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H: Can I test the EM shielding of a material with a spectrum analyzer?
I am a hobbyist and I am intersted in testing some materials for their ability of reflectance/absorbance/transmittance for microwave radiation and I want to know if a spectrum analyzer is suitable for this task and how to do this (setup)?
I posted this question in engineering site and I was recomanded to post here.
AI: Shielding effectiveness is tested by measuring the gain between two antennas, then inserting the shield material between them and measuring the gain again. The measurement must be done in an area where backscatter is strongly suppressed or avoided, or such that the shield seals the two antennas from each other (splitting one anechoic chamber in two).
To perform such a test, a spectrum analyzer, tracking generator, RF power amplifier, and pair of antennas would be needed at minimum. Technically the TG isn't required, and an arbitrary source could be swept, or a noise source used even, with the spec in peak hold mode; the TG is merely faster. A VNA could also be used, which is basically the same thing but with phase reference included. (TG+spec can also be described as a SNA, scalar vector analyzer, so this is no accident.)
The amplifier is required to account for antenna factor, which will most likely be pretty significant to begin with, and obviously goes up by even more with the shield included. A very high SNR TG+spec could also be used; just to say that, you need enough signal level to account for losses, so that you have something to measure above the noise floor even in the shield-in-place condition.
Regarding anechoic chambers: they're rarely perfect, and more likely the reflection is on the order of -20dB, significantly attenuated but hardly an "optical black". The reflected wave (plus whatever scattering occurs) bounces back towards the shield and gets a second go at it, and so on and so forth; thus, there will be a small correction factor due to backscatter. This might be ignored entirely as it most likely amounts to a couple dB; or it might be ignored by specifying the test environment, backscatter included; or the chamber might be measured separately, then calibrated out of the result. With scattering a factor, this isn't the easiest thing to calibrate for (the speckle pattern of the scattered radiation will manifest as random peaks and dips -- speckle in the spectrum, as it were).
Looking up standards on these materials, and their measurement, likely sheds some light on the methods to resolve these issues (and, no, afraid I don't have any offhand, sorry). |
H: RCA coax end termination
I am building an amplifier that takes in an RCA output from a TV. From my general understanding, RCA coaxial cables typically have an impedance of either 50 Ω or 75 Ω and a small amount of capacitance in the pF range, and require end termination to prevent signal 'bounce' (or reflection).
A couple of questions:
Is series or parallel termination with resistors (no isolation transformers used) better in terms of signal quality and latency/propagation?
What value resistor do I use for terminating 50 Ω or 75 Ω cables (single end termination only)?
Is it possible to design a termination circuit for both 50 Ω and 75 Ω cables? This is especially since my local store does not list impedances for generic RCA coax cables, nor have I found a reliable way to measure said cable.
AI: "RCA output" is a bit vague but this answer assumes you're trying to handle video signals. The answers are mostly valid for audio signals as well, it would just be unusual or audiophile-grade weird. If your TV has audio outputs through RCA connectors they are likely not designed for a terminated receiver!
Since you only have control of the sink (as opposed to the source; you're at the receiving end) you can and must do parallel termination. Video transmission depends on it, and the source assumes it.
The values must equal the impedance. (but don't obsess about sub-percentage tolerances)
To my knowledge there are no consumer or professional video applications with 50 Ω impedance. It's all 75 Ω, so don't bother trying to do both. If you don't know, assume it's 75 Ω.
If you have an oscilloscope you should be able to measure the impedance (and length!) of an unknown cable. |
H: What does really happen when you connect a capacitor in DC series?
To specify what I am asking: How do we even get a current for a moment when the circuit technically never is completed? And what is it that does this? Here I am thinking of a very simple circuit where there is a DC power source and a capacitor (if we want to be accurate, I'm sure we'd need a resistor as well).
So, when we connect the DC power supply to this circuit, what happens inside? I understand that plates connected to a power supply will be charged according to the charge coming into it, so if the negative terminal of the battery were connected to a plate, it would cause that plate to be negatively charged as well. However, why does this happen and how?
(Just realizing this is only indirectly correlated to a capacitor, but whatever.)
First, I thought maybe the charge density in the two plates of the capacitor were changed according to the voltage applied, so that there would be a voltage across, but that also confused me and didn't make sense, because (unless I'm wrong) for there to be higher charge density in one part of the wire, wouldn't the charge density have to decrease at other places in the wire? And wouldn't this have created a local electric field through this wire?
I tried to draw a picture to visualize what I'm thinking (the stronger the color, the greater the charge density):
This just didn't feel right, so what is it that actually happens? Also, how does this give us a current for a second before it flattens out?
AI: Your reasoning of a change in charge density is about right.
If you push an electron onto one plate of the capacitor, the charge density there will necessarily rise, but in doing so, the increased negative potential of that plate will repel an electron out of the other plate, thereby reducing the charge density on the other side of the dielectric.
Said another way, if an electron enters one side of a capacitor, an electron must leave the other side. Repeatedly forcing an electron onto one plate leaves you with an accumulation of electrons on one plate (the plate became negatively charged), and a deficit of the same number of electrons on the other plate (positively charged). To an external observer, Kirchhoff's current law (KCL) is not violated; current in both capacitor terminals is equal.
That's how a current can flow "through" a capacitor. There appears to be a current flow, equal on both sides in accordance with KCL, but in reality no electron ever crosses the dielectric barrier.
Also to an external observer, the net electric charge of the capacitor is still zero. There are still equal numbers of protons and electrons throughout the material of the entire capacitor but locally to the plates there is a charge imbalance, immediately either side of the dielectric, an imbalance which gives rise to the potential difference (voltage) across the capacitor.
Current stops flowing altogether when the potential difference of the voltage source (battery) equals the difference in potential of charges on the capacitor plates. That's logical, considering that to push an extra electron onto a plate, that electron must have greater potential energy than those already there. This means that current flow will only be momentary, until the source and capacitor potential differences equalise.
As far as the wires are concerned, it's true that for there to be a current flow along a wire, there must be a potential difference between the two ends, suggesting that there is a place of lower potential energy that a charge can/will move towards, and this would imply a charge density gradient along the wire. Initially, immediately after the capacitor is connected to the source, this will be the case.
If current were unconstrained by resistance in the loop, it would be infinite, but that's never the case. Everything in the system has resistance, including the battery and wiring, so current will be capped. Initially the voltage source has some potential difference, and the capacitor has zero potential difference, and by connecting them together you produce a situation in which charges must move from one to the other to make them equal. The imbalance must be resolved. Charges will move through whatever resistances are present in the loop, and it is those resistances which initially have the greatest potential difference across them, bridging the gap between source and capacitor voltages, but limiting the current that will flow according to Ohm's law. If those resistances are small, with a large potential difference across them current will be huge.
So, yes, there will initially be a potential gradient along the wires, and along resistances inside the voltage source itself, which is what is meant by an "electric field", and that gradient is what defines the forces on charges that find themselves there, and the potential energy they possess. That's what propels the charges, always to a place of lower potential energy.
Explicitly placing a resistor in the loop is a measure to control initial current, and the greater that resistance, the greater its initial share of the potential difference, meaning that the source and wiring resistances share a much smaller fraction of the overall potential discrepancy.
Eventually though, capacitor voltage will have risen enough to equal the voltage source, and there will be a state of equilibrium, in which there's no current because no conductive element in the system has any potential difference across it; all charges at all points along any conductive part of the loop have the same potential. The two wires each have the same electrical potential along their length, so there's uniform electron distribution (thermal and quantum fluctuations notwithstanding) along them. The charge imbalance is constrained to the vicinity of the capacitor plates.
All this is a macroscopic understanding of the system, in classical terms. Depending on how deep you want to go down this rabbit-hole, what really happens is less prosaic. Here's another answer I wrote addressing these mechanisms in terms of potential waves and, to some extent, quantum principles. |
H: Can you identify these connectors (JST or Molex type)?
I am trying to identify these board-to-wire connectors found on a PCB for a garage opener. The pitch is 2mm. At first I thought they were JST PH, but the two slots are spaced a little too far apart, and a JST PH cable will not fit. The unit is sold in the US and the PCB has the year 2017 printed on it.
They are not latched, and fully shrouded, with openings only on the sides.
I have tried web searches and the Identiconn utility but failed to find any matches.
Ideally I would like to find a ready-made cable for the 10-pin connector.
AI: It looks like a TE AMP CT header. Part numbers probably 292133-10 and -6. |
H: How much more energy-efficient is it to remove the magnetic ballast for an LED tube?
The diagrams for the GE T8 tubes have these alternative retrofit circuit diagrams:
On this website, 9W in Starter, it states:
....with a further ~9 Watts lost in the magnetic ballast...
I have magnetic ballasts and a starter, so the retrofit does not require a rewire, but I will leave the magnetic ballast in the circuit. However, I could rewire and remove the ballast.
My question is: will it actually save 9 W, or is that loss only when a fluorescent tube is in the circuit?
I have found that the Philips installation instructions match the GE ones, but I could only found the GE circuits.
I have read this SE.EE question, but it really does not have an answer.
AI: A magnetic ballast is no different than a choke, so it has an inductance and a series resistance (DC resistance - DCR).
If you leave it in the circuit it'll be in series with the entire circuit. Depending on the current drawn, its DC resistance will dissipate some power.
...with a further ~9 Watts lost in the magnetic ballast...
This looks like an exaggeration to me. Normally, a 36W tube along with a magnetic ballast (the entire circuit), draws an RMS current of ~0.5 A when supplied from 230V grid and with no power correction at all. A dedicated magnetic ballast is usually an iron-core choke with an inductance of usually 1~1.5 Henries and a DCR of 30~40 Ohms. Now if you work out the current and DCR the real dissipation of the choke should be about 6-7 Watts. But maybe, in some circumstances, the dissipation may increase so the author of the reference might have taken those into account.
EDIT: My "~0.5 Amps" above was a crude guess from memory. The RMS current drawn by a magnetic ballast and tube network should be around 0.3 ~ 0.4 Amps from 230 V if there's no power correction which is usually done with a parallel capacitor across the L-N.
Now, for an uncorrected setup, the total apparent power will be around 80-85 VA but let's be pessimistic and take 90 VA. For an uncorrected PF of 0.5, the real power will be 45 W, therefore leaving 9 W of real power dissipation to the ballast/choke if we assume the tube will dissipate 36 W (usually they are overdriven slightly but let's be optimistic here). So a pessimistic approximation verifies that the ballast choke may dissipate 9 Watts, but the real numbers usually hover around 6-8 W. I also took the cold resistance of a typical 36 W magnetic ballast, so it may increase a little when hot and increase the dissipation further to approach 9 Watts. |
H: The purpose of resistor and capacitor in parallel in MCU reset circuit
I'm designing a device based on Prolific PL2773 and in reference PCB schematic there is an unusual reset circuit.
VCC3 is 3.3V and reset input (RESETB) is active low 3.3V TTL input with Schmitt-trigger (Vt- 1.1V, Vt+ 1.6V). What is the purpose of R32 resistor? And, if I want to make a manual reset, is it ok to place a button in parallel with capacitor?
AI: What is the purpose of R32 resistor?
R32 does two different things: -
It discharges the capacitor should the main power (VCC3) be removed
It elongates the time the device remains in reset when VCC3 is applied
Personally I wouldn't recommend designing it for (2) because there is a danger that it never comes out of reset should the Schmitt trigger threshold not be reached. Instead just make C7 a larger value.
if I want to make a manual reset, is it ok to place a button in
parallel with capacitor?
Yes it's OK but you need a small value resistor in series with the switch to avoid a high impulse discharge current. Maybe 10 Ω to 100 Ω. |
H: Why won't this differential amplifier work without these two additional resistors?
I'm trying to measure a differential voltage in LTSpice using the following structure:
The simulation results are correct, the voltage at the output is very similar to the differential input voltage. My question is: Why are the 40k resistors R7 and R9 needed? When I remove them, the results are not correct (see second picture), and with some OpAmps the simulation will even fail.
These are the waveforms when I remove R7 and R9. But that is the typical differential amplifier structure. Can anyone tell me why this happens?
Edit: this is the full schematic including Vcc and Vss:
AI: Can anyone tell me why this happens?
You are applying a sinewave of amplitude 48 volts p-p to a differential amplifier that has a gain of unity powered from a +/- 3.3 volt supply. The fact that you have "limited" the sinewave with an attenuator and diode doesn't automatically mean that the offset seen on Vdiff+ isn't going to push the op-amp into excessive saturation.
You need to make R8 and R10 at least ten times lower in value. The clue is that your output waveform is badly clipping.
To prove this to yourself, view the waveform Vdiff+. |
H: Would reducing the air gap between the rotor and stator to micrometer/nanometer scale distance proportionately increase a motor's power?
I'm asking this because the smaller the distances between electromagnets, the stronger the electromagnetic field.
I don't know if this would mean a proportional increase in power, be it either torque or RPM.
The closest I could find to this was this article which explores the use of ferrofluids to reduce the gap. It was able to double the torque, but nothing in a proportional sense.
AI: For a constant field in the gap, the motor armature and running characteristics are independent of the size of the gap.
However, to get that constant field, a wound motor needs excitation current more or less proportional to the gap, so the gap affects efficiency. A permanent magnet motor needs magnet strength proportional to the gap, so the gap affects cost.
If the gap is already running at the maximum flux it can, so the pole piece material is at saturation, then reducing the gap further will not increase the field or armature efficiency. It will however reduce the MMF required to maintain that field.
Achieving a very small gap needs high mechanical precision, which is expensive. It's therefore a reasonable tradeoff for the manufacturer to aim for a larger gap, and spend more on magnetic field and less on mechanical precision, when trying to cost-optimise a motor. A larger gap also means slightly less 'windage' loss, from shearing the air in the gaps. |
H: AC delay circuit question
Sorry if it's a stupid question but.. is there any AC delay circuit? All I know is RC and RL circuits but are based on DC current
For example, to delay for a second the triggering of an AC relay
thanks
AI: Yes, there are AC time-delay relays, usually based on heating a bimetallic strip at a controlled rate.
Other methods for more complex sequencing include motor-driven cam switches, etc.
I can't think of any purely electronic methods that don't involve converting the AC to DC at some point. |
H: I2C circuit resistance value
I have a question about I2C and esp32. when I started long time ago making circuit based on esp32 I think i have made a mistake on the pullup resistor by putting 4.7K on 5 volt on i2c lines.
the I2C lines work well I want to know why i did not burn esp32 SDA and SCL pin witch are 3.3 volt logic ?
if I am wrong what is the value for the resistor on 3.3volt.
Last question about slaves modules with are 5v do I need a level shifter on the i2c line in case of 3.3 volt resistor line.
AI: Yes, it was a mistake to pull 3.3V IO pins to 5V with resistors.
It does not blow up because you can assume that MCU IO pins have built-in protection diodes to supply voltage to clamp excessive input voltages from doing damage.
And since you have resistors, they limit current which allow the protection diodes to work instead of damage due to excessive current.
Assuming the MCU supply is 3.3V, and the protection diodes start to conduct at about 0.5V above that, there will be 3.8V on IO pin and clamping current is only about 0.25mA. So this is why it did not burn.
There is no one true and correct value. There is a suitable range which depends on many things you are not mentioning. It depends on which speed you want to communicate, and how much total capacitance there is on bus, which depends on your wiring/cabling, bus length, number of devices on bus, how fast they are capable of communicating and in which mode, and in which limits they need to operate. So there are I2C specification limits, and limits imposed by the chips which may be within or outside I2C specs, and there is the limits imposed by the environment the chips are in, and that may also be within or outside of I2C specs.
So 4k7 to 3.3V could either work or not. It will be somewhat slower bus than 4k7 to 5V. It depends if any other module already has pull-ups or not. The total pull-up resistance must not be too strong or too weak, so that all chips are capable of reliably communicating at the speed you want to use, and you may need to do some tradeoffs.
And yes if you have devices with 5V I2C bus which are not compatible with 3.3V I2C bus then you need a level shifter.
Even if you have slave modules that take in 5V supply, it means nothing how they work internally, as they might still have internal 3.3V regulation and must not be connected to a 5V I2C bus. |
H: What are the potentials at front and back of inductor when primary DC supply is disconnected in LR circuit?
I have been learning the basics of inductors in simple LR circuits with DC currents. I think I get most of it but I am a bit unclear about the secondary EMF produced by the inductor after the primary supply is disconnected. In particular I am unsure about the actual electric potentials at the ends of the inductor just after the primary battery supply has been disconnected.
In the QUCS (v. 0.0.19) transient simulation image below, the maximum voltage drop across the inductor (L1), at time > 0.150 s is about -5 V (see the graph for voltage probe PR2). (I would like to plot the value of potential around the circuit at different times - but I have not found a way to do this with QUCS).
I assume that this means the potential at the front is 0 V and at the back it is about +5 V. This would make sense to me in that the inductor subsequently acts like a dying battery - pushing a dwindling current around the secondary loop in a clockwise direction.
But I am also aware that the -5 V voltage drop across the inductor could also be caused by potentials of -5 V at the front and 0 V at the end. It seems to me that scenario would also drive a current around the secondary loop in the clockwise direction.
So my question is: which of these two scenarios is correct?
Update after Question Answered
Following the receommendation by u/RussellH I probed the potentials at points A,B,C,D with the following (QUCS) graphical results (below). I also plotted (in Excel)(further below) the profile of Potential vs. X (relative position along the section of the circuit containing R1,L1,R2).
AI: So my question is which of these two scenarios is correct?
Neither.
At the instant that S3 shorts, there is still 1 amp flowing through L1, R1 and R2.
Given that R1 (2.5 Ω) is grounded to 0 volts by S3, the top node of R1 has to be at -2.5 volts.
I think you can work out the other voltage now. |
H: Calculate the inductance voltage from its current time function
I have a solution for the following exercise, but I'm not sure if it's a good one.
This is the current time function of the inductance:
$$i(t) = 2-2e^{-t/T}\text{ (A)}$$
$$L = 10\text{ mH}$$
$$T = 2\text{ ms}$$
Calculate the inductance voltage at \$t = 3\text{ ms}\$.
