a machine. Others allow a few inputs to be used as interrupts so that program execution can jump to a subroutine and execute immediately. |
H: What are promising battery technologies for longer battery life?
Most electrical devices lack the capacity to work for longer time (e.g. smart phones work only for a day) without recharging. This is mostly as a result of low energy storage capacity in current chemical cells. I am not an expert in solid state physics nor am I an expert in battery technology. I am just curiosity driven simple guy in the corner who wants to know why we could not have small size batteries that last longer than aday in smart phones, more than a week in PC's and months in electric cars. So, my question is, what is the current status of battery technology? What options do we have in parallel with chemical solid state batteries?
AI: Due to the need to permanently contain both the reactants and the reaction products (except for the special-case "zinc-air batteries"), batteries cannot be more that a certain amount efficient in terms of mass, or volume, power density.
Because of this fact, researchers have been working hard for many years on portable (micro) power generation ideas, and alternative energy storage devices.
One alternative storage device recently developed/released is the "EDL Capacitor"/"Supercapacitor" with an order of magnitude higher power density than any previous capacitors.
A new (small) generation device that has come from this pursuit is the Hydrogen Fuel Cell, along with the closely related NiMH battery technology, and hydrocarbon fuel cells.
Finally, there are several "over the horizon" technologies still being developed/pursued, the most promising (imo) of these is the pairing of a new generation of super-small microturbines with similarly miniscule microgenerators. |
H: LiPo Charging Ciruit - Switching Help
I do understand that similar questions and maybe even the same question has been asked before, I've found them but unfortunately as a result of inexperience on my part I have been unable to adapt the posted solutions to my design specifically.
I am currently working to build the power side of my overall circuit that uses an 18650 (3.7v, 2600mAh) battery that is charged through an MCP73833 IC. The source power for charging will be a 5V wall wart power supply. What I am trying to accomplish now is to create a path or switch that disconnects the battery when the 5V power supply is connected, while preferably allowing the the 5V to be used to power the project and charge the battery at the same time.
Based on prior answers here on EE, I've heard that two Schottky diodes can be used with a little forward voltage drop; I've also seen recommendations for FET's. I'm trying not to add any more IC's to the project if I do not need too especially since I'm hand soldering all of this (not meaning FET's, but actual IC's).
Forward voltage drop is not critical since I'll need to step-up the battery voltage anyway after this circuit (to 5V for MAX7219). So it wouldn't be a big deal to run both power sources through the same booster.
My problem is that I have zero experience with Schottky diodes and even less with Mosfets. I've been reading and trying to understand but just when I think I do, I look at my circuit and my brain turns to mush.
Obviously if you see any errors within the circuit and wouldn't mind pointing them out, I'd appreciate that as well.
AI: Microchip's AN1149, "Design A Load Sharing System Power Path Management with Microchip's Stand-Alone Li-Ion Battery Charger", describes how to use 3 components to allow switching between off-line and on-line operation without interruption of power. |
H: Kirchhoff law in Electronics Workbench
I have a task to check Kirchhoff laws in this circuit.
If I understand everything right, I3 + I2 should be equal to I1, but it isn't. What's wrong?
AI: 9.649mA + 32.16mA = 41.81mA Your I3 meter needs to be flipped around, and you'll get the result you expect. Take this as a lesson in understanding both the simulator and circuit. If you don't understand both, interpreting the results is an exercise in futility. |
H: Eight bit ALU with Overflow in Verilog
I have the above assignment and here's what I have so far in verilog:
module eightbit_palu(input [7:0]a,
input [7:0]b,
input [1:0]sel,
output [7:0]f,
output ovf);
reg (f, ovf);
always @ (a, b, sel);
case(sel)
2’b00: f = a + b;
2’b01: f = ~b;
2’b10: f = a & b;
2’b11: f = a | b;
endcase
endmodule
I'm new to verilog so I'm not sure if this is right or what I should do about the overflow value. Any tips/suggestions?
AI: Given that you have shown effort and done most of it, some hints:
ANSI module declaration:
The standard way of declaring modules is as follows:
module <name> (input <wire> <width> <name>,
output <reg or wire> <width> <name>
);
So for example:
module test (input [7:0] a, //Input 8 bits wide
input wire b, //Input 1 bit wide - 'wire' is optional
output reg [2:0] c, //Output register 3 bits wide
output reg d, //Output register 1 bit wide
output wire [4:0] e, //Output wire 5 bits wide
output [4:0] f //Same as above - 'wire' is optional
);
The way you have done it, using the extra reg directive in the body of the module may work on some synthesizers, but it will definitely not work on all - some will complain about redeclaration of ANSI ports.
Getting the overflow from an Addition
There are several ways of getting the overflow from an addition. Firstly, create a module which does addition with overflow. Secondly, create a function which does the same thing. Thirdly, and probably most simple, use concatenation.
For the third option, something like this is quite possible. This is an example for a 4 bit adder with carry out.
wire cout;
wire [3:0] sum;
wire [3:0] a;
wire [3:0] b;
assign {cout,sum} = {1'b0,a} + {1'b0,b};
I will let you convert that into your code. But essentially what it does is make the inputs a and b 5 bits wide by concatenating 1 bit of 0 as the MSB. The output of the addition will then be 5 bits. The output of this addition is then fed into both cout and sum. Because the two are concatenated together, the sum gets the lower 4 bits of the addition, and the cout will get the upper bit. |
H: calculate encoder period from motor rpm?
Let's say I have a stepper motor which rotate at a speed of 1000 RPM and I have an incremental encoder attached directly to the shaft with 4096 PPR (pulses per revolution).
How I can measure the period of the square wave of channel A or B of the encoder?
Thanks
AI: Pulse frequency, f, is given by:
$$ f = motor~speed \cdot pulses/rev = \frac {1000}{60} \cdot 4096 = 68,267~Hz$$
(Divide by 60 to convert RPM to rev/s.)
$$Period = \frac {1}{f} = \frac {1}{68,267} = 14.6~us$$
Homing routine
From some of the OP comments it appears that you are using the encoder to detect home. The incremental encoder won't tell you where you are on power on. You'd need an absolute encoder or encoder battery backup for that.
The usual approach (e.g., in inkjet / dot-matrix printers) is to run a homing routine. e.g., run left at low speed until the home sensor turns on. (Allow mechanical over-run to allow motor/load to decelerate to a stop.) Then move at low speed to right until home sensor just turns off. Stop.
Now you just have to keep track of steps. The move-right step ensures that even if you power-up with the actuator left of the home position that it will always initialise to the same position.
An encoder can be used if there is a danger of stall and this needs to be detected. Alternatively you could re-home at every move left. |
H: Resetting a microcontroller with a button connected to Vcc
I would like to have a way to reset my PIC by removing power from the chip. My hacker self thinks that the simplest way would be to add a resistor between the output of my LDO and Vcc on the PIC, and then add a button between GND and after the resistor. To avoid any potential issues with switch bounce, I figure I could add a cap in parallel with the button and a resistor in between, like this: (ignore component values, those are the defaults from the editor)
simulate this circuit – Schematic created using CircuitLab
Is there anything seriously wrong with doing it like this? I'm trying to keep it simple and not require a bunch of components.
AI: You'll suffer variable voltage drop depending on load current. It doesn't look too elegant.
simulate this circuit – Schematic created using CircuitLab
Figure 1. Power cut button.
How about using a normally closed pushbutton? Switch bounce isn't a problem. The circuit will just reset again. |
H: What type of connector use smart phones adapter?
What is this type of connectors?
http://www.aliexpress.com/item/Universal-Qi-Wireless-Charger-Receiver-Charging-Adapter-Receptor-Receiver-Pad-Coil-For-Xiaomi-THL-OnePlus-Honor/32334054624.html?spm=2114.01010208.0.472.8hRw0V
AI: The connector at the top of the main pic. is micro USB. If you mean how does the charger work, it's a "wireless charging" inductor that couples with a "base" inductor to create a transformer to convert ac mains voltage down to 5V for usb charging. |
H: Relay for Intermittent Wiper Function
I want to add an intermittent function to the wipers on my 40-year-old car without any additional switches. The details of the wipers can be found here [http://www.globalsoftware-inc.com/coolerman/fj40/5G.htm]. The wiper knob closes the path to ground allowing the motor to begin sweeping the wipers. (There is another ground at the motor that mechanically closes when the wipers are not parked. This allows the wipers to park regardless of when the wipers are switched off.) It seems that all I need to do is complete the path to ground whenever I want to sweep the wipers, but it seems too easy so I’m second guessing myself. I’ve drawn out a small circuit using a ATtiny85 to detect if the wiper circuit is closed (not shown on the diagram). If the circuit is closed (knob pulled out) for longer than one second, the wipers operate as long as the knob is out. If the knob is pulled out and returned in less than one second, the uC activates the relay long enough to start a sweep and repeats this at a predetermined interval.
Am I missing something? Is this a reasonable approach? I chose an automotive relay I’ve used before, datasheet here [http://www.farnell.com/datasheets/1809465.pdf] with a diode and RC snubber across the contacts.
[Edited from original to reflect accurate wiring.]
simulate this circuit – Schematic created using CircuitLab
To be completed by OP.
| A | B | C |
+-------+-------+-------+
Everything off |12.9 V |12.9 V |12.9 V |
Bulb in, WIPER_LO closed | 0 V | 0 V | 4.9 V |
Relay energised, WIPER_LO closed | 0 V |12.9 V | 4.9 V |
simulate this circuit
[Second schematic illustrates switch and two ground paths.]
AI: [Update: major rewrite after additional information.]
simulate this circuit – Schematic created using CircuitLab
Figure 1. Redraw of wiring based on Toyota wiring mystery.
Figure 1: how it works
In OFF position the RUN switch will keep the motor running at LO speed until it reaches the PARK position.
In the OFF / PARK position the motor is shorted out. This causes dynamic braking of the wiper motor and will stop the motor abruptly preventing run-on into the RUN position.
In LO speed the off contact is open, the LO is closed and the motor runs at low speed.
In HI speed the other two contacts are open and the high-speed winding is energised.
Note that + to GND short-circuits should never occur with this arrangement as the wiper switch contacts will be break before make.
Figure 2: replacing switch with relays
Figure 2 shows the rewiring for intermittent add-on control. I am recommending this approach rather than the single-relay approach of earlier edits as it provides complete isolation between the logic and the power wiring, gives you great flexibility in the logic and is simple to wire and understand.
There is one danger to be avoided in the circuit of Figure 2: when RLY2 is energised and high-speed is selected the LO wire must never connect to the PARK switch as it will toggle between + and GND while the HI winding is energised.
simulate this circuit
Figure 3. (Almost) full circuit.
Figure 3: the full circuit - how it works
The original wiper knob has a few limitations: there are fixed internal links which cannot be broken. Disconnecting all the original wires - except the GND wire - allows us to monitor both LO and HI settings using the logic circuit without interference from the +12 V supply.
R1 and 2 pull up the /LO (not LO) and /HI (not HI) lines when not selected.
Selecting LO or HI will pull the corresponding line low. Both lines high indicates OFF position selected.
The ATiny executes the logic for the relays.
Q1 and 2 are open-collector drivers for RLY1 and RLY2. D1 and D2 provide inductive kick-back protection for Q1 and Q2 when switched off.
As required in the Figure 2 comment above we need to ensure that if RLY2 is energised then RLY1 is too. This should be done in software but D3 ensures that if Q2 (and RLY2) is switched on then RLY1 will be energised regardless of the quality of the software. ;^)
Software
As Dwayne Reid points out in his answer there are some neat tricks you can do with this to use an on-off-on sequence to set and modify the delay time. This was in my mind as I wrote my original because I remember reading an Elektor article (April 1980) on the subject. Simple micros such as the ATiny weren't available then and the design used some tricky logic to make the timer 'memory'. The Elektor design is referenced in US patent 4388574. |
H: What am i missing for this to work? (Zener Diode)
Since the zener breakdown voltage is set to 5v, i actually thought that the voltage would be 0, since it goes directly to ground, I just wanted to try a "simple overvoltage protection". How is it that the voltage is 5v instead of 0? Without the diode, the voltage is around 22 volt or so.
Can anyone explain why this is and maybe give a few examples on how to fix it?
(This is my first time working with Zener Diodes)
Thanks!
AI: You seem to be confusing the Zener action with a 'crowbar' circuit. A Zener will limit the voltage whereas a crowbar circuit will short out the supply in the event of an over-voltage. The crowbar short can usually only be removed by cycling the power.
The Zener diode will start to conduct in the reverse direction when the voltage across it reaches about 5 V. From then on very small increases in voltage will cause it to pass an exponentially greater current. At the same time the voltage will increase slightly above 5 V.
The problem with your arrangement is that if the regulator can pass, say 1 A, before going into limiting then your Zener has to handle all that power. \$P = V \cdot I = 5 \cdot 1 = 5~W\$. It will get hot.
Crowbar circuit
simulate this circuit – Schematic created using CircuitLab
Figure 1. Crowbar circuit.
This is a basic crowbar circuit. When the supply voltage rises above 4.7 V the Zener starts to conduct. When the voltage across R1 gets high enough SCR1 will be triggered and pretty-well short out the supply. The thyristor has to pass all the current (1 A in our example above) but this time the voltage across the device is only about 0.5 V so \$P = V \cdot I = 0.5 \cdot 1 = 0.5~W\$. Cool!
The thyristor will stay on until the current though it is reduced to zero by interrupting the supply.
