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H: Edison effect in vacuum
I was reading this page which says:
What Edison discovered (and it was promptly dubbed the "Edison
effect") was that electrical current doesn't need a wire to move
through. It can travel right through a gas or even a vacuum.
Maxwell introduced the term displacement current for explaining how the capacitor is passing an alternating current through its plates.
But what I understand is that the displacement current does not move in the form of electrons, but it moves/passes in the form of circulating magnetic fields up to the other side of the plate.
The paragraph claims that the electrical current can move in vacuum. But it doesn't state the form of the current.
For the Edison effect in vacuum, is the current moving in the form of actual electrons or magnetic fields?
AI: In Edison effect in vacuum, is the current moving in the form of
actual electrons or magnetic fields?
The current that Edison was observing was due to an electron current with electrons moving through a vacuum. The effect in vacuum tubes is from one side of the tube being heated to thermally excite electrons off of a metal plate, where they can transfer charge to another metal plate. So to answer your question the charge was due to actual electrons.
This is different than the fields Gauss was noticing, because he was experimenting with capacitors, the displacement current is due to electric and magnetic currents changing in the capacitor. |
H: Is it possible to charge a li-ion battery with brushless motors?
Can I charge a li-ion battery pack with the help of extra brushless motors(with a rectifier) attached to the front wheels of my electric skateboard. Will it work? Anything that I should know before doing this? And also if I sent the power coming from the motors into the battery through a BMS, will it charge the batteries properly?
Thanks!
AI: If you're thinking about "free energy", or, collecting "wasted" energy from the movement of the board, forget about it, because it's impossible.
If you're thinking about collecting some energy (even if it actually brakes the board's movement), the answer is theoretically, yes.
But I'm sorry to say that building this would be very difficult, or even impossible. The mechanical assembling would be critical; you would have to use gears to adjust the rotation of the motor. It could make harm to your equipment. And you would collect so little energy at all that it would be practically no gain.
There's no way for an amateur building a KERS system.
Anything that I should know before doing this?
Yes, there are a lot of things that you can learn about this, the internet is full of tutorials about electronic, free energy, etc. I think that as soon as you learn more things, you'll become suddenly aware that this project is not worth your efforts. |
H: Which relay will last the longest?
I'd like to control a 20 amp, 240 volt AC resistive load with a heavy duty contactor.
The contactor I have is a Siemens 3TX7130-0DB03 rated for 30 amps, and DPDT. I was originally going to switch both lines on this relay. But I got to thinking...
Would it be easier on the contactor to use both poles in parallel to switch the same line, and leave the other line always powered?
What theory or research would support that conclusion?
In short...
simulate this circuit – Schematic created using CircuitLab
Which of the above relays will last longer, why, and what other pros and cons are there?
AI: Interesting question.
In terms of switching there is no real difference. When switching a contactor/relay one contact will always make or break first/last. Or to put it another way, only one contact actually does the switching. That is why you should never double up contacts to try to turn 30A relay into a 60A unit.
So, what else is there?
Contact resistance. When closed the circuit on the right shares the load current equally through both contacts and will therefor drop half the voltage and heat less. In the left circuit, both contacts must carry the full load so you end up with twice the voltage drop compared to a single contact.
There is one more factor. Isolation. In the left circuit when the contactor is off you have full isolation between the load and the supply. The right circuit you have only disconnected one line. The left circuit is therefore safer.
So in other words it comes down to do you prefer isolation versus quarter the voltage drop and a cooler contactor.
ADDITION:
Since folks brought it up in the comments, here are the fault scenarios, not including coil failure.
From this it is evident the series connection is possibly better if you need it to fail off. |
H: Understanding Audio Jack Connection
I have circuit that has an audio connector that has the following diagram.
Audio connector is one that I am not familiar with. What I gather is that the rectangle on the left is the sleeve (entry point).
The tip is pin 2.
The ring is pin 3.
The sleeve is pin 1.
Is this correct or did I misunderstand how to read this symbol ?
Added:
Link to product page
Datasheet
AI: TL;DR; You are correct in your determination of how the contacts will make connections to a TRS jack (see diagram below).
Based on the datasheet, the jack is indeed designed for 3-pole TRS jacks. However on examination we can see that the shell of the connector is plastic. We can also deduce from the measurements that the contact for pin 1 is not technically the sleeve contact.
Instead pin 1 is equivalent to the second ring contact on a 4-pole TRRS socket. This explains why in the connector symbol they show pin 1 connected as a third arrow as opposed to a wire joining to the sleeve (the rectangle). From the diagram below of a TRRS jack, each of the arrows on the symbol corresponds to one of the Tip, Ring 1, and Ring 2. The box corresponds to the sleeve.
If we examine the difference between the various types of jack (specifically TRS and TRRS ones), it can be seen that the connector will still work fine for a TRS jack. This is because the contact at the second ring position will in fact contact the sleeve on the TRS jack.
Note:
This type of connector could be used fine in audio for headphones that use a TRS jack - i.e. just headphones, and no microphone.
The connector could also be used with TRRS jacks using the AHJ or CTIA standard where the ground is located at the second ring.
You could not use this connector for TRRS jacks wired for the OMTP standard where the ground connection is the sleeve, because pin 1 would not make contact. |
H: Classify Signals (Static vs Dynamic) Homework Assignment
For the following assignment, I only want to see if I am on the right path when answering the question.
The following question ask:
let x(t) be the system input and y(t) be the system output. Determine whether the system is static vs. dynamic and causal vs. non-causal.
1) y(t) = 2x(t-1) + 4x(t-2) + 3x(t)
2) y(t) = 10\$x^2\$(t)
I started off by plugging t = 0 in equation 1, which results in:
y(0) = 2x(-1) + 4x(-2) + 3x(0)
From this the first two terms on the left side of the equation depends on past inputs. For this reason, I believe the system is dynamic and casual as the output depends on the past and present inputs.
For the second problem, I also plugged in t for 0 for:
y(0) = 10\$x^2\$(0)
I believe this is static and casual(output depends on only the present value).
AI: A system is static iff the output at time t depends only on the input at time t
A system is dynamic iff the output at time t depends on the input at time t as well as the input at other times (i.e. in the past or future)
A system is causal iff the output at time t depends only on the the input at times at or before time t (i.e. not future values of the input)
A system is anti-causal if its output depends on the input at times after time t. |
H: AVR MCU frequency vs VCC
Why is the max frequency of AVR MCUs a function of VCC? Maybe they have constant current sources that starve as the power demands go up?
Is this relationship unique to the AVR line, or is this typical?
Edit: I am not looking for a quantitative figure or equation. I realize the function exists, and have the datasheets. I'm asking why it exists.
AI: Generally CMOS circuits are not as fast when the Vdd is reduced (within design specifications). The drain current goes down with less gate voltage and a given Vds, so the load capacitances cannot be charged and discharged as quickly. So it is not unique to the AVR, although it may not always be shown explicitly since the specifications usually show only the guaranteed operation region, not the typical. If you plotted the typical region of operation (assuming you had some way to actually fully test the chips) you would see a smooth curve rather than the typical piecewise linear relationship.
For example, here is the specification for another Microchip (this one a PIC) part:
That is assuming the limitation is based on switching speed. It's possible there are thermally limited chips that cannot be operated at 100% duty cycle at higher voltage even if they can switch that fast- because the heat cannot be removed fast enough. Power consumption (to be specific, the dynamic portion of power consumption) increases with the square of supply voltage, all other things being equal. A little 8-bit CMOS processor is not going to be thermally limited though. |
H: Single supply OPA445?
So basically I'm trying to make a circuit for a piezo-electric actuator. I only have access to a single supply at +70V for my op amp, an OPA445AP. Vin varies between 0 and +3V. I've looked at the help given by TI in this PDF and have come up with the following circuit. I've implemented it but when I power it up, Vout saturates at around +68V regardless of Vin. When I've simulated the circuit it indicates that it should work as expected.
Have I messed something with my circuit design up or is the OPA445AP not suitable for use with a single supply? Thanks in advance for your help.
Matt
simulate this circuit – Schematic created using CircuitLab
AI: Figure 1. From the OPA445 datasheet page 2.
This is telling you that the inputs don't work within 3 V of the supply rails. |
H: How to turn an SPST momentary push button into a momentary DPDT?
How to turn an SPST momentary push button into a DPDT? I would like to close 2 circuits at the same time. (From comments: I want to ground the reset circuit while getting a pin to HIGH on an Arduino like board.)
SPST
DPDT
Image source.