This is my solution:
$$UL = L \times \frac{di(t)}{dt}$$
$$UL(t) = 10\text{ mH}$$
$$UL(t) = 10 \times e^{-t/t}$$
$$UL(t=3\text{ ms}) = 10 \times e^{-3/2} = 2.23\text{ V}$$
Is it right?
AI: Well, notice that we know that:
$$\text{V}_\text{L}\left(t\right)=\text{L}\text{I}_\text{L}'\left(t\right)\tag1$$
So, when:
$$\text{I}_\text{L}\left(t\right)=2\left(1-\exp\left(-\frac{t}{\text{T}}\right)\right)\tag2$$
We get:
$$
\begin{alignat*}{1}
\text{I}_\text{L}'\left(t\right)&=\frac{\partial}{\partial t}\left(2\left(1-\exp\left(-\frac{t}{\text{T}}\right)\right)\right)\\
\\
&=2\left(\frac{\partial}{\partial t}\left(1\right)-\frac{\partial}{\partial t}\left(\exp\left(-\frac{t}{\text{T}}\right)\right)\right)\\
\\
&=2\left(\frac{\partial}{\partial t}\left(1\right)-\exp\left(-\frac{t}{\text{T}}\right)\cdot\frac{\partial}{\partial t}\left(-\frac{t}{\text{T}}\right)\right)\\
\\
&=2\left(\frac{\partial}{\partial t}\left(1\right)+\frac{1}{\text{T}}\cdot\exp\left(-\frac{t}{\text{T}}\right)\cdot\frac{\text{d}}{\text{d}t}\left(t\right)\right)\\
\\
&=2\left(0+\frac{1}{\text{T}}\cdot\exp\left(-\frac{t}{\text{T}}\right)\cdot1\right)\\
\\
&=\frac{2}{\text{T}}\cdot\exp\left(-\frac{t}{\text{T}}\right)
\end{alignat*}
\tag3
$$
So, we get:
$$\text{V}_\text{L}\left(3\cdot10^{-3}\right)=10\cdot10^{-3}\cdot\frac{2}{2\cdot10^{-3}}\cdot\exp\left(-\frac{3\cdot10^{-3}}{2\cdot10^{-3}}\right)=\frac{10}{\exp\left(\frac{3}{2}\right)}\approx2.2313\space\text{V}\tag4$$
So, you're right! |
H: integration of multiple microphones in same ESP32-S3 I2S interface
I'm currently working on a graduation project where I am tasked with building an embedded system using an ESP32-S3 and four MEMS microphones capable of detecting ultrasonic waves. The goal is to understand the sound specifications of an industrial machine. The specific microphone model I am using is the SPH0641LU4H-1. You can find the datasheet here
(https://mm.digikey.com/Volume0/opasdata/d220001/medias/docus/930/SPH0641LU4H-1.PDF).
In the ESP32 documentation, particularly in the ESP-IDF programming guide, I've learned that the ESP32 has two built-in I2S (Inter-IC Sound) interfaces, namely I2S0 and I2S1. However, I noticed that while I2S0 can support up to 8 multiple sound sources in mono mode, I couldn't find clear information on how the microcontroller distinguishes between these data sources. Unlike I2C, the I2S bus frame doesn't have a specific addressing mechanism.
AI: I2S bus does have an addressing mechanism. Otherwise it could not know the difference which sample is left or right, or even which bits belong to which sample.
That sync signal can have multiple names but it is called WS in ESP32 MCUs.
And there are multiple methods how to transfer 8 PCM channels on I2S. The ESP32 supports 8 channels as four data wires sending 2 I2S channels each, or in TDM mode with up to 16 channels on single data wire.
But the mic sends PDM bits, not multi-bit PCM samples. There can be only two mics per data wire, and the correct clock to use for the mics is WS because also the ESP32 has to be in PDM input mode. So the bit clock is not needed. |
H: Would this sensor get damaged without a relay?
I'm just learning and looked at this circuit to implement. I have read the infrared sensor could be damaged by the larger current requirement of the light/siren without a relay. Is that true?
If so, why doesn't the light/siren just not operate right due to the sensor only outputting 100 mA vs the need to have 300 mA to operate?
AI: If you exceed the sensor's current specification limit for an extended time, it will heat up and eventually fail.
You can boost the siren current by using a transistor buffer. For example, a 2N2222 is good for about 600mA.
Such a setup would look like this (simulate it here):
You could also use the sensor to activate a relay, assuming the relay coil current is low enough. |
H: Why is the right half of the circuit irrelevant?
A follow-up this post regarding the following circtuit:
I continue to struggle to understand why the right half of the circuit is irrelevant when computing the voltage between \$A\$ and \$B\$. One answer to the previous post reads
The \$500Ω\$ resistor is not short-circuited, but there's effectively only one connection from the left half of the circuit to the right, so the current source and the \$28V\$ voltage source can have no effect on the voltage from \$A\$ to \$B\$.
Aren't there two connections from the left half to the right?
Even if there is only one connection, why does that imply the sources in the right half of the circuit cannot have an effect on the voltage from \$A\$ to \$B\$?
Another answer reads
The right half of the circuit only connects to node \$B\$, there is no path for any current from the \$28 V\$ or \$10 mA\$ sources to flow to node \$A\$, so that whole part of the circuit is irrelevant to the voltage from \$A\$ to \$B\$.
It seems to me the right half connects to \$A\$:
AI: For a current to flow you would need two independant nodes between the left and right. The right side only connects to the left at node B. They may have drawn it to look like there are two nodes, but the connections from the right are in fact both to node B, so only one node is connected.
Note that I use the word node. In a schematic a node is a part of the circuit where two or more circuit elements connect, and it is assumed that this connection has zero resistance. It is considered to act as a single point even though it is drawn as lines between the components. So node B includes all of the lines connected to where it says B, up to where they attach to a component.
Here I have highlighted node B, all of the red is considered to be at the same voltage potential, and all of the currents entering and leaving it will sum to zero.
So as you can see, current coming from node B can't flow around to A and through the 500\$\Omega\$ resistor back to B.
For any current to flow, some part of the right side that's not connected to node B would need to connect to some part of the left side other than node B.
That is why I said in the previous question that the right side of the circuit is irrelevant to the voltage from A to B. |
H: How to properly calculate the run time of a battery using a power converter?
I want to buy a 12v 100ah lithium battery to power a 750 watts power converter that provides 120V.
I would like to know how many hours of energy will I have with this information. I tried some online calculators, but it's not clear if I'm doing it right.
Why?
Well, the 750 watts is related to the 120V part of the power converter, so I don't know how to calculate on the 12V part of the battery.
AI: Generally, a 12 volt 100Ah battery has an equivalent watt-hour rating of 12 x 100 i.e. 1200 watt-hours (Wh). So, if your load needs 750 watts then the battery will typically be able to supply that power for 1 hour and 36 minutes.
However, there is not much info about the battery and, it may droop its terminal voltage both under load and, over time so, I wouldn't bet on getting anything more than 1 hour from the battery. |
H: Why is the bottom half of the circuit shortcircuited?
A follow up to this post, regarding this circuit:
Below \$V_1\$ applies a rectangular wave of amplitude \$24V\$. For this post I'm only concerned in the case \$V_1 = -24V\$. In that case, according to the teacher, the bottom half of the circuit is short-circuited i.e. need not be taken into account. Why is this the case?
AI: For charge to flow there must be a closed path around the voltage source V1. If V1 is negative D1 is reversed biased so can be removed for the sake of analysis. So the circuit can be redrawn as in Figure 1.
For charge to flow in the lower circuit, there must be a closd path from node A through V1, through the lower circuit to node B, then return to node A through an additional path, which does not exist.
Rx in Figure 2 is an example of an additional path from node B back to node A, closing the circuit.
In figure 1 Rx is infinite, thus opening the circuit, so there is no current in any of the elements in the lower circuit, making them irrelevent to circuit operation when V1 is negative.
...according to the teacher, the bottom half of the circuit is short-circuited...
The circuit is not shorted, but open.
simulate this circuit – Schematic created using CircuitLab |
H: Does coiling a transmission line affect its characteristic impedance?
If I take an advertised 50 ohm impedance cable and turn it into a coil will this change its characteristic impedance from 50 ohm to something else? My intuition tells me it should as if we look at one section of the coil the inductance per unit length should be higher as we still have the wires self inductance but the magnetic field produced now interacts with the neighbouring wires and induces a current in them which resists the initial current which sounds like the inductance per unit length would be higher which would increase the characteristic impedance. Is this correct?
EDIT:
For context I'm working on an RFID project which has an coil design similar to this
Essentially I'm wondering if I set my trace impedance to 50 ohm and make this coil will the characteristic impedance of the trace be altered?
AI: That's not a transmission line, that's a trace in space.
It would be a transmission line if the trace were over ground plane, and spaced adequately from nearby traces (microstrip geometry).
Follow standard calculation methods for square spiral inductors.
To explain in more detail:
A "normal transmission line", would be a well-defined pair of conductors, of constant cross-section, dielectric inbetween, and well-defined impedance and propagation velocity.
For the flat spiral geometry, you can think of (for very short time scales i.e. ≪1ns), one trace, with respect to its neighbor, as an edge-coupled flat (planar) differential pair; which will be a fairly high impedance, 100-200Ω say. But after the wave propagates once around the loop, now what was the signal becomes the ground, and vice versa.
The structure can't be modeled as a simple normal TL.
We further have the problem that, for each pair of traces, it's not against ground at all, but the companion of one pair, is in series with the next, and so on until the end of the winding. The transient impedance is actually a whole bunch of TLs in series (at the "start" port), then looped around in parallel (at the "finish" port, skewed by one index, except for the start and finish which connect to nothing else).
And furthermore, all the TL lengths vary, because the length around the spiral varies. (They're equal when it's e.g. a cylindrical helix, as in a single-layer solenoid coil.)
For more information along those lines, you might want to research the helical waveguide: this is one model of the helical solenoid winding, used to describe its impedance. The overall result is a much higher impedance (Zo in the kohms is easily achieved), and dispersion (propagation velocity varies with frequency). Similar analysis applies to the spiral case, with much stronger dispersion I suppose, and there are other known periodic structures with peculiar frequency response, meta-materials, etc.
Also consider the converse:
A transmission line is a transmission line, specifically because it is not an antenna. That is, EM fields are confined within the structure, rather than propagating out into free space.
Microstrip is a fairly poor example, as TLs go -- one whole side is open to space, and indeed radiation is noticeable from such structures. But it's usually low enough to be an acceptable compromise; for example, a 4-layer PCB with logic-level signals (e.g. LVCMOS at 3.3V and ~ns edge rate, or LVDS-style signals at ~10mA and ~Gbps data rates) can pass commercial emissions limits without much difficulty.
This is in direct contradiction to the purpose of an RFID coil: to couple to free fields.
A transmission line, as a trace over ground plane, is precisely the wrong thing to use here; the ground plane closes the loop local to the trace, greatly reducing the coupling to free fields. Rather, a loop antenna is normally used, with huge open area and no ground plane, specifically so that wireless communication is possible.
As above, we can still make some kinds of transmission-line-based observations -- but they become much more hand-wavey in nature, harder to scope out and model and analyze, and pretty quickly our one-dimensional analysis method (the transmission line) fails and we must submit to full-field 3D analysis.
Conversely, when we don't need the full-field analysis, as say at microwave frequencies -- it suffices to use low-frequency approximations. Hence the above suggestion to follow standard calculation methods. The most common of which I believe is due to Wheeler (sadly, often repeated but rarely cited!). Approximations by the way, that include ignoring various aspects such as wire shape, diameter, spacing, material (conductivity) and frequency; so expect some error (easily >10%) between calculation and the real article. This is worse than the tuning can be for an RFID gadget, so allow a step in the project plan to tune it -- for example, design for a low target value then increase load capacitance until on frequency; or iterate the layout until the self-resonant frequency is correct.
Finally, to answer the title question as curious readers may find it:
Does coiling a transmission line affect its characteristic impedance?
No, or not very much. If you wind up some coax on a spool for example, as long as the radius of curvature is much larger than the outer diameter, nothing much happens. Do observe that the cable is made of a range of materials -- the polyethylene or whatever dielectric is much less stiff and strong compared to the copper (or plated steel) conductors -- and so the cross-section becomes deformed when bent around, more and more as the radius of curvature approaches the outer diameter.
The resulting displacement, for tight bends or kinks, still doesn't amount to much for most signals -- it's a brief interruption where the impedance changes, and such changes are only sensible by waves by their size in relation to the wavelength. A deformation 1cm long is hardly sensible by a wavelength of 1m, etc. Such a kink can be detected by TDR (time domain reflectometry), where the wavefront length is comparable to the defect size.
Similarly, crushed cables can be detected by TDR; such is the basis of some traffic measurement systems for example.
The most important consequence from bending, is probably the reduced breakdown voltage, if the line is operated at high voltage (including at DC), or near the maximum [apparent] power rating. The core conductor being squished towards one side within the dielectric, causes the dielectric to thin, reducing its breakdown voltage. In the extreme, the core conductor can indeed get pulled all the way through the dielectric, shorting out the cable entirely.
Hence, you're advised not to bend such cables tightly. Always read the datasheet and supporting information from the manufacturer! |
H: Will this electric guitar wiring scheme be noisy?
I want to wire my guitar using two SPDT switches for my pickup selector. This is so I can have four sounds: lower, upper, both in parallel, both in series. I'm concerned that when it's set to use just the lower pickup (as shown in the schematic) that it will cause some noise/hum to be added to the output signal since the upper pickup still has one wire connected to the output jack with the other wire left floating. If this does add noise, will it help much if I keep the wires from the upper pickup very short (an inch or so)?
AI: This will be fine.
The inductive pickup of noise from the pickup that is connected with only one wire will be entirely prevented. Its extremely large series resistance (of the open switch) prevents any signal contribution.
It will pickup a slight bit of electric field noise as will any dangling conductor that is connected to your circuit, such as the bridge, the strings, even yourself. But this noise is common-mode noise. As opposed to a differential signal between the two poles of your guitar cable, the common-mode noise is usually not amplified, so it doesn't matter either.
Another intuitive way, you can see why this will be fine is the following: With a standard pickup selector, the unselected pickup is still connected with one side to the "ground" node of the guitar circuit, but doesn't contribute noise. Since the guitar output is a differentiel voltage between its two output poles, anything dangling on the "ground" pole is no better or worse than dangling on the "signal" pole. |
H: How does a power meter differentiate between three-phase and single-phase loads?
How does a three-phase power meter (for example the one used by my power company) know the difference between a single three-phase load and multiple single-phase loads?
For example: a three-phase heater at 4 kW (400 V * 10 A) vs three single phase heaters also totaling 10 A (230 V * 3.333333… A * 3 = 2.3 kW).
What am I missing here?
AI: The difference is phase voltage vs line voltage.
Consider the following:
simulate this circuit – Schematic created using CircuitLab
In the 3P arrangement above, R, S, and T are "phases" with an RMS voltage of 230V separated (delay / phase difference) by 120 degrees each, N is the common (neutral). Above is a balanced system so the neutral current is zero.
For each phase, "phase voltage" is 230 Vrms. Each phase supplies a 100R load, therefore phase current for each is \$I_R=I_S=I_T=230/100=2.3 \ \text{Arms}\$, which makes ~530 Watts each and ~1.6 kW total.
4 kW (400 V * 10 A)
Now 400V here is the "line" voltage which is basically \$\sqrt3\$ times the phase voltage: \$V_L=230 \ \sqrt3\approx 400 \text{V}\$. This is the voltage across each phase i.e. R-S, R-T and S-T. So the phase voltage (i.e. R-N, S-N, and T-N) is 230V but the "line" voltage is 398V.
And 10A is the total RMS current of the network which is
$$
I_{tot}=\sqrt{I_R^2+I_S^2+I_T^2}
$$
where \$I_{R,S,T}\$ are "phase" currents. For a balanced network the phase currents will be equal so \$I_{tot}=I_R \sqrt3\$. And for the example circuit above, it's
$$
I_{tot}=2.3\sqrt3\approx 4 \ \text{Arms}
$$
Now if you multiply the "line" voltage with the total RMS current above, you'll get
$$
P_T=V_L \ I_{tot}=1.6 \ \text{kW}
$$
three single phase heaters also totaling 10 A (230 V * 3.333333… A * 3 = 2.3 kW)
The currents do not sum up to 10 A. The powers can be summed but the currents cannot because they are out of phase. All you can do is to take the square root of the sum of squares which I tried to show above. |
H: Why are a transformer's high and low impedance reported directly?
For context I've looked at DI boxes for audio applications. Passively operated ones use a transformer to step a source voltage down as well as provide an impedance bridge for a potentially high-impedance source to a low-impedance input for maximum voltage transfer efficiency - desirable in audio applications. A bridging impedance is Z_load >> Z_source, usually a ratio of 10 or so.
The DI circuits are basically equivalent to only a transformer (see the "block diagram" section here and below). However, products in this space often report input and output impedances. One reports 140 kOhm in and 150 Ohm out for a 1 kHz signal, with a turns ratio of 11.5 (see the transformer specs here and below).
To my knowledge, transformers only have an impedance ratio which is the square of the turns ratio. So in a voltage divider, a transformer would only shift the apparent impedance of the load, unlike say when using an active buffer amplifier that has set input and output impedances. But then the ratio of the reported impedances of the transformer (140k / 150 = 930), does not match the square of the turns ratio either (130).
Here is a simple schematic of the system:
simulate this circuit – Schematic created using CircuitLab
So above, the load impedance of 1 kOhm would ideally present as 100 kOhm to the source, providing a bridging impedance.
Is there something else at play or have I misunderstood the issue? Is it only a matter of marketing and the transformer not being ideal? Thank you.
AI: Is there something else at play or have I misunderstood the issue?
I don't see any problem with the data sheet for the transformer: -
12 squared is 144 and, for an ideal transformer with a 12:1 step down ratio and a 1 kΩ load, the input impedance would be 144 kΩ. I have no idea where the 150 Ω load comes from that you mention: -
One reports 140 kOhm in and 150 Ohm out for a 1 kHz signal
No, you are misreading that...