History lesson: the term crowbar protection comes from the railway third-rail electrification. If the lineman saw a problem and needed an emergency isolation of the track power supply he would throw his crowbar to short the third rail to one of the running rails and trip out the breaker. |
H: Buck Converting Voltage, to Switch Original Voltage
I'm learning electronics, so I want to sanity check this approach in case I'm missing something.
I have an arduino (at 5V) controlling power to a set of lights at 12V, via a mosfet. This is all fine, using a separate 5V and 12V power supply.
Is there anything to stop me using 12V for both in parallel, taking one of the parallel branches, passing it through a Buck converter to 5V, and supplying the Aruino, then using the other branch, raw 12V to power my lights, with the arduino switching the 12V via the mosfet.
I've plenty of available power in the 12V supply, but is that in-parallel structure what I need.
I feel like it is an obvious question, but I was unable to google it. Examples seem to use different power supplies for the control + switched circuit. Is my suggestion the way complex circuits use a single power supply for different voltages?
AI: Yup that's fine. Just make sure the output of the buck converter to 5V is regulated well enough to not have too much ripple to cause the Arduino problems.
It's not recommended to control the buck by the Arduino because if you don't do it right, you could fry the Arduino or cause it to power itself off. It is possible to do that though.
Easiest setup is to just get a separate buck regulator IC and have it create your 5V from the 12V line. |
H: 100nF MLCC bypass capacitor spontaneously exploding on 3.3V bus, how does this even happen?
A 1608/0603 multilayer ceramic chip capacitor was functioning as a bypass capacitor for the 3.3V supply of a InvenSense MPU-6050 6 axis (gyro + accelerometer) MEMS sensor. It somehow exploded forming an open circuit, without damaging the sensor IC or any other component on the board, and without affecting its operation.
Illustration of a typical circuit from the InvenSense datasheet:
Some parts of the ceramic material are still attached to the PCB and bits of the wreckage are scattered all around. The cap was reflow soldered with lead free solder by the manufacturer and the PCB has not been reworked. I am pretty sure that the component was intact when I first got the board.
I am curious about what failure mode this was, and how to prevent such failures in boards of my own design. Is this a common ocurrence in mass production? I thought these were fairly reliable components.
AI: Could be high ripple currents (or poor cooling and moderate ripple currents) causing overheating. Or it could be one or several of the mechanical stress/defect sort mentioned in the comments, or a combination of such factors. Without knowing if the capacitor is prone to blowing up on more than one copy of the board it's hard to say for sure. Post-mortem analysis of a single failure without extensive supporting data is not definitive.
Linked below is a TDK app note hosted at Digi-Key which pretty much explains that while MLCCs are often not exactly ripple-current rated, ripple currents do affect them - but it's more of a temperature thing than a specific current. They also mention that when ripple currents are measured on MLCCs it's typically at room temperature - so it's quite possible that some lazy designer found a number, ignored the "25°C" associated, and said whoo-hoo, here's my ripple current - let her rip - at 65°C.
http://www.digikey.com/en/pdf/t/tdk/ripple-current-mlccs
The "4V capacitor on 3.3V supply" (also mentioned in comments after I wrote the above) is likewise a poor design choice that may well contribute, but it's unclear if the spec is what the actual capacitor is, since you indicate a source that is evidently not the manufacturer of the board (chipmaker? in which case they are nuts to spec that) for the spec. |
H: Disagreeing 3-phase power
So this feels like a silly one but I'm absolutely stuck. I'm trying to balance the power between a 3 phase source and two 3 phase loads in parallel. One is a Wye and the other is a Delta:
So I figured out that the single phase source current is 5A per branch, so the source complex power is (3*120V*5A) = 1800 W and 0 var.
The problem comes when I try to find the load draw. It's easy enough to find the delta line-to-line voltage and the current going through the phases of the loads:
When I go to sum the powers, though (including line impedance), I get a strange answer that doesn't balance:
Does someone see my obvious mistake?
Thanks!
AI: Never mind, answered my own question. My line impedance power calc should have been 3*I^2*Z. |
H: Dangers of IR LED
I would like to build a DIY laser tag system.
The website milestag recommends a Vishay Tsal-6100 as IR-LED.
Here is a datasheet: http://www.mouser.com/ds/2/427/tsal6100-279822.pdf
Since I want to focus the light (maybe one or two degrees divergence after the lens) , I started thinking about potential dangers to the eye. Reading up on the topic reveals that IR light can be very problematic. The eye doesn't see the IR wavelengths and can't react to overexposure. Therefore it is crucial to be well within the limits of safety standards.
Such as IEC 62471 which seems to be the applicable norm.
In fact Vishay has publicated a document with data about their LEDs referenced against this norm: http://www.vishay.com/docs/81935/eyesafe.pdf
The Tsal6100 is described as 400mW/sr in a worst case scenario. This is supposed to mean that the LED is "exempt" = less dangerous than class 1.
But what does that mean ? The specs say 230mW/sr so it seems like they already included some kind of safety margin. I couldn't find the distance at which this intensity is reached. If the specs record 230mW/sr at 1m distance and the worst case according to the norm is 50cm then a focused beam (diameter 10cm^2, a few angles divergence) might have a much higher intensity.
My question : How do I calculate the intensity of my beam ? How do I know if the LED is safe to use ?
Update:
I read up on the unit mW/sr and found this definition:
phi as intensity
omega as angle
The half angle before focusing the LED is 10 degrees, after the lens I hope to get something like 1 degree. So the factor is about 10.
I applied some math:
If I understand this correctly, a focused beam of light is 10 times as powerful, as the unfocused one.
Is this right ?
A new problem is: How do I get the source area? Should be the surface of the lens.
I tried to enter the values in this calculator: http://www.intersil.com/en/products/optoelectronics/ambient-light-sensors/eye-safety.html
(many thanks to Dave for the great link)
but there are many fields that I just don't know how to fill.
An LED with 2300mW always turns out to be lethal, and that doesn't seem right.
As a sanity check I tried to just copy the values from the spec sheet into the calculator. Turns out that the LED is dangerous, even unfocused. Now I'm sure that I made some mistake, since Vishay said that this product was "exempt".
What is the LED type for my setup ? I chose "Lensed"
If I choose lensed, what exactly is "extended source area" ? The surface of the lens ?
Could you maybe help me and try to enter the values yourselves ?
The calculator is an Excel spreadsheet.
I copied it into my dropbox, so you can just use Microsoft Excel online.
Here is a link: https://www.dropbox.com/s/r28n3p6bdf5m7hs/exposure-calculator.xlsx?dl=0
Again, the link to the spec sheet is: http://www.mouser.com/ds/2/427/tsal6100-279822.pdf
Big Update:
To be more precise about my problems with the calculator: In order to get interesting limits such as exposure limit and safety factor it is necessary to choose a "intersil proximity device type". For this field you can choose between "standard proximity senser" or "long range proximity sensor".
If no choice is made, the fields for safety factors remain empty.
Maybe we can solve the problem without this calculator ?
I thought about this unit mW/sr. Apparently it doesn't depend on distance. So in order to get the effect on the eye it is probably necessary to determine the fraction of the area that actually "hits" the eye.
At a distance of 10m from the IR LED, the half angle of 10° has produced a circle with radius 1.73m and area 9.4m^2. The eye (not the pupil, I'm not completely sure what can be harmed) maybe has an area of 3cm^2.
That's a very low percentage, surely harmless.
So I could assume that the lens creates a perfectly parallel beam of light and then create the fraction between lens surface area and eye surface area.
This makes my question simpler: What power is harmless for the eye. Assuming a fixed area for the eye, can you compute a lens diameter which makes the LED harmless ?
Is this approach OK ?
I checked against the Vishay safety document:
In case of IR - Emitters the dominating limit is the cornea/lens risk
in the wavelength range from 780 nm to 3000 nm. This limits the
irradiance to E_e= 100 W/m^2 which is expressed as intensity a value
of I_e= 4 W/sr with the measurement condition of that standard with
0.2 m distance in mind
This specifies: A viewing distance of 0.2m and a limit of 4 W/sr.
According to my thinking above they probably calculated the cone at this distance and then determined the percentage of the surface area of the eye. Then you could get to a concrete value for I_e = 4 W/sr.
Which means that I could get values for other distances. Power per area is max 4W/sr for the 0.2m distance. At 0.1m the cone area is 1/4 of that area, so I would get a maximum of 1W/sr for I_e - Thinking: The cone is 1/4 of that area -> percentage of eye surface is 4 times as high -> Power per area must be 1/4 the reference value.
And at 0.05m only 250mw/sr would be allowed.
For TSAL-6100 the document says: "maximum intensity at absolute max. ratings" 400mW / sr.
Therefore I believe that I could use an LED with f > 0.063m.
Calculation behind that: Maximum power per area of TSAL is 400mW / sr. This is 10 times lower than the reference power per area. The cone-bottom area decreases quadratic with the cone height. Hence I can decrease the distance by sqrt(10). This leads to a viewing distance of 6.3cm.
I believe that 6.3cm is the safety limit for looking directly into a TSAL-6100.
Could you check my calculations.
If I would mount a lens at exactly this spot with the exact focal length 6.3xxx cm, then from any distance it would be as though I would look at the LED from 6.3cm distance. Which is the exact safety margin.
Something that still bugs me: Different LED's have different half angles. How come they can specifiy one concrete I_e for all their IR-LEDs ? The cone of a TSAL-6200 (20° half angle) is much bigger than that of the TSAL-6100. Therefore the fraction of the light that enters the eye should be smaller. Therefore I_e should be bigger.
Maybe my whole approach is broken ?
AI: Le Goog has procured a document on "IEC 62471" from Intersil with the relevant equations. Thank you, Le Goog.
Please try Google-ing the safety standard. Here's Intersil's walk through.
Edit: Okay, I really should have been straight forward and just told you to do the math. Ee = Ie/(d^2) = (400mW/sr)/(0.2m^2) = 10 W/m^2.
According to the definitions in the intersil sheet that I provided you above, if there is no time limit or if the time limit exceeds 1000 seconds (which you shouldn't need to transmit the hit data for laser tag) The sum of all Ee wavelengths should be less than 100 W/m^2 (which you would get).
The reverse of which is that the minimum safe viewing distance for your IRLED would be (0.4W/sr)/(100W/m^2) = d^2 = 0.004 -> sqrt(0.004)= 0.063246m. So yes your math for min safe distance was correct. But again I'd just like to point out that your laser tag system doesn't take 1000 seconds to send the pulsed light. More likely you could get closer and not suffer irreparable harm (seriously, most laser tag systems do about 0.1 secs of pulsed light). Anyways, Please use the math. (not calculator)
Edit 2: A concrete example about laser tag.
Hacking the LightStrike
The link provided has some basic info on a version of laser tag that exists on the market (existed, not sure if sold anymore). Using his decode, the worst case time for transmitting "hit data" is 6750us + (32interbit markers * 900us) + (32bits * 3700us(for a one)) = ~4 seconds. No one should ever wait that long and this example is just a worst case example using a known product scheme. Going off that 4 seconds, it's less than 1000 seconds. Using Eq.1 from the intersil info sheet that I linked. Ee <= 18000*(4sec)^-0.75 or Ee <= 6363.96 W/m^2. Again applying the the reverse math to obtain min safe distance we would get (0.4 W/sr)/(6363.96 W/m^2) = d^2 = 6.28539e-5 -> sqrt(6.28539e-5) = 0.007928m or around 8mm. So your minimum safe distance is 8mm if you were to use this with a lightstrike laser tag marker. Results will vary based on encoding of course. I hope this has provided you with more than enough information to judge the safety of your device (which is the first question). If you happen to have any more questions, please ask them in a new question so that others have the opportunity to help you. |
H: Is it possible to eliminate output ripple of SMPS's by using two parallel SMPS's to produce outphase outputs to each other?
simulate this circuit – Schematic created using CircuitLab
Obviously it can not be that way. But why? Imagine one converter with a phase detector of its switching frequency, outputs a flag when the phase is 180° and other one synchronises its own oscillator to output an outphase ripple to other. Then it will be like below:
Green one is the output sum of two AC ripples and the common DC. What are the reasons why this can not be done?
AI: The voltage ripple on the output is the integral of current in the output capacitor(s). If we assume the load draws a constant current then for the ripple voltage to be zero the current delivered from the switch mode converter(s) must also be constant.
If the ripple current from the switchers was a nice clean sinusiod at the switching frequency then what you propose would be possible. Instead as the name suggests it's a waveform resulting from switching, sometimes the current may be zero (bucks running in "continuous mode" do deliver current all the time, boosts, flybacks and converters running in discontinous mode don't) and when current is being delivered it's not quite at a constant rate (how close to constant it is depends on the value of the inductor). The exact shape of the waveform will depend on many factors including supply voltage and load current.
Having said that while your method won't completely eliminate ripple it will significantly reduce it. We call this a "polyphase" converter and it can be done with any number of phases from 2 upwards.
For now lets assume that the waveforms from all the converters are time-shifted copies of each other and are periodic at the switching frequency (these are approximatations of relality but good enough for now). We can view any periodic waveform as a series of sinusiods at harmmonics of the switching frequency.
If we have n perfectly matched phases then we eliminate any harmonics that are not a multiple of n. In general lower harmonics tend to be larger than higher ones for most waveforms and on top of this the output capacitor is acting as an integrator which means high frequency ripple currents have much less impact on ripple voltage than low frequency ones.