AI: Based on the info in your comments. You do not want a DPDT switch, but a resistor and an invertor.
simulate this circuit – Schematic created using CircuitLab |
H: Calculate Collector Voltage In Collector feedback bias Configuration
I've tried using KVL and KCL but I always end up with two or more variables, and I've got another question, how can I know if this transistor is in the saturation region?
AI: Call the collector node \$C\$ and call the voltage there \$V_C\$. Call the base node \$B\$ and call the voltage there \$V_B\$. We know that \$V_B=700\:\textrm{mV}\$, by definition. We also know that \$\beta=315\$ and therefore that \$I_C=\beta \:I_B\$, by definition.
Direct Route:
You can immediately compute \$I_8=\frac{V_B}{R_8}\$. That current, plus the base current must flow through \$R_7\$. So \$I_7=I_B+\frac{V_B}{R_8}\$. That current, plus the collector current must flow through \$R_6\$. As \$I_C=\beta \:I_B\$, so \$I_6=\frac{V_B}{R_8}+\left(\beta+1\right)\:I_B\$. The sum of the voltage drops across the three resistors must be your voltage source, \$V_{CC}=9\:\textrm{V}\$. So it must be the case that \$I_6\:R_6+I_7\:R_7+I_8\: R_8=9\:\textrm{V}=V_{CC}\$. From this information we have:
$$\begin{align*}
V_{CC}&=\left(\frac{V_B}{R_8}+\left[\beta+1\right]\:I_B\right)\:R_6+\left(I_B+\frac{V_B}{R_8}\right)\:R_7+\frac{V_B}{R_8}\: R_8\\\\
\end{align*}$$
Solving for \$I_B\$ you should get:
$$I_B =\frac{V_{CC}-V_B\left(1+\frac{R_6+R_7}{R_8}\right)}{R_6\left(\beta+1\right)+R_7}$$
That method is pretty straight-forward.
Using KCL:
This is using nodal analysis. It will get you to the same place, but through a slightly more complex route.
Then, by KCL at each node I get:
$$\begin{align*}
\frac{V_C}{R_6}+\frac{V_C}{R_7}+I_C&=\frac{V_{CC}}{R_6}+\frac{V_B}{R_7}\\\\\frac{V_B}{R_7}+\frac{V_B}{R_8}+I_B&=\frac{V_C}{R_7}
\end{align*}$$
The first equation is just putting all the currents "spilling away" from node \$C\$ on the left and all the currents "spilling into" node \$C\$ on the right. The two must equal each other, of course.
The second equation is just putting all the currents "spilling away" from node \$B\$ on the left and all the currents "spilling into" node \$B\$ on the right. The two must equal each other, of course, again.
The above solves out easily as:
$$\begin{align*}
V_C&= \frac{V_{CC}\:R_7+V_B\:R_6\left(1+\beta\left[\frac{R_7}{R_8}+1\right]\right)}{R_6\left(\beta+1\right)+R_7}\\\\
I_B &=\frac{V_{CC}-V_B\left(1+\frac{R_6+R_7}{R_8}\right)}{R_6\left(\beta+1\right)+R_7}
\end{align*}$$
As you can see, the equation for \$I_B\$ works out the same way. It's just that this also gets you \$V_C\$ along the way.
Since you already know that \$V_B=700\:\textrm{mV}\$ and all the resistor values are known as is \$V_{CC}=9\:\textrm{V}\$, I think you should be able to make the calculations here. You should also be able to come up with those equations.
In the end, I think you will find that \$V_C\$ is large enough that the transistor cannot be saturated. |
H: MCU + LED + SSR + (how to determine resistor value)
I am pretty new to electronics and trying to learn some basics. What I am having trouble with is how to figure what resistor value to use. I'll put links in to the data sheets of the parts I am using. If someone could explain the formulas I'd greatly appreciate it. Here is the setup I have... I have an MCU that produces 3.3V and max rating of 20mA. I really want to stay between 5mA - 10mA on each pin. What I would like to do is use the MCU + tinyled + solid state relay (which has small led in it) + resistor (or maybe no resistor?).
So we have 3.3V to work with.
The led I was thinking of using 550-1304 on this datasheet.
Then in series with this solid state relay.
So here is what I got to but I don't know if it is correct or not:
3.3V - 1.8V (Vf of the led) - 1.15V (Vf of led in SSR) = 0.25V (So my resistor needs to handle 0.25V?)
3mA is needed to activate the SSR, and it looks like 2mA is needed to make the led light up, so I am guessing 5mA.
I then took that for the Ohm's law formula, R = V / I which I did this: R = 0.25 / 0.005, R = 50. Is this the correct way to do it? And I am hooking these up in series, how do I know that the led is going to get 2mA and the SSR 3mA and that they don't share it equally i.e. both get 2.5mA? How do I determine how much each component uses? Also is this the correct formula to figure out what I need? I was trying to keep my part count down and not have to throw in a transistor if possible since the pins seem capable of driving the 5mA load. Some insight and general how you figure it out info would be great to see. Thanks!!
AI: 3.3V - 1.8V (Vf of the led) - 1.15V (Vf of led in SSR) = 0.25V (So my resistor needs to handle 0.25V?)
3mA is needed to activate the SSR, and it looks like 2mA is needed to
make the led light up, so I am guessing 5mA.
Current is the same for all the elements that are in series, but voltage is dropped at each element. (In parallel the voltage is the same, but the current is shared/divided).
Assuming your LED is rated for 2mA and that your SSR requires 3mA to turn on, you cannot put them in series without:
Exceeding the recommended current of the LED, or
Not be guaranteed to
turn the SSR on
Both are unwanted scenarios, so you have to go with another route.
Considering that a GPIO on your MCU can source 20mA, even if we stay within the desired ~10mA, you could wire them in parallel:
simulate this circuit – Schematic created using CircuitLab
Calculating R LED:
Vled = Vtotal - Vforward
3.3V - 1.8V = 1.5V
U = R * i
1.5V = Rled * 2mA
Rled = 1.5V/0.002A = 750R (Which is a standard value, so you can use this or use the next higher one)
Calculating R SSR:
If we aim at 5mA (3mA + 2mA margin), datasheet says about 1.1Vf.
So, same thing:
(3.3V - 1.1V) = Rssr * 5mA
Rssr = 2.2V / 0.005A = 440R (which isn't a standard value, so you can use 420R which should give a bit more than 5mA, or 470R which will give a bit less)
So, in total, you have 2mA for the LED + ~5mA for the SSR, so about 7mA coming from your pin.
simulate this circuit
So, in summary, you got Ohms law right, but you got the series current wrong. :) |
H: Why can't Hall effect sensors detect electromagnetism?
I read that no matter how powerful the electromagnetic field, the HE sensor won't budge.
I'm trying to develop a way to sense neodymium magnets from 12" away, and understanding the differences helps with that. If it doesn't, then delete the post.
AI: Hall Effect (HE) sensors don't detect a lot of things, electrical field, smell, sound, and plenty of others. What they do detect is the magnetic field component normal to the element averaged over the response time of any limited bandwidth amplifier that follows the Hall element.
I'm not entirely clear what you mean by 'electromagnetism', but if a HE sensor doesn't respond to it, then I'm guessing it's something whose magnetic field component has a DC value which is zero when averaged over a suitable time.
Commercial IC Hall sensors tend to have a bandwidth in the kHz ballpark, so should be able to respond to mains and audio magnetic fields as well as DC fields, and not to those of radio waves and light. |
H: Wireless switch using ESP8266
I am a software developer, sorry if my question may sound dumb.
I want to make my home smart by controlling lights through a centralized smart home web page. So rather than buying ready-made wireless unit I want to make it by myself (it will be a fun and satisfying).
A centralized Web page will contain buttons for lights in my hall/kitten/bedroom. So while search for this I come know about ESP8266 wifi module which works on STA mode. Also, I may need a relay which will take signal from wifi module and on/off the light but as I am a novice in this I have some questions:
Which ESP module need to be selected? there are two variants I see on net one is NodeMcu and another is ESP-01, which one I should go with?
If I want to control 2 lights from a light switch panel, then is it possible control it by using 1 wifi module and 2 channel relay module or I have to buy 2 wifi module (no. of lights in a panel = no. wifi module) one for each light?
Both relay and wifi module works on 5v, What should I use to pull 5v from the 250v power supplier? On net I've seen switching power supplys but these are all bulky and contain transformers. I was hoping that there was a smaller solution.
AI: p.s.: From the level of your questions I assume you are starting with electronics, and I should say this is not recommended as a first project because mains voltages can kill or severely harm you either due to direct shock or fire, etc. Please consider doing a similar project controlling lower voltages first (12VDC for example).
1.NodeMCU will give you much more GPIO and is arguably easier to develop on due to the onboard USB to serial converter. It is also easier to plug into breadboards.