The 150 Ω actually comes from an analysis of test circuit 1 using the 6.8 kΩ drive resistor and the series 4.6 kΩ primary resistance. When converted to the secondary (by dividing by 144), the impedance is 79 Ω. Then, add this to the secondary resistance and you get 142 Ω. Consider also that the primary and secondary have a little bit of leakage inductance and you can readily see that the output impedance of test circuit 1 is nominally 150 Ω.
Relevant info: - |
H: Battery consideration and topology choice
I have a system with a battery. I do not know what will be the battery at the moment. And so I can select the voltage battery in function of my system. I need to regulate a voltage (15 V) thanks to a converter. The output current will be around 10 A, i.e 150 W.
The question is it better to use a boost converter or a buck converter for doing the job as I can choose the voltage level of the battery? Also I think that If the voltage level of the battery increase, it will also increase the output impedance of the battery as more cell will be put in series and so when some current will be drawn by the load, the voltage of the battery will be lower. Also I think that the battery will have a lower energy density if I increase the voltage than if I took a battery with a lower voltage. So I think that it will be better to use a boost converter. But as I do not know really much battery, I would appreciate to have your opinions.
I put away the non inverting buck boost topologies as it would be less efficient.
AI: This is not a simple thing to be answered in a few sentences. Different things should be considered together such as failure modes, environmental conditions, cost, etc. I'll try to cover only a limited section of these.
Whether you go for a buck or a boost, the complexity, overall efficiency, and material cost will be pretty much the same for both.
Lower battery voltage and boost converter require thicker input wires, but higher battery voltage and buck converter require good interconnection of the batteries/cells. This will make you consider the overall build of the system. And, for now, we can assume the power loss at input section to be equal for both buck and boost.
Practically speaking, 10 Amps is not a low number for an output current. It's high enough to give you headache due to the losses and heat.
If you go for a boost converter with, say, 12V battery voltage, the rectifier diode will be in conduction for almost 80% of a switching cycle (assuming CCM). This means that the diode dissipation will be substantial. You can go for a synchronous boost converter and replace the rectifier with a MOSFET but it's not a common thing these days. So you can select a higher battery voltage and go for a synchronous buck instead.
You should also consider the potential failures:
In a boost converter if the switch fails first it'll short the battery. But if the rectifier diode fails first your load will see the input voltage, 12V.
In a sync buck converter, if the top switch fails first the load will see the full input voltage, presumably 24V. But if the bottom switch fails first the battery will be shorted.
There are also other possible failure modes. If you want to deep dive into the details google "FMEA".
Consider the potential danger of the load being exposed to higher or lower voltage.
Do a rough calculation/estimation of the losses and cooling of the critical components, and take the complexity and implementation difficulties along with cost and availability, and decide which is acceptable for you. Since there's no such thing as "best" in engineering, it is not to make the best but it's to make the "good enough".
PS: Personally, I'd start with a synchronous buck. |
H: How can I find out what the value of the capacitor on an Intel Core i5-4690 processor?
How can I find out what the value of the capacitor shown in the photo is on an Intel Core i5-4690 processor?
AI: Find an identical CPU with identical capacitor.
Remove the capacitor from a working CPU and measure it.
Then you can put the capacitor back, and buy a new one for the CPU that is without the capacitor.
Please understand that this may not be the best way, but will give you a reasonable guess for the capacitor.
However, you can only measure the parameters of the capacitors your meter can show you, not all.
You can see the physical size, and you can measure capacitance. Good meters show multiple things that might be important in addition to the capacitance, such as ESR and ESL. Or an impedance curve to find out which capacitor will be at least as good as the original.
But there are some parameters that cannot be measured. You can't know if that is a 10V, 16V, 35V, or 50V capacitor.
As you see, the specs of the cap is not known, so even if you replace it with some random capacitor with same capacitance, there is no guarantee any more if the CPU will work as specified.
While I did mention a procedure how to figure out the capacitance, I cannot recommend doing it, as you may end up with two broken CPUs and removing and measuring and re-attaching the capacitor properly requires quite a lot of good equipment and skills. |
H: Control chip with a GPIO pin using a transistor and a push button switch
I need some help regarding a small circuit that I'm currently working on.
I have a microcontroller (ESP32) that I want to use to control a chip with a GPIO pin of my ESP32 as output, and at the same time, I want to be able to also use a physical push button to do the same job.
The chip is a RF module with a power on/off pin. When this pin is held low (sub 1.5 V) for 1 sec the module boots (or reboots). Otherwise it's internally pulled high to 1.8 V during normal operation.
datasheet
The datasheet suggests to use a NPN transistor for this purpose following this figure:
I used Falstad circuit simulator (first time using it) to try to design a good circuit but I have to admit that I'm struggling. I came up with this circuit. It seems to work fine without the push button, the RF pin goes high when ESP pin is low and goes low when ESP pin is high.
However, I would like this push button to also reboot the RF module. And I don't know how and where to place it. I tried to connect the base with the emitter and the collector like this and now the RF pin follows ESP pin (RF high when ESP high and low when low) but now of course it's as if I didn't have a transistor. Moreover, the voltage is now 3.3 V at the RF pin and I think it might damage the RF module given that its internal voltage is 1.8 V. What's the best and simple way to deal with this? Just for info, I will be using this transistor:
Datasheet
Another question: What's the purpose of the 47 k resistor? It doesn't change anything when I remove it and the simulator says that it's a bad connection. I don't understand.
AI: There's a few options for the pushbutton. You could simply parallel the C-E of the transistor with your pushbutton. Especially in this case, the TVS (ESD protection device) shown is highly recommended since ESD could work its way from the finger to the switch to the input.
If you make the pushbutton drive the transistor you could also do this:
simulate this circuit – Schematic created using CircuitLab
The 47kΩ resistor deals with any leakage in case the input is left open (for example if the ESP is unplugged or in a state where the GPIO is high-Z). It will probably work okay with it removed. |
H: How to bind a module in system verilog, with parameters not from the target location
I would like to bind a module, and pass a parameter from the module I declared the bind in, rather than them all coming from the module I am binding to.
module dut(input logic A, output logic B);
assign B = A;
endmodule
module dut_verif_bind#(
string IF_PATH = "a_string"
) (
inout A,
input B
);
import uvm_pkg::*;
AB_intf vif();
assign vif.A = A;
assign vif.B = B;
initial begin
uvm_config_db#(virtual AB_intf)::set(null, IF_PATH, "vif", vif);
end
endmodule
The bind module would be bound from some higher level (a testbench, or another bind module) to dut. The fact that this can be another bind module is why I want to be able to bind the parameters like this.
module tb#(string IF_PATH = "uvm_test_top*") ();
dut my_dut();
bind my_dut dut_verif_bind#(.IF_PATH($sformatf("%s.dut*", IF_PATH))) dut_if_binds (.*);
endmodule
The above however, gives errors, because in the parameter binding of the bind module (#(.IF_PATH($sformatf("%s.dut*", IF_PATH)))) it is searching for the IF_PATH parameter in the module being bound to, not in the scope of the bind declaration. This is somewhat intuitive, though I hadn't considered that issue, but it also seems like there should be a way around it.
For those wondering, this structure is something I am exploring because it should wind up making mid-level testbenches and tests nearly trivial to run if you have a unit and system test anyways. Just need a couple extra sequences and a couple more agent instantiations, but you don't have to start over with declaring interfaces and tossing them in the database, and shouldn't have to move them around the database at all.
How to bind a module in system verilog, with parameters not from the target location?
AI: Limit the use of parameters to places that require constant expressions. In this case, there is no need to make IF_PATH a parameter; it can be a string variable.
module dut_verif_bind (
inout A,
input B
);
string IF_PATH;
import uvm_pkg::*;
AB_intf vif();
assign vif.A = A;
assign vif.B = B;
initial begin
wait(IF_PATH !="")
uvm_config_db#(virtual AB_intf)::set(null, IF_PATH, "vif", vif);
end
endmodule
module tb#(string IF_PATH = "uvm_test_top*") ();
dut my_dut();
bind my_dut dut_verif_bind dut_if_binds (.*);
initial
my_dut.dut_verif_bind.IF_PATH = $sformatf("%s.dut*", IF_PATH);
endmodule |
H: USRP B200 sample rates
What are the possible sample rates for both transmit and receive on the Ettus USRP B200 mini ? This SDR uses the AD9364 chip.
Does the radio use integer divisors of the master clock rate? Or does it support arbirtrary sample rates as suggested here?
AI: What are the possible sample rates for both transmit and receive on the Ettus USRP B200 mini ?
Too many to explicitly list. Whatever clock rate the synthesizer on the ad936x can generate, divided by an integer between 1 and 512.
Does the radio use integer divisors of the master clock rate?
Yes
Or does it support arbirtrary sample rates as suggested here?
That page does not suggest that. |
H: Large current flow through SMD shunt resistor
What is the best way to large current flow through SMD resistor?
As you know, latest FPGA like ULTRA series or Versal series need large Core current almost over the hundred ampere.
It is not problem, if i not use shunt resistor or use expensive DC/DC regulator that has power management function(can get measuring output voltage or current), because they connected through just copper plane from voltage source to FPGA.
If i use shunt resistor and that has not enough big smd pads, i think it will be make problem.
Because i can make big enough copper island but current must flow through shunt resistor smaller and narrower than island.
is there any good way?
AI: You can put multiple shunt resistors in parallel or you can use a copper trace as a shunt resistor. |
H: Half-bridge driver ground loop elimination
I built this half-bridge driver using a generic Schmitt trigger/hex inverter IC and a 555 timer as the square wave generator. By varying the value of the resistors, the dead time can be adjusted and it works perfectly!
The outputs are connected to the input of a TC4420 MOSFET driver IC. One for each output. I thought this would work by using two (galvanically-isolated) power supplies for each of the TC4420s but then realized that a ground loop still exists between the input signals.
Is there a way I can eliminate this ground loop without optocouplers? Optocouplers are slow and even the fast ones might draw more current than the hex inverter can output.
AI: Your circuitry isn't going to work. Even with isolated supplies. The fundamental issue is that the logic input reference is common to the MOSFET gate return. What you need is a high-side gate driver with bootstrap to drive an N-channel MOSFET. What you really need is a half-bridge driver that includes the bootstrap diode. There are so many to choose from including high-voltage versions (up to ~600VDC). Some even include adjustable deadtime. TBH, the whole left side of you circuit (everything except you power stage) could be replaced with a single 8-pin gate drive and boot capacitor.
You probably want something like this if you have no idea what I'm talking about: https://www.ti.com/lit/ds/symlink/ucc27712-q1.pdf?ts=1706252591500&ref_url=https%253A%252F%252Fwww.ti.com%252Fproduct%252FUCC27712-Q1
I don't know your requirements, but there are plenty of options available: High-voltage, built in shoot-through protection, single input, independent gate control, fixed dead-time, adjustable dead time, programable slew rate if you want to get fancy, ect. All the big names make these: TI, ADI, Infinium, Microchip Tech, OnSemi, ect. |
H: Which pin is pin #1?
Where is pin #1 on this IC?
It seems to have a line and a dot. The dot doesn't seem to line up with a corner pin though.
Chip is a INA210AIDCKR.
AI: Generally, when the text/print is the correct way up, bottom-left is the pin-1. This applies to QFP/LQFP, QFN, SOT, DIP, SOIC, SOP and maybe some other packages. For your question, it's the top-left according to the picture. |
H: How to supply a data logger with electricity inside a tumble dryer at 80 °C?
I have constructed a data logger which will be used in a tumble dryer. During the drying process the temperatures inside are 60-80 °C. The data logger is designed to move freely inside the tumbler. The data logger is very compact. How would you supply it with electricity (3 V, 50 mA)?
AI: There are some specialty lithium batteries (LiSOCl2) that can take up to 125°C.
For an example see Tadirn TLH batteries.
Yet another manufacturer using a similar Li chemistry claims a 165°C limit. |
H: MOSFET biasing and Rg value
I have a doubt and can't find relevant information in books/internet about setting a suitable value of Rg resistance during MOSFET biasing using a large drain-to-gate feedback
resistance. Here is a visualization of the problem:
The current of Rg is 0 amperes, meaning that we cannot use Ohm's law here. But as far as I understand, the voltage at the gate is the same as drain Vd (the upper half of the MOSFET). It seems a bit not very intuitive to me, but I will get it.
Now, the common knowledge says that the value of Rg must be "sufficiently large" around some Mega Ohms, but still, I had not seen any formula or a way to find its minimum value or something. Is there a formal way to determine it resistance, or just say that it should be "large" around some Mega Ohms?
AI: as far as I understand, the voltage at the gate is the same as drain Vd
In your circuit that is correct because the gate to source resistance can be massive and, that means virtually no DC current passes through \$R_G\$.
I had not seen any formula or a way to find its minimum value
Its minimum value depends on what you are trying to use the MOSFET for i.e. what is the application you are trying to fulfil. In some applications the minimum value could be zero. In other applications it could be tens of mega ohm. It's application dependent.
Is there a formal way to determine it resistance, or just say that it
should be "large" around some Mega Ohms?
Yes, but it's application dependent. |
H: High R for low frequency op amp integrator
I have to design an integrator for very low frequencies (corner frequency around 0.5 Hz).
Are there any reasons why I cannot choose a really high resistor value (say 5M Ohm) to keep the capacitor value reasonably low?
Consider that the signals at play are also quite small.
AI: 5MΩ is not a very high value if your accuracy requirements are modest and the environment is benign. You normally don't start to run into problems with long time constants until the times get into the minutes. I have made analog products with time constants as high as 30 or 60 minutes. But that was then, now we would generally do that digitally once it gets beyond seconds.
You should pick an op-amp with a guaranteed low input bias current and a capacitor with guaranteed low leakage (neither of which are much of an issue for the numbers involved). And get a reasonably stable resistor.
I suggest also that you at least consider a digital solution, which can largely avoid the relatively large (relative to the accuracy of a crystal) temperature dependency of R and C values and several other sources of error. |
H: A bit tricky exercise that I'm currently looking for help to solve it
simulate this circuit – Schematic created using CircuitLab
The exercise:
We switch 200 V as a resulting voltage on two capacitors in series.
The charge of the two capacitor is 10 µAs (Q1=Q2).
What is the capacitance of the two, if C1 = 4C2 ?
The capacitative voltage divider would be an option, but we do not know either capacitance.
AI: Let X be the total capacitance, then
Q = U * X = 200V * X = 10 uC
X = 1/20 uC / V = 50 nF
This is the total capacitance of the two series caps. Now figure C2 in the task:
50nF = C2 * 4C2 / (C2 + 4C2) = 4C2^2 / 5C2 = 4/5 * C2
C2 = 5/4 * 50nF = 62.5nF
C1 = 4*C2 = 250nF |
H: Same MOSFET component but reversed arrow
I'm trying to implement a USB PD controller, STUSB4500, but the typical application example shows 2x P-Channel MOSFETs STL6P3LLH6 with different arrow indications. I'm now confused about the pins, T4 has the same MOSFET drawing as in the STL6P3LLH6 datasheet, so the left pin is the drain (5,6,7,8) and the right source (1,2,3). But for T1, which is reversed, is the right side pins 5,6,7,8 or 1,2,3? Any further clarifications on this will be appreciated.
Image Sources/Datasheet links:
STUSB4500: https://www.st.com/en/interfaces-and-transceivers/stusb4500.html
STL6P3LLH6: https://www.st.com/en/power-transistors/stl6p3llh6.html
AI: The two sources (1,2,3) are connected together. The grey thick line has obscured the internal connection from body to source.
This is a standard back-to-back MOSFET connection to make a bilateral switch. |
H: Choosing a supply voltage for an instrumentation amplifier
I am using a Wheatstone bridge to measure a resistance and an instrumentation amplifier to amplify the signal. Specifically, I am using the INA121. According to the amplifier datasheet, the input voltage can be in the range of ±2.5V and ±15V. In addition, it says that
Typical performance curves, “Input Common-Mode Range vs Output Voltage” show the range of linear operation for ±15V, ±5V, and ±2.5V supplies.
Therefore, my choice would be one of those voltages. However, which one should I choose? Does a high voltage offer advantages over a low voltage (or vice-versa) in instrumentation amplifiers?
AI: With instrumentation amplifiers, in addition to input voltage range and output voltage range you would look it for an op-amp, you also have to pay attention to the "Diamond" diagrams to ensure you have high enough supply voltages that internal nodes don't saturate.
From your datasheet link:
Analog Devices has an excellent "Diamond Plot Tool" which can simplify this analysis with their products, however the part design you have is owned by TI (née Burr-Brown). |
H: Should you include multiple low-pass capacitor circuits in integrated IC PCB design?
I have two circuits, one is a POE to 5V, another is an IC for leds. They both include capacitors between 5v and gnd. Should I include the first 5V circuit POE capacitors since it should already be doing the filtering?
Transformer for POE:
https://www.coilcraft.com/de-de/products/transformers/power-transformers/power-converter-transformers/fct1-xxm22sl/fct1-50m22sl/
POE datasheet:
https://www.monolithicpower.com/en/documentview/productdocument/index/version/2/document_type/Datasheet/lang/en/sku/MP8009GV-Z/document_id/4873/
LED Current Regulator:
https://www.digikey.com/en/products/detail/lumissil-microsystems/IS31FL3236A-QFLS4-TR/14308348
Or is there something I am missing. Here are the schematics:
[![enter image description here][2]][2]
AI: The capacitors for the LED driver are necessary for stabilizing the chip - they decouple the power supply and provide local noise immunity. If you want a robust design, keep the capacitors. Also, they prevent disturbing the VCC plane that other IC's are using especially if this LED driver modulates via PWM.
Will the circuit work with out them? Probably. But it's bad design practice - it's a lot of risk for practically no cost or effort. |
H: Custom made NetTie in KiCAD7 can't be routed
I have created a footprint for NetTie of custom size (I wanted 0.2mm pad/connection), as the library-provided one has the size of 0.5mm and is too big.
I have tried two ways of routing:
from other pad to NetTie: trying to route it from outside (of NetTie) the track stops before reaching NetTie's pad
when trying to start the track from NetTie, when clicking pad of NetTie, this green block shows up, the screen blinks and no track appears.