This is done in practice, for example http://www.linear.com/docs/4166 describes a converter IC that is designed to drive two phases directly and provides functionality for syncronising multiple chips to build converters with up to 12 phases. |
H: Sensitive AC detector circuit
I built a field mill based on this video link. This device generates a sine wave, where I am interested in the amplitude.
Once i have the peak, or average value (does not matter, but average would be better), I will send it over to an arduino data logger.
What circuit could I build to accomplish this task. This will be on a high altitude balloon launch in a few weeks, and need to find a circuit quickly for a high school high altitude balloon launch in less than two weeks.
The AC wave peak ranges from a bit less than one hundred millivolts to a few volts, and the data logger will read values from 0 to 5 volts.
AI: So, as far as I understand the field mill (not familiar with it, looks neat), you have a sinusoidal voltage at some frequency (couldn't read the time division on the scope), and you want to convert it to a value that an Arduino can read.
First off, it'd be helpful to know the absolute peak voltage this thing is going to generate in a thunderstorm or something. In the video, it looks like it goes above 5V.
Here's how I'd go about it:
First, clamp the input to a sane range. Let's say -0.7V to 15V. This will require some diodes (ideally two). This eliminates troublesome voltages that will ruin the rest of the circuit.
Next, run the signal through a precision peak detector (a regular peak detector outputs one diode drop below the input, and won't work for your low input voltages). This will eliminate the sine wave, and give you a slowly-varying (almost DC) voltage. Here is a helpful document for designing precision peak detectors. Googling will give you many more resource for designing precision peak detectors.
Feed this signal through a voltage divider (and maybe another low-pass filter to clean it up), and then into your datalogger's ADC. This assumes that the field mill output voltage range is actually more than the 0-5V ADC range (because that's what I saw in the video). If this is not true, you should add a small gain stage to scale the voltage appropriately.
Edit: WhatRoughBeast raises a good point: motor noise can disturb the peak detector circuit. So, an alternative:
Clamp the input, as before. Run the signal through a precision half-wave rectifier. Low-pass filter this signal, and run it into the ADC, amplifying or attenuating as needed. |
H: Splitting AC voltage based on phase
Is it possible to separate the positive and negative portions of an AC voltage wave into two separate lines, where one line has the positive voltage portion (and is neutral when the source is negative), and another has the negative voltage portion (and is neutral when the source is positive)?
Here's a diagram to try to clear up any misunderstandings:
simulate this circuit – Schematic created using CircuitLab
So if VMI's output looked like this:
VM+ and VM- would look like the following:
If it matters, this is purely hypothetical; I don't plan on trying to build said circuit anytime soon.
AI: I think a Half-Wave Rectifier is what you mean. |
H: What is a startup code how it is related to bootloader or bootloader code?
I am doing a project on building a bootloader for a kinetis E series controller (MKE02Z64VLD2). My mentor told me to have a thorough knowledge about Startup code and linker files before starting the project. How are they essential for a bootloader code ?
AI: You have to understand how the processor boots, understand the state of the chip on boot, no doubt the boot code is in a flash/rom somewhere. You have to know where. Then with that knowledge you have to place the vector table or boot code at the right address per the rules of how the processor boots, which means you need to have complete control over the compiler, assembler, and linker in order to do that. You are not compiling an application running on linux or windows where the default linker script knows how to place things, this has to be place per the rules of the hardware not the rules of some operating system. |
H: Proper Arrow Directions for Off Page Connectors / Symbols...
Good day everyone,
As I understand it, there are two kinds of objects that can be used on a schematic to indicate that the Net to which they are attached is connected to a location elsewhere on the schematic (maybe the same or a different sheet). The reason is to allow the flow of the schematic to be clear and / or to allow a signal path to be continued on another page.
I have heard / seen these objects referred to as off page connectors (or symbols), ports, etc. I am guessing that they may (or may not?) have different names depending on whether or not the signal appears elsewhere on the same sheet. In that case, I have often seen the Net line simply end, with a little alias or name at the dead end that can be found elsewhere on the same sheet.
The real question I have is this: I am trying to understand the correct direction of the arrow to select for the off page (or on page for that matter) connector symbol. I gather that, for logic signals, gate outputs and the like use the arrow direction arranged such that the outputs are pointing to some other gate's input. That other gate input has a corresponding incoming arrow. But what if there is a control voltage / current to be applied to an external load and there are two wires, a send and return. I can imagine that there would be one arrow out (the outgoing driving current (conventional)) and one arrow point back in (the returning current (conventional)). But, I wonder if both of these arrows should point outward (towards the load), since they work together to act as a 'data signal output' (and what if the output pair is an AC voltage?). In summary, what is a good, consistent, standard practice to assign directional arrows for on page / off page connector symbols? Do I follow the conventional current, or the flow of data? Or does it depend? I know there are bidirectional arrows available.
This is my first question. I do believe I searched thoroughly enough to be sure it is not a redundant question.
Thank you very much for reading! Any advice will be very much appreciated! Best,
Eric
AI: An off-page connector for an output should be drawn as an out arrow even if the output sinks current. Here's an example.
Block diagram perspective:
Detailed perspective (contents of the green block): |
H: Power supply filter choke failure - how and why?
Trying to repair a non-working Fiber to Ethernet Converter. When connected to power, the unit made a hissing/sizzling/scratching sound and the all LEDs glowed dimly. I isolated a small surface-mount choke on the input power side, filtering 12VDC input from wall-wart to input of main 3.3V voltage regulator. (Schematic below.) Removed this choke and replaced with a straight wire shunt. The unit powers up and works perfectly.
I don't have an inductance meter, but the choke measures about 0.1 Ohms, so it's not burnt out.
The voltage regulator is a LM2591HVT-3.3 (a 150kHz Buck switching regulator). I see a clear 150kHz signal at the input after replacing the choke with shunt. Before replacing, there was a very noisy signal at this point, almost impossible to see with my scope.
Three questions:
1) What actually failed inside this choke? Why does it "sizzle" when powered?
2) What possibly caused this failure?
3) Is it OK to replace this choke with a shunt? What problems may arise from this modification?
Output of power circuitry is now stable at ~3.4VDC and unit powers up normally.
Thanks!
Schematic:
simulate this circuit – Schematic created using CircuitLab
AI: There are many options.
First of all, as said in the comments it's possible it wasn't the choke, but the caps that have gone bad. It's possible it was the choke after all.
The choke is there to get out the buck converter's switching noise spikes, so that:
The adapter isn't overstrained (adapters are cheap and designed on the edge of well)
To conform to EMI/EMC standards.
The second may not be an issue for you, the adapter will probably filter that out before it gets to the AC and possibly it will not couple from the DC wire onto anything else.
If it was the choke its inductance may have increased due to a mechanical default in the core. But you should also make sure the caps aren't dry by looking at the ripple if you put in a small extra resistance in the supply path.
Or just replace them anyway with a couple fresh ones. |
H: What does "V.A." mean in datasheet dimensions?
I'm looking at a datasheet for an LCD panel (pdf link to datasheet).
I don't understand these two encircled dimensions:
What does the acronym "V.A." mean for these dimensions?
I have referenced a couple of other questions here on EE, but they don't cover this:
Abstract Datasheet Dimension Notation
Reading dimension in datasheet
AI: After looking into the linked LCD datasheet and answers here, I looked into other display products of the same company NEWHEAVENDISPLAY. And the interesting thing I found is apart from V.A. they use another therm called as A.A. Below is an example. Adding the datasheet link as well, if someone more interested can look into.
V.A. can be Viewing Area or Viewing Angle(if mentioned in degrees)
A.A. is Active Area also called as effective area which the area for active pixels or icons or segments or patterns.
These links say all the same link1, link2 and link3.
Just hope ambiguity stops here. |
H: Finding the equivalent Resistance of acircuit
simulate this circuit – Schematic created using CircuitLab
I am a little confused on how to find R equivalent of this circuit because of the wire.
Right now I have the following expression:Req= (R1//R2) +(R3//R4)
Is this correct?
AI: Redrawing the circuit should hopefully make the solution obvious:
simulate this circuit – Schematic created using CircuitLab
So yes, you are correct. |
H: Do I need a galvanically isolated step-down DC-DC converter to power my digital electronic from a battery pack that powers also the electrical motor?
I have a battery pack with lithium ion cells. The battery pack has a maximal voltage of 84V (it goes from 84V to ~50V -> depending about the current state of charge).
I use this battery pack to power all sort of things (lights, horns, MCU(BMS), inverter - that then gives power to the eletrical motor...). The inverter draws about 60A of current continuously from a battery pack when the motor is rotating.
For my concept to work I must have different DC voltages on a single PCB:
12 VDC (for lights and horns)
5 VDC (for some sensors...)
3.3 VDC (for MCU, operational amplifiers...)
The idea is to use some sort of step-down converters on the PCB to convert from 84V to 12VDC and to convert from 84V to 5VDC. And then to use a LDO to give me 3.3V from 5V.
The 12VDC must have a power of ~40W, and the 5VDC must have a power of ~10W.
The question is regarding if the step-down converter should have galvanic isolation (like a flyback) so that I can have different grounds for high voltage and low voltage? Or is OK to use a "normal" buck regulator?
I think that for 12VDC that supplies the horns and lights is not so necessary to have different grounds as for 5VDC that supplies most of the electronic ?
Any suggestion?
AI: The inverter draws about 60A of current continuously from a battery
pack when the motor is rotating
Not a small amount of current and, if you don't get your grounding sorted out correctly you might find tens of amps trying to flow through the wrong ground return wires and hurting your MCU.
Therefore there is a case for providing galvanic isolation to your sensors (5V) and MCU (3V3) supplies because, although your MCU may need to connect to ground at the motor controller, having isolation in a potential ground return path could save you lots of heartache and dollars. |
H: how to remove this error "ERROR 152: Port on instance does not exist in entity declaration for instance"?
I am new to Cadence Allegro HDL for schematic design and for my first design i got this error.
In 728 page document provided by cadence its is mention as follows,
"Open the symbol for the component in Concept HDL or Part Developer and delete the port
on the symbol, if it is not required. Then save the symbol. This will bring your symbol and
entity ports in sync."
but I am not understanding what to do,please help.
AI: Use of the words "Port" and "Entity" suggests that you are working in the VHDL language, perhaps your schematic editor is a tool that allows the visual creation of VHDL designs.
The actual meaning of the message is clear : you are trying to connect a signal to a pin that doesn't exist. For example, take this AND gate
entity AND2 is
port (A,B : in std_logic;
O : out std_logic);
end entity;
If you try connecting 3 inputs (A,B,C) and one output to it, this will fail with a similar error message about port C.
Now the specific cause in your environment is not so clear to anyone who doesn't use that environment.
It might be that you have created a schematic symbol for a 3-input AND gate
simulate this circuit – Schematic created using CircuitLab
and as part of that symbol, associated it with a VHDL entity called "AND2" which of course represents a 2-input AND gate.
Now either you need to find the correct VHDL entity (presumably called "AND3") or delete the extra input pin (and make sure the other two are labelled "A" and "B").
But you must end up with a schematic symbol and a VHDL entity that match. i.e. both the graphic view and the VHDL textual view must describe teh same thing.
If you don't know how to do that, you need to find and study tutorial material for the specific tools you are using. |
H: Formula for electromagnet strength parallel, outside the coil
I am working on a electromagnet project to collect metal shavings off the floor after machining.
I have a formula for the magnetic strength at a distance from the ends, but not parallel to the coil. How do I calculate F parallel to the coil?
The formula I have now is: $$F = \dfrac{(NI)^2 μ_0 A}{2 g^2}$$
[ Edit - formula from http://depts.washington.edu/mictech/optics/sensors/week2.pdf ]
Here is a visual example of what I am looking for:
AI: How do I calculate F parallel to the coil?
It doesn't have to be an electromagnet to show that the force halfway along is zero: -
Note the iron filings at the centre of the bar magnet are not attracted to the centre - they form lines (with other iron filings) that attempt to bridge the ends of the bar magnet. |
H: Does AM Broadcast Use Sideband Modulation
Does AM broadcast radio in the United States use a sideband modulation, or are both sidebands and the carrier present in the signal? Please provide a reference.
AI: Commercial AM stations in the US use AM modulation. AM modulation carries the signal in sidebands on either side of the carrier. If you were to suppress the carrier or either sideband, you'd no longer have AM modulation. |
H: Ideal transformer output voltage
I know that the output voltage and current of an ideal transformer depends on the number of wounding of the coils. It means that no matter how many resistors I put in a series in my secondary loop, the current would not change and it depends only on N (Number of turns in coil).
I have been thought that resistors define the current that flow through them and it is hard to look at the current independent of the load.
Could someone explain the physics behind v1/v2=N1/N2
So is it safe to assume that the current in secondary coil acts as an ideal current source because if I know the source current I can get the secondary loop current no matter what the load of the secondary circuit is
AI: Could someone explain the physics behind v1/v2=N1/N2
In a transformer electric energy is converted into magnetic energy and then back to electric energy.
The formula V1 / V2 = N1 / N2
applies to this process without any losses.
Losses can occur due to:
Resistance of the transformer wires
Saturation of the magnetic core of the transformer
Inductance of the transformer (a transformer is also a coupled inductor)
frequency of the AC signal fed to the transformer
So the formula V1 / V2 = N1 / N2 does apply but it only applies to an ideal transformer.