2.You can control 2 relays with 2 outputs from the same microcontroller (with the appropriate driving circuitry, unless you are using modules like this or similar, which already contain the driving circuitry). No need for one ESP for each relay.
3.You can probably use a 5V phone power supply ("charger"), provided it can supply enough current for all the relays you are using.
simulate this circuit – Schematic created using CircuitLab |
H: How can the CMRR of an opamp be negative?
I was reading a paper from 1977 regarding the analysis for the CMRR of a three opamp instrumentation amplifier, I got the paper through my Unviersity but I cant post the paper since its copyrighted, however here is the link to download it legally through IEEE Xplore: C.M.R.R. analysis of the 3-op-amp instrumentation amplifier
The circuit im refering to is as follows
Basically the short (1 page) paper describes the procedure to obtain the CMRR of an instrumentation amplifier. The formula is the following:
$$ CMRR=\frac{Ad\cdot cmrr_o}{1+Ad\cdot cmrr_o(\alpha_2-\alpha_1)}$$
Where $$Ad=\frac{(R1+R2+R3)}{R2}$$
is the differential gain of the buffer stages (both opamps at the input)
$$cmrr_o \text{ is the CMRR of the differential amplifier stage}$$
and
$$ \alpha_i=(cmrr_i-1)/cmrr_i $$
where
$$cmrr_i $$ is the CMRR of the ith opamp
The paper follows the equation with a simple example:
An instrumentation amplifier was designed and built
with the following parameters: Ad = 100, R4=R5=R6=R7= 10k (nominal value), with R5 adjustable. 741-type
op-amps were used with the following measured cmrr's:
$$cmrr_1= -111\times 10^3$$
$$cmrr_2= 40 \times 10^3$$
$$cmrr_3= -52.5 \times 10^3$$
R5 was used to obtain $$cmrr_o = 100 dB$$
The measured CMRR of the IA was $$-2.4 \times 10^4$$
in reasonable agreement with the previous equation which predicts a CMRR of
$$CMRR=-2.9 \times 10^4$$
I can do the math and arrive at the same result as the paper, but my question is this: how is it possible to obtain negative CMRRs (when not stated in decibels) such as the values of cmrr_1 and cmrr_2, does it mean the differential gain or common mode gain is negative in that particular opamp? in that case what does it mean that an opamp has a negative differential or common mode gain?
AI: Neil hit things on the head. I'll try and write it up a little.
The paper you are looking at is a little aged and the terms they use and the way they use them may be a little unfamiliar. First off, take a look at the wikipedia page on the topic: Common-mode rejection ratio. There you will see the following equation:
$$V_o=A_d\left(V_{\left(+\right)} - V_{\left(-\right)}\right)+\frac{1}{2}A_{cm}\left(V_{\left(+\right)} + V_{\left(-\right)}\right)$$
Let's fabricate a somewhat fuller schematic:
simulate this circuit – Schematic created using CircuitLab
(To get their nominal gain of 100, \$R_1=R_3=49.5\cdot R_2\$. And yes, I did take note that they adjusted the gain towards its nomimal value by making changes in \$R_5\$.)
If you take some measurements of the inputs and the output of each of the three opamps while setting \$V_{CM}=0\:\textrm{V}\$ and \$V_D=1\:\textrm{mV}\$, and do this a second time with say \$V_{CM}=100\:\textrm{mV}\$, then you can solve for both the differential gain values as well as the common mode gain values for each of the opamps. And these values will not only be different, they will also differ in sign for \$A_d\$ for each.
The following solution set to use for each opamp would be:
$$\begin{align*}
A_d&=\frac{V_{O_1}\left(V_{\left(+\right)_2}+V_{\left(-\right)_2}\right) - V_{O_2}\left(V_{\left(+\right)_1}+V_{\left(-\right)_1}\right)}{\left(V_{\left(+\right)_1}+V_{\left(-\right)_1}\right)\left(V_{\left(-\right)_2}-V_{\left(+\right)_2}\right)-\left(V_{\left(+\right)_2}+V_{\left(-\right)_2}\right)\left(V_{\left(-\right)_1}-V_{\left(+\right)_1}\right)}\\\\
A_{cm} &= \frac{V_{O_2} \left(V_{\left(-\right)_1} - V_{\left(+\right)_1}\right) + V_{O_1}\left( V_{\left(+\right)_2} - V_{\left(-\right)_2}\right)}{V_{\left(-\right)_1} V_{\left(+\right)_2} - V_{\left(-\right)_2} V_{\left(+\right)_1}}
\end{align*}$$
In the above pair of equations I used the subscript of '1' to indicate the first measurement with, say, \$V_{CM_1}=0\:\textrm{V}\$ and the subscript of '2' to indicate the second measurement with, say, \$V_{CM_2}=100\:\textrm{mV}\$. I suspect they made these kinds of measurements in order to arrive at their values.
Given the above solution equation set and the circuit arrangement, then it's actually true that there will be a negative value for \$A_{d_{A1}}\$ and a positive value for \$A_{d_{A2}}\$ and a negative value for \$A_{d_{A3}}\$, with this arrangement. Nothing magical here.
Yes, this doesn't comport with the usual meaning for the computation of decibels today, where the differential gain is always taken as positive. But in this case, I think they were actually making voltage measurements using a voltmeter and using the above purely mathematical solution approach to computing \$A_d\$. In that case, you can and will get negative values for \$A_d\$, given this topology.
I think that's all it means.
I don't want to try and evaluate a paper I haven't read, though. But at least I can see how they were able to arrive at values such as those you mentioned. The sign is all about the topology here.
To get the equation solutions I gave earlier, let's say we assume there exists a value \$A_d\$ known as the differential gain and that there exists a value \$A_{cm}\$ known as the common mode gain. Let's say that we want to work out what those values are.
Well, we have two unknowns so we will need two equations:
$$\begin{align*}
V_{O_1}&=A_d\left(V_{\left(+\right)_1} - V_{\left(-\right)_1}\right)+\frac{1}{2}A_{cm}\left(V_{\left(+\right)_1} + V_{\left(-\right)_1}\right)\\\\
V_{O_2}&=A_d\left(V_{\left(+\right)_2} - V_{\left(-\right)_2}\right)+\frac{1}{2}A_{cm}\left(V_{\left(+\right)_2} + V_{\left(-\right)_2}\right)
\end{align*}$$
If you simultaneously solve those two equations for \$A_d\$ and \$A_{cm}\$, treating everything else as measurements you made, then you will get the solution equations I provided earlier. |
H: Can i charge 18650 single cell using buck converter?
I get some 18650 cells from old laptop battery. Cell voltages are good and seems healthy. I want to use this cell as a battery pack for some of projects. I add protection circuit to prevent short-circuit undervoltage. But i dont have proper charger for this.
Now is it ok to by charge from other sources. I have buck converter with CC/CV pot. Can i charge 18650 cell by set th CC(~4.2V) and CC(0.4C ~1000mA)?
Do i have to take extra safety steps while charging or is it ok if i can just control Voltage and Current?
AI: It should work, if the settings are accurate enough. You can set it to 4.1V for extra safety, you won't lose much capacity. Check voltage and current with a multimeter...
You will need to monitor the voltage to know when the charge is finished though. It is not recommended to let the battery charge at 4.2V for too long (most charge control ICs include a timer and various other safety checks).
From mkeith:
The standard method of charge termination for lithium ion (and lithium polymer) chargers is by current. During the CV stage, if the current drops to some low level, the charger stops charging. The timer would be a backup termination method in case the primary termination condition never occurs. For a single 18650, the charger might terminate at like 50 mA or something.
Also make sure the cell is within the acceptable temperature range. If this happens indoors, it should be.
This starts to sound like a lot of work.
You can buy a 2-cell 18650 charger for little money these days, so that would be a better and safer solution. |
H: Know processor speed of our application
I am working on a project where i have to log data of many Analog and Digital sensors, GPS with lots of processing Algorithms. I am in a hunt to select a Microcontroller for my application.
My question : How do i know, how much processing speed do i require for my application.? as well as How much RAM do i need for it as well.?
anybody help.
thank you
AI: What I did for my project (for from unfinished):
For memory usage:
Count for each item (global variables):
How many you need to store in SRAM
How many bytes each type cost
Add some extra for stack, and temporary variables
Check what you want to optimize
For processor speed, it's a bit trickier. You should decide what is the most performance and critical part of your program, than decide how much instructions you need to perform it against the speed you need. You could analyze this by writing the critical part and check in the assembled code the exact amount of instructions, calculate it by the clock cycle used for that instruction and sum all up. |
H: Silk screen layer over pad resist cutouts on pcb
I am designing a SMD PCB and have noticed that some component silk screen outlines extend over the pad resist cutouts. Is this acceptable? If the resist cutouts are made after the silkscreen process I can see this working.