Most probably there is a DRC rule on the way. To prevent it, I ensured that during footprint creation all layers, polygons etc. were the same as the working NetTie from library. Unless there are some additional settings invisible in KiCAD of which I don't know.
What to check to make this custom made NetTie to be possible to route?
AI: @Hearth It worked, thanks!
To make it clear: Footprint needs to be opened in Footprint Editor, next click on Edit Footprint properties (yes, there are more than one 'Properties')
Next, go to the 'Clearance Overrides and Settings', click small '+' and place desired pads (in my case: 1, 2) |
H: Why do I see unwanted current through this BJT switching circuit?
I built the following circuit but am getting a result I wasn't expecting.
Here is a link to the CircuitLab model (I don't know how to upload a saved circuit to StackExchange):
LED_module_circuit_diagram
The circuit design was supposed to work as follows:
The bank of LEDs labelled FWD are three white LEDS, the bank of LEDs labelled REV are two red LEDs. These represent the lights at one one end of a model locomotive which may move forward or reverse and the lights respond accordingly. The lights on the rear of the locomotive would behave the opposite way.
The four inputs on the left are logic high (+5V) or low (0V) (from a MCU), except Vss which is tied to the positive supply (+5V). For the purposes of this question the input nINTEN may be ignored.
A logic low on input nEXTEN is supposed to enable use of the FWD and REV banks of LEDs. A logic high should switch off both banks. While nEXTEN is logic low the input FWDnREV should dictate which bank of LEDs is lit. With logic high the white LEDs should light (red LEDs off), with logic LOW the red LEDs should light (white LEDs off).
The circuit works as desired except in one case. With nEXTEN low, and FWDnREV low, the red LEDs light as desired, but the white LEDs are dimly lit as oposed to fully extinguished.
What is the cause of this and what should solve it?
AI: To turn on Q4, there must be a very small current through white LEDs to base of Q4. The current is small but it is enough to light up the LEDs so that they are visible.
You could try to put e.g. a 1k resistor over the white LEDs to keep the voltage over LEDs low enough while the 1k resistor passes current instead of the LEDs. |
H: 2nd-order Sallen-key low pass filter with unity gain outputs rail voltage when input is 0V
I was designing an active low pass filter for simple sigma-delta modulation signals. I found that the op-amp will output 3V when the input is 0V given that the power supply is 0~3.3V.
I simulate the circuit and find the same result. The circuit and the result are shown below. The blue line is the output of the op-amp.
According to my analysis, it is essentially a unity gain buffer with 10k2 input resistance in this scenario and the output should be 0V.
I noticed that the voltage will follow when V2 is slightly greater than 0V.
Why does that happen?
AI: You are exceeding the input common mode voltage range of the opamp. The AD712 is designed to operate with a minimum of 10 Volt (±5 V). power supply.
You also have an input biasing problem. To get it to work with a single power supply, you need to bias the +input at half the supply. You can do this by adding a DC voltage source in series with V2.
For your experiments, use a ± power supply perhaps at ±5 V. |
H: Ultrasound cIeaning of PCBs with isopropyl alcohol
I have tried to clean the flux remnants on a high-voltage PCB using isopropyl alcohol in ultrasound. The flux remnants were visible around the big SMD components such as transistors or in the creepage-distance cutouts. The main idea of the cleaning was to prepare the PCBs for potting.
The result was that after applying ultrasound for one hour the flux remnants are still there but seem broken up. The broken up flux remnants still stick to the PCB even though they can be manually removed relatively easily.
Is there any chance I could completely clean the PCB in ultrasound? Would increasing (how much?) the isopropyl alcohol temperature help?
AI: Is there any chance I could completely clean the PCB in ultrasound?
I would switch to a better solvent, if that is something that is tolerable and still PCB compatible. I usually recommend some of techsprays stuff but it wouldn't work well in a bath. IPA is a weak solvent:
IPA isn’t a great cleaner. It has a Kb (Kauri-butanol) value of about
50, meaning it’s not strong enough to tackle the tougher-to-remove
lead-free and no-clean flux residue commonly used today. It also isn’t
very good at removing greasy or oily fingerprints.
Source: https://www.microcare.com/en-US/Resources/Insights/June-2021/Stop-Using-IPA-to-Clean-Your-PCBs
Also make sure you are using 99% IPA and not something from the local store that has a lot of water in it.
So try and find something higher on the Kb scale that will work for your application.
Would increasing (how much?) the isopropyl alcohol temperature help?
Yeah, but it will also cause your IPA to evaporate much faster as the boiling point is at ~80C, lower than water. Increasing it beyond 40C gets it into the air in a hurry.
Another thing would be to switch the flux, I really like chipquick 291 (no clean), and maybe they have something that would be easy to clean in a water-soluable or rosin based that would work better with IPA |
H: Using impedance to get RLC Transfer Function
I am trying to get the transfer function G(s)=Vo(s)/VI(s) of this circuit and I am lost. I know that in the s domain, L=sL and C=1/sC. I tried to calculate it already and got R+(Ls)/((Ls^2)*C+1). But I don't know if that is correct because I didn't really do anything with current?
I think I am confused on the overall method of getting a transfer function for a circuit. I thought the steps were get Vi, manipulate it until the right is in terms of Vo, and then do Laplace, and then put into Vo/Vi form. But I am seeing around this forum another method where you use impedance somehow? That is what I attempted to use to get my top answer, impedance of R + Xc || XL. I tried looking at other RLC circuit transfer functions on here where that method was used and I did not understand what the process is.
So how am I supposed to calculate this?
AI: I am confused on the overall method of getting a transfer function for
a circuit
You analyse it like you would a potential divider made from resistors: -
The lower resistor is, in effect, the parallel impedance of the capacitor and inductor. The upper resistor is \$R\$ in your circuit.
Calculate \$\dfrac{1}{sC}\$ in parallel with \$sL\$ (call it \$Z\$ if you want)
Use the potential divider formula to get the TF = \$\dfrac{Z}{R+Z}\$
how am I supposed to calculate this?
Use the above method and drill down to you get a solution. A quick calculation tells me it is: -
$$TF = \dfrac{s\frac{1}{CR}}{s^2 + s\frac{1}{CR} +\frac{1}{LC}}$$ |
H: How to connect to these terminal blocks?
I have this power supply (actually part of a control box for a robot arm) which has terminal blocks, but most of the screws seem complete inaccessible due to how closely the blocks are placed together. Here is a picture of the one semi-reachable screw in the whole device.
Here is a top view of where I want my wires to go in.
I thought maybe the connection pads were spring loaded but I could not pry them open with a tweezer.
Taking a picture from one side, I noticed there is a label! "DORABO DB2EK-3.5"
AI: Here is a Dorabo terminal block PDF close to the P/N that you have. It is the top half of the terminal block assembly. As you can see there are also key guides that help locate and seat the halves.
Not sure if there is a locking tab that first needs to be disengaged.
You might try prying very carefully at the first plastic to plastic interface above the PCB surface to separate the two sections. Once separated you would loosen the terminal screw, insert the wire, re-tighten the screw, repeat for all other wires on that terminal block, then reinsert the upper block into the lower section. |
H: Is GND connection required for two independent modules to communicate in MIL STD 1553 protocol?
I use a Holt IC 2130 MIL-STD-1553 transceiver. When I connect only the A and B differential lines in two independent modules, I can't communicate between the two modules, but when I connect the GND pins, the communication is completed successfully.
Is GND connection required for two independent modules to communicate using the MIL-STD-1553 protocol? Or is there another point I missed? As far as I know, GND connection is not required for differential lines.
Source: Holt IC AN-550
AI: Is GND connection required for two independent modules to communicate in MIL STD 1553 protocol?
Because the physical layer of MIL STD 1553 requires transceivers to be connected to the bus via isolation transformers, no common ground connection is required between transceivers.
The Holt IC application note AN-550 referenced in the question shows a center tap on bus-side of the isolation transformer, which it says is "often" not connected.
Data Defense Corporations Review and Rationale of MIL STD 1553 A and B has a similar schematic which shows no bus-side center tap on the isolation transformer.
If however, the bus-side center tap of the isolation transformer is grounded on one module, the differential pair will not "float", but will be referenced to that ground. If more than one module has the bus-side center tap connected to ground, and the modules are direct-connected to the bus, then there is a ground path through the differential pair. If the ground potentials are different at the different modules, this ground current could be significant, and might possibly interfere with communication. Therefore, the bus-side center tap of the isolation transformer should not be connected to ground if the modules are directly connected to the bus.
Rather than connect the bus-side isolation transformer center-tap to ground, it might make sense to connect it to the shield of the twisted pair cable. Whether or not shields should be connected on both ends has been debated. I will only say that "both ends connected" might work in some circumstances, and "one end only" might work in other circumstances. I will leave it to others to comment further on this topic if they choose to.
If the modules you are using require a common ground connection in order to communicate, they are failing to meet the standard. Something is wrong. The schematic for the Holt IC module provided in the question shows such an isolation transformer. Can you check the modules to see if they do indeed have isolation transformers? Can you verify that the bus-side center taps of the isolation transformers are not connected to ground? |
H: My MQ6 gas detector stops working properly after some time when using a battery
I've connected an MQ6 gas sensor to the output of a 78L05 5 V regulator, with its input connected to the + terminal of a Duracell 9 V battery. The DOUT pin of the MQ6 is connected to the - terminal of a buzzer.
After a while, the green light on the MQ6 indicating that it is operational goes out, the red light indicating that there is a gas leak lights up very dimly and the buzzer starts to sound even though there is no leak.
When I disconnect the whole thing and let the battery rest for a while, the sensor starts working again until it stops working properly.
can anyone tell me what's wrong with my project ? I was inspired by this project https://www.youtube.com/watch?v=wq1bM2EpRdI but I had to add a voltage regulator because the MQ6 requires a voltage of 5 volts.
Datasheet : https://joy-it.net/files/files/Produkte/SEN-MQ6/SEN-MQ6_Datasheet_2023-09-26.pdf
AI: The MQ6 has a heater inside it that draws about 200 mA @ 5 V. This is far more than a standard rectangular 9V battery can supply for any length of time. |
H: Input impedance of instrumentation amplifier
I have a problem understanding which resistors set input impedance of an instrumentation amplifier. What is the input impedance of the circuit given and what is the purpose of R250 (10k to GND)?
AI: R250 is needed to allow natural bias currents from the op-amp inputs to flow to local 0 volts. If it were not there, the inputs would float and likely cause common-mode input range problems.
The differential input impedance (S2A and S2B in current positions) is R163 + R164 = 1022 Ω. This increases by 28 kΩ when S2A and S2B are changed over. |
H: Looking for circuitry to save power on an I2C sensor
I have an I2C sensor connected to a microcontroller. The whole system is battery powered, so I need to find ways to switch off power from devices when they are not in use.
For the sensor, I am thinking of using an IO Expander or something similar to disconnect the sensor from Vcc. But what to do with the I2C pins? They are pulled up with 10 kΩ resistors.
Even though I am disconnecting the sensor from the power, there will be still current flowing through the pull-up resistors. Is there a way to also disconnect SDA and SCL pins, so that the sensor is not drawing any current at all when not in use?
I am considering using MOSFETs as switches, but wondering if there are any standard solutions or best practices.
AI: If you are going to turn the sensor supply on and off, put the I2C pull-up resistors on the sensor supply so they turn on and off with the sensor.
This will prevent leakage through resistors to sensor.
It also prevents problems of bus voltage ever being higher than sensor supply, during switching things on or off. |
H: DIY guitar amp cabinet selector switch
Problem statement:
I have 2 guitar amplifiers and I have 2 speaker cabinets that normally connect via speaker cables (1/4" connectors). I would like to create a passive device that lets me switch which guitar amplifier to connect with which speaker cabinet.
What I think want/need:
I think I could do this by wiring together 2 identical switches; one for the amplifier selection and one for the cabinet selection (the cabinet selection wired in reverse of the amplifier selection since the signal would be coming IN from the select amplifier yet going OUT to the selected cabinet)
I think each switch would need 6 connections (4x2)
2 sets of 2 connections connected via switch to a single pair of connections
For context:
I've done some soldering for car audio applications and an old synthesizer chip replacement, and done some LED projects with breadboards and microprocessors but do not have an incredible amount of experience in this space.
After some research:
Within each of the 2 sections, the components must be completely isolated (the amplifiers speaker outputs do not share terminals) - therefore, I think I'm looking for DPDT (Dual Pole Dual Throw) switches
In an effort to super-simplify the project, I looked up DPDT "light switches" and found these.
Would that work for what I'd like to do? I have a few concerns:
that switch is very expensive for a light switch, and I'd need 2!
what if I wanted to grow this to handle 3 or 4 amplifiers, and maybe another cabinet or 2? I'd need to handle more than 2 sets of inputs/outputs.
So after some additional digging, I came across this video. My main question is around how he selected the right switch at this point in the video. The switch he selects is rated at 500 mA and I think that is too small. According to this question, even an 8 Ω speaker rated at 25 W could peak at just below 2 A.
My concern is slightly reinforced here where it is mentioned:
Let’s say you’re listening at loud levels and hitting 100W peaks, and you have 8-ohm and 4-ohm speakers. Using the equations above, you can calculate that the 8-ohm speaker will demand 3.5A of peak current, while the 4-ohm speaker will demand 5A of current.
And yeah, guitars can get loud so I think I'd like this device to suffice for a max. 100 W speaker.
All that to say, if I go the Mouser route, I think I'd be safest with a slightly beefier switch, so I tried to tweak the Mouser search and much to my surprise, the results of that search are listing up to almost $100.
So - thought I'd ask for input from anyone that might be able to temper my logic and help make sure that indeed that is the type of switch I'd need to consider - or, if you'd offer a different suggestion. At this rate, I'm starting to cost more than buying something off the shelf.
AI: Since you want to be able to expand the system you might as well start with a system that can grow with your needs. Whatever solution you choose, you need to ensure that you can't short the outputs of two amplifiers together so switches would need to be break-before-make type.
I suggest that you use a patch panel in conjunction with a 2-input / 2-output selector box similar to the off-the-shelf unit linked in your question. Using the patch panel removes any limits on the number of devices that can be connected.
simulate this circuit – Schematic created using CircuitLab
Figure 1. A combination of two patch panels and a 2-in / 2-out selection box gives a good low-cost solution. As shown here, either AMP1 or AMP3 can be connected to SPKR2 or SPKR4.
Notes:
The patch panels could be made with standard 1/4" jacks or Speakon(?) connectors. Make sure that the jacks can't short sleeve to tip during insertion.
You now have the ability to do A-B testing between any two amplifiers and any two speakers.
You need to be sure that the amplifiers are OK running open-circuit.
Watch out for amplifiers with DC blocking capacitors on the output. These may result in a thump from the speakers when switched in.
You need to be sure that you can't switch, say, a 4 Ω speaker into an 8 Ω amplifier.
Check your local car parts stockist for heavy duty toggle switches. |
H: Input stage for sampling audio sine wave with ADC
I'm building a prototype, checking the potential for a project.
I want to be able to sample a sine wave from a power amplifier using a Teensy4 ADC.
I have a Teensy 4.0 powering an Evisun isolated DC-DC module to generate +/-15v
I have a full wave precision rectifier to eliminate any negative voltage on the output and increase digital range on the ADC (I don't need to recreate the sound, just need to capture the half cycles)
My missing part is a differential input stage, where I can connect either a single ended output or a differential output amplifier, then feed the unbalanced output into the precision rectifier >> ADC and sample the half cycles.
I need to have a voltage divider connected on each input to reduce up to 200v peak-peak down to 6v peak-peak.
I'm looking for suggestions on a balanced to unbalanced circuit that is capable of handling frequencies of 10Hz to 20kHz.
Also, can I connect the input ground of the dc-dc converter to the com output? I don't really need an isolated supply and used it just for low component count, and I need a ground reference for the ADC anyways.
AI: Looking for suggestions on a balanced to unbalanced circuit that is
capable of handling frequencies of 10 Hz to 20 kHz
I could easily suggest a differential amplifier but, given that the power output might have considerable common-mode signals associated with it, I would favour an audio transformer from people like Hammond.
Also, can I connect the input ground of the dc-dc converter to the com
output?
You won't need an isolating DC-DC converter with this method.
If you insist on using op-amps, this is a decent mixer input for a balanced signal from Rane. The following is an image of the specific section (microphone amplifier with an added 30 dB attenuator): |
H: Will this battery to USB auto switching circuit work?
I want to have a project be powered by six AA batteries or USB. When the USB cable is attached the circuit should automatically switch off the battery input and start using 5 V from the USB.
This is the circuit I've come up with by copying component setups from the Arduino Uno R3 and an Adafruit Feather, the idea here being that when the USB power is connected it applies voltage to the MOSFET that then disconnects the battery from the voltage regulator.
The Zener diode that goes between the USB voltage input and the voltage regulator output protects the USB from backfeed. In total my entire circuit that this supplies power to will use less than 500 mA.
My understanding is that it is typically not recommended to have another voltage input attached to the output of a voltage regulator without a protection diode, but I assume if the Arduino has gotten away with it then maybe it will work in this case and save me a component as well as the voltage drop over that diode.
If, however, I do need a diode there, what would be a good one to use?
AI: When the USB cable is attached the circuit should automatically switch off the battery input and start using 5 V from the USB.
Your circuit will not work like that.
When VIN is available, regardless of whether VUSB is available or not, the 5V regulator will see VIN through the PMOS DMG341 (either its body diode or its "on" resistance, RDS-on) as per the connection shown:
When VUSB is unavailable whilst VIN is available, the PMOS will be on so the regulator will see VIN.
When VUSB is available whilst VIN is available, the PMOS will be off but the regulator will see VIN through the PMOS's body diode.
The Zener diode that goes between the USB voltage input and the voltage regulator output
It's not a Zener diode, it's a Schottky barrier diode.