Also an ideal transformer does not lose any power nor will it add any. So that means:
P1 = P2 and also V1 * I1 = V2 * I2
If you would connect a current source to one side of the transformer then the relation will still hold however you must make sure that the current on the other side of the transformer can flow so you must apply a load or short it.
This is the opposite of connecting a voltage source at the primary side, you are expected to have a load at the secondary side or leave it open, but you're not allowed to short the V2 side ! |
H: DIT-MCO hardware with Rotary switch & Nixie Tubes?
I really needed some help. I can't for the life of me find any information pertaining to this hardware manufactured by the DIT-MCO Electronics Division. The most information I can glean is that It's a Type 45 switch and was made in Northlake Illinois and that it's attached to a rotary dial. Any information or applications for these pieces of hardware would be greatly appreciated!
One more bonus, my grandfather also gave me a Nixie Tube display and I have yet to test it out. I'm not sure any specifications or what equipment to use with it to make sure I don't blow it up (figuratively); any information on the tubes would also be appreciated. Thanks!
Here are pictures for the hardware!
AI: It's a stepping switch. I refer you to the complete engineering information that an engineer would have used in 1964- "How to use Rotary Stepping Switches" from Automatic Electric.
It's an 63MB 81-page book, scanned and linked above in it's entirety. If it disappears, look for the file named "10539-how-to-use-rotary-stepping-switches-automatic-electric-1964-ocr-r.pdf" |
H: What does a black bar across transformer symbol designate?
What does a black bar across a transformer symbol designate?
This appears in a power electronics context, and in particular in this case with substituting a DC transformer as an equivalent circuit for the guts of a DC-DC converter.
AI: As you (and others) have said, it's the symbol for a DC transformer, which is an ideal model used to transform impedances for calculations involving switching DC-DC converters. The only source I can find for the symbol is Robert W. Erickson's book Fundamentals of Power Electronics, so it seems likely that Erickson himself invented it. I took his Coursera class a couple years ago, and there wasn't any special meaning associated with the bar. The bar is visually distinct from the (many) other transformer symbols and solid horizontal lines are often associated with DC.
UPDATE: As mentioned in his comment, Scanny tracked down Professor Erickson on Coursera and asked him about the symbol. Erickson's response:
This symbol was invented by Profs. Middlebrook and Ćuk in their classic papers on converter modeling from the 1970's. It denotes that the transformer is not a physical magnetic transformer, but rather is an ideal DC transformer. Many of us in the power electronics field have adopted this symbol.
The earliest Ćuk paper on record is "A General Unified Approach to Modeling Switching-Converter Power Stages" (Middlebrook and Ćuk 1976), an IEEE Power Electronics Specialists Conference presentation from June of 1976. This paper introduces the idea of using idealized transformers to model switching action. The IEEE copy of the paper (linked above) uses plain transformer symbols without the horizontal lines. The reprint at Scanny's link (possibly from the International Journal of Electronics in 1977, Vol. 42, Issue 6) uses both a straight line and a wavy line together -- more on that in a moment.
At the next PESC in June of 1977, Ćuk and Middlebrook presented a follow-up paper covering the model's application to discontinuous conduction mode. This paper seems to be the first use of the horizontal line by itself. It's behind the silly IEEE paywall, so I've reproduced the part that introduces the symbols here:
So the straight line (bar) indicates a DC transformer, the wavy line (sinusoid) indicates an AC transformer, and both together indicate an AC+DC transformer.
Fun fact: At that same conference, Ćuk and Middlebrook also introduced the topology known to modern engineers as the Ćuk converter. |
H: Does ground suffer from a voltage drop like effect?
I am currently working on a project where I need some very long wires (about 20 meters, and that is 40 meters both ways), that will be connected to a button which will be used to trigger a pin on the controller (ATmega8).
Due to expected voltage drop problems, I have chosen that I will pull the I/O pin high, and run ground through the button (button pulls I/O pin low and triggers it).
Hence my question: Will there be any problems using such long wires when I run ground through them, instead of Vcc voltage level (5V)?
Does ground suffer from "voltage drop" like problems?
AI: No, it doesn't (but...)
Ground, by definition, is the zero point in a circuit so it can't experience "drop". Ground wires (e.g. connections to ground) are subject to Ohm's Law like any other wire.
This is your circuit as best I understand from your description:
simulate this circuit – Schematic created using CircuitLab
From the Arduino's perspective the ground is a large sheet of copper foil buried inside the circuit board. All determinations of voltage (and therefore logic levels: high/low, 0/1, true/false, etc) derive from comparing the signal potential energy to the potential energy of this foil sheet (which is usually connected, ultimately, to a battery/power source's negative terminal).
Long wires are ok in your application because...
In your question you are concerned with the voltage losses in a ground wire (the wire connecting the leg of the switch to ground). This wire can (and will) develop a voltage as current flows through it (Ohm's law) and so "drop" in your understanding, but this drop is not significant enough to cause problems due to the way the switch circuit is designed:
R3 is typically three orders of magnitude greater than the resistance in the wire pathway through the switch. When the switch is open the resistance is almost infinite and the voltage at the Arduino GPIO node is equal to V1. When SW1 is closed, the resistance between the Arduino GPIO node and ground is now the resistance of the two wires to the switch and the switch itself.
24AWG wire (the kind used in network cables and other small wire systems is about 0.085 Ohms/meter). You could go more than a kilometer before you reach 100 Ohms! Even at these great resistive values, the total resistance in the wire path would be less than 250 Ohms and therefore represent only 2.5% of the total voltage (e.g. still almost 0 and certainly low enough to be read by the Arduino as logic 0).
Not all "grounds" are the same...
The concept of ground is defined for the system. If you have multiple systems there can be differences between their respective grounds.
@Techydude points out several interesting examples of this problem:
long wires in the example above, the ground plane(s) of the PCB, the
ground pins of chips, the bond wires between the pins & the silicon
die, and the silicon pathways themselves.
This relative relationship to ground occurs because voltage itself is relative. Voltage is the difference in potential energy between two points. "Ground" is just the name given to the second point when all voltages in an analysis are sharing this same second point. If they do not, you will experience (and have to account for) drop in your ground connections. |
H: How to choose a bypass capacitor for a common emitter amplifier?
This site says: "To increase the gain for AC signals the emitter resistor bypass capacitor is added. This should be calculated to have a reactance equal to the emitter resistor at the lowest frequency of operation".
I assume that I have 1K ohm emitter resistor. If I added a parallel capacitor of reactance 1k ohm. The total impedance would be 0.5 K ohm.
In this case,
Should I increase the emitter resistor to be 2K and increase the reactance to be 2K so that I get 1K ohm (Which is the original value of emitter resistor before adding the bypass cap)?
Or it is ok to have an impedance of 0.5k ohm although the original value of emitter resistor is 1K?
Will a bypass capacitor affect the bias voltage of the biasing resistors?
Thank you very much,
AI: It depends on the bandwidth you want the amplifier to achieve. A simple thumb rule is that a capacitor is seen as a "short-circuit" at higher frequencies and an "open-circuit" at DC.
Therefore, the gain of a common-emitter is about the load connected to the collector divided by the load connected to the emitter. So, you can change the AC gain of the amplifier using a capacitor at the emitter in parallel without changing the bias point of the transistor. A .1 uF in parallel with a 1k resistor will put a low cut frequency of about 1.6 kHz. Therefore, higher values should be used for lower frequencies. |
H: PWM ~100mA LED's
If someone knows of a prototyping solution for driving several LED's at >75mA I'd be grateful.
I've taken 8 LED's out of an existing product. They appear to expect 70mA to 100mA for max brightness. I've looked around for breakout board solutions to dimming or driving these LED's but most breakout boards for multiple channels that I can find either have a max of around <200mA for all channels together or around 35mA max per channel. I have yet to find something with a LT3760 (for example).
AI: Use several mosfets.One for each led circuit . Here is an example of one circuit for one led: MOSFET LED CIRCUIT |
H: Propagation across a surface
How do stud sensors actually sense current, what would cause the effect to propagate across an 8x8 foot surface area and remain strong enough to be detected?
I asked another group "what is happening", now I would like to ask your group "why". This is a question about electrical sensors included on some stud sensors used in home repair to find wiring within walls. The garage for my home was built in 2000, the garage interior is naked particle board directly nailed to wood studs, and the insulation has paper backing (not foil). When using the electrical sensor, most of an entire wall surface shows having current instead of indicating specific wire routes. Other walls in the garage do not show this effect. When I turn off the breaker for the outlets, the current disappears (so the reading is related to current flow, not a bad sensor). What is the effect the sensor is designed to detect, and why might that effect propagate across the surface of one wall and not the (similarly constructed) other walls? Thanks.
AI: So there is a magnetic stud finder and an electric one. I will assume you are speaking of the electric kind. In these, there is a capacitive plate in the front end which is driven using two multivibrators. As a multivibrator is driven, the voltage on the plate drops to zero and then as it goes low, the plate charges back up. The rise time/slope determines the dielectric changes between your plate and the plate in the wall. This dielectric change is what we detect.
Since an the electromagnetic lines are disrupted by the secondary plate, this causes a change in flow of the field lines. Thinking of this in terms of your sensor, there is something blocking the field lines over the whole wall. This means that the thickness of the wall is proper in all areas. So your whole wall is a stud.
As for when the power is turned off, that's a great question. My thought would be that the wood is a bit wet. This damp wood is then carrying current. You did say it was bare, and it may be damp. Might want to look into waterproofing. |
H: What is the pole frequency of a three-phase LC filter?
Suppose I have this configuration:
Assume L1=L2=L3 and C1=C2=C3. What is the pole frequency of this filter?
Edit: As mcmiln says, this is not identical to the question here:
Cut Off Frequency in three phase LC filter
I want the closed form (and preferably derrivation) of the filter pole.
AI: Ground is defined as the common node of the 3 voltage sources. There's no load attached.
I threw this into a symbolic math solver and get the following relations (a.k.a. sorry, no derivation):
\begin{gather}
V_{R1} = \frac{2 LC \omega^2 - 1}{3 LC \omega^2 - 1} V_1\\
V_{R2} = \frac{V_1 (4 LC \omega^2 - 1) + j \sqrt{3}}{2(3 LC \omega^2 -1)}\\
V_{R3} = \frac{V_1 (4 LC \omega^2 - 1) - j \sqrt{3}}{2 (3 LC \omega^2 - 1)}\\
I_{V1} = \frac{j C V_1 \omega}{3 LC \omega^2 - 1}\\
I_{V2} = \frac{-C \omega (j V_1 - 3 \sqrt{3})}{2( LC \omega^2 - 1)}\\
I_{V3} = \frac{-C \omega(j V_1 + 3 \sqrt{3})}{2(3 LC \omega^2 - 1)}
\end{gather}
I'm relatively certain voltage source currents are assumed to flow into the source to ground.
Note that this assumes 120 degree out of phase sources (\$V_2 = \frac{V_1}{2} - j \frac{\sqrt{3}}{2}\$, \$V_3 = \frac{V_1}{2} + j \frac{\sqrt{3}}{2}\$)
I'm unfamiliar with how to compute 3-phase filter frequencies, but hopefully this should be enough that someone else more knowledgeable can use these relations and find a filter frequency.
I do see a common factor of \$3 LC \omega^2 -1\$ in the denominator for all of these equations, so my best guess is there's poles at
\begin{gather}
\omega = \pm \sqrt{\frac{1}{3LC}}
\end{gather}
I did a quick simulation with LTSpice, and it appears that this analysis does correctly predict the pole for \$V_{R1}\$. |
H: Physics explanation of charging time by increasing dielectric constant
I am having a problem understanding in physics of the relationship between charging time and dielectric constant of a parallel plate capacitor.
I'd like about physics aspect not based on time constant.
If dielectric constant increases then capacitance also increases and so the time constant RC.
However, how would you explain it in physics?
With dialectic material between two plates of capacitors, the electric field is reduced. But how that affect the time constant?
AI: Capacitance of a parallel place capacitor is related to the dielectric constant, area, and distance between plates by a very simple equation:
$$C = \frac{\epsilon_r \epsilon_0 A}{d}$$
Where \$C\$ is capacitance, \$A\$ is plate area, and \$d\$ is the separation between plates. \$\epsilon_0\$ is the permittivity of free space and \$\epsilon_r\$ is the relative permittivity.
The voltage across a capacitor at any given instant is given by:
$$V = \frac{Q}{C}$$
Where \$V\$ is voltage, \$Q\$ is the charge on the plates, and \$C\$ is the capacitance.
So clearly as capacitance increases, the amount of charge required for a given voltage across the plates will also increase proportionally.
If you require more charge, it will take longer to reach a given voltage if you have the same initial current - which in an R-C circuit is governed by the resistor. In essence at time \$t=0\$, if there is no charge on the capacitor, all of the voltage is across the resistor, so this sets the maximum current. As the capacitor charges, the voltage across the resistor will decrease (it increases over the capacitance), so current decreases (Ohms law).
Given that current is the rate of change of charge, then with the same limitation on current, it will take longer to charge the larger capacitor to a given voltage. |
H: Need help in biasing TIP31 in common collector mode- edited
I am using an Arduino Uno board and R-2R DAC to obtain a signal which should vibrate a transducer connected across RL of common collector amplifier using TIP31C.
I need around 300mA of current to vibrate the transducer which is much like speaker (resistance is about 10 ohms).
I have calculated the biasing resistor values, by using the link from guitarscience.net,(sorry forum is not allowing to insert the link) as R1=39ohm and R2=82ohms with Vcc=5Volts for all the components.