AI: The silk screen is the last step in PCB production. The solder mask is the penultimate step.
If your silk screen is over (or too near) a pad, that should throw an error from your DRC. Most likely the silk-pad spacing in your DRC settings is greater than the solder mask expansion so you should not have the silk screen going over the edge of the solder mask unless it is too close to the pad (or unless you are doing something special on the solder mask layer such as removing it over a large area for some reason).
Keep in mind that the registration of the silk screen layer(s) may not be as good as the other layers so you may wish to set conservative DRC rules for the spacing, lest it end up on a pad and affect the soldering.
If the solder mask cutout is unrelated to a pad, I don't think running it over the edge of the cutout will cause serious problems but it might affect readability if your designators are very small. |
H: Computing output resistance for zener power supply
I have a simple zener power supply, the schematic is below. My question is how to find the output resistance of the circuit?
I also have small signal model given, but I cannot get a proper result.
AI: If you calculate the impedance seen across \$u_s\$ by shorting \$u_i\$, clearly R comes in parallel with \$R_z\$. Thus, potential at the node connected to current source becomes,
$$V = -i_b(R_z||R + h_{ie})$$
Also total current will out from that node will be \$-i_b(1+h_{fe})\$.
The impedance becomes the ratio of these two i.e. \$(R_z||R + h_{ie})/(1+h_{fe})\$. This actually comes in parallel with the \$R_s\$. |
H: Charging and using 3 separate Li+ batteries in parallel (using 1 or 2 or 3 together)
The idea is to make a battery setup in such a way that the user can use 1 or 2 or 3 18650 Lithium cells. The circuit would draw power via USB to charge the setup as well as run the application circuit after the boost section. The load is around 500mAh.
An p-Channel mosfet is used to act as a switch for the final output. USB-5v is connected to the gate of the mosfet, the output of the UV protection circuit is connected in parallel to the drain of the mosfet. So as soon as USB is shut off, the circuit would draw power from the battery pack that is connected. Circuit reference from ESE answer.
The main reason to use the setup was to allow the user to use 1S by default. If the lifetime of the application circuit was to be extended, the user attached another 1S to the setup. So the boost section ( 3.7v-4.2v to 5v) remains the same.
Questions
Is it safe to use 3 separate charging ICs and 3 separate 1S battery protection ICs to charge 3 separate Li+ batteries?
Is is safe to draw power by connecting the output of the protection circuits in parallel?
I came across TP4056 modules on Amazon. So looking at the reference design, I was wondering if the above setup would be possible and safe to execute?
Update:
Added diodes to the output of the main block.
AI: No, this is not safe.
Edit: Your updated version, with the diodes, should be safe.
The problem isn't with the charging/discharging of the batteries, but with connecting their outputs in parallel. If you put two batteries in parallel when they're not at exactly the same voltage, a huge current will flow between them, possibly damaging the batteries, wires, and protection circuits. One of the batteries will basically charge the other at the maximum current it can supply. In the best case, the protection circuit will just switch off until the batteries are separated again; in the worst case, it'll catch fire.
It will work fine if you put a diode after the output of each battery so that the batteries can't charge each other when they're paralleled. You will lose some capacity due to the voltage drop across the diodes, however. A high-power Schottky diode in series with the "OUT+" terminal of each protected battery should do the trick.
You could also use a MOSFET-based "ideal diode" circuit which is a little more complicated but avoids the voltage drop. |
H: Reversible Constant Current Source (CCS)
I need a current source, that can work in both directions.
To make it simple, I need to generate a square signal in current.
I am currently using a simple H-bridge, but without any current regulation. All the current regulators that I found don't work for both negative and positive parts.
voltage range: up to 30 - 40 V
current range: few mA
period range: few ms
Do I need to use diodes and use separate regulators for the negative and positive part of the signal?
Can I find a H-bridge controlled by PWM that can deliver a regulated current?
Any other idea to solve this problem?
AI: You're on the right track regarding the use of an H-bridge.
Put some inductor (a couple of mH) in series with your "load" and just use PWM and be fine and dandy with that, as an open loop system.
If you want to be certain that the current is what you want it to be, then use closed loop system. So get a current sensor, put in series with the inductor and the "load" and feed the output of the sensor to your µC (MicroController).
I'm 99% certain there's current sensors that can give out a value in both directions, if not then make one yourself with a differential amplifier (op-amp + resistors + shunt resistor), bias the signal with 2.5V and then convert it to digital (ADC), that way 2.5V will mean 0A, above 2.5V will mean current in some direction, less than 2.5V will mean current in other direction.
Then make an interrupt happening at regular intervals and set up a PID system in software using the sensor values as input and PWM as output.
If you want to know how to implement a PID then I can tell you that it's easy and you have youtube / wiki for that. Or just make another question.
This is what I had in mind:
The diodes on the gate in parallel with 10 Ω is just to remove the ringing of the gates. I'm aware you're going to use like 40-50V, so use appropriate voltage shifter and appropriate high voltage op-amp.
I've explained pretty much everything in the image.
Here's the link in case you want to mess around.
If you're going to try this out, then keep in mind that you will definitely have to mess with the PWM (if the resistance of your load is very low, as in the simulation), because right now it's duty cycle is at \$\frac{0.06}{5}=1.2\%\$. Scale that up with 255 = 100% and 0 = 0%, then 1.2% translates to 3.06 = 3. That's not enough resolution, 1 will be 2 mA, 2 will be 4 mA, 3 will be 6 mA, as in the simulation. But that will depend heavily on the resistance of your load, in the simulation I used 10 Ω because whatever, but if I'd used say 100 Ω then the binary value 30 would have given me 6.3 mA as in the image. Oh well, there's always some drawbacks.
Another way to increase resolution would be to change the size of the inductor to.. say 500 mH and use a 16 bit PWM, then all your problems will most likely go away. You can probably find a couple of Henry in transformers, look in a garbage dump for the transformer in a microwave oven or something.
But with large inductances like that, it would result in long delay changes, say you want 1 mA, maybe 2 seconds later you'll get that, then you want 2 mA, 2 seconds you'll have that. I'm just speculating. |
H: Altium Wire vs. Bus entry
I have following situation in Altium Designer:
Now, AUDIO_INPUT_SELECT[0..3] are connections to Bus, made using Wire command. However, after researching Altium, I've stumbled upon Bus Entry command and the question is, do I have to change these Wire entities to Bus Entry entities or not and if I must, how do I transform Wires to Bus Entries in most elegant way? I know I can delete Wire and place instead of it Bus Entry, but I have several such connections and I want to know the most elegant path of executing such task.
AI: You can connect wires directly to a bus; bus entries are not required.
A bus entry allows you to connect to both sides of the bus without the wires from both sides being shorted together. |
H: How is the 220V line in US different from other parts of world? Does a US 220V line need two switches for each wire?
Today my A/C compressor had given up its ghost and when I tried to disconnect it from the wall switch after turning OFF the wall circuit-breaker,I assumed like most 110V lines where only one line (Live-L) lights with neon tester.
When the circuit breaker was turned OFF the neon tester on A/C switch line turned OFF and I assumed the power was out. But while disconnecting I accidentally touched the other wire and I received a shock. On testing with neon tester I noticed the light was glowing for the other line also and I later discovered it had in fact two breakers(no handle tie) for the A/C circuit. When I checked on my normal 110V lines the neon test light glows only on one wire i.e., Live wire.
Even most of my google searches showed a 220V line with just L and N. So,doesn't the US 220V have a neutral?
AI: US residential (and most commercial) uses split-phase electricity, therefore both wires of a 220V outlet are live; the neutral is used to provide two 110V connections, each 180 degrees out of phase. |
H: How far should a TEC1-12706 peltier module be away from the heat source to prevent damage when using it to generate thermoelectricity?
I am working on my science fair project and only have 5 peliter modules. In order to build my project I need to know how far it must be to keep from being damaged because I can not replace these once I have begun.
AI: I recommend that you add a piece of aluminum block around 3-4 mm thick with good quality thermal paste and then bring it close to the candle at a distance and at the same time measure the temperature of the aluminum block. Make sure you stay below the limit around 55-65C since 75C is max temp. Because too much heat you will melt the solder connections inside and ruin it. Good luck! |
H: Mosfet controlled LED-strip is dimmed
I'm gonna start to excuse me for my bad soldering/layout, but this is .