My understanding is that it is typically not recommended to have another voltage input attached to the output of a voltage regulator without a protection diode
I don't know the 1117's design but I personally wouldn't apply an external voltage across a regulator's output, regardless of whether through a diode or directly. If 1117's series pass element is a PNP transistor (I really don't know, I just assumed so), for example, which means the output is taken from the collector, then the external voltage applied across the output will simply forward bias the CB junction when the regulator is off (i.e. no input). This may or may not result in an IC failure, I don't know. If you want to have the regulator output and the VUSB OR'ed then the simplest way is to connect them with OR'ing diodes:
simulate this circuit – Schematic created using CircuitLab
... but I assume if the Arduino has gotten away with it then maybe it will work in this case and save me a component as well as the voltage drop over that diode.
I've just checked the R3's schematic. When there's no power applied from the power input connector shown as X1why would a designer name a connector with X anyway?, the USBVCC (USB 5V) will pass through T1 PMOS (FDN340P) body diode 1 and power LP2985 to generate 3.3V, also will be used as +5V for the rest of the circuit. Now the potential problem here is that the USBVCC might be supplying current into the regulator's output when the regulator is off. If this usage is allowed by the design/designer then that's fine, but I wouldn't do that.
1 The T1 is simply there just as a reverse-polarity protection. If the main intention is the disconnect the USBVCC when a DC input is available then simply it won't work like that. |
H: Running traces across THT footprints
Will this work, knowing that I'm running traces across THT footprints?
Kicad lets me do it, but will PCB manufacturers be able to print the board?
On the other hand I had no other choice as I might also want to etch boards myself, so I had to keep it single layer.
AI: It is perfectly normal to run tracks through THT footprints, and even between IC pins, as long as you leave sufficent clearance between the tracks and component pads. |
H: UC3842 optocoupler feedback confusion
I am having a slight confusion regarding the selection compensation pin of the UC3842 IC and the current mode control system given in the datasheet.
Why is there an RC combination at the feedback of the error amplifer? Can I simply use a resistor at that pin?
Secondly as far as I have seen, the optocoupler will allow a current to flow in the right side only if the voltage gets below the threshold. But this will only give either a high or a low voltage at that pin. It wount give the amount by which the voltage got decreased or the real reference as is done in the TL494 IC where the IC gets a propper reference from the voltage divider as shown below. But here from getting either a high or a low how can we figure the measure of change of pwm duty cycle??
UC3842:
Kindly help me out in understanding these two concepts and the working of this IC with optocoupler. Considering I am using the voltage mode of the UC3842 IC. Here is a scehmatic of the changes made in the current and voltage mode control:
AI: The TL494 PWM controller hosts two op-amps for controlling current and voltage. However, if the vast majority of CC-CV controllers (constant-current and constant-voltage) use ORed op-amps wired in an inverting configuration, the TL494, for an unknown reason, has its two op-amps configured in a non-inverting configuration which makes compensation an extremely unpleasant exercise. It is part of the response I gave on SE and I released a PDF you can download from my page. The \$RC\$ network you see on the schematic is linked to compensation of this converter but I suspect trial-and-error is behind this process.
Regarding the UC384x and an optocoupler, I recommend to stay away from the application circuit you reproduced. The most reliable way is to disable the internal op-amp by grounding its FB pin. You then connect the COMP pin to the 5-V VREF pin via a given pull-up resistance - say 10 kOhms - and you connect the optocoupler from COMP and the GND pin of the IC:
It important here to care about the PCB layout - sorry for the crapCAD schematic : ) the two copper traces from the optocoupler (FB and GND) go side-by-side and the emitter connects to the IC ground which is quiet: do not connect to a noisy point or the converter won't be stable. Same for the collector, go to the COMP pin and have the filter capacitor which introduces a wanted pole, wired very close to the IC. This is an industry-standard configuration that I have seen many times in ac-dc converters. |
H: Shadow-triggered LED fade circuit
I'm making an interactive art project and I'd like to have an LED that turns on when someone puts their hand in front of a light sensor (photoresistor). I got that working. Next step is that I'd like to have the LED fade back to off after a second after it turns on.
What I'm after is:
When photoresistor goes from uncovered (low resistance) to covered (high resistance), turn LED on immediately, then fade to off over ~1 second.
When photoresistor is not covered, the LED should be off.
I don't care too much what happens if the photoresistor becomes uncovered while the LED is fading, but the best would be if it keeps fading to off.
Here's my circuit attempt. It works the first time but there's no path to discharge the capacitor so it won't turn on a second time.
Can someone point me in the right direction?
AI: Circuit below:
Discharging path ensure two 10k resistors.
The right PNP creates a monostable circuit to not stop fade if hand is out of LDR.
470k trimpot sets the LDR trigger point according what LDR is used and light conditions around.
Better use lower resistance LDR like 10k. |
H: Complex-conjugate poles from transfer function
This comes from Sedra's Microelectronic Circuits, example 17.1.
If I have a transfer function with a pole \$p_1 = \omega_0(-0.1736 + j0.9848)\$.
The example talks about "combining \$p_1\$ with its complex conjugate", becoming \$s^2+0.3472s + \omega_0^2\$.
By trial and error I found that this "combining with its complex-conjugate" assumes an expression \$s^2 + (p_1 + \overline{p_1})s + p_1 \overline{p_1}\$. But I don't get what this combination means exactly and why it needs to happen here.
AI: For realizable systems, complex poles always occur in pairs. For example, a second order transfer function can be written:
$$
G(s)=\frac{N(s)}{s^2+a_1s+a_0}
$$
Or factored into poles as:
$$
G(s)=\frac{N(s)}{(s-p_1)(s-p_2)}
$$
If \$p_1\$ is complex then so must \$p_2\$ be the complex conjugate of \$p_1\$.
If a pole is complex, then its conjugate must exist, so the denominitor of the transfer function can be found as is shown in the OP. That's all this means.
For higher order systems, each pair of complex poles will reveal a quadratic factor in the denominator. |
H: ESP32 max input voltage UART
Before doing any mistake, can you confirm me that it is not possible to connect directly UART pins of Arduino UNO on UART pint of ESP32 ?
I see on the datasheet a voltage of 3.3V. I would like to make communicating both boards by TX/RX pins.
Thank you,
AI: You should use a level converter (also called level shifter) from 3.3V to 5V (and around). These devices are bi-directional.
You connect:
3.3V to the LV (Low voltage) pin
5V to the HV (High Voltage) pin
RX of the ESP to A0, A1, A2 or A3, e.g. A0
TX of the ESP to A0, A1, A2 or A3, e.g. A1
RX of the Arduino opposite of the TX of the ESP, e.g. B0
TX of the Arduino opposite of the RX of the ESP, e.g. B1
Connect the GNDs, for clearity better put them on the same row (ESP next to A4, Arduino next to B4).
See bi-directionional-logic-level-converter-hookup-guide for more details. |
H: Changing noise level with 24-Bit A/D Converter
I work on a 8 Channel 24-Bit A/D-Converter. I have an issue with the noise level of this converter. There are some "quiet" periods where the noise is low and acceptable and there are other times where the noise increases by a factor of 5.
Y-Axis is in mV, X-Axis in minutes, sample rate is 2Hz, the pcb board electronic is in a aluminum case to reduce the general noise. The A/D conversion part of the electronics looks like this:
The AD converter is the AD7190 from analog devices. Prior to the A/D conversion part we use the following electronic to scale the incoming signals to the A/D converter range:
I tested the electronics connected to a function generator, with open inputs, in a case and without a case several times at different locations and i always get a behaviour like that...
My Question is what is the reason for this behavior and how can i "fix" it? many thanks in advance!
AI: There could be many sources of noise, I'll cover a few:
1) Temperature. Temperature changes in the electronics can create noise in electronics, especially those of low level measurements. With ADC's you can find the temperature coefficents in the datasheet listed as ppm or in a graph like this:
Source: AD7190 datasheet
Usually it's a good idea to temperature control low level analog electronics
The other problem is air on chips, take a cover and put it over the board if you haven't done so already.
2) Power and conducted emissions
One of the biggest problems is common mode noise (noise getting in through the ground), if you have other loads that are switching these can cause problems (especially if the return currents for those loads cross the analog subsection of the PCB)
DC to DC converters can wreck havoc on low level measurements, make sure your supplies are noise free, and you use good regulators for the power.
3) Radiated noise. Remember that 5/2^24 is 63nV, it's not that hard to get 63nV of noise from radio sources on your PCB (and that means even less before the amplifier, because the amplifier gains up noise). Make sure your inputs to the amplifiers and electronics are properly shielded.
Source: Art of Electronics 3rd ed |
H: How can I calculate length of wire, given its gauge and resistance?
As the title suggests; I am trying to calculate the length of a wire given its gauge and resistance in ohms per 1000 feet. I also have the amount of voltage drop, which is 1.1 V.
Specifications
Gauge: 14 AWG
Resistance (per 1000 feet): 2.525 ohms
I researched briefly and found this equation:
Vdrop = IR
However, I don't have an amperage value, so is it possible to calculate cable length without amperage? This is a DC system.
EDIT: I may be confusing some people. I'm looking for Total Length of the Cable.
The ohms/1000 feet value came from this table here, not from me.
AI: $$ \Delta V = \frac {R_{1000}}{1000} l I $$
where \$ \Delta V \$ is the voltage drop, \$ R_{1000} \$ is the resistance per 1000 feet, \$ l \$ is the length in feet and \$ I \$ is the current.
There are four variables. You can find any one of them if you have the other three. You only have two variables, \$ \Delta V \$ and \$ R_{1000} \$ so you can't solve for \$l\$ or \$I\$.
One other thing: don't forget that the loop resistance will be twice that of a single core. |
H: What is meant by the Sending end and Receiving end in 3 phase electrical circuits?
I do not understand the what sending or receiving end means in 3 phase circuit analysis.
Aside:
Also I am struggling to understand 3-phase just in general with respect to circuit analysis so if there are any links to helpful sites, please add them.
Thank you
AI: In three-phase power analysis it is more common to refer to the supply and load rather than the sending end and the receiving end. The supply end is usually the sending end, but is possible for the load end to return power to the source. If someone has solar panels on their house or some other renewable energy system, they could send power to the source if they use less power than their renewable energy system supplies. In that was people often refer to their electric meter "running backwards" or their power use being negative. The source is still the considered the to be the normal "sending end," but some customers are exceptions, so their power flow is "negative" or the opposite direction of the standard.
Unless other wise stated, the voltage of a three phase system is the voltage between any two phases. Equal or balanced voltage among the phases is assumed unless otherwise specified. For a three phase, 4-wire, wye source, both the line-to-line and the line-to-neutral voltage should be specified, such as 380/220 V. If only one voltage is specified it is normally the line-to-line voltage. Just the line-to-neutral voltage would be specified only if referring a to a single-phase circuit.
Reference material links are not usually given in this forum because there is no system for repairing dead links. You will need to search for educational material yourself. It may help to use advanced search tools such as limiting the search to *.edu sites or searching for *.pdf documents. |
H: Bode Phase Plot of RC High-Pass Filter
I have been trying to understand Bode Phase Plot of RC High pass filter.
I know that from my engineering that voltage lags behind the current passing through the capacitor. So, I assume the current through capacitor will be in-phase with the input voltage waveform and that same current will pass through the resistor. So, output voltage should be in-phase with input waveform.
Can someone explain me the Bode phase plot of the filter when frequency is swept from zero to cut-off frequency via time domain analysis?
AI: Note that current is actually common to both C and R, not really as shown in the schematic.
If you compare the input voltage (Vin) with the output voltage (Vout), the input amplitude is always >= output amplitude. At low frequencies, output is much smaller than input.
And at low frequencies, phase of Vout leads phase of Vin:
At high frequency, well into the passband, Vout amplitude is nearly equal Vin amplitude, and phase of Vout approaches that of Vin: |
H: Correct charging current for lithium-ion batteries
I am trying to replace a lithium-ion battery for my Bose QuietComfort 35 headphones. I cannot find the datasheet for it. The battery is an AHB110520CPS (AHB110520) by Synergy. It is supposedly an "Advanced Hybrid Battery" which uses thinner materials than LiPo or something.
Its specs are 495mAh 3.7V, max 4.2V (I think).
I could not find a datasheet or anything about it on the net, including at archive.org. The battery model is probably about 2 years old and could be obsolete by now. The manufacturer's website has some AHBxxxxx.. battery models listed but not this one.
Other numbers found on it:
1ICR11/53
SYNERGY MH10048-E7
(chinese writing)
The battery has 3 wires labeled T (temperature), B+, and B-, so I don't think it has anything sophisticated inside it.
I would just replace it with a drone battery of similar capacity and voltage but I'm concerned about the charging current used for the battery. Do I have to find a battery with the same or more max charging current? I suppose I can measure the existing battery's charging current but what I'm curious about is what specs I need for the replacement battery in terms of charging current. I've read that lithium charging circuits are constant current/constant voltage (which is it?). So if the internal resistance of the battery is lower than the spec would there be a problem or not?
AI: CC/CV charging means that it begins by providing a constant current until the voltage hits a preset level, then switches to that constant voltage until the current trickles down to something marginal.
The internal resistance of the battery doesn't affect the charging routine, although the charging efficiency might change.
This target charge current is relative to the battery capacity ("C"). For standard Li-ion or Li-polymer batteries, chargers often target 0.5C charge current. In other words, if the battery is rated at 500 mA-h, the target current is 250 mA.
It is not unusual to charge at 1C (500mA), but this compromises the battery's capacity over time.
I don't know about Synergy's "Advanced Hybrid" technology, but the voltages match standard Li-Po batteries. I would choose a new battery of (at least) the same capacity.
Hopefully you can find one that fits! Obviously, if you have to reduce the capacity to reduce physical size you risk charging the new battery too fast. |
H: 60 Hz signal between thumb and forefinger?
I was playing around with the oscilloscope and noticed when I measure between two points on my hand, I see about a 1 V signal at about 60 Hz. Do I oscillate at 60 Hz? Does this have anything to do with lighting or main's hum?
Where does this 60 Hz signal come from?
AI: The wiring in your walls acts like one plate of a big capacitor, with your body being the other plate, picking up mains-frequency noise and coupling it to the oscilloscope probe. It's completely normal. |
H: Dual DC output voltage power supply made from two power supplies?
I need replacements for power supplies which output two voltages. For example, a Nintendo 64 needs 3.3V DC 2.7A and 12V DC 0.8A. There are readymade options but they seem to be of questionable quality. Can I combine two single output power supplies? Do I connect the output GNDs? What are the disadvantages to this approach; apart from maybe being overkill, cost- and space-inefficient? The advantage I see is that 100-240V AC to 3.3V, 5V, 12V, ... DC power supplies are readily available from reputable sources.
As an alternative, I could probably use an ATX power supply and just use some of the output voltages. Is that a better idea?
AI: This is generally perfectly acceptable. The only major concern is that of power supply sequencing, meaning ensuring that the various power supplies:
start up in an acceptable order
reach their respective target voltages within an acceptable period of time
start up within a suitable window of time before or after each other.
This is important because an improper power sequence can cause disruptive or damaging currents to flow through the signal lines that connect components that are connected to different power supplies.
Sequencing is more difficult to accomplish with two different power supplies versus one integrated system. You would have to carefully examine the requirements of the various loads to really determine if this is a problem.
As an aside, I'm not sure why an N64 would need 12V, unless it has a fan, which I don't recall.
Generally you would need to tie the negative outputs of the power supplies together if you need two positive voltages. It is also possible to use separate power supplies to produce positive and negative voltages by tying the positive output of one power supply to the negative output of the other, provided the negative power supply has no internal connection between its negative output and its chassis/equipment ground terminal.
A major advantage of combining power supplies in this way is that it's easy to create arbitrary sets of voltages and current capacities without incurring the substantial cost of designing and certifying a custom multi-output power supply. This makes it a very attractive solution for low volume products with high power requirements or requiring nonstandard voltages.
An ATX power supply could work, however given that you need less than 20W, even the smallest ATX supply is going to be tremendous overkill, and will not likely demonstrate good efficiency or regulation. Also, since most modern PCs derive most of their operating power from the 12V (via point-of-use step-down converters located near the CPU, GPU, and other low-voltage/high-current components), some power supplies sacrifice the efficiency and stability of the lower voltage rails to save cost while providing improved performance and efficiency on the 12V rail. |
H: Registering IOT products across Europe based on NodeMCU
We've been designing some smart home devices and softwares in past few years and they've been sold in some countries in middle-east. Basically, they were combination of ESP, Arduino and NodeMCU boards which we've added some relay board to them and make them a product since they came with a UI and web service we had provided.
At the moment, I am planning a company in Poland to start my own IOT service and my goal is to sell some NodeMCU based devices with a hosted software, such as temperature control or GPS tracking services.
I would like to import relay boards and temperature sensors and so on from Aliexpress dealers, directly from China to Poland and Estonia.
What I am aware of, at least in paper work level I do need certifications from someone to cover my ass when I am selling these products, since they can make problems if they are overheated or higher current provided to them.
The question here remains is, which certifications the new product that I am designing must pass, in order to be installed on premises legally. Just in case, these devices make a fire, explosion or anything, what would be my escape goat, so I can say I have passed all tests before selling them to customers.
AI: There's different things involved here:
To introduce an electronic device to the EU market, you will need some certifications. I'm not an expert, and you will have to see an import attorney or something similar to get the legal stuff sorted out, but the things from the top of my head are:
CE certification. That's essentially a document that says that you, yourself, guarantee the device adheres to standards. Depending on the type of device, these are different standard. You kind of picked the worst kind of device, it falls into a few of the hardest categories:
consumer device: Needs to be generally safe to handle.
attaches to the power outlets: Most device suppliers elegantly work around this by not building their own power supplies, but buying CE-conformant supplies that someone else builds. For you, that's not an option, because the core of your application is switching outlets. So, make yourself familiar with the regulations that CE has for such devices. I'll be honest here: most pictures of China-sourced relay boards that I've seen probably cannot be certified legally; clearances and creep distances just seem too small.
integrates transmitting radio hardware: Your things are based on WiFi and/or other wireless standards. Thus, the transmitters need to adhere to the radio regulations. You will need to know how to bring your devices into a test mode, and you will need a certified RF lab to test them for conformity with radio regulation. That's costly. You can avoid that by only using non-modifiable, pre-certified radio modules only.