But I dont get the waveform shown in the fig (1) instead output is always HIGH.
I have measured the voltages for the above values using multimeter and they are, VBE = 3.7V, VCE = 1.8V, VRL(voltage across the load) = 3V
Then used the link available from hyperphysics (sorry forum is not allowing to insert the link) and used R1=31 kilo ohms and R2=55 kilo ohms
upon measuring VBE = 0.5V, VCE = 4.8V, VRL(voltage across the load) = 0.4V and this time output is very much suppresed but has the same shape of the input.
Kindly help me to bias the bjt properly. The images are as below:
[![enter image description here][2]][2]
My first constraint is I have to work with 5 Volts for all the components.So arduino, LM358 and TIP31 are all at 5volts supply.
The previous ckt was incomplete and the complete ckt is as below (Fig 3), here the R1 to R11 = 20K ohms and R12 to R18 are 10K ohms and
R19 is the load (of about 10 ohms). Upon measuring without connecting the transistor, i.e. with only arduino, R2R and LM358, I am getting 2.5 Volts at the point 'A' as well as 'B'.(seems to be working properly).
But when BJT anlong with the load is connected across TIP 31C (removed biasing resistors after the previous reply), the output at 'A' = 2.5Volts and But at 'B'=0.5 volts, VBE=0.5 Volts, VCE=5Volts, VRL=0 Volts.
Kindly tell me the mistake
AI: Assuming you are using multiple Arduino outputs to drive the ladder, the ladder output is attempting to drive a current into the LM358. This ensures that the - input will always be larger than ground, and the op amp output will always be zero, or whatever your - supply is (I'm assuming you're powering the LM358 from ground and +5). Your amplifier is faithfully trying to drive the load to ground as a result.
In addition to this, you have no feedback around the opamp, so there is no way for the load to follow the ladder in any proportional manner.
Since you have an op amp, there is no need to try to bias your output transistor. Instead, add another op amp and do something like
simulate this circuit – Schematic created using CircuitLab
And yes, this requires that you add a second, negative power supply.
Also, an opamp like an LM358 will not work well at +/- 5 volts, as it is only guaranteed to drive something like +/-2 volts on the output. You need to get some rail-to-rail op amps. |
H: Current flow from positive to positive between batteries?
If you have two batteries with different voltages, let's say 12 and 9 V, and you connect the negative terminals while you put a LED or whatever between the positive terminals, there will be current flow. I reason that is because there is a potential difference even though both terminals are positive. However, I'm not sure I understand how electron current would work here.
If there is current then that must mean that electrons actually flow out of the lower positive terminal into the higher one? Is that what's going on? If not, could someone clarify it?
AI: It sounds from your question that you aren't asking about the effects of electron flow in the "wrong" direction on battery chemistry, but rather asking if a "positive" voltage can "act like" a negative or a ground. If so, then the answer is yes, it can.
All voltage is relative, so your reasoning is correct. If you connect two positive but non-equal voltage nodes together, current will flow between them.
Calling something "positive" only means that it has a higher voltage potential than something else which you are using as a ground reference.
You can think of it as pressurized cans of air. Let's call 1 atm (which is air pressure at sea level) our reference pressure. If you pressurize one can to a pressure of, let's say, 2 atm, you might say it has a "positive pressure". If you then pressurize another can to 3 atm, it also has a positive pressure. Connecting either can to a 1 atm pressure (i.e. by opening it) will cause air to flow out of it (this is analogous to electrons flowing in the wire) until the pressure equalizes. But if you connect the two cans to each other, air will flow out of the 3 atm can and into the 2 atm can until the pressure equalizes.
So it is with voltages. The electroncs that are under more "pressure" (voltage) will push harder against the ones that are under less pressure (but still "positive", vs. some arbitrary reference). |
H: Powering and operating dual electric linear actuators
I am in the process of building a height-adjusting desk. I am pretty good with computers (I'm a programmer), but I never learned much in the way of electrical engineering.
I recently purchased two Eco-Worthy brand linear actuators. They each have an 18-inch stroke and a 320 lbs. static load bearing capability. They just have two wires, the positive and the negative wires. They are supposed to operated at 12 volts with a 1 amp minimum and 3 amp maximum.
I intend to operate them both at the same time in the same direction. So what I am unsure of is primarily the power source. My understanding is that I would still use a 12 volt power supply to power both at the same time.
But will I need it to have a 3 amp output or more than that? Will the amperage be reduced as it is divided among devices? Am I misunderstanding how this should work?
Also, if I get a momentary rocker switch will I wire both together and then to the switch? Should I get a relay to run this through? Any recommendations on how I hookup the A/C power to it? Since it won't have any plug inputs.
Here is a link to the info about the actuators I purchased. Any help in figuring out the electrical setup is welcome.
AI: You want to buy a switching power supply with a 12V output capable of at least 6A if you want to be able to operate both struts at the same time under full load. The amount of current drawn will depend on how much force is pushing against each strut. If you only need to operate one strut at once, then a 3A supply will probably do.
You could, in the simplest case, wire them directly in parallel. However, it is unlikely that they will move synchronously unless they are directly bolted together, e.g. if there's one at each end of the desk they may become misaligned and jam or break the desk. If one is under more load, it will move a little slower. So probably you want a separate switch for each strut so that they can be controlled individually.
You will also need some means of reversing the polarity in order to change the direction of travel. The usual means is a DPDT switch or relay; you can google up a schematic pretty easily.
If you buy switches rated for at least 3A (one strut) or 6A (both struts) then those switches can directly control the motor current. More likely is that you might buy smaller/cheaper switches which actuate the coils of large (10A) relays, which are responsible for enabling and reversing the power.
Edit: you might want to look into building your desk so that it has a single central strut to perform the lifting work, supported by stabilising arms/scissors/whatever at each side of the desk. That way, there are no alignment issues and no need to synchronise two struts. If your lifting geometry results in more than 150kg load on the single strut, you could probably put the two right next to each other and they would be mechanically synced and provide up to 300kg of support together. |
H: Pull up on output pin of Buffer SN74HCT245
I am using SN74HCT245 Buffer operating on VCC=3.3V. So, I am getting 3.3V at each output pin of buffer (B1 to B8).
But if I want to drive a load with 5V input, what options are available to me by keeping VCC=3.3V?
Is 5V pullup on output pin of buffer is valid option?
Please refer below image where I have connected pull up resistor R1 (10K) to 5V.
AI: You Have a few options here. I'm assuming you are driving a logic input since if you simply need to switch a higher voltage load (e.g. LED) then you should probably use a transistor instead of an octal buffer (OR - octal buffer with OC outputs, see below).
The first is the simplest
It is safe to drive 5V inputs with 3V3 logic levels
3V3 Output will drive a 5V logic level correctly on most logic families.
Most IC's today will output rail to rail to handle the less sensitive 5V inputs. You must make sure that the buffer or the higher logic family can only be configured as input.
You can use an "Open Collector" device
There are open collector octal buffers that will work as you want them to, by having a pullup to 5V on the output side will allow you to use a basic IC as a level shifting tranceiver
The device to look at is the SN74621A, but there are many other families
You can use a dedicated level shifting buffer chip
This is pretty easy if you only need one direction. Multiple manufacturers offer similar parts, searching "ic level shifter" or similiar should give you a lot of options
There is a "3V3 to 5V" octal tranceiver in the same "74xx" family - the SN744245
Addendum
All 5V systems will tolerate a 3V3 input safely, however certain logic families will have higher input thresholds for 5V logic. The last letter of the middle numbers being - *T for TTL typically refers to TTL compatible CMOS subfamilies. LVTTL (3V3) and compatible CMOS families (including LVCMOS types) will work with TTL (5V), but not with vanila CMOS logic.
Note: there are many other subfamilies, (the middle letters), so care must be taken when mixing jelly beans, for a somewhat comprehensive list you can checkout this list from Texas Instruments |
H: Bipolar Voltage to Current Converter
I need a circuit or component which converts bipolar voltage to bipolar current. I have found the XTR111 from Texas Instruments and the 1B22 from Analog Devices. However, they only convert 0-10 V to 0-20 mA current.
I to need convert -10V-0V to -10mA-0mA too (the opposite polarity range). How can I do that? Or can I add a circuit to work with 1B22 or XTR111?
AI: Try the Howland current pump: -
Here's an article that gives a lot of mathematical insight into it but is not inaccessible to the equation-hater.
AN-1515A is an article by TI on the subject. |
H: Car battery capacity in colder countries
The capacity of the battery is affected when the temperature drops, particularly Lead-acid batteries. So how in cold countries this situation is solved.
In this PDF about automotive batteries at low temperature, they have mentioned the specific gravity of the electrolyte is affected when the temperature dips proportionaly the charge as well. This also observed from the graph.
And also is there is any possibility, I have load of 2 amps apart from starter will it function when the battery capacity is reduced to certain low percentage(eg. 50%) at the time of temperature drops.
AI: "The capacity of the battery is affected when the temperature drops"
That is incorrect, it is not the capacity (amount of energy stored) that gets lower at lower temperatures. It is the internal resistance of the battery which increases and this lowers the current capability of the lead-acid battery. So you will be able to draw less current (and power) from the battery at low temperatures. The amount of energy stored does not change.
If a lead-acid car battery is in good condition it will still be able to start a car even under cold conditions. A workaround could also be to turn on the lights of the car before starting, now a current will flow warming up the battery and increasing it's current capability. Of course you should only do this when you are sure that the battery has enough stored energy. |
H: Choosing voltage divider resistor values
Above is the circuit in question.
R1 and R2 forms a voltage divider to supply desired current for the load.
On the left side there is the original circuit and on the right side its Thevenin equivalent.
Lets say the ideal supply voltage is known and we want 0.01A current through our load RL which is 100 ohm.
So we can write the given and desired as:
Vcc = 5V
Ith = 0.01A
RL = 100 ohm (load resistance)
Here we can employ the Kirchhoff's law for the Thevenized circuit as:
Vth = IthRth + IthRL
Vth - 1 = Ith*Rth
5*R2/(R1+R2) - 1 = 0.01 * R1*R2/(R1+R2)
R1 = 400*R2/(R2+100)
So now we have a relationship between R1 and R2.
For example:
if R2 = 100ohm then R1 = 200 ohm
if R2 = 10k ohm then R1 = 396 ohm
So far so good. If we plot R1 versus R2 for 1 to 100ohm and 1 to 10kohm we obtain the following plots:
It seems like for R2 at around 10k, R1 goes to a limit like 400ohm.
I think one can choose here like 400ohm for R1 and any resistor above 10k for R2.
I also noticed if R1 and R2 is low resistances such as R2 = 100ohm then R1 = 200 ohm, the current flowing through R1 and R2 is much higher.
Here is my question:
Does that mean we should avoid choosing low resistors for R1 R2 just not to drive more current?
And if so, what is the rule of thumb if we only know the exact value of RL and the relation R1 = 400*R2/(R2+100) ?
I mean is there a rule of thumb between RL and Rth or RL and R2? Would you have a rule of thumb between RL and Rth or RL and R2?
AI: In the situation you outline, the rule of thumb is to eliminate R2 and determine the value of the voltage required at the \$ R1Rl \$ junction by using Ohm's law:
$$ E_{J} = IR = 0.01A \times 100\Omega = 1 \text{volt} $$
Since the current in a series circuit is everywhere the same, the 0.01 amperes through the load resistor must also go through R1.
Then since the supply voltage, Vcc, is 5 volts and you want the motor to see 1 volt, Ohm's law dictates that the value of R1 must be:
$$R1 = \frac{Vcc - V_{J}}{I} -\frac{4V}{0.01A} = 400 \text{ohms} $$ |
H: Adaptation of the ICD2
I'm trying to adapt the ICD2 so that it can be used with a microcontroller family (dsPIC33F) that has a 3.3V supply.
Projects using the ICD2 must be 5V. How can I do this?
AI: According to this from Microchip the ICD2 works across 2.0V to 6.0V. http://www.microchip.com/DevelopmentTools/ProductDetails.aspx?PartNO=dv164005
However, AFAIK the ICD2 is no longer supported by Microchip, and you should upgrade to an ICD3. |
H: Short circuit when probing live ATX power supply primary section with voltmeter
I was probing around an old ATX PSU to inspect voltage levels in the primary section. The first 15 minutes went smooth (bridge rectifier inputs, big caps, Power IC Vcc, etc..) but as soon as I probed the solder pads of the large coil circled in red in the picture, a short happened with the usual bang, sparks and smoke.
First question: what is that large coil (Hi-Pot) connected to the primary side? Is it part of the EMI reduction circuit?
Second (probably stupid) question: my DMM has 10Mohm input impedance when used as voltmeter. How can I cause a short like that if I am careful to ensure that meter probes do not touch one another and that I do not short any PCB tracks with the probe tip?
I am extremely careful when I work on energised circuits, so I am rather annoyed by what happened as I may make the same mistake again if I do not understand exactly what I did wrong.
A last question related to general safety: is there any difference between using a 1:1 (230VAC -> 230VAC) isolation transformer vs connecting the appliance to be probed directly into the mains using a lead without the earth wire (or with such earth pin isolated/disconnected)?
Thanks to the community for your precious help.
AI: I'm not sure what the big inductor is for. It doesn't look like part of an EMI filter to me, since that would typically be a common mode choke with four leads, and this apparently has only two. (That's probably a common mode choke on the circuit board next to the the torroidal inductor, labeled FL3, with the white bobbin and the ferrite core.) It's also rather large for that. My guess (and it's only that) is that it may be part of a power factor correction circuit. BTW, "HI POT" is short for "high potential", and generally just means that it has passed an insulation breakdown test.