I am using a mosfets (IRF520) to which I have connected a 12V power source to the drain and source. They are connected up to a led strip. The gate I am controlling with my arduino MKR1000 (3.3v outputs). The circuit does exactly what I expect from it except for one thing: the leds dont burn as bright as I would expect. When I connect a 12v input to the gate they do burn as bright as I expect. My guess is that it is because the voltage I put on the gate is too low, and the mosfets are acting as a variable resistance.
The store I bought them though tells me that they should burn as bright with 3.3v as with 12v. Am I doing something wrong, or am I using the wrong mosfets?
Schematic I used:
Link to datasheet:
https://www.vishay.com/docs/91017/91017.pdf
AI: I think your Vgs is too low. You could easily check this by connecting the gate to VUSB or even the 12V as you did. I used IRF520 with 3.3V logic and had to use a higher voltage transistor coupling to the gate to increase Vgs. |
H: Mounting SIP Power Module with 1.7mm Pin Pitch @ 8 Amps
I'm attempting to mount a GE NQR010A0X4Z power module (datasheet here) on a prototype board or custom PCB and am not sure what I can do besides directly soldering wires to the pins (which seems very hacky).
At first, I was looking to create a simple breakout board but the narrow pin pitch restricts me from creating traces wide enough to handle the current and creating heavier traces also get's too expensive.
What are my options?
AI: One option I have seen done many times:
Have the traces soldered over to thicken without the extra thick copper board cost. This is popular with budget ATX computer power supplies and small AC to DC converters. You can usually at least triple the thickness. |
H: What is an "Ni-Cr" battery?
I have encountered a few references to Ni-Cr batteries, such as a Charger for NiMH and Ni-Cr batteries. I think it may be a synonym for NiCd. If you can shed any light on this, it would be most welcome. If it is just a synonym for NiCd, how did it come about?
I also found several alibaba listings for "NiCr" batteries. They appeared to just be NiCd, but I am not sure.
AI: There is no such battery called Ni-Cr. It is likely just a typo.
As other commenters have mentioned the 'r' and 'd' are close on the keyboard.
The only common nickel based battery chemistries are Ni-Cd (or Nickel Cadmium) or Ni-MH (Nickel Metal Hydride). They operate similarly with Ni-MH being the more recent with better capacity and not using toxic Cadmium. They both have terminal voltages of ~1,2v which is similar to zinc-carbon (manganese alkaline) primary batteries. They are relatively safe and do not usually suffer from explosive or thermal failure in the way that Li-Ion batteries do.
They both suffer from high self-discharge rates of many % per month and have been supplanted in modern equipment by Lithium based batteries (e.g. Li-Ion). |
H: Is the dynamic input resistance zero for constant voltages?
This might sound a stupid question but I'm confused with the following excerpt:
If Vbe is constant 0.7V which means ΔVbe is zero, does that mean the dynamic resistance is zero?
I think I have problem with interpretation of the graph above. If there is no change in the Vbe signal and Vbe is kept always constant at 0.7V, can we still talk about any kind of resistance here?
AI: If Vbe is constant 0.7V which means ΔVbe is zero, does that mean the
dynamic resistance is zero?
All that means is that you can't experimentally measure it, since you are refusing to vary \$V_{BE}\$. On the other hand, you can compute it from theory (with or without the Early Effect -- which I see in your chart.)
For a BJT, the following equation is broadly true over quite a range of collector currents (often 4 or 5 or even 6 orders of magnitude):
$$\begin{align*}
I_C&=I_{SAT}\left(e^{\left[\frac{V_{BE}}{n\:V_T}\right]}-1\right)\\\\
I_B&=\frac{I_{SAT}}{\beta}\left(e^{\left[\frac{V_{BE}}{n\:V_T}\right]}-1\right)
\end{align*}$$
Now, there is also an Early Effect that can be approximated this way:
$$\begin{align*}
I_C=\frac{I_{SAT}}{1+\frac{V_{BC}}{V_A}}\left(e^{\left[\frac{V_{BE}}{n\:V_T}\right]}-1\right)\\\\
I_B=\frac{I_{SAT}}{\beta}\left(e^{\left[\frac{V_{BE}}{n\:V_T}\right]}-1\right)
\end{align*}$$
But even if you don't actually vary \$V_{BE}\$, it can be computed from theory.
Dynamic resistance is the local slope of a curve where the y-axis is voltage at the x-axis is current; or else it is the multiplicative inverse of the slope where the y-axis is current and the x-axis is voltage.
[Slope calculations are the meat and potatoes of 1st year calculus, starting with the limit theorem definition of the derivative of a function (grounded rigorously in mathematics by Dedekind and Weierstrass in the mid-late 1800s and impressively complemented by Abraham Robinson's "Non-Standard Analysis" work in the early to mid 1960's.)]
Just take the derivative and you are done. Using the version without the Early Effect and applying the D() operator (see Heaviside):
$$\begin{align*}
\operatorname{D}\left(I_C\right)&=\operatorname{D}\left(I_{SAT}\left(e^{\left[\frac{V_{BE}}{n\:V_T}\right]}-1\right)\right)\\\\
&=I_{SAT}\operatorname{D}\left(e^{\left[\frac{V_{BE}}{n\:V_T}\right]}-1\right)\\\\
&=I_{SAT}\operatorname{D}\left(e^{\left[\frac{V_{BE}}{n\:V_T}\right]}\right)\\\\
&=I_{SAT}\:e^{\left[\frac{V_{BE}}{n\:V_T}\right]}\operatorname{D}\left(\frac{V_{BE}}{n\:V_T}\right)\\\\
&=\frac{I_{SAT}\:e^{\left[\frac{V_{BE}}{n\:V_T}\right]}}{n\:V_T}\operatorname{D}\left(V_{BE}\right)
\end{align*}$$
At this point, it's pretty easy to see that for almost all useful, realistic values in life:
$$I_C=I_{SAT}\left(e^{\left[\frac{V_{BE}}{n\:V_T}\right]}-1\right)\approx I_{SAT}\left(e^{\left[\frac{V_{BE}}{n\:V_T}\right]}\right)$$
So:
$$\begin{align*}
\operatorname{D}\left(I_C\right)&=\frac{I_{SAT}\:e^{\left[\frac{V_{BE}}{n\:V_T}\right]}}{n\:V_T}\operatorname{D}\left(V_{BE}\right)\\\\
\operatorname{D}\left(I_C\right)&=\frac{I_C}{n\:V_T}\operatorname{D}\left(V_{BE}\right)\\\\
&\therefore\\\\
\frac{\operatorname{d}\left(V_{BE}\right)}{\operatorname{d}\left(I_C\right)}&=\frac{n\:V_T}{I_C}
\end{align*}$$
Which is the slope of the curve at any given value for \$I_C\$. (Note that \$V_T=\frac{k\: T}{q}\$, which is approximately \$26\:\textrm{mV}\$ at room temperatures. Also note that \$n\$ is the emission coefficient and, for small signal BJTs, is almost always very close to 1. [It is never less than 1.] It is almost always more than 1 when discussing diodes and LEDs, but can also be more than 1 for BJTs. But usually just take it as 1, unless there are reasons to say otherwise.)
This value is assigned to the emitter and not the collector, so the value you want to compute is:
$$\begin{align*}
r_i&=\frac{n\:V_T}{I_E}
\end{align*}$$
At \$I_C=0.5\:\textrm{mA}\$ (and \$n=1\$) you'd expect \$r_i\approx 52\:\Omega\$ (as \$I_E\approx I_C\$.)
Note that this is in the range that they said. No surprise. It's based on the physics going on. So it pretty much has to be that way. Do note, though, that their range is mentioned in a narrow context. But the above equation applies over a very, very wide context they do not discuss. So their range is not a limitation. You should use the equation and not use their range. The equation tells you the value. Their range does not. Their range is was given because of the context of their discussion.
Take careful note that \$r_i\$ is heavily dependent on the collector/emitter current and that changing the quiescent point changes this dynamic resistance. Also take careful note about the fact that it is also dependent on temperature. (Finally, take careful note that if your emitter resistor is small-valued, then \$r_i\$ in the common base mode matters more and is temperature dependent and this leads to a circuit whose behavior is too easily varied as ambient temperatures vary.)
There are adjustments made due to the Early Effect (modeled with \$V_A\$ parameter.) But I think we can avoid this for now, as it is very simple to include in the above calculations, if needed.
If there is no change in the Vbe signal and Vbe is kept always
constant at 0.7V, can we still talk about any kind of resistance here?
There is always a slope to a curve, even if you aren't changing the point on the curve that you are sitting at and can't yet tell what the slope might be. If you are sitting on the road up a steep hill and not moving, you might not notice the slope. But the slope is still there and if you start moving up or down the road you will soon find out about it, too. Kind of like that, I suppose. |
H: Why use an opto-coupler?