Handling of electronic waste: If you want to sell a device in the EU, you have to pay for the disposal of electronics. It's a rather complex system, but in essence, for every ton of devices you sell, you'll have to pay for the disposal of a ton of devices. You also need to document that all your device's components adhere to RoHS guidelines. That, for example, means that your chinese boards must not be soldered with lead solder, and that the manufacturer guarantees that.
Aside from these mandatory certifications, you said you wanted someone to certify your device is especially safe. That's a good idea! There's TÜV Rheinland which does device testing and certification. The "GS" (for "geprüfte Sicherheit", tested safety in German) logo is a popular one and acts as advertisement logo, too.
As a general note: Generic IOT devices are a hard market where you compete with much larger companies that have very much lower legal and waste handling costs per device than you have. You have no distribution channels – while these devices are sold by the millions through supermarket chains. More than the legal and engineering problem that you'll be facing, I'd be worried about the economic aspects of this. Can you really sell your devices at even remotely competitive prices, competitive reliability and service? Consumer electronics is a hardly-fought battleground with low profit margins, and unless you have a unique value proposition for that market, I'd not try to enter it with something that established manufacturers can produce and sell by the millions. |
H: Fan speed switch: why OFF :: HIGH :: MED :: LOW?
Many AC fans, including inexpensive floor models, have a rotary speed control which rotates from OFF to HIGH to MED to LOW. Since the switch does not rotate 360°, you have to rotate it backwards to shut it off. This has always slightly annoyed me, because it appears to make no logical sense to have to speed the fan up to turn it off.
Why are the speed settings from fastest to slowest? This design seems counter-intuitive, but I assume there is a good reason because the design is so common.
AI: Fans of that type have induction motors with two windings with a capacitor in series with one of the windings. For every individual motor design there is a certain capacitor value that allows the motor to develop maximum torque and operate at the maximum corresponding speed. Smaller capacitor values are used to reduce the torque so that the load overcomes the motor's torque and slows the motor down.
When the motor is turned on, the motor must overcome the static friction of the bearings and get the fan moving. It must then accelerate the motor inertia. That means that using the highest torque setting is desirable for starting the motor. Once the fan is moving, the torque can be reduced for lower speed operation. |
H: Voltage to PWM Circuit, need to understand frequency
This is an off shelf component that might fit my need. Unfortunately the documentation is terrible. It should convert a 0-5v or 0-10v analog to PWM (0-5v/0-10v is a jumper). The analog voltage signal controls the duty cycle. The potentiometer controls "Precision" per the description. I think they mean it can offset the duty cycle from the control signal. I don't care about that, I can program the control signal. This should be perfect for me if the PWM frequency is 21-28KHz.
I can't find/get any info about the frequency. It was cheap enough that I picked one up and started mapping out the traces. I've been trying to learn for a few hours now and I can't quite figure it all out. I think the 555 is controlling the frequency, but this doesn't look like any examples I have found. I want to know specifically which resistors/capacitor are controlling frequency and what it calculates to. I have the ability to rework the board and change them if needed. I don't have an oscilloscope right now, but I'm working on borrowing one. I'm hoping to figure this out and get any possibly needed components before I borrow the scope.
The picture should show connections, sorry if it is ugly. I will gladly clarify anything if confusing. FYI, I just used purple when it would cross over another red line.
AI: Your circuit will look something like this:
simulate this circuit – Schematic created using CircuitLab
Help understanding how to use a SG3525
And the PWM frequency will depend on \$C_2\$ and \$R_6\$ values.
\$Q_1\$ together with \$R_6 , D_2 , D_3\$ forms a constant current source.
And this current will be around \$I_S = \frac{0.66V}{R_6} = 140\mu\textrm{A}
\$
And this current will charge the \$C_2\$ capacitor. And if the voltage across the capacitor reaches the NE555 threshold value (2/3 Vsup). The 555 will quickly discharge the capacitor to 1/3Vsup.
So, the equation for PWM frequency will look like this
\$\Large F \approx \frac{3 \cdot I_S}{C_2 \cdot V_+} \$
So for 5V supply, as you have in your circuit the equation becomes
\$\Large F \approx 0.6 \cdot\frac{I_S}{C_2} \$
I'm trying to simplify all the resistors and jumper leading to U2+
(R9,R10,R1,R3,R4). Suppose I don't need the option to jumper to a 0-5V
input and only using 0-10V input. Also would like to eliminate
adjustability with R1. From testing I know that for best function R1
is set at 8k Ohm between jumper above R3, and 94k Ohm to ground below
R3.
In theory, this voltage divider at the input should do the job: |
H: Contradictory information in LM339 datasheet
I am inspecting the LM339 datasheet (link) and I find contradictory information regarding the operation principle.
Page 10 states:
The output consists of an open-drain NPN (pulldown or low-side) transistor. The output NPN sinks current when the positive input voltage is higher than the negative input voltage and the offset voltage.
But page 12 says:
If IN– is higher than IN+ and the offset voltage, the output is low and the output transistor is sinking
current.
Which one should be considered?
AI: The text on page 10 is incorrect, that is clear when you look at the schematic:
The output NPN sinks current when the positive input voltage is higher than the negative input voltage and the offset voltage.
No, when the IN+ has a higher voltage, more current will flow through the right side of the differential pair (the side connected to IN-), and pull the voltage at the NPN current mirror up, that will switch on the NPN that has the 80uA source connected to its collector and then means that the output transistor will be off. So that NPN cannot sink current at all.
Amazing that there are still errors like that in a TI datasheet. And talking about a drain, hmm. Someone wasn't paying attention. |
H: Refractive index of Infrared light through air and perspex
What will the refractive index of infrared light traveling to air and through perspex be?
So n1 is air and n2 is the perspex, I am using Snell's law to prove that as thetha 1 decreases so will thetha 3. But I need to the refractive indices of IR though air and Perspex first. Are the values for n1 and n2 correct?
AI: According to this site, the index of refraction of acrylic is as follows:
Of course the index of refraction of air is 1.
Edit: By 1, I mean very close to 1 (as it is in a vacuum). If you're interested in the tiny variations from that this NIST site has some equations, but the variations will be much larger in the index of refraction of acrylic, so that will dominate the relative change when the light enters the acrylic from air (or vice versa). Not all acrylics are exactly the same. Wikipedia gives a range of 1.490–1.492 for acrylic, which is around an order of magnitude greater. |
H: Buck converter inductor calculation doesn't agree with typical application schematic
Im trying to play around with some buck converters and have found one that I might want to make a circuit with. The IC I want to use is the the TPS53125 in QFN package. It's a dual output buck converter IC which takes between 4.5 V and 24 V.
Data sheet
When I use equation 3 from the datasheet to find the suggested inductor value for regulating 1.8 V with a max input of 24V and a output correct of 4 A I end up with the value of 4 uH for the inductor.
Working:
4uH is somewhere in the range that I would expect and sounds correct.
My problem is that in the example circuit above they use a 1.5 uH inductor although they try to regulate 1.8 V at a max of 4 A which should use a 4 uH inductor if you follow equation 3, but instead they use 1.5 uH:
So now I'm very unsure who is wrong? Is the diagram using an inductor value which doesn't conform to the equation or is my calculation faulty?
AI: TI provides the sample design targeting 12V Vin. From the datasheet:
This explains the discrepancy between formulas (that assume 24V Vin) and the value of inductor in the example design. They should however mention this fact in their schematics, which boldly says "4.5V to 24 V", and you can try to advise them to correct the schematics.
However, I think the conversion from 24V to 1.8V is unwise, it will suffer in terms of efficiency.
CORRECTION: It is true that 12V V in doesn't change much the result. However, the whole selection of inductor uses just 30% ripple current at max load, and the datasheet has another note:
Note:The calculation above shall serve as a general reference. To
further improve transient response, the output inductance could be
reduced further. This needs to be considered along with the
selection of the output capacitor.
So if you (I mean quality of your output capacitor) can afford 75% current swing in the inductor, then all numbers will align.
I would say this: don't confuse yourself with formulas too much, all simplified formulas are derived under many assumptions, which may or may not be exactly true. Always use the manufacturer's reference design as the main guidance. They build it, thoroughly characterized it, and sell it. More, if you want to avoid surprises, follow their recommended (tested!!!) layouts, and get their particular inductor from their BOM - not all inductors are made equal. |
H: I have a H07RN-F 3G cable which is rated at 400V/16A, can it be used for a normal 220V schuko connection without consequences?
Basically making it a 2.5mm gauge cable, will it have any negative effects on appliances that run on 220V - I am not sure if a 400V/16A rating is the max or the supposed to be used at/for rating?
AI: Voltage and current ratings on cables are always maximum ratings. You can freely use the cable for any current and voltage below its rated values. |
H: Amplifier Output Clipping
When an amplifier output clips at a certain voltage, does that mean the maximum voltage that its output can provide is the voltage value where the clipping occurs?
AI: When an amplifier output clips at a certain voltage, does that mean
the maximum voltage that its output can provide is the voltage value
where the clipping occurs?
It's load dependent. The no-load clipping point is usually the maximum voltage swing but, if you place a several mA load on the output, it will clip at a lower voltage level. Be also aware that feedback resistors form a small output load and, values that "low" means clipping occurs at a lower level. |
H: Z80 RD and WR to RD/WR?
This may sound like a very stupid question. However, I am new to the Z80 stuff. I am planning on how to connect the Z80 control signals to a SRAM. But the Z80 has seperate RD and WR, while my SRAM has them combined. My issue is that the Z80 when is reading for example, instead of just low, it makes a square wave. Any ideas?
SRAM Model: TC551001BPL-70
Datasheet: http://pdf1.alldatasheet.com/datasheet-pdf/view/31687/TOSHIBA/TC551001BPL-70L.html
READ IS HIGH
WRITE IS LOW
AI: Assuming timing conditions are satisfied (it has been two decades since I touched a Z80).
You can connect the ¬WR signal directly to RD/¬WR in the SRAM
You need to connect the ¬RD signal directly to ¬OE in the SRAM
You need to generate the SRAM ¬CE signal for read access AND for write access. Which, requires anding ¬RD and ¬WR together and combining it with address decoding to provide ¬OE to the SRAM (a single multiplexer IC can accomplish all of this).
If there is no ¬CE in the SRAM, that probably means that the RD/¬WR and ¬OE signals need to be generated from address decoding AND both ¬WR and ¬RD.
I believe that covers most cases.
As @Janka points out, the Z80 is one of those processors that has different instructions for addressing memory snd peripherals. This requires an extra line (¬MRQ) to be added to the memory decoding. |
H: What is an expression for the output voltage?
simulate this circuit – Schematic created using CircuitLab
Is this just a voltage divider principle and it boils down to sin(1000t)+sin(5000t) ?
AI: The circuit is linear with ideal sources, so the output voltage is the sum of the contributions from all the different sources.
To calculate the contribution from 1 source, replace all voltage sources, except the one from which you want to determine the contribution to \$V_{out}\$ by a short circuit, and all current sources (I know, there are none, but this is the general procedure), except the one from which you want to determine the contribution to \$V_{out}\$, by an open circuit.
In your example this boils down to first (1) short circuit \$V_4\$, then $$V_{out_1}=\frac{1}{3}V_3$$Next (2), short circuit \$V_3\$ and see that $$V_{out_2}=\frac{1}{3}V_4$$.
Now the result is the (linear) sum of both: $$V_{out}=V_{out_1}+V_{out_2}= \frac{1}{3}(V_3+V_4)$$.
The reason that voltage sources are replaced by short circuits and current sources by open circuits, is that an ideal voltage source has zero impedance, so in fact it would already be a short circuit if the voltage were zero.
An ideal current source on the other hand has infinite impedance, so that if its current is zero, you're left with two open terminals: an open circuit.
So in fact you just summate the effect of each source on the variable you want to evaluate while keeping the circuit exactly the same: all but one source values are zero for each contributing source. |
H: Function of the Third Pin Inside Laptop Barrel Connector
So I've been doing research on barrel pin connectors, and have been trying to apply this knowledge to understand laptop barrel connectors, specifically. My confusion relates to the function of the third pin that is recessed within the "tip" of the connector.
To give some context, I am using my laptop power connector as an example. It runs 2.1A of current at 19.5V. According to the markings on the adapter, the sleeve is positive, while the tip of the connector is negative. But there is an extra pin that is recessed within the tip. What is the function of this pin?
So far, what I have gathered is that there may be a few possibilities:
First, I have seen some diagrams which seem to suggest that some barrel connectors can use this third pin as a mechanism in a switch. So, would that mean that current only flows to the positive sleeve rather than being rerouted back to the ground when contact is made with that third pin? Could this be a sort of safety feature?
I have also seen some explanation in another post (What to do with third contact in DC barrel plug with only two input contacts) involving a DC barrel plug that mentioned the possibility that this third pin may involve some "swtich" behavior not on the end of the barrel connector itself, but instead on the end of the component which it has been "plugged into". As in, is it possible that contact with the third connector inside of the jack causes a switch behavior that rerouts current within the laptop to indicate that it is "plugged in."
Thank you for your help. Unfortunately, I have not been able to put together a clear understanding from what I have been able to find.
AI: Standard DC barrel female jack 3rd pin is a detect or switched pin. When the male plug is not inserted, the 3rd pin is connected with the center pin. This is used to route a signal around the jack, like a battery pack, or to signal to a controller that nothing is inserted. While I have not seen all laptops ever designed, I have seen many and I have never seen a laptop use the 3rd pin of a standard DC connector. The question you link to is talking about those connectors. Which is not the connector you are asking about.
Newer laptops that use different 3 or 4 pin cylindrical connectors are not using old standard DC connectors. These actually use the extra pin for other signalling like power supply identification (Apple, dell, Lenovo use 1-wire or similar standards for this) or maybe dual voltage inputs. These 3 conductor connectors have a center pin, and the outer barrel is two non-connected sleeves, providing two conductors. |
H: What's the frequency used by the remote control of this RC car?
I'm trying to find out the frequency used by my RC car. In particular, what band it runs on (27MHz or 49 Mhz or something else).
It looks like this:
And the back view:
AI: One can assume that the frequency is 2.4 GHz looking at the "2.4G" written on the board. A quick google on he "XL932R4" confirms this. |
H: Identify audio plug
I got an old Siemens RW666 LP player, but the cable is a little short... What plug is this, so I can get an extension cord?
AI: That looks like a 180 degree 5 pin DIN they were fairly common a few decades ago. less so recently.
180 degree describes the angle populated with pins |
H: Printer Data Cables with *suspected* IC boards
I've one found of those printer data cables that have some kind of pair integrated circuit along them, one at each end just before the plugs, I was wondering if there is a less invasive way of determining the specs for them other than slicing their enclosure.
I'm sure it will be fine in doing so, and I know I'm over thinking this just a tad but I just feel uncomfortable with a pair of IC that isn't detected by windows, and if it were some kind of spying boom doogle it would probably have some kind spring/pressure based mechanism that wipes the microcontroller's programming or something of that nature
complete novice in this field apologies for any misuse of terminology and or incorrect terminology.
AI: Those black lumps are noise supression beads especially if they are heavy and stick to magnets.
eg: https://nz.mouser.com/ProductDetail/Fair-Rite/2643540202?qs=sGAEpiMZZMsuct6UGZJC7V3VDI5SybAbaF8BfY3CDFQ%3d |
H: Rough estimation of power consumption in an electronic design
I want to get a rough estimation of the power that my design is going to need. Is it sane to calculate the power of every IC component based on values in DC Characteristics in the corresponding datasheet, add them together and get a rough yet reasonable value or will I be totally off?
EDIT:
DC Characteristics have 3 columns minimum, typical and max. I will use various combinations based on the use of every IC in a given state. I thought it was clear in the original question that I will use values (which include min, typ, max) from DC Characteristics table, not the max values per se, but apparently it is not.
AI: Adding the maxima (from the normal operations sections) of your ICs will yield a (very conservative) upper bound on your current consumption.
To get at a more realistic value you can either measure, or figure out how the actual values vary with the circumstances. Power voltage, temperature, operating mode, and clock frequency often have a big impact.
Start with the list of the power consumptions of all your componets, sorted from high to low. Concentrate of the top entries, you can probably ignore the tail of the list. The 'attack' on the individual components requires experinece, skill and insight, and depends a lot on the type of component.
In a comment you mention the AT25SF041. Its current depends on its mode of operation (power-down, stand-by, active). The conservative estimate is that it is always active (max from the table: 16 mA). If this puts it on top of the list, you need to get more info about how it is used.
If you have a lot of components that attribute roughly the same current, and you can live with < 100% working products, you could use the typical values instead of the max ones, or do a distribution-based addition (assuming some distribution..).
Another interesting aspect is whether you need the average of the peak power consumption. A good design probably needs to take both into account.
PS thumbs-up for using the normal-operation values, be be aware that we see too many questions (and sadly even answers!) here that use the "absolute maxiuma", which are almost never relevant. |
H: Why is earth at 0 V with respect to phase?
In 3 phase transmission lines we have different voltages depending on how we measure it. Phase to phase voltage is going to be 1.73 times more than phase to neutral. So far I understand everything that I've just said. Now, phase to neutral has the same voltage as phase to earth. This means that phase has a higher potential than earth and that earth has a lower potential than phase which would in theory be 0 V. Why is earth at 0 V with respect to phase and not floating?
The only reason I think that earth is at 0 V (with respect to phase) is because at the power station there is a metal pole stuck to the earth which is connected to neutral thus allowing current to flow through the earth and up the neutral pole. If there weren't a pole, current wouldn't flow through earth and earth would not be at 0 V it simply would be floating. Right?
Any help is greatly appreciated.
AI: Yes, in some grounding schemes (eg. TN) N is grounded at the power station end and that is what makes N = E.
In some other systems like IT, N is not grounded and the Ls are floating.
Look up grounding schemes/grounding systems. |
H: How to get the MSB into a logic gate which will check if a number is negative or not?
I need to create a logic gate which will find out whether a number is negative or not.