I don't have any really great theories about the short, except to say that, especially if you're running on 230VAC mains, the peak voltages in such circuits can be quite high, especially in circuits with big inductors. So it's possible that you didn't even make true metal-to-metal contact but simply decreased the air gap just enough to cause an arc.
Re the isolation transformer, YES, there is a difference! Please use the isolation transformer! Disconnecting the safety ground lead on the equipment under test doesn't remove the potential between the hot circuit and earth ground, so you are still exposed to danger. Even with a well-insulated, floating DMM, you can still have an accident. Sometimes engineers will float the ground on oscilloscopes so that they can probe line-side circuits, but this is still dangerous. An isolation transformer is the only reasonably safe way to work on these things. |
H: Stabilizing the 12 volts from a car battery
I'd like to put a small gadget that I made into my car and power it from the car battery. The gadget works on 12 volts and the car provides 12 volts also. And the consumption would be somewhere around 2-4A.
However as I understand those 12 volts from the car are not very stable and can have short spikes up to 60 volts. Which would be a problem for my board.
Could you recommend a good way to stabilize those 12 volts from the car?
I'm wondering between maybe buck converter, linear stabilizer and overvoltage protection.
Thanks!
AI: I think the linear stabilizer will be acceptable solution, if your gadget may eat lower voltage than 12V (about 10% lower). If yes, then you may use LM-338 for example. This stability margin (10%) is needed because the every linear stabilizer has its own voltage drop near 1V or higher. And, of cause, follow the recommendations regarding capacitors - their capacity must be enough for expected load, and their voltage must be 50-60V at least to respect your remark about real voltage pikes.
If your gadget is pretty simple and has some resistance to voltage overload, then you may use only LC-filter based on ferrite-ring and a combination of big capacitor and ceramic capacitor (purpose of second one is to eat the noise of generator's rectifier and ignition of brushes).
If the voltage precision is a key, then your choice is a double convert 12 to ~220 and then ~220 to 12, by using a car power adapter (inverter). These adapter should be able to give you much more than 2-4 Amps. |
H: Humbucker-style microphone interference cancelling
For a recent experiment, I decided that it would be a good idea to try to add some electric guitar distortion effects to my otherwise Classical flute playing.
In order to do this, I attached a small electret microphone to my flute, which is powered by a phantom power unit and provides an audio signal to a guitar amplifier. A rough idea of my setup is shown in the following circuit diagram:
simulate this circuit – Schematic created using CircuitLab
(I did not include component values as I don't think they are that important in this problem, but I can add them if required.)
This setup seems to work quite well, I am able to add as much distortion as I want using the amp. However, there seems to be a bit of interference (mains hum, etc.) present, as there probably would be in proper electric guitar setup. My friend, who plays electric guitar, immediately noted that I should try to implement something similar to a Humbucker pickup, which is a type of guitar pickup that cancels out most interference, but with microphones instead of pickups.
My understanding of how a Humbucker pickup works is that instead of having just a single pickup, there are two, wired in series or parallel. The pickups have opposite magnet configurations (ie. one has all magnets with the north poles pointing upwards, and the other has magnets with the south poles pointing upwards) and wire coils wound in opposite directions. Any interference induced in the coils is then cancelled out because the interference from each coil is equal and antiphase in relation to one another. The signal from the guitar string remains unaffected because the although the coils are wound in opposing directions, the magnets are also in opposite polarity.
Based on that understanding, I came to the conclusion that it isn't possible to replicate the Humbucker setup with my microphone, because there is no equivalent to "winding the coils in opposite directions" or "reversing the polarity of the magnets" in an electret microphone. Simply connecting two of my microphones in series or parallel (which I believe is what my friend was thinking) would increase the audio signal but also the amount of interference, due to the microphones (and therefore interference) being in phase with each other.
So my question is, is it possible to replicate the concept of a Humbucker pickup but with microphones instead of guitar pickups?
(And if so, will it actually make a difference? Is the perceived interference I'm getting actually from the microphones themselves, or another source such as wiring?)
AI: is it possible to replicate the concept of a Humbucker pickup but with
microphones instead of guitar pickups?
A humbucker guitar pickup is intended to directly cancel the induced voltage caused by magnetic fields. The magnetic fields that a guitar is susceptible to will be barely a problem for a microphone because the coil is normally much smaller and shrouded by ferromagnetic material that would cause external alternating magnetic fields to bypass the coil.
A guitar pup on the other hand, has an "open" sensitive area in order to project a static magnetic field that can be "modulated" by the movement of the "iron rich" strings. It is much more susceptible to hum because of this.
If a microphone is picking up a lot of hum then, the first question to ask is if the cable feeding the microphone is balanced and is the microphone driving a balanced signal. I suspect that the answer here is no because clearly, an electret is a polarized microphone and one end needs to connect to ground whilst the other end feeds the signal to the amplifier.
I'd try using a conventional moving coil microphone and a balanced cable and balanced input module. A guitar produces an unbalanced signal and the amp is also unbalanced (because it doesn't need to be balanced).
If the magnetic field hum were picked up by the electret - you'd have to use another electret and invert its signal in order to cancel the main microphone's hum but, the problem you have here is that both microphones need to be in close vicinity to "receive" the same level of hum and this means acoustic cancellation and therefore poor levels for the signal you want to keep. |
H: Ways to use electromagnets to pick up metal shavings at a distance
I work at a rotor machining plant and there is an issue for shutting a line down to clean up metal shavings, causes too much down time. I was thinking of using a strong electromagnet to have a "pick up" radius under the machining cells, say 12". After thinking about creating my own and a failed prototype and also looking online to buy one I am not seeing any that have a effective radius. Is it even possible to achieve what I am trying to do? Any suggestions?
Thanks
AI: If you are trying to pick up shavings 12" from the magnet, that's probably not going to work, even with absurdly strong magnets. If it did work, what would you do with them THEN?
I'm having a hard time visualizing exactly what your setup looks like, but something that moved a magnet around the area so it was closer to the chips (and then moved it over a hopper, bin, or belt and turned off the magnet) might be more effective. Then again, ignoring the "magnet" aspect completely and mechanically sweeping, or vacuuming (as in a large air-transport chip collector, not a shop vac) might work (will depend somewhat on the nature of the chips - long, stringy, curly, sharp ones are much harder to handle than short chips.)
If you have issues with build-up IN the machines, you may need to arrange air-blast or something to get them to the floor or collection system. Depending on what exactly the constraints are, something like a wide belt under each machine section that either dumps the chips where they can be safely collected as operations continue, or onto another belt that takes them away might work. |
H: Driving a 5V IR led with a 5V charger?
I'm trying to build a remote control with IR led.
I'm experiencing low range with my 1.5V, If=50mA IR leds, so I looked for and found a 5V, If=100mA IR leds.
twice the current, hence probably better transmission capabilities.
The problem is, I'm not sure what to do about the current limiting resistor.
I know that its important to limit the current so that the led wont burnt,
and so with the 1.5V led I used a 35 Ohm resistor.
But now voltage drop is not needed, because the source is the same as the consumer.
Do I still need to worry about current limiting resistor or do I simply plug it directly?
AI: You've misread the datasheet. The reverse voltage is 5V (which is not uncommon for small LEDs), but the forward voltage is 1.6V. You will still need a current limiter. |
H: Would you prefer a voltmeter or a scope when measuring average voltage precisely?
As far as I understood the DC setting of a voltmeter or a multimeter is measuring the average value of a periodic voltage signal(not rms but the average value).
On the other hand, an oscilloscope as well in DC coupling mode also measures the average value of the signal.
Which one in general measures more precise? Why?
AI: Scopes are designed to plot waveforms so you can see what the wave-shape looks like. While modern scopes have the ability to do a lot of other things with the data gathered from the ADCs they are not designed as precision measurement devices.
The downside of a basic multimeter is you typically don't have control over the averaging length. If the averaging length is too short compared to the waveform you are measuring then your result won't be an accurate represention of the long term average (and your display will jump around). If it's too long then the reading on your meter will take an annoyingly long time to settle. The more accuracy you need the longer the averaging period you need.
Looking at Why do DMMs have so low update/refresh rates? it seems that ordinary multimeters make about 4 readings per second. High end bench meters often have settings available to adjust the averaging behaviour. |
H: Probing with Oscilloscope Stops ADC Fluctuation
I've gone back to my constant current project to fix an annoying fluctuation in the required current display. See schematics below.
The entire circuit consists of a CCS section with signal outs for both the requested current (represented by Vreq) and the actual current (represented by Vact developed across the sense R).
These signals are sent to an amplifier which scales them up x10 and the amplified signals are sent to a uP ADC with an external 2.5V ADC reference.
I'm using a star-ground approach and I keep digital (uP, display driver) and analogue (CCS, amplifier) grounds separate, with a separate ground point for the load return.
Despite this, the required current selected by the 100k POT moves up and down a mA or two on the display. In preparation for some low-pass filtering, I took some readings with my oscilloscope probe (set to x10) and to my astonishment, the display was rock solid. No fluctuation, perfectly smooth with the POT turns.
It didn't matter which "Probed" point I tried (see diagrams below), the effect was the same. Realizing that the probe contains some capacitance and a large resistance, I tried placing a 22pF and 10M in parallel at one of the points probed. So, far I haven't managed to replicate the "fix" brought about by the probe. I'll get a chance to do some more this evening, but in the meantime: can anyone tell me why the probe has had such a "beneficial" effect on my circuit?
Thanks!
UPDATE: I should have pointed out that the oscilloscope is mains connected and its ground lead is connected to earth ground. Also, the circuit itself is floating, that is it's connected to a 5V wall regulator that doesn't use earth ground. One other thing, in all cases when I probed the circuit ("probed" points), I connected the oscilloscope earth lead to the local ground at that point (not the star ground point) - not sure if that makes any difference.
While I'm updating, another intriguing thing is that if I attach a load (say a 100 ohm resistor) the readings seem to be quite stable. Not sure if this is related, but I thought it was worth mentioning.
ANOTHER UPDATE:
Well, I had a chance to test things a bit further. Seems like it's the connection to earth ground that fixes the issue, so @WhatRoughBeast is on the right track. Unfortunately, the twisted pair trick didn't work.
FINAL UPDATE:
I finally got the circuit to work reasonably well. I changed the 100k multi-turn pot for a 10k (and the 1M5 resistance to 150k) and the increase in current seemed to cure the flickering, except when the pot resistance was quite high. I also shortened the signal wires and implemented a simple digital low-pass. If I had this to do again, I'd use a proper PCB with a ground plane.
AI: This may sound odd, but I suspect it is your grounding philosophy which is getting you in trouble. Star grounding is the standard approach for power wiring in order to avoid ground loops. It is not appropriate (usually) for signal circuits. The problem is that, for high impedances and long(ish) wires, it invites the formation of large-area ground/circuit loops, which will respond to varying magnetic fields by acting as antennas and injecting noise into the signal path.
Try connecting your Vact to the A/D input via a twisted pair, with the other wire connected to ground at both ends. Yes, this will produce a ground loop, but with very low currents the imposed noise should be very small and dominated by the reduced pickup on the microphone signal.
If that doesn't work, try disconnecting the twisted pair's ground wire at the A/D side, but leave it connected at the sensor side.
EDIT - Looking further at your circuit, there are a number of issues. If this is all there is, you are getting yourself in trouble with your grounding philosophy. Unless there are other loads involved, what you are doing is simply not worth a star ground. Your CCS only produces a maximum current of about .6 amps, assuming your 1k load resistor is an error. This is not an enormous current level.
You have not described your circuit partitioning, but if all 3 circuits are part of the same PCB, then you should simply use a single ground plane for all grounds. The low resistance produced by the ground plane will overwhelm any other effects. As a further tip, place the ground power connection close to the bottom of the sense resistor.
If the first section is physically removed from the other two, and you are uncertain of coupling via twisted pair, the standard approach is to use a difference/instrumentation amplifier
simulate this circuit – Schematic created using CircuitLab
Use twisted pair for the connection from the sensor. C2 and C3 should be ceramics. They filter out any pickup around the loop produced by the separation of the grounds. The op amp responds to the difference between the two sensor points, and since it draws virtually no current in its inputs, it rejects the effects of any current flow between the two ground points.
Twisted pair acts as a poor man's shield. The ground wire acts to intercept radiated energy, and because it is in close proximity to the signal wire it tends to shield it. For really high-sensitivity applications, you use coaxial cable, which has an inner conductor completely surrounded by wire mesh or metal foil. |
H: AL value: nH/N² vs. nH/T²
I'm in the process of sizing some toroidal ferrites. My problem is that I am seeing the AL value on the datasheets shown two different ways as nH/N² or nH/T².
Are these two units identical? I believe they are, being that N and T both stand for "turn".
AI: You are correct, both N and T stand for turn. |
H: Power in AC circuits
When I look at the graph of the power in AC circuits versus time (not the average power I am talking about the instantaneous power) it goes from a negative value to positive. On the other hand, I know that a positive value of p(t) represents dissipation and negative power represents generation. How could I link this concept to a sinusoidal form of power? For example, for a resistor that is connected to AC voltage source, the positive power describes the heat that is dissipated, what would negative power in negative cycle represent?