A side note: please be forgiving of me because I'm just starting out in Electronics so I'm still confused about certain terms.
Why are optocouplers used?
Yes I know they link a circuit together since there is an LED inside and a photoresistor. I also read that it prevents high voltages. But why use an optocoupler over another component? For example, to reduce high voltages we can use step-down transformers but why, in a certain scenario, we would use an optocoupler over a step-down transformer?
In what scenarios are optocouplers appropriate?
Hope my question is clear enough for anyone who can help.
AI: Both transformers and opto-couplers can provide isolation between 'hot' and 'safe' regions of a circuit.
The difference is that opto-couplers are very small, cheap, and can work on simple DC, so can shift a logic signal from one side to the other with no fuss and no or few other components.
Transformers are big, expensive, and need AC to work, they cannot simply be inserted into a logic line. When you need the thing that a transformer does well, moving power from one side to the other, then you have to use a transformer. Otherwise, you use something smaller, cheaper and easier. |
H: Low current divider for battery voltage monitoring
I’m working on a low power sensor and would like to measure its battery voltage.
The battery voltage is too high my ADC input so I have been dividing it with a pair of 1M resistors and a 100n cap so that the ADC’s input capacitance doesn’t load it too much while sampling and affect the results.
The trouble is that the resistor divider drains a few uA all the time and this is far from ideal. Is there a common solution with a low BOM cost to this problem?
I thought of switching an N channel mosfet to the base of the divider but it would allow the ADC input to float too high. Putting a P channel on the +ve rail would be difficult to drive.
Do special ADCs exist that can measure beyond the +ve supply rail?
Any suggestions welcome.
AI: Use a simple circuit like this. The P-Channel MOSFET can be some other low current device - The one shown is just the one that showed up in the schematic tool.
simulate this circuit – Schematic created using CircuitLab
The idea is to connect the GP_OUT signal to your MCU and have the software drive that high at the time an A/D measurement is to be made. I show using 100K resistors for the divider but these may need to be scaled down depending on the input impedance of the A/D input pin. It's been my experience that 1Meg resistors used for this purpose lead to too much measurement error.
Some delay from setting the GP_OUT high will be needed till the A/D conversion is commanded to allow for settling of the A/D input.
Note that when the NPN transistor is off the only load on the battery is leakage current. |
H: speed of rotation in rads/sec
Can anyone help me with the speed of rotation in rads/sec.
I am trying to find the equation for it.
I have 730 rpm as my motor speed. It is a DC motor.
So far the equation i have is: 2pi x 730/60
And the answer is 76.445.
Is this the correct equation to use tho find speed of rotation in rads/sec?
Thanks.
AI: $$\Omega=2\pi f= \dfrac{2\pi N}{60} = \dfrac{\pi N}{30} $$
$$\Omega=\dfrac{3.14\cdot 730}{30}= 76.4 [rad/s]$$ |
H: Problem with chaining many small DMX cables
In DMX cables (used for lights), all cables are daisy-chained.
In my band we are normally on small stages, but occasionally on bigger.
To prevent buying very long cables, I wonder if it is a good idea to buy multiple 5 m (or yard) cables and just connect them together (without a DMX device in between) in case we need longer length.
I assume it is not a problem, but just wondering.
There is some 'common' used maximum of like 10-15 fixtures, but does this count the number of DMX devices or the number of 'cable connections'?
AI: There's a maximum rule of 32 devices and 1800 feet (548m) of cable in a single chain, though generally people seem to aim for about half that number of devices in practice. Each connector you add in will have an impact on the maximum length because of the resistance it adds in.
I doubt that you'll have any problem with multiple extensions unless you're talking about hundreds of metres. Even so, I'd be inclined to go with longer extensions (10m or even 25m) and less of them - there will be fewer connections to fail. |
H: What switch would replace a series of 6 pins linked by jumpers?
I've bought a small electronics kit - a function generator which produces a 1kHz signal in various waveshapes - square, sine, triangle and integrator. It's a basic kit on a single PCB (555 timer based) and you select the waveform you want using a jumper between 4 pairs of pins.
I would like to put this kit in a small box to use as a more robust tool for testing equipment. I need a chassis mounted switch which could essentially imitate the action of moving a jumper between these four sets of pins. What sort of switch would I need, and how would I wire it?
AI: It's a bit hard to answer without a schematic, but a rotary switch is a typical solution. You can get them in many different configurations. For example, a four position double pole switch as shown below. |
H: Find the largest value of \$K_1\$ and \$K_2\$ for instability of the system
A cascade control system with proportional controller is shown below
Theroritically the largest values of the gain K1,and K2 that can be set withot causing instability of the closed loop sytem are
Given \$G1=\frac{1}{(s+1)(2s+1)}\$ and \$G_2=\frac{1}{(3s+1)}\$
\$(A)\$10 and 100 \$(B)\$100 AND 10 \$(C)\$10 AND 10\$(D)\$\$\infty\$ and \$\infty\$
The closed loop T.F will be
\$\frac{C(s)}{R(s)}=\frac{K_1K_2}{(s+1)(2s+1)(3s+1)+K_2(3s+1)+K_1K_2}\$
Now C.E will be
$$6s^3+11s^2+(6+3K_2)s+1+K_2+K_1K_2=0$$
Now after applying R.H criteria I got two conditon for stability
that is $$20+9K_2-2K_1K_2>0$$ and $$1+K_2(1+K_1)>0$$
Now after satifying Options one by one all options becoming unstable for this equations.
Is the options are wrong or I am doing it wrong?
AI: You seem to be doing things right. I verified them using Mathematica. So either you have \$G_1\$ and \$G_2\$ wrong, or the question is wrong.
There are two possibilities:
The max value of \$K1\$ is \$\frac{9}{2}\$, and the max value of \$K2\$ is \$\infty\$.
\$K1\$ is some finite value greater than \$\frac{9}{2}\$, and \$0<\text{K2}<\frac{20}{2 \ \text{K1}-9}\$. |
H: Electromotor + transformer poor output
I kindly ask you for help in my project.
Outline: Electromotor used as generator, connected to transformer does not provide expected output; output is lower than input. When connecting the joule thief with a power supply to the same transformer, the voltage amplification is as expected.
Diagram:
Primary winding of transformer is also connected to the ocilloscope, channel 1.
Picture:
A picture of the experiment. The generator is rotated with the battery drilling machine.
DSO output:
There is more voltage on the primary output than in the secondary output, as can be seen on the picture.
Now, the problem is clear - no voltage transformation, which makes me a very sad puppy, since I will not get rich by manually spinning the generator.
I have discarded the generator and replaced it with a powered joule thief, just to test the transformer connections and setup.
Picture:
Joule thief + transfomer + DSO, 2 channels.
Diagram:
Output:
This is a good output, the signal gets clearly amplified from the transformer.
Questions:
The transformer is from ATX power supply, MP-EI-33-2
What is the cause of the poor tranformation of signal from the electric generator?
How can I improve the signal transformation?
Isn't the varying signal the best thing you can provide to the transformer????
Thank you in advance,
Kindest regards.
I think I have found an answer to the problem:
I have opened the electromotor and performed a test with a metallic screwdriver to see if it will jump to any side of the stator.
What I have found out is that the magnetic force is very weak. The screwdriver barely goes to left / right side.
I would conclude that the stator does not consist of a permanent magnet and it requires AC input to switch polarities, which does not happen with the motor being disconnected.
Would this be the correct problem?
Thank you,
Boris.
AI: It appears that you are trying to generate electricity using a universal motor.
Figure 1. A universal motor features series connected field and armature windings. There is no permanent magnet involved. Source: Electrical Easy.
When connected to DC power the poles will be magnetised as shown (or reversed depending on polarity) and the motor will run as a DC brushed motor.
When connected to AC the polarity of the magnet reverses on each half-cycle but so does that of the rotor so it continues to spin in one direction.
To reverse the direction of rotation for DC or AC the connections to the brushes are swapped.
As you have discovered, this motor cannot be used as a generator as it has no built-in magnet. You could split out the field windings and apply a little power to them to generate some magnetism. The brushes would then output DC and be useless for your transformer. |
H: How do I measure internal resistence of voltmeter without ammeter?
I know how to do it with ammeter, but how can I find internal resistence without ammeter?
AI: simulate this circuit – Schematic created using CircuitLab
Figure 1. DC voltmeter impedance measurement technique.
For a DC meter such as a typical digital multimeter you can try the scheme of Figure 1.
Use a very stable reference voltage. (OK use a PP3 9 V battery.)
Take a voltage reading from your reference voltage with the meter in question (Figure 1a).