The input is 8 bits and the output is 1 bit, and if the input is 1 (i.e. negative number) then the output should also be 1, but if the input is 0, then the output should also be 0, so it kinda works like a NOT gate without the inverter.
I've read about the MSB and how it works and that I'd need to feed it in to the input but I'm not entirely sure how to write it. I'm going to be testing this on a HDL, so I'll need to write it a bit like:
Not(in=in[1], out=out[1];
and so on.
Any ideas?
AI: In Verilog you'd just assign out = in[7] and be done.
There might be a gate available in your HDL that does this in one step. It might be named something like Buf or Buffer.
But if you don't have the documentation for your HDL, or there really is no way to simply tell it that one signal is an alias of another, you could cascade two inverters.
simulate this circuit – Schematic created using CircuitLab
As an aside, the language you're using looks more like a netlist language than a full featured HDL. It's probably been designed solely to support your class, and isn't used in industry at all. Your professor or TA is probably the best one to explain the syntax and available gates to you. |
H: How to fix this wifi FPV antenna
I do not know much about antenna's. A wire broke off of a Wifi antenna, being part of a video transmission unit of a quadcopter.
The short grey colored wire on the right side of the picture broke off.
I simply tried soldering it back on. But that does not seem to work. The range of the antenna now is not even 2 meters. The thick shrinkwrapped black part of the antenna is a holow metal tube.
I assume that the center of the antenna is going through the tube, while the outside of the antenna should be attached to the tube. But this is just a guess of mine, I have not much knowlegde about antennas.
The black antenna wire is so small that I can not see a distinct core or shield.
Is this a coaxial wire? How should it be attached to the metal tube? In other words: How could I fix this antenna?
Thanks.
AI: Frequency is not given, but this looks like it could be based on a quarter-wavelength antenna main element (your broken wire), and a quarter-wave sleeve balun element:
Coax cable extends into the left end of the metal sleeve, which is hollow. The left end of the sleeve is electrically floating, unattached to anything. The right end of the sleeve is attached to the coax shield.
Coax centre conductor is attached to the quarter-wave antenna wire, where it appears at the right end of the sleeve. |
H: Identifying strange burnt resistor color code
I'm trying to figure out the resistance of a burnt resistor in the PCB of a Hoover FD22G 011 (a cordless vacuum cleaner, with rotating brush; should be the same as the Deik ZB1516, which is the actual code on the PCB).
The resistor is marked as R27 (so it should indeed a resistor and not a similar-looking component), is between the negative battery terminal (B-) and the negative brush (not vacuum!) motor connector (SB-). It's ~4.5 mm in diameter, so it should be a 1 W resistor.
Despite being burnt (it left some horrible black marks on the PCB, plus the usual burnt resistor smell), some color bands are still visible. I'm quite certain about the black on the left, the red on the right, and I'm decently confident about the gold, second from right.
However, to me this doesn't make much sense, as AFAIK black cannot be neither the first nor the last band in a resistor! Same as the gold one - how can it be second or second to last?
To try to investigate this further, I tried to scratch a bit over the wire to measure the resistance between a terminal and several different windings. Bizarrely, I found very small resistance (under 0.2 Ω per winding, and I don't even know if I should trust my multimeter for such small values), which seems to me quite strange - such a big beast of resistance to introduce a minimum resistance on a DC motor? Doubly bizarre if we think that the crude contacts that bring the current from this PCB down to the brush will introduce more resistance than this...
Any idea?
Update: forgot to say, there's a similar, non burnt resistor in the circuit:
Sorry for the much lower quality photo, I don't have my camera with me now; the colors are a bit altered by the phone camera, but here the rings are definitely black, yellow, violet, silver and (most probably) brown.
The position in the circuit seems similar (it's between the negative battery terminal and the negative fan motor wire), so we can suppose it's a current sensing/limiting resistor as well, but unfortunately it isn't the same resistor - the power rating is lower for sure, as it's a little shorter and thinner (it's ~4 mm of diameter). However, this still sports:
the initial black ring (so, it's not that it's charred in the burnt resistor) - so, maybe it's actually a leading zero?; in that case, this would make it a 0.47 Ω ± 1%;
the shining ring, that here is clearly silver;
the final ring, which is most probably brown, and it may be compatible with the orange-ish tint of the last ring of the burnt resistor.
AI: That resistor is toast for sure, but the spiral 'bump' indicates it is a wire-wound resistor of at least 1 watt. The first band is either red or orange, so if it is 0.33 ohms or 0.22 ohms that would make sense as a series current limiting or current sensing resistor. It may be so damaged that the nickle-chrome wire is no good, or its bond to the end caps has gone bad.
If you look at the top of the burnt resistor the nickle-chrome wire appears to be burnt away, creating a gap, so this resistor should measure as being 'open' or infinite resistance. It could also be in the category of "Fusible" resistor or at least "Flame proof".
If installing it as it is now causes problems I would try both a new 0.22 ohm 1 watt and a 0.33 ohm 1 watt to see which works best. The higher value is best as long as the motor runs at expected speed and torque. Note that this was a 1% tolerance resistor. Such a tight tolerance is normally used for accurate current sensing. |
H: What distinguishes an ordinary thyristor from a GTO thyristor?
A thyristor, I know, is a four-layer PNPN structure, with an anode on the first P section, a gate on the second P section, and a cathode on the second N section. This simple structure suggests that any thyristor ought to be possible to turn off, by routing all of the anode current out through the gate, making the cathode current go to zero, thereby unlatching the thyristor.
In a simulator, a two-transistor model of a thyristor as shown below does indeed turn off when a sufficiently low-resistance path to ground is provided.
simulate this circuit – Schematic created using CircuitLab
And one can purchase thyristors specifically designed to be used like this, called GTO (gate turn-off) thyristors.
So my question is this: What makes a GTO thyristor special? Is it just an ordinary thyristor but with specified characteristics for this mode of operation? Or is there some different silicon structure inside of it that makes it work fundamentally differently?
AI: Interesting question!
Let's start with how we typically use a Thyristor. The Cathode will usually be connected to Ground and the Anode to supply via the load:
simulate this circuit – Schematic created using CircuitLab
So the electrons enter at the Cathode and travel to the Anode.
In the drawings below, the Cathode is at the top! So the electrons flow from top to bottom (only in the doping profiles, not in the schematic above)!
After some searching I found these two drawings of the doping profiles of both devices.
This is the doping profile of a "normal" Thyristor, from this site.
And here is the doping profile of a GTO (same source as above, press Next a few times).
The main difference that I see is that the GTO has an additional P+ region (highly doped P-region) for the Gate contact. Such a highly doped region is used to make a "better", more low-ohmic contact to that doping region.
According to Wikipedia:
Turn off is accomplished by a "negative voltage" pulse between the gate and cathode terminals. Some of the forward current (about one-third to one-fifth) is "stolen" and used to induce a cathode-gate voltage which in turn causes the forward current to fall and the GTO will switch off (transitioning to the 'blocking' state.)
For me that could explain why the GTO can be turned off while the normal Thyristor cannot. In a normal Thyristor the gate doesn't have such a good contact to the top P region which prevents it from diverting enough of the electrons to make the Thyristor turn off.
In a GTO the contact to that P-region is much better so many more electrons can be removed (via the Gate) from that P-region. Also the voltage of this P-region can be controlled much better through the low-ohmic contact. That also allows the Gate to pull down the voltage of this P-region relative to the Cathode which will bias the Cathode (N+) to Gate (P) junction in reverse and blocking the Cathode current. |
H: How can I solve for impedances if given RMS currents?
Given a very simple circuit such as this.
If I where given the RMS values for the three currents, how could I go about solving R and Xl? (the resistor and inductor in paralell)
I know I can not apply KCL using the RMS values, even if I tried the sum of the three currents is not zero (as in total current is not equal to the sum of the currents of the two branches).
I tried getting the peak value from the RMS values and then apply KCL but I'm stuck because the total current and the current that goes thru the inductor will have some unknown angles.
What would be a good strategy to solve this?.
AI: Well, we know that:
$$\underline{\text{Z}}_{\space\text{in}}=\frac{1}{\frac{1}{4}+\frac{1}{\text{R}}+\frac{1}{\text{j}\omega\text{L}}}=\frac{\frac{1}{4}+\frac{1}{\text{R}}}{\left(\frac{1}{4}+\frac{1}{\text{R}}\right)^2+\left(\frac{1}{\omega\text{L}}\right)^2}+\frac{1}{\left(\frac{1}{4}+\frac{1}{\text{R}}\right)^2+\left(\frac{1}{\omega\text{L}}\right)^2}\cdot\frac{1}{\omega\text{L}}\cdot\text{j}\tag1$$
So, know you know that:
$$\left|\underline{\text{I}}_{\space\text{in}}\right|=\frac{\left|\underline{\text{U}}_{\space\text{in}}\right|}{\left|\underline{\text{Z}}_{\space\text{in}}\right|}=\frac{\left|\underline{\text{U}}_{\space\text{in}}\right|}{\sqrt{\left(\frac{\frac{1}{4}+\frac{1}{\text{R}}}{\left(\frac{1}{4}+\frac{1}{\text{R}}\right)^2+\left(\frac{1}{\omega\text{L}}\right)^2}\right)^2+\left(\frac{1}{\left(\frac{1}{4}+\frac{1}{\text{R}}\right)^2+\left(\frac{1}{\omega\text{L}}\right)^2}\cdot\frac{1}{\omega\text{L}}\right)^2}}\tag2$$
And:
$$\left|\underline{\text{I}}_{\space2}\right|=\frac{\left|\underline{\text{U}}_{\space\text{in}}\right|}{4}\space\Longleftrightarrow\space\left|\underline{\text{U}}_{\space\text{in}}\right|=4\cdot\left|\underline{\text{I}}_{\space2}\right|\tag3$$
And we can write:
$$\left|\underline{\text{I}}_{\space1}\right|=\frac{\left|\underline{\text{U}}_{\space\text{in}}\right|}{\sqrt{\left(\frac{1}{\text{R}}\cdot\frac{1}{\left(\frac{1}{\text{R}}\right)^2+\left(\frac{1}{\omega\text{L}}\right)^2}\right)^2+\left(\frac{1}{\omega\text{L}}\cdot\frac{1}{\left(\frac{1}{\text{R}}\right)^2+\left(\frac{1}{\omega\text{L}}\right)^2}\right)^2}}\tag4$$
And remember that:
$$\left|\underline{\text{I}}_{\space1}\right|=\sqrt{2}\cdot\text{I}_{\space1\space\text{rms}}\tag5$$
And that:
$$\omega\text{L}=\text{X}_{\space\text{L}}\tag6$$ |
H: PID Controller Subsituition
I just have a quick question. In my lecture notes, he defined the PID controller as
And then he said that this is equal to the following
How did he do that? What is k, z1 and z2 equal?
AI: Okay:
$$\begin{align*}
k_p + \frac{k_i}{s}+k_d\:s&=\frac{k\left(s+z_1\right)\left(s+z_2\right)}{s}\\\\
k_d\:s^2+k_p\:s + k_i&=k\:s^2 + k\left(z_1+z_2\right)s + k\:z_1\:z_2\\\\\therefore\\\\k&=k_d\\\\
k\left(z_1+z_2\right)&=k_p\\\\
k\:z_1\:z_2&=k_i
\end{align*}$$
If you accept that \$k=k_d\$ solves for \$k\$, then that's two remaining equations and two remaining unknowns:
$$\begin{align*}
z_1+z_2&=\frac{k_p}{k_d}\\\\
z_1\cdot z_2&=\frac{k_i}{k_d}
\end{align*}$$
Can you move forward from here using the quadratic equation? |
H: Why are there so many vias on this board?
I was looking at the MMZ09312BT1 development board layout, and I was curious about all the holes they have on the board. Are these vias? What is their purpose (I heard somewhere that they are meant as a filter)?
Also it doesn't say explicitly, but is it possible to tell if they have a ground plane on the bottom layer?
Datasheet: http://cache.freescale.com/files/rf_if/doc/data_sheet/MMZ09312B.pdf
Development board on page 8
AI: This is generally referred to as via stitching, and it's generally used to reduce either the high-frequency electrical impedance or the thermal resistance between layers. It can also be used to provide a low DC resistance path between layers for high current pathways. In this case the reason is certainly RF impedance, however the level of stitching shown is probably overkill even for a 900MHz RF part. However it's easy to do, and doesn't generally hurt anything on a board as sparsely populated as this one.
You would need to consult the design documents to determine the stackup details if the layers aren't clearly visible. Often for dev/eval boards the manufacturer will provide a full package of manufacturing documents. |
H: How MCU provides an output to pins
I would like to know how does MCU provide it's output current. here is a little drawing of my assumption (simplified) of how a pin is functioning when it is set to output.
Basically it is a transistor that allows current to flow or it doesn't. it cant be something else, but what actually bothers me is V in to the mcu. Does MCU provide an output using Vcc input directly or this input is going through some other internal components.
If this is a transistor and 5v goes through it, do i get that Vce drop ? which is 0.2V. and therefore i actually get 4.8 volts on my pin ?
If i am performing some calculations how would apply a Kirchhoff rules to a circuit with micro controller if i don't know nothing about its internals, only datasheet data.
AI: Somewhere in the literature for most processors there will be representative schematics for the pins. They can be complicated, but usually with a pin set up as an output it boils down to a push-pull CMOS output stage (sometimes open-drain).
Usually this is found in the MCU's user's manual -- the data sheets are usually at a higher level.
The reason for no drop is -- as stated -- because microprocessors these days are CMOS. I can't think of one that's not. |
H: Attaching a decoupling capacitor to PCB. What should go where?
I'm nearly finished with the recreation of a PCB that'll be used in creating a modchip for my PlayStation 1 that is gathering dust. I'm basing it off a modchip I bought last year but accidentally threw out while cleaning and contacting the seller to purchase a new one now is difficult.
Right now I have the capacitor connecting to the pads that are connected to where pin1(VDD) and pin8(VSS) of a PIC12F508-I/SN will be mounted. I have absolutely no idea if this is correct and if it is not where should I be connecting it to?
AI: You have the decoupling capacitor connected to the correct pins on that IC, that's for sure. However, it is a good rule of thumb to have a decoupling capacitor as close as possible to the pins on the IC it is decoupling (for one source, look here http://www.capacitorguide.com/coupling-and-decoupling/, though nearly every article you read will talk about how they need to be close to the IC).
However, if your board needs to be single layer and/or it needs to be that size, I wouldn't worry too much.
What capacitance have you chosen? |
H: Power Dissipation of Op-Amp
If we look at the output stage of the MP111 op-amp it's just 2 MOSFETs in series. I'm curious why the power dissipation of the device isn't simply \$R_{\text{ds(on)}} \times I_{\text{out}}\$, but rather \$I_{\text{out}} \times (\text{supply voltage} - \text{output voltage})\$
If we had a MOSFET in a box, the power dissipated by that 'box' would still just be \$R_{\text{ds(on)}} \times I_{\text{out}}\$ wouldn't it?
Datasheet: https://www.apexanalog.com/resources/products/mp111u.pdf
AI: \$R_{\text{ds(on)}}\$ is relevant when the MOSFET is fully on, in which case the channel resistance can be considered to be constant, but in that case the power dissipation would actually be \$R_{\text{ds(on)}} \times {I_{\text{ds}}}^2\$, which you can derive from Ohm's law and \$P = IV\$.
However, this is not relevant when the MOSFET is operated in the linear region, especially when there is feedback that regulates the voltage across the MOSFET, as in an op-amp output stage. Here channel resistance is a limiting factor in the maximum output voltage range with respect to the power supply as a function of output current, but it is not a useful parameter in determining power dissipation in the part. This is because the equivalent channel resistance changes as the surrounding circuity manipulates the gate voltage to hold the circuit's output voltage (which is directly related to the MOSFETs' \$V_{\text{ds}}\$) constant. This is not really an accurate way to understand MOSFET operation, but it's good enough to understand why \$R_{\text{ds}}\$ is not useful here.
Since \$V_{\text{ds}}\$ is controlled, we can simply multiply that by \$I_{\text{ds}}\$ and get the power dissipation in each MOSFET in the output stage. Note that this isn't quite the same thing as saying that the power dissipation is \$I_{\text{out}} \times (\text{supply voltage} - \text{output voltage})\$, because we have to consider the direction of current flow. If the circuit is sourcing current into a load, than the current is flowing from the positive rail through the high side transistor, and so \$P = (V_{S+}-V_{\text{out}}) \times I_{\text{out}}\$. Conversely, if the circuit is sinking current, then current is flowing out of the load, through the low side transistor, and into the negative rail, so \$P = (V_{S-}-V_{\text{out}}) * I_{\text{out}}\$. Thus if the output voltage is close to the positive supply then the device will dissipate more power when sinking current than when sourcing current. |
H: How to determine op-amp gain with active feedback?
The Howland current pump uses an op-amp in the configuration below with a resistive feedback network which gives me the gain show below:
But if I decide to swap the feedback resistor for an instrumentation amplifier for less noise and better resolution, what will the new gain be? I've tried searching but can't seem to find anything on this.
simulate this circuit – Schematic created using CircuitLab
AI: What you need to do is add a couple of resistors
simulate this circuit – Schematic created using CircuitLab
If the instrumentation amp has a gain G, then, since the current through R1 must equal the current through R2,
Vin/R1 = G iL Rs/R2,
where iL is the load current.
Rearranging the terms gives
iL = Vin(R2 /R1 G Rs)
Note that, strictly speaking, an instrumentation amp is not required, since Rs is grounded, and a simple non-inverting op amp would do the job. In practice, an instrumentation amp would be a good idea, since tiny differences in ground resistance will have a noticeable effect due to the large gain of the amp.
Also note that this configuration will almost certainly oscillate like crazy. The phase shift caused by the instrumentation amp will need careful compensation. |
H: Choosing the right transformer to replace batteries?
I have an idea for a hack I'd like to build, but I need some help figuring out the right transformer. Let me know if there's a better place to post this sort of request.