AI: If an AC waveform is connected to a pure resistance, the delivered power never goes negative. The current is always in phase with the voltage. When the voltage is +ve, so is the current. When the voltage is -ve, so is the current. In both cases, the power, the product of voltage and current, is +ve. While the power varies from large to zero, it never goes -ve.
If the load has an inductive or capacitive component, then the current waveform will lag or lead (respectively) the voltage waveform, and at some points in the cycle they will have different signs. When the signs are the same, the component is storing energy, and the generator is delivering positive power. When they are different, the component is returning energy to the generator, which could be mathematically consistently described as negative power.
The power due to this storage of energy is called reactive power, usually written VAr. |
H: Why is the power supply terminal is not connected to ground in this bridge rectifier?
Above is two identical bridge rectifier circuits. Input is a 12V sinusoidal voltage source. Red is the source. Blue is the output from the circuit on the right side. Green is the output from the circuit on the left side. It seems like the left circuit is wrong and the one on the right side has true output. But I don't understand why it is not the other way around.
The only difference is that the circuit on the right side has a ground between the load and two diodes.
On the other hand, the circuit on the left side has a ground connected to the negative terminal of the supply.
Normally I connect the negative terminal of the power supply to the common ground.
An example to this is the circuit on the left side.
But when I plot the output for both circuits, the circuit on the left side acts weird for negative alternating input.
But my logic says:
When V2 goes positive the current must flow through D7 R2 D6 and finally to the supply terminal.
When V2 goes negative the current must flows throug D8 R2 D5 and to the supply terminal.
So I would expect the same output as in the circuit on the right side. Hence I would expect a rectification for negative halves as well.
What is wrong in my logic here?
Regarding the working circuit in the right side:
I dont understand the circuit on the right side by the way. Because:
When V1 goes positive the current must flow through D3 R1 and finally to the ground or to D2 ???.
When V1 goes negative the current must flow through D4 R1 and finally to the ground or to D1 ???.
Why do we need ground here? Current still should flow to the terminal of the supply without a ground there.
My question is:
Why is the common ground connected like in the circuit on the right?
AI: Both circuits output rectified unsmoothed DC, and if you plot the voltage across R1 and R2 you will find that both waveforms are absolutely identical, looking like this:
However you are not plotting the voltage across R1 and R2. You are plotting the positive output of the rectifier against ground. In the first case ground is tied to the incoming AC, causing the weird output you see. In fact, grounding the other side of the AC supply would result in the exact same effect:
If you are ever building a real world power supply, ground the negative DC output of the diode bridge (like on the right of your schematic), as grounding the input of the bridge is not only useless but will short out a diode if you also ground the negative output of the bridge. |
H: Is an ideal transformer an ideal current source?
I define every circuit element of a primary coil of a transformer that is connected to a voltage source and calculate \$I_1\$ (current of the primary loop). I then connect it to a secondary coil with a known number (\$N\$) turns.
Regardless of the values of the secondary coil, I always can calculate \$I_2\$ (current in secondary coil) by using $$I_2=\frac{N_2}{N_1}I_1$$ Does it mean that an ideal transformer is an ideal current source?
AI: Does it mean that an ideal transformer is an ideal current source?
An ideal transformer will have infinite magnetization inductance and, because of this, it will take zero primary current when there is zero secondary current. This doesn't make it an ideal current source.
Look at the equivalent circuit: -
An ideal transformer doesn't have any series elements so basically they become shorts. It doesn't have any parallel losses so Gc becomes open circuit and the only contentious thing left is the primary magnetization inductance shown with a blue square surrounding it. For an ideal transformer this is infinite in value and no current can flow into the primary if there is no load on the secondary. If there is load on the secondary then the primary impedance looking in will have an impedance of: -
\$P_Z = (\dfrac{N_1}{N_2})^2 \times S_Z\$
You could argue that a CT (current transformer) might be an ideal current source but it isn't. With a high primary current of (say) 100 amps it still has a primary magnetization inductance that will naturally limit the open circuit secondary output voltage despite rumours of thousands of volts being talked about on some websites. |
H: What does FG stand for?
I have an old 12V cooler master fan that I am converting into a fume extractor for soldering and am connecting it to a standard 12v computer power supply. I am trying to wire it and am confused about one thing.
On the fan's label it has a coding for the different wire colors (red, black, and white) and it says:
red +
black -
white FG
I have looked up FG on google all over the place but can't seem to find anything that says what it stands for. I was thinking that it might be "Ground" but then what would the black negative wire be? Maybe I just have a fundamental misunderstanding (pretty new to all of this electronics stuff)
I did connect the black wire on the fan to a black wire on my power supply and the red wire on the fan to a yellow wire on the power supply (white wire connected to nothing) and it turned on and seemed to be working fine but I want to make sure that this is actually a correct wiring and isn't going to cause any problems in the future.
AI: FG stands for Frequency Generator (or Feedback Generator), it has an output with a frequency proportional to the fans speed. It is used by the CPU to determine the fans speed.
Some (older) fans have an extra winding internally and the FG signal is a sinusoid with both amplitude and frequency proportional to fan speed.
Modern fans almost exclusively use a Hall-Effect sensor and the signal is an open-collector square-wave signal where the frequency is proportional to fan speed. Peak voltage is determined by the magnitude of the power supply that feeds the pull-up resistor. |
H: Should serial triac dimmers work for resistive load?
I have a cheap 120V hot plate that I assume regulates its temperature using a triac dimmer (because the heating element vibrates/hums at lower settings). But at its lowest setting it still gets too hot for my application.
So my first reaction is to put another household dimmer between the device and the power. But I'm not sure how the second triac in the series will behave if it's seeing chopped AC from the first.
I think the capacitors in the second triac are going to hold their charge between "cuts," and so the cumulative effect of the dimmers will be multiplicative as expected: I.e., if both are set to "half" power the device will see "quarter" power. Is that correct? Or is it not that simple?
AI: Stacked dimmers are probably not going to behave the way you want. Household dimmers expect a sine wave input: they look at the zero crossing and fire a triac at some fraction of the cycle. When the second dimmer receives the chopped up sine wave, it might try and fire at the same time, or it might get confused by the non-sine input. It may even have undervoltage trouble, depending on how the low-voltage DC for the control circuitry is controlled.
Instead of two dimmers, here's a couple different things you could try to accomplish your goal:
Feed the hot plate from a variac. However, this could cause the hot plate to malfunction (again, feeding it a lower voltage might make the control circuitry unhappy), and large variacs with large power ratings are expensive, heavy, etc.
Control power to the hotplate with an SSR. Because the hot plate has a slow thermal response time, you could basically "PWM" the power to the hot plate with a very low frequency (1Hz or lower).
Break open the hotplate, and bypass the built-in temperature control (if possible, leave the internal temperature cutoff for safety), and control the hotplate with an SSR. This eliminates all uncertainty with the built-in control, and you have direct PWM control. Again, because it's a hot plate, you don't need high-speed PWM control, 0.2Hz or something will be fine. The nice thing about SSRs is that you can buy a cheap SSR with screw terminals, and you don't have to deal with any mains voltage other than wiring the thing properly.
The above suggestion might also work with a household dimmer. However, make sure the dimmer is rated for your hotplate wattage, and keep in mind that some household dimmers don't go close enough to zero (you might have the same problem). |
H: How do I open a grill ignitor?
My grill rusted apart, so I decided to salvage the spark circuit.
But there's a problem - I can't find the polarity of the ports, and I can't open up the blackbox. https://i.stack.imgur.com/Q2Ku3.jpg
How can I crack this open without breaking it, or alternatively, is there a way to figure out the polarity? All the exposed pins are outputs, and the wires lead to a switch.
AI: If it's a standard "push button, goes clunk, get spark" type setup (can't tell from that picture), there's just a striker (somewhat like an automatic centerpunch) and a piezo disk inside it. You push the button, the striker strikes the piezo disk, the piezo disk makes volts (might well be AC as the crystal "rings" when struck, but I don't know, and I don't hate my oscilloscope enough to try and find out.) No electrical hanky-panky to speak of in one of those.
If it's some type of powered (110VAC - 230VAC - 9V?) unit, you'd have to supply more helpful information about how it WAS wired and powered than "all the pins are outputs and the wires go to a switch" which is pretty much no information at all.
If you wish to test the polarity, you'll need a resistor string able to take the output voltage, and you measure across one resistor on the ground end of the string - a voltage divider. That either needs some high voltage resistors or enough not-so-high voltage resistors in series that they add up to a high voltage. Odds are the resistors needed cost more than a 14 year old ignitor is worth. |
H: Need Help In Creating and Debugging a 8Mhz CLK Circuit
I want to create a CLK generator circuit that will work at 8Mhz.
I'm following this design (from http://www.electronics-tutorials.ws/oscillator/crystal.html):
My crystal is 8Mhz HC49/US, with 20pF capacitance as described in the website I bought it from.
I'm using R1 = 10Mohm , R2=1Kohm , and as for C the smallest I had was 150pF so I used 6 of them in series which should give 150/6 = 30pF .
My inverter is a 6 input buffer NOT gate from TI (cd40106b) , I'm using two of its gates for the circuit (AMP + buffer).
After connecting it all, and hooking up my (PC based) oscilloscope, I saw I'm only getting a 1.5Mhz signal , which was very far from a square wave I was expecting - more of a triangle kind of waveform.
Then I removed the crystal from the circuit, and the waveform stayed the same, so I'm obviously doing something wrong here as I'm just seeing the RC circuit oscillating.
I've replace the crystal with two others and got the same results.
The NOT gate should just fast enough for this circuit ,working at VDD=5V , transition time is typically 100nS and up to 200nS and I'm no way near this with my results.
I think I'm using the correct values for R2 and C but I'm not sure.
How do I calculate the correct R and C values to use for my circuit? Could it be the reason for this behavior?
I've also tried to assemble a circuit using CD4060 (counter) like the one from the comment.
to see if the problem is coming from the inverter but I still can't get the input frequency to be 8Mhz.
What can I do to further debug this?
AI: The usual recommendation for the Pierce oscillator is to use un-buffered standard inverters, ie not schmitt trigger input inverters. The inverter needs to work in its linear region which is almost impossible with a schmitt trigger input. Try TI's LVC1GU04. |
H: Analog PID controller (transferfunction questions)
I have to build an analog PID controller (with op-amps), I've made a schematic:
My questions:
Can I build my PID controller this way?
My PID tranferfunction (in the s-domain) looks like:
$$\text{H}(s)=\text{K}_{\text{r}}\left(1+\frac{1}{\tau_is}+\tau_{d}s\right)$$
But what in my schematic represents the variables from my transferfunction?
AI: You need 3 more resistors of value R3 to make this add the components equally. This is a textbook parallel PID controller. Assuming R1 = R2**
\$\tau_i\$ is R3*C1
\$\tau_d\$ is R4*C2
While you can certainly build this, it has a number of problems. The derivative will cause noise problems. Usually a filter has to be placed in front of the differentiator if you really need the derivative function (it's often not necessary and causes more problems than not).
You also have to consider saturation in each of the three terms as well as the output. As part of that consideration of saturation, if the integrator continues to integrate while the output is saturated you'll have integrator windup, which will usually cause a lot of over/undershoot, especially at start-up.
** If they are not equal, or if the summing resistors are different from R3 then it will effectively scale each of the three terms. You can probably see that that's (ideally) the same as scaling the time constants. In practice, you cannot use an integration capacitor 1/100 the size and use a large summing resistor because the op-amp will saturate before it can apply enough integral action to force the output over the whole range. |
H: What is the difference between high precision resistors and current sense resistors
Actually, I am trying to build a current monitor for a wireless sensor network. I am going to use a current sense resistor as the current will pass through the resistor. By measuring the value of the voltage, the current value can be calculated.
My application is very sensitive as the current that will be measured is in the range of mA, 30 mA maximum.
When it came to buying the resistors I got confused. I found a type of resistor called "precision resistor" that has a very small error, and another one called "current sense resistor".
Here is a picture of a current sense resistor:
So my questions are:
What is the difference between these two types current sense and precision resistors?
How does this current sense resistor differ from the ones we normally use, such as as these 5.6k Ohm, 5% resistors?
AI: Resistors are resistors. They only see the current thru them and the voltage across them.
However, specific models can be targeted to specific applications. What you show as a current sense resistor looks like it's designed to dissipate significant power. However, it would work for lots of purposes within its power and voltage limits.
Current sense resistors tend to:
Be run with little voltage across them.
Be fairly accurate, since being used for measurement.
Drift little due to temperature.
Have low values.
Sometimes have 4 leads so that you can use a Kelvin connection. Two leads carry the current, then the voltage is measured across the two other leads. This keeps the voltage drop due to current in the leads off the measurement. |
H: How to correctly wire this transformer for 240V?
I have an old transformer (without any details about brand and model) with dual primary windings. I need to set it up to take 240V, but I am unsure about the correct way to connect the windings in series.
Here is a photo of the primary side:
So what is the correct way to connect these for 240V input?
Do I connect the two 0 connection points and supply 240V at the two 120 points or cross over a 0 to a 120 pin?
Note: the windings is on each side, i.e. there is no connection between 0 and 0, and 120 and 120. On each side I have about 8 Ohm between the top 0 and the bottom 120 pin.
AI: The primary windings go in series, but get it wrong and the transformer turns into a (bifilar wound) resistor. It is important to not connect them in "anti-series". It cannot be seen from the outside how the windings are internally connected. You 'll have to wire it, then carefully test if the transformation ratio primary-secondary is about right.