Use a precision reference resistor. (OK grab a 1M resistor out of your bag of resistors.)
Take a second voltage reading from your reference (Figure 1b). If R1 is significant relative to the meter the reading will drop.
e.g. If the meter reading falls to half when using a 1M resistor then the resistor must have an input impedance of 1M.
The easiest way to do it is to keep adding or subtracting resistance until you get a reading of half of the battery voltage. That resistance is equal to that of the meter.
Update
VM2/9V = (R1+Ri)/R1 Is this correct?
No. \$ \frac {meter \ voltage}{total \ voltage} = \frac {meter \ R}{total \ R} \$. So,
\$ \frac {VM2}{9} = \frac {Ri}{R1 + Ri} \$. |
H: What is the purpose of these elements in this circuit?
I am analyzing this continuous waveform amplifier circuit for a Microwave Motion Sensor. This high gain amplifier is recommended in the datasheet as the signal output of the sensor is extremely small (~1uV). I understand that they are using an op-amp integrator and an op-amp differentiator configurations, but what are the roles/purposes of the circled parts? Could someone please kindly explain it to me as my circuit analysis is a little bit rusty.
AI: Figure 1. Waveform amplifier.
The resistors form a half-supply reference for the single-rail powered op-amps. With a 5 V supply this reference voltage will be 2.5 V. The 100 uF capacitor stabilises this voltage. As @Trevor points out in the comments, it also prevents any AC input signal from (3) affecting (2).
The non-inverting input is held at 2.5 V. With no signal the output should go to 2.5 V as well.
This stage is a non-inverting amplifier. Without the 330k resistor the bias current of the op-amp would charge or discharge the 4.7 uF capacitor until it reached +5 V or 0 V. Providing a DC path to the 2.5 V reference prevents this.
Difficult to know what the 12k is for without a schematic of the motion sensor innards. The capacitor means that the rest of the circuit will only respond to rapid changes from the motion sensor.
The op-amp is a non-inverting one. The gain is given by the standard formula \$ A = 1 + \frac {R_F}{R_G} = \frac {1M}{10k} = 100 \$. (Note the sloppy units on the schematic. It should be 'k' for kilo and 'M' for mega.) The capacitor blocks DC again as the DC path for the bias is provided by the 1M feedback resistor. It could probably have been omitted and (5) connected to the 2.5 V reference but it may be doing some high-pass filtering too.
The capacitor again blocks DC from reaching the next stage. The 8k2 resistor is the input resistor of an inverting op-amp. |
H: Slew Rate Controlled GPIO
What is the purpose or advantage of using Slew Rate controlled output capability for GPIOs?
AI: Slew rate control limits the production of high frequency components in a transition and reduces the potential 'noise' produced by any given circuit. This becomes ever more important as PCB track length gets longer or you transition into wires to go offboard. |
H: What's wrong with my logic in my adder-subtractor?
I'm trying to create an adder-subtracter using only one adder, not gate, and multiplexer. When A = 1000, B = 0001 it results in Q = 0111. If my math is right, that is 8 +(-1) = 7 and 7 = 0111 which is indeed what Q outputs. Now, if I try to switch the values in my circuit as I do in the circuit below I get Q = 1001. Going over my math again A = 1, B = 8 so 1+(-8) = -7 and -7 = 1111 which does not equal the Q output. I'm really confused as to where I went wrong here.
AI: -7 = 1111
In 4-bit two's complement, -7 would be found by inverting 0111 to get 1000. Then add 1 to get 1001.
So 1001 is indeed the representation of -7 in your system.
1111 is the representation of -7 in 4-bit sign-magnitude so that might be where you got mixed up. |
H: How can I connect multiple 24LC04B's on the same I2C
I'm trying to store some data on a EEPROM, but unfortunately I cannot fit it into a single 24LC04B, therefor I want to connect another 24LC04B to the I2C line to compensate for the lack of memory.
The 24LC04B has only two actual addresses/pages (0x50, 0x51), but the problem is that it's also using another 6 addresses/pages as a mirror of the first two for some reason..:
i2cdetect -y 1
0 1 2 3 4 5 6 7 8 9 a b c d e f
00: -- -- -- -- -- -- -- -- -- -- -- -- --
10: -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --
20: -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --
30: -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --
40: -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --
50: 50 51 52 53 54 55 56 57 -- -- -- -- -- -- -- --
60: -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --
70: -- -- -- -- -- -- -- --
For 24LC128, I can change the I2C address of the EEPROM by connecting A0, A1 and A2 pins to the VCC pin, which gives me an address selection in a range of 0x50 to 0x57.
For the 24LC04B this seems to have no effect and I cannot find any information on how to disable the address mirroring ether. Even if the 24LC128 was big enough, I still can't add it to the I2C line, because 24LC04B is using up all the address space with the mirrored addresses..
Is there a way to connect two 24LC04B's on the same I2C line? Thanks!
AI: The datasheet I checked (Microchip) explicitly states that the A0-A2 pins are not connected, hence there is no way to have more than one such chip on a pure I2C bus.
You could restore to tricks like using two separate (bit banged?) I2C busses - the I2C protocol is not that complex - or swithing one or both of the lines, using some other GPIO pin. That could be as simple as a few resistors and maybe a diode, or as complex as using a dedicated IC. |
H: Using Octave instead of Matlab in control engineering
I would like to know if in a first course in control systems at the undergraduate level I would miss much by using Octave instead of Matlab? I understand that Matlab has simulink but regarding basic control system design, is Octave enough?
AI: I've finished my masters degree in electrical engineering one month ago and throughout my entire studies, i only used octave instead of matlab. I also had some control systems classes, which can be perfectly done with the control toolbox for octave (it is not built in). However, as you mentioned, octave has nothing like simulink, so if you need something like that, you don't really have a choice. For the matlab part and the usual undergraduate control systems classes, octave should be perfectly fine! |
H: Safe load on a 40 VA Transformer
I apologize ahead of time for this elementary, and quite possibly poorly posed, question, but I am a coward and want to make sure that I am understanding everything correctly, and not blow four motors...
We have a hydronics heating system, and we are adding a fourth heating zone. Each zone is controlled by a motorized zone valve (https://forwardthinking.honeywell.com/related_links/water/5000_series/install/95c_10932.pdf).
The (above) manual says that the motors have, as
Actuator Electrical Rating:
24Vac 60 Hz, 0.30 A Current Draw,
5 W, 7.2 VA maximum
[The label on the actual device says 0.32 A; in any case, I note that 0.3 X 24 = 7.2 above]
Now, these are on the 'secondary' side of a transformer; the relevant portion of the transformer label says
40 VA
PRI 120 V
SEC 24 V
My practical question is:
Since
$$ 40 > 4 \times 7.2, $$
am I reading things correctly, and it is safe to add the fourth motor?
I also have an 'enquiring minds want to know' question:
Are the numbers above enough to know what is the power draw on the 120 Vac side, when no motors are running, and when the 4 motors are running?
Thank you very much...
Peter
AI: As pointed out in comments this will be fine.
The VA rating of the transformer is the maximum you can take from the secondary: This is literally just output current times output voltage so you want 28.8VA which is less than 40 so OK.
Note its VA not watts because power also depends on phase and shape of the current taken by the load. Transformers like this are typically better than 95% efficient so the input VA will be slightly higher than this but not much. We can estimate this \$ VA_{input} \approx \dfrac{28.8}{0.95} = 30.3 VA\$ About 0.28 amp from your 110V mains.
Without any load there will be some current in the primary. This is called the magnetising current as is a result of the inductance of the primary winding. The more inductance you have the less magnetising current there will be. |
H: Is it okay to connect a European 240V transformer to a US 240V outlet?
I've an old Bosch German made automotive charger transformer with ratings 14V/9V, 220-250V @ 50/60Hz 132VA. Can I use it on the 220 to 240V outlet here in the US which uses a split - phase system as the transformer was intent to be used on Europe style 240V with a Live and Neutral system?
Will there be any difference in its performance?
Note: I'd initially asked this question in the comments section of "How is the 220V line in US different from other parts of world? Does a US 220V line need two switches for each wire?". But has been removed by moderators and was asked to make separate thread so this info is seen by everyone.Thanks.
AI: Yes, you can connect that to a US 220-240 V outlet. Just connect to the two hot lines and connect nothing to the neutral. However, there are normally only 30 to 60 amp 240 volt outlets available in US residences. They are dedicated to clothes dryer cooktop and oven use. To add a 240 volt circuit requires two adjacent 120 volt circuit breaker spaces in the distribution box. You will need a 15 or 20 amp double breaker whichever is the smallest available. You can connect additional 120 volt circuits to the same breaker, but they must not be circuits like bathroom and kitchen outlets that can not be mixed with outlets elsewhere.