I have some exterior LED path lights that work on 4 C-Cell batteries each. There are 6 of them, so having to replace the batteries is relatively expensive and a bit of a pain. Looking at the lights, I see that I could easily solder wires to the battery contacts and replace the batteries with a transformer. I could run a low voltage wire indoors to a transformer and never have to worry about batteries again. My problem is that I'm not sure what transformer to get.
I did some simple math but I have no electronics background, so I would appreciate getting guidance from folks that have more knowledge about this stuff.
Here's what I know:
Each light takes 4 C-cell batteries in line. That is they are stacked one on top of the other with positive terminal touching negative. I think that means I'd need to add the voltages together meaning 4 x 1.5V = 6V required.
I have no idea how much power each light draws (specs don't say anywhere) and the math I did doesn't seem to make sense. I know that alkaline batteries are normally rated for about 8000 mAh. The manufacturer of the light claims that the lights last about 50 hours before needing the batteries replaced. 8000 / 50 = 160 mAh draw. That seems really low to me. Is 160 mAh a reasonable amount for an LED light (supposedly 80 lumens output if that helps)?
So if I multiply that out 6 * 160 = 960 mAh required. I want to be able to add a few lights so let's double that (so just under 2A).
Does that mean I need a 6V transformer that can do 2 amps? That seems really low to me, so I'm hoping someone with electronics knowledge can provide some help figuring out what I really need.
Any guidance or nudges in the right direction would be greatly appreciated.
Thanks.
AI: 160mA at 6V is about 1W that would be easily suffucuent for a fairly bright LED path light.
If you're not equipped to measure the current consumption of the lights
maybe go for a 2A 6V supply that should be plenty if you estimate is in the right ballpark.
You could probably even use a 5V 2A "phone charger" as battery powered devices are usually designed to operate from as low as 1.1V per cell so 5V should be plenty. |
H: How can relays be rated at 10A for both 240VAC and 120VAC?
I have a small W1209 DC12V thermostat PCB that's had a small 12VDC relay attached.
The print on the relay says that it can do 5A @ 240VAC and 10A at 120VAC.
That makes sense, as both of those equal 1200 Watts.
However I'm wanting to wire it to a 2000 watt load at 240VAC. So have been looking at other relays I can replace the one on the PCB with.
And I'm seeing a fair number of relays which rather than having an 10A @ 240AC & 5A @ 120VAC rating they have a 10A @ 240VAC & 10A at 120VAC
So how is that possible? Is the relay that has 10A for both built for 10A @ 240, and 20A @ 120 too much for it, or is the rating fudged somehow?
AI: There are a large number of design elements that affect the current and voltage ratings, technically, and then there are the approvals they choose to get. Many relays have multiple ratings, depending on the safety standard to which they are tested.
Generally "watts" is not the limiting factor, it's amperes for heating, and the voltage for contact opening. The wear on the contacts can also be limiting and that's affected by factors such as the inrush current of the load and any reactance (eg. inductance) in the load.
Generally small relays are rated for some number (100,000 is typical) life at full load, and you can derate that (eg. operate a 20A relay at 5A) to get a better lifetime.
It's not that common to find a drop-in relay that will give you substantially more life or current rating with the same load and contact metallurgy-- relays are fairly well optimized. You'll usually find the higher rated relay will draw substantially more coil current (perhaps that your existing circuit cannot safely handle) or will be substantially larger or both.
Your best bet may be to operate a larger relay or contactor with your small relay. |
H: Arduino MKR Vidor 4000
Can it capture the image in full Rpi camera resolution and save it somewhere?
Thank you for your time.
AI: The Raspberry Pi camera is an 8-megapixel sensor.
The Arduino MKR Vidor 4000 has 8 MB of on-board SRAM.
So the answer is no, with only 8 bits per pixel, the board cannot store an uncompressed image from the camera. It would normally require 2× (YUV 4:2:2) to 3× (RGB 4:4:4) that amount of memory. The best you could do would be to store a "raw" (Bayer-encoded) 8-bit image for post-processing elsewhere.
BTW, while the board does contain a relatively powerful SAM D Cortex-M0+ based microcontroller, the primary feature that gives it its real power is the FPGA — this is going to require knowledge of hardware design (and HDL) in order to make use of it.
What are you trying to do? For most applications, it would be easier (and cheaper) to plug the camera into an actual RPi. |
H: Calculating input impedance of 3 port network
I am trying to derive a expression for calculating the input impedance of a 3 port network to use as direct calculating code and avoid SPICE/simulator solving of the same.
I am able to solve the input impedance of a 2 port system with a load, \$Z_{load}\$, connected to port 2. The impedance looking into port 1 would be (by solving basic 2 port theory):
$$
Z_{in} = Z_{11} - \frac{Z_{21}Z_{12}}{Z_{22}+Z_{load}}
$$
If I have a 3 port network with 2 ports connected to different loads, \$Z_{load1}\$ and \$Z_{load2}\$, how can I generate an expression for \$Z_{in}\$ looking in from port 1 starting from only the Z-matrix, \$Z_{load1}\$ and \$Z_{load2}\$?
AI: The 3-port can be described in 3 equations, using
$$\left(\begin{matrix}
V_1 \\
V_2 \\
V_3
\end{matrix}\right) = \left(\begin{matrix}
Z_{11} & Z_{12} & Z_{13} \\
Z_{21} & Z_{22} & Z_{23} \\
Z_{31} & Z_{32} & Z_{33} \end{matrix}\right)\left(\begin{matrix}
I_1 \\
I_2 \\
I_3\end{matrix}\right)$$
I will now load ports 1 and 2, while using port 3 as the input. Adding two loads adds two new equations
$$\begin{align}
V_1 &= -Z_{L1}\cdot I_1 \\
V_2 &= -Z_{L2}\cdot I_2
\end{align}$$
This can be inserted into the matrix notation:
$$\left(\begin{matrix}
Z_{11} & Z_{12} & Z_{13} \\
Z_{21} & Z_{22} & Z_{23} \\
Z_{31} & Z_{32} & Z_{33} \end{matrix}\right)\left(\begin{matrix}
I_1 \\
I_2 \\
I_3\end{matrix}\right) = \left(\begin{matrix}
-Z_{L1}\cdot I_1 \\
-Z_{L2}\cdot I_2 \\
V_3
\end{matrix}\right)$$
Which is the same as
$$\left(\begin{matrix}
Z_{11} + Z_{L1} & Z_{12} & Z_{13} \\
Z_{21} & Z_{22} + Z_{L2} & Z_{23} \\
Z_{31} & Z_{32} & Z_{33} \end{matrix}\right)\left(\begin{matrix}
I_1 \\
I_2 \\
I_3\end{matrix}\right) = \left(\begin{matrix}
0 \\
0 \\
V_3
\end{matrix}\right)$$
Since we want to solve for \$Z_{in} = \frac{V_3}{I_3}\$, we can use Cramer's rule to find
$$
I_3 = \frac{\left|\begin{matrix}
Z_{11} + Z_{L1} & Z_{12} & 0 \\
Z_{21} & Z_{22} + Z_{L_2} & 0 \\
Z_{31} & Z_{32} & V_3
\end{matrix}\right|}{\left|\begin{matrix}
Z_{11} + Z_{L1} & Z_{12} & Z_{13} \\
Z_{21} & Z_{22} + Z_{L2} & Z_{23} \\
Z_{31} & Z_{32} & Z_{33} \end{matrix}\right|}\
$$
Or also
$$
Z_{in} = \frac{V_3}{I_3} = \frac{\left|\begin{matrix}
Z_{11} + Z_{L1} & Z_{12} & Z_{13} \\
Z_{21} & Z_{22} + Z_{L2} & Z_{23} \\
Z_{31} & Z_{32} & Z_{33} \end{matrix}\right|}{\left|\begin{matrix}
Z_{11} + Z_{L1} & Z_{12} & 0 \\
Z_{21} & Z_{22} + Z_{L_2} & 0 \\
Z_{31} & Z_{32} & 1
\end{matrix}\right|}
$$ |
H: How does Ethernet to CAN conversion work?
I am wondering how an Ethernet-to-CAN converter works. CAN is a multi-master broadcasting network where messages are associated with IDs. Each message type has a unique ID. Devices listen to all messages on the bus and decide whether an ID is of interest. Ethernet, however, uses point-2-point addressing, meaning that sender and receiver are specified. These information cannot be mapped on CAN, because there is not sender or receiver field.
Let's say, device A sends message m to device B. Device B is connected via CAN to the converter. Device A communicates over Ethernet.
How does the CAN-to-Ethernet converter select the message ID when converting an Ethernet frame to a CAN fame? Are the same CAN IDs used for packets to the same recipient?
How does the CAN-to-Ethernet convert a CAN message to an Ethernet frame? A CAN message has no recipient, but an Ethernet frame knows the recipient address. How does the mapping work?
AI: It varies as per product manufacturer but generally speaking,
CAN-to-ETH converter reads all the messages from CAN Bus, packs every message as payload of ethernet datagram and sends it to computer. This payload contains CAN message ID, DLC and data.
On Computer, you can configure to filter out messages as per your requirements. Usually a DLL is provided and you can build your own application. Some even provide application where you can monitor the bus.
We use CAN-ETH converter from ProconX. Their user manual contains this simple diagram which will help you to understand:
Again, it depends on manufacturer and you can find details in the manual. |
H: Question related to the interference seen at the spectrum analyzer's receiver
I am preparing for the RF course exam. I have an old question on hand that I am trying to solve. Please review my attempt.
Question:
A simple spectrum analyzer with superheterodyne receiver having 10.7 MHz intermediate frequency is given. It works well unless the analyzer is used to analyze a signal on 318.6 MHz by setting the analyzer to 318.6 MHz center frequency, 5 MHz span, and 10 kHz RBW. Before we switch on the desired transmitter, the analyzer detects a clear signal at the center frequency. Later, when we learn that there is a 340 MHz transmitter in the same antenna tower you immediately realize what happened.
1) Explain the "issue".
2) Improve the design of the spectrum analyzer that protects against this kind of problem.
My attempt
Does the below arguments make sense to you? Thanks.
a) The adjacent channel interference generated by the carrier at 340 MHz frequency could be leaked to the spectrum analyzer receiver at 318.6 MHz.
b) The IF bandwidth of the receiver can be narrower in order to reject the adjacent channel interferences.
AI: Neither (a) nor (b) are correct as far as I see it.
It's highly likely that the question is trying to get you to realize that the local oscillator is set at 329.3 MHz (or 10.7 MHz above 318.6 MHz) and, will be precisely 10.7 MHz below an interfering signal at 340 MHz i.e. both 318.6 MHz and 340 MHz will be down-converted to a signal at 10.7 MHz and you will not know whether you are measuring 318.6 MHz or 340 MHz.
This is why decent spectrum analyzers uses several tiers of frequency-decreasing IF stages.
For instance, if your spectrum analyzer was designed for measuring input signals from DC to 100 MHz, the first local oscillator frequency might be set at 250 MHz. This results in an up-conversion of wanted signals into the range 250 MHz to 350 MHz.
The impact of this is that a potential interfering signal of (say) 150 MHz would produce a difference frequency of 100 MHz and would be outside the wanted up-conversion range. In fact, only when an interferer of 500 MHz was present would the difference frequency be at the edge of the desired range (250 MHz).
So, the spectrum analyzer would apply a reasonable low pass filter on the signals it measures such that anything above 100 MHz would be attenuated and, by the time 500 MHz is reached, the attenuation would be over 100 dB typically. This could be an 8th order filter having an attenuation of 48 dB/octave; 200 MHz would be attenuated by 48 dB; 400 MHz would be attenuated by 96 dB etc.. |
H: USB hub and upstream/downstream speeds
I am designing a USB hub that has a 2.0 upstream connection (pc is only 2.0). To one of the downstream ports I am connecting a Ethernet to USB IC capable of USB3.0 speed and Gb ethernet.
My question is...does it make sense to have a USB3.0 hub IC or a 2.0 will be more than enough ? Will I get any boost in performance with a 3.0 hub ic ?
AI: No.
The 2.0 upstream connection to your PC will be your bottleneck. Even with 3.0 hardware downstream, the devices will only be able to transfer data to the laptop at a 2.0 rate. Don't spend extra money or time implementing USB 3.0. |
H: How is a load cell made?
Someone know how a common cheap 4-wire load cell is made?
I'm referring at one that it is in the cheap Chinese scale, like this.
In particular, using a multimeter, I would try to identify the structure of load cell for simulation purpose.
I know that there is the Wheatstone bridge under the load cell, but I'm unable to identify correctly the shape of it (eg how many strain gauges are in the load cell) and the resistor values.
What kind of measures is useful to identify this kind of load cell?
AI: A typical 4 wire load cell has 4 resistors in a bridge.
They are wired so that opposing resistors are to experience the same strain and adjacent resistors are to experience the opposite strain. This results in the excited load cell being forced out of a balanced condition under load and a voltage to appear on the sense pair under load.
When measuring a load cell you can find opposite terminals by measuring the largest values between two terminals. This will be equal in value to the element resistance R=(R+R)||(R+R). If you measure across two adjacent terminals you should read a value of R'=R||R+R+R=0.75R.
If the top and bottom pairs are not equal in value you will have a much harder time determining the exact values as you will have to take more readings and possibly short out other terminals to gain enough known values to calculate the unknowns. It is certainly possible but less fun.
Once you have found the opposite pairs of terminals you need to provide a excitation current and then see which polarity the sense pair supplies and determine the gain of the bridge with applied maximum load. If you plan to use some cheap devices with little documentation then I would purchase a bunch, and test them to load and current and temperature destruction (dielectric insulation too), check the gain and resistance tolerance and yield and THEN if they meet your need I would order a lifetime supply hoping to receive the same type and redo the tests in case they are from a different batch.
If you need to reorder do so from a reputable source with a data sheet and pay for them from the profits of your first successful run. |
H: Matlab Three Phase System Model
I have a question regarding MATLAB Simulink if anyone is familiar with the software.
I usually have no problems creating basic models like this one below:
I know the image is rather small.
It is basically 3 phase generator with internal series impedance > VI series measurement block > series ammeters > RL series transmission line impedance > RL series load with load volt meters.
But however, sometimes when I run my circuits and enter my oscilloscope, I don't see a pure sine wave. It appears rather choppy. I get the correct voltage and current values in my displays as I have calculated them also. I am nearly sure all my parameters like frequencies, base voltages etc are all typed in correctly.
This is what my scope looks like:
Any tips or ideas?
Thanks.
AI: The issue is the final display not the actual data.
If you are using a fixed step solver then use a small timestep. If however you are using a variable timestep then you need to inform the solver to render/calculate additional datapoints.
Simulink is aware what hte actual signal is as it is an integration engine and is sensitive to zero-crossing. In rendering a plot it has deemed this is all the information that is needed to represent the data.
This does produce unexpected results.
Consider a 100Hz sinewave as below. Not very sinus
What is needed
What is needed is to change the default refinement factor IF you are using a continuous solver.
The defaults is 1 so change it to 100
You now have the expected waveform. What this does is requests the solver to generate additional points between the datapoints it needs to realise the waveform.
A good explanation is found at "guy on Simulink" blog. https://blogs.mathworks.com/simulink/2009/07/14/refining-the-output-of-a-simulation/ |
H: How to add the TVS protection?
I'm still busy with my DMX splitter (soldered everything), however, I made a mistake with my TVS protection diode. I put it in series with 220 V AC, and nothing happens (no current flows).
I expect I have to put this diode between the two AC wires, but before I mess up with 220 V AC I rather hear some feedback.
What I have now is:
simulate this circuit – Schematic created using CircuitLab
(I could not find all symbols, so I misused a few):
V1: 220 V AC (Dutch power socket)
F1: Now fast blow, later slow blow 0.25 A
F2: Resetable temperature fuse, 73 degrees
D1: Bidirectional 440 V TVS resetable diode
Load: 5 x 600 mA 5V AC/DC converters
I expect I need to use the following circuit:
simulate this circuit
Can this be confirmed?
Also, does the order of F1, F2 and D1 matter?
AI: Your second circuit is right. The bidirectional suppressors are really two diodes back to back so that one has to break down and the other forward conducts to pass current during an overvoltage.
The order of the fuses doesn't matter. |
H: Why does this power opamp have MOSFETs at the output stage
I saw a post a few days ago.
Power Dissipation of Op-Amp
where I was actually curious about the circuit where they use BJTs all over the circuit but use 2 MOSFETS for the output stage.(also why are they using JFETs?)
In most designs that I have seen online, only BJTs are used.
Is there any advantage of using MOSFETs instead of BJTs for the output?
Are MOSFETs not less linear than BJTs?
If so what is their advantage?
It would be great if someone can shed some light on it.
AI: The reason could be partly due to the absence of second breakdown. In BJTs the power dissipation capability is often reduced at high voltage as the current tends to be concentrated in localised hot spots. Some MOSFETs do not suffer from this. With this device the safe operating area graph shows a straight line up to 100 V when the device is dissipation limited. |
H: Where does the negative sign come from in this transfer function?
I have a transfer function that was given to me. It's been a really long time since I've done block diagram math, so sorry if this is a newbie question.
Based on the block diagram below (I forgot to edit the image, but Tin is the output y, that last arrow)
I am give the equations
$$T{_i}{_n} = T{_d}{_i}_s{} - T{_m}{_o}{_t}$$
$$G{_r}{_e}{_f}{\rightarrow}{_i}{_n} = \frac{ -C(s)M(s)}{1 - C(s)M(s)}$$
I can't for the life of me figure out where the two negative signs come from in these equations. Can anybody point me in the right direction as to where these come from?
AI: The output signal: \$\quad y = d-C(s)M(s)e\$
The error signal: \$\quad e = r-y\$
$$
y = d-C(s)M(s)[r-y]
$$
$$
(1-C(s)M(s))y = d-C(s)M(s)r
$$
From superposition principles, take \$d\$ as 0 and you get the transfer function from \$r\$ to \$y\$:
$$
y = \frac{-C(s)M(s)}{1-C(s)M(s)}r
$$
As can be seen, the minus sign comes from the first equation. If you were to redefine \$y = d+C(s)M(s)e\$, then the following equation would hold:
$$
y = \frac{C(s)M(s)}{1+C(s)M(s)}r
$$ |
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