I think it'd be connected as below, but you'll have to carefully verify the result before using it at full power.
As @jms suggests, add a (regular old fashioned incandescent) 240V light bulb in series when testing. If it lights up while the transformer is unloaded, then you've wired it incorrectly.
~ ---o 0 o
/
/
/
/
/
/
o 120 o--- ~
I understand from your comments that the transformer is rated for 240V, but do be careful, a new transformer with decent spec sheet is cheaper than rebuilding a burned down lab. |
H: How to overcome transistor heat in slayer exciter?
I build this circuit and it works but I have two problems. The major problem is that the transistor becomes very hot quickly. I burnt my transistor due to excessive heat. How can I get rid of the very high temperature of the transistor.
The second problem is that the circuit does not provide enough range. When I approach a florescent lamp, It does not light up. I brought a wire and modified its shape to be round (a circle). It looks like a coil that have one turn. I connected the high voltage terminal to one of the coil's terminals. The other terminal of the coil is not connected (open circuit). When I put the florescent lamp inside the coil it lights up. This is a very small range. I need a good range, something like 5 or 10 cm.
I need to know how to achieve a good wireless range? and What controls the range? Is it supply voltage, the base resistor or something else?
My supply voltage was 10 volts because I don't have a battery of 9 volts. The number of turns of the coil is 300 instead of 275.
Thank you,
AI: Have you tried finding the current through the coil? Find the current through the coil. You can either simulate it with a spice package or place a meter between the coil and transistor. Once you know the current use the equations located in this online textbook in chapter 11 Inductance and Magnetic energy (Yeah it requires math).
Then find the magnetic field through the coil (you'll need to know the area of the coil and the turns.) Then calculate what kind of energy you need at the input coil. I'll give you a hint as to what you'll find. Your coil is generating a magnetic field at a certain frequency, keep in mind the magnetic field is going to 'attenuate' by the the inverse cube of the distance (magnetic field=1/(distance^3)). Meaning if you had 1 unit of magnetic field at 1cm, at 2cm you will have 1/8th of the field, at 3cm you will have 1/27th and so on. By the time you get to 10cm you will have 1/1000 of your original field so you will need ~1000x the current or you will need to redesign your coil.
This is also why there will forever be companies that promise wireless power but you never see any products beyond the charging pads. The lightbulb demonstrations use phenomenal amounts of power even with a directed antenna(coil). The best one I've seen is this one Wireless power transmission. You can changed the antenna/coil however and it can help. Look up Helmholtz coil if you think that might work for your application. If you want you can parallel several BJT's if you want them to get less hot. |
H: Why do we need to rectify or amplify the current/voltage?
I am a beginner in Electronics. What i read now a days in my text book is about diodes,transistors etc. They are used for some rectification and amplification etc ? I wonder why do we need to give different shapes to voltage and current. For what these different wave forms of current and voltage are used ?
AI: A power company wants to generate electricity, but they don't want to waste it on power transmission.
Power loss in a wire is given by
$$P = I^2R$$
but you can also calculate power as
$$ P = IV$$
where I is current, V is voltage, and R is resistance.
So power companies use very high voltages, as this lets them use very low currents, which means (relatively) little power loss in the transmission lines.
You don't want to use very thick insulation on your electrical cables in your house, though, which is what's required to prevent you from getting shocked by the 100,000 volts the power company produces. You would like to get the voltage down to, maybe, 120V or 220V.
So the power company makes alternating current, or AC power, because this creates a time-varying magnetic field around the wire. If you get a bunch of this wire together, then you can use that magnetic field to induce electricity in another bunch of wire.
The neat thing is that the voltage output of that cluster of wire, called a "transformer", is related to the ratio of input to output turns. This means you can now step down from 100,000 volts to 120 volts.
But, your computer is a digital device. It likes things to be "ON", at a constant voltage, or "OFF". If the voltage is time-varying, you don't get to have nice things like logical TRUE or "ON". So you would like to rectify the AC power into DC power.
You do that with an arrangement of diodes called a bridge, which technically creates ripple DC, which you can smooth with capacitors to approximate a steady DC supply.
Then your computer works.
Also, you might want to shape electricity to make interesting shapes with it, like the PAL signal to make images on TV, or to reproduce audio, or any number of other amazing things that you can do with analog electricity. |
H: Finding L and R in AC circuit
Ammeter readings:
A1 = 4A
A2 = 2A
A3 = 3A
Voltmeter reading 24V.
Circuit is in AC.
I have to find R1,R2 and L.
While calculating R1 (R1 = 24V/2A= 12 Ohm) seems to be easy, next steps are not particular clear to me. I know that ammeter measure magnitude of complex current, but I cannot utilise this fact.
edit: f=50Hz
AI: Here's a hint (or two): -
Draw a triangle with sides of length 4, 3 and 2
Each side represents the angle and magnitude of the currents relative to each other
Calculate those angles using trig (here)
The side that is length 2 is resistive (R1)
The angle of side "3" relative to angle of side "2" is due to the presence of the inductor in series with R2.
After that you are nearly there. |
H: What happens if I connect a solar panel and an AC/DC PSU in series?
I'm thinking about connecting the DC12V output of an ATX PSU with a solar panel in series.
I want an always-on power source with as much power coming from the Sun as possible.
AI: If they acted like ideal voltage sources, you could put them in series and they would additively combine their voltage (19V from the panel and 24V from the supply = 43V). However ideal voltage sources can supply infinite amounts of current, your supply and panel cannot do this. The main problem is the current, your panel has an IV curve, the more current you pull from it the less voltage you will get. They have circuits\IC's that optimize the power being pulled from the cell (solar MPPT tracker). You may be able to use a topology like this, you would have to keep the MPPT voltage right above the rail of the supply by tenths of volts and a topology like this would not be easy to get working.
simulate this circuit – Schematic created using CircuitLab |
H: USB powering batteries without OS support?
On a bare-metal project I have in mind, I would like to know if I can avoid to implement the multiple USB specs (O/E/xHCI) and still benefit from usb charging my battery. It will run on a almost broken/useless flashed smartphone with one mini usb port.
I literally don't want any usb port for media storage or hid or anything for that project. There will only be one app running.
I was thinking that the already existing hardware usb controller(s) would allow me that, but.. on the other hand, if it is not activated by software, it may be an issue.
Question:
Can I power up my batteries via a usb cable, without software usb implementation? Is there a shortcut avoiding the full usb implementation?
Thanks
Update, to make things clear:
My goal was to use one of my cables, without modification made on it, plug it into my phone, and let it charge. With that information, would you consider that it is still a "yes" to that question?
AI: Answer: Yes.
Maxim Integrated Guide to smart battery devices. Or if you're a "dumb device" kinda person, the device being charged can tie their data lines (D+ and D-) and receive maximum power from an adapter. Vice Versa YOu can always charge a USB device 5V @ 500mA regardless of what is present on the data lines (tied up or down or floating or anything). Anything past that would require one of the chips similar to the link's that intelligently (and without programming, i.e. you just solder it in and it works) charges the device that follows USB charging spec.
On the off chance you're talking about using your phone to charge something (still not quite sure, but adding for completeness), that's called OTG and from what I know you'll only be able to provide around 8mA from your phone to the device.
Addendum: Apple doesn't follow USB Charging spec. Ask apple why that is. |
H: Output State of comparator circuit for AC signal when open?
I am trying to design a circuit which would interface AC signals from pick-up coils.
Below is the circuit.
R4, R3 are for DC bias, and R2 is for feeding that bias to inverting input of the OP amp.
R5 and R8 is for hysteresis regarding noise immunity.
V2 is the input signal.
V1 is power rail.
My question is that how should I determine the state of Vout(18V or 0V) when V2 is open?
I simulated it on CircuitLab. and below is the result
V+ = 8.946 V
V- = 8.955 V
Vout = 1.783 mV
My concern is that V- is higher than V+ only by 9 mV, which I believe is leading Vout to near negative rail(0V).
Can I rely on current circuit for the output to be low always?
Or should I add some kind of pull-up or pull-down resistor for circuit stabilization?
I tried adding pull-up or pull-down resistors to V+ or V- but it prevents the signal from making Vout with 18V peak to Peak.
Below is the link to the circuit i made public
https://www.circuitlab.com/circuit/74bt8h/comparator/
AI: To start, eliminate V2, C1, R2 and R5. Then the junction of R3 and R4 will have a Thevenin equivalent voltage of 9 volts, and a resistance of 5k. Adding R5 provides a hysteresis of $$ \Delta v = 18\times \frac{6k}{1.006 M} = .107\text{ volts} = 9 +/- .053\text{ volts} $$ this will also provide a hysteresis at the R2/R3 junction of $$ \Delta v = 18\times \frac{5k}{1.005 M} = .089\text{ volts}= 9 +/- .045\text{ volts} $$ With V2 still unconnected, the - input will be forced to the value of the R2/R3 junction. Since this is undetermined, the output voltage is undetermined.
Effectively, with V2 disconnected the comparator can be in either state, and there is no a priori way to determine which state that will be. |
H: Using Dither to achieve finer PWM resolution
In practical industrial applications (regardless of any particular industry), is there any use for a dither module that goes along with a PWM module considering HRPWM's are available. With my limited experience I can justify using the former as it's most likely cheaper than the latter. However, this is a rather weak reason, and I was hoping someone can share their experiences/thoughts on the usefulness of dither?
My intention is to keep this question as far as possible from being opinion-based or too broad, but just to make sure - I am looking for an extended 'yes' or 'no' answer to whether a dither module may be useful in practical applications.
AI: Lower cost is hardly a weak reason.
Using a more expensive part than necessary, without good reason, is hardly a mark of a good design, and specifying a high-res type may constrain the design in other ways (for example, forcing the designer into a vendor with inferior pricing, availability or performance).
Power consumption may be less too, though it's probably not a huge difference overall. |
H: What voltage is a 3-LED light from a LED-TV?
I have some "NDV rev1.1 4248 back lights" from a television.
There is a print that sais "94V" on the strip, and on the wire supplying the strips it said "96V".
Every strip has 9x buttons which have 3 LED's in them. So i devided 94 by 9 and figured that every button should have 10.4V roughly, which makes sense, because a white LED is normally 3.5 volts.
I have a 12v Lithium battery which measured 11.1V so i tested one of the LED's with it and it lasted less than one second.
Are the LED's are rated for 10.4V roughly and not more?
AI: If you blew them out that quickly, it is most likely because LEDs need controlled current. Controlling only voltage to them doesn't work because they are diodes and have a "non-linear response" to voltage.
Try connecting a potentiometer to the + of your battery, turn the pot to about 1Kohm rssistance, then connect the 1 pod of leds & turn down the resistance until they light well without blowing.
Once you find a good point with the pot, measure voltage across your pot, then disconnect the battery & measure resistance the pot is set at. dividing the voltage (in volts) across the pot by the resistance setting (in ohms) will give you the current (in amps) that your leds need. |
H: Finding Vout of a Transformer
I am a little confused on how to read the ratios of a transformer
Suppose there is a ratio of 1:6, is this read as the following?
Number of turns on Primary: Number of turns on Secondary?
so suppose a question asked that there is a voltage of 20 voltages rms and it enters a transformer of 1:6, is the Vout rms calculated as the following:
Vs/Vp = Ns/Np,
so Vs= VpNs/ Np
so Vs= (20*6)/1 which yeilds 160 rms?
does the distance and type of gauge wire effect the answer?
AI: I'll assume that multiplying by 8 was a typo, and that you meant 6, as your original ratio was 1:6.
The usual way to read it is the first number represents the primary, the second the secondary, so in your example you would have 120v output.
However, I know many people who describe all transformers in the style N:1, where N is larger than 1, and then describe which way round it's connected.
When it matters, you need to double check what the ratio means. It matters in simulators, check the simulator is using the convention you expect. It matters if you are plugging the transformer into the wall, and then accidentally going to touch the secondary!
The length and thickness of wire will not change the ratio, but they will affect the output voltage slightly when on load, due to differing voltage drops in the wire with the load current (and primary current!) |
H: Flash Magic for code dumping (LPC1778)
I was reading a book called "Insiders guide to the Phillips ARM7 based microcontrollers".
There it says "The easiest way to program the Flash is by the built in bootloader which allows the code to be downloaded via UART 0 into RAM and then programmed into FLASH".
In Flash magic when we dump the .hex file via the PC's COMM port to the LPC1778 UART 0 port, does the .hex file go into the RAM and then does the bootloader inside place the code into the flash?
If so,the onchip RAM is 64K in size and the flash size is 512k in size. What if the .hex file size i want to dump is greater than the size of the RAM?
AI: In general, firmware images are not transferred over the UART verbatim; instead you communicate with a bootloader that provides in-system programming (ISP) functionality.
Section 37 of the LPC1778 user manual discusses the ISP commands and functionality.
A tool like Flash Magic sends commands to the bootloader to reprogram the chip incrementally. The particular algorithm will vary, but it probably looks something like:
Transfer a sector's worth of data to the LPC1778's buffer in RAM.
Erase the next flash sector to be reprogrammed.
Copy the buffer from RAM to flash
Compute the checksum of the newly written flash sector
Compute the checksum of the data in RAM to verify
Repeat until the entire firmware image has been written.
Accordingly, the amount of RAM available doesn't limit the bootloader's ability to reprogram the flash, though having more RAM available can speed up the process by reducing the amount of data transferred back and forth between the bootloader and the PC. |
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