It is not a good idea to connect to a higher current circuit because the higher rated circuit breaker might not prevent a fire if you have a short-circuit in the charger or in the charger cord. If you want to do that you should have smaller fuses near the outlet. I don't know if that is permitted by code.
You should use a US receptacle and plug rated for 240 volts and the fuse or circuit breaker current rating. You probably will not find a suitable converter for a 240 volt US receptacle.
There could be a difference in performance related to the actual voltage in the previous vs. present locations. However the transformer will be operating within the specified voltage and frequency range, so that is nothing to be concerned about. If the new voltage is higher, the charging current may be a little higher and the charging time a little quicker and the opposite if the voltage is lower. I assume an old charger may not have any voltage regulation. The operating temperature may be lower or higher.
A 15 or 20 amp 240 volt receptacle installed for a window air conditioner is an excellent choice for this application. |
H: How do I use fuse bits on AVR to configure frequency
I compiled a hex file where the controller is supposed to blink 1 time per secound. The frequency was set to 1mHz (it is the internal generator). I programmed an Atmega 8a using avrdude 33 on default fuses and now the LED is flushing very fast. I think if I rise the frequency to 8mHz it will work ok. But I need the Atmega running as slow as possible to reduce power consumption. Because if i rise the frequency to 8mHz it will increase the power consumption 8 times! So how sould I configure the fuses?
P.S. I tried to use someone else's fuses and killed a few microcontrollers so I want to be sure that it will work 100%.
AI: You can use this website to calculate what your fuses should be:
http://www.engbedded.com/fusecalc/ |
H: P channel mosfet as a switch different setups for USB and battery switching
Recently I came across the below 2 examples of using a p-mosfet as a switch. Running a simulation leads to the following observations.
Circuit 1
Reference 1
OUTPUT is 4.951
Circuit 2
Used this design in some previous circuits
OUTPUT2 is 4.379
The output needs to be 5v is 5v is available. If 5v is disconnected, the output has the be the voltage of the battery that is connected.
Could someone please explain the difference in both the circuits.
simulate this circuit – Schematic created using CircuitLab
AI: This circuit should give the full 5V on its output when the 5V power supply is connected at the cost of about 1mA idle supply current when it's operating on battery.
simulate this circuit – Schematic created using CircuitLab
The circuit compares the voltage of the power supply to that of the battery. If the power supply has a higher voltage than the battery, it switches on M1, otherwise it switches on M2. |
H: Unknown ticking sound on bad ite power supply
I have a power supply with the model number NU70-1120520-l1. It is supposed to have an output of 12vdc@5.2A. When i test the voltage i get a reading of 1-2vdc so i opened it and i hear a ticking with the frequency of about 2×/second. After removing the heat sinks and testing the voltage again i got a proper reading. So whatever the problem is its intermittent. But as far as i can see there is no relay or anything else i can see that would make such a noise. When i unplug the supply the noise decreases in frequency and then stops. What could cause such a noise in this video. https://www.dropbox.com/s/qrl4aqnctn70doh/20171029_222750.mp4?dl=0 It would be much easier to replace the power supply but i am interested in the knowledge i will gain from diagnosing the problem with the circuit. I can identify certain parts of the circuit and have a basic understanding of most electrical components.
AI: First, a warning: Power supplies like this one contain high-voltage capacitors that will hold a potentially deadly charge even after you unplugged it from the mains. So if you unplug the power supply and then start to work on it, you might still get a pretty dangerous electric shock.
The problem with this power supply is most likely caused due to the controller chip's feedback/bootstrap circuitry not working.
Most power supplies have a controller chip (and switching transistors) on the primary side which creates the waveforms needed to drive the main ferrite transformer. This chip needs to be powered as well, so most manufacturers of such power supplies put an additional low-voltage winding on the transformer which will power the controller chip. This creates a chicken-and-egg problem, however: The controller chip can't start before it has power, but it doesn't get power before it starts and drives the transformer. In order to solve this problem, there's a large-value resistor that slowly charges a low-voltage capacitor from the ~300V line voltage. When this capacitor has enough voltage, the controller IC can start the power supply using the energy in the capacitor and then sustain its own operation using the auxiliary transformer winding.
If the auxiliary winding does not work for some reason, the chip will use up the capacitor's energy and then shut down again which results in a clicking noise from the transformer. This repeats every time the capacitor has reached the start-up voltage of the controller chip, causing rhytmic clicking.
If I had to take a guess, I'd say there's a dry solder joint or a dead diode somewhere in the thing. |
H: Width of Antenna Feed Line for Chip/PCB Antenna?
I am designing a board having nrf52832 and a chip antenna (2450AT18B100 Johanson Technology) for 2.45 GHz (BLE basically). But I have few questions regarding standard 50 ohm characteristic impedance of Antenna Feed Line.
In my PCB, antenna feed line will be coplanar waveguide with bottom ground plane. As far as I know transmission line will have capacitance, inductance and resistance. But the standard suggests to keep impedance of antenna feed line to be 50 ohm (which is resistive).
Does that mean the transmission line will also resonate at RF frequency (2.45 GHz in my PCB) and it will be resistive at that frequency only?
The chip antenna (2450AT18B100 Johanson Technology) states that it has impedance of 50 ohm. Does that mean if my antenna feed line is of 50 ohm, then I won't need any matching components?
AI: If the chip antenna has an input impedance of 50\$\Omega\$, and the line has the same impedance, then the line loaded by the antenna will also look like 50\$\Omega\$ impedance looking in. If that's what your amplifier is expecting to drive (usually is), then you won't need any matching.
If you place a line with the correct geometry, that is correct width on a substrate of specific thickness and dielectric constant, with correct gaps to the topside ground plane, then it will have a broadband 50\$\Omega\$ impedance.
There are many online calculators you can use to give you suitable dimensions. Coplanar with ground has too many variables to keep dimensions in my head, but note that with ordinary microstrip (ie coplanar with ground with very big gaps!) on FR4, you get roughly \$50\Omega\$ with a line that's twice as wide as the substrate thickness. A coplanar line will be narrower than this, as there is some field coupled to to the top metal.
Although the line has capacitance to ground, and you can define an inductance per unit length, they are both uniform. The effect is that when a voltage signal passes along the line, a current signal runs along with it, and they are in the ratio 50 volts to 1 amp, or \$50\Omega\$. This impedance is a nominally constant 50 ohms from DC up to frequencies where the imperfections of the line geometry or dielectric uniformity start becoming significant, often many GHz even for plain FR4, and 10s of GHz for 4350. |
H: Current Draw with Stepped Down Voltage
If I have a 24V source, and then step down the voltage to 5V. Then at this 5V level I draw 1 Amp, what current draw will this be on the 24V source. My gut tells me it's 5V/V24 * 1 Amp mathematically but is this true in real life?
Edit:
DC/DC converter
AI: It depends on what you use to step down the voltage. If you use a switching regulator, you are mostly correct, less inefficiencies in the regulator.
On the other hand, if you use a linear regulator, you’ll see a 1 amp (plus inefficiencies) draw on the 24 volts.
In either case, the losses will show up mostly as heat. With the linear, the regulator will put out 19 watts of heat. |
H: How to determine how long a relay can stay engaged?
I have a device where I want to use relays to run some important equipment. I want to use monostable relays because if power is lost or if there is any sort of system failure I want the connected device to shut off. This means I'll probably be running the relays continuously or nearly continuously for up to a few hours. I'm concerned about coil heating or otherwise damaging the relay.
I've been looking at some datasheets like these G5Q Datasheet, G5LE Datasheet and I'm having trouble determining how long I can safely keep the relay engaged. I don't see anything indicating a max time or max duty cycle. Does this indicate I can simply leave the relays engaged indefinitely or is there a proper way to go about calculating this?
AI: The air temperature around the case of the relay should not exceed the published rating. The referenced data sheets show a maximum ambient temperature of 85C or 105C for one variation. There also derating curves for operation above 40C. That information shows how the relays are intended to be used. If the relays are in a confined space or in a space that has other heat producing items, you need to test or otherwise verify the air temperature around the relay.
The relays should be ok kept operated permanently unless they are in a very confined enclosure. It is unlikely that they would overheat unless the temperature outside their enclosure is also high. |
H: Vbe vs Ic characteristics of NPN transistor at different Vce in active region
Below is an NPN transistor characteristics for Vbe versus Ic at different Vcb or Vce:
It seems like in the active region, Vbe vs Ic curves gets steeper with an increasing Vce.
The following equations relates Vbe to Ic in detail:
But from the above equations how can we conclude that the above curves become more steeper with increasing Vce?
The first part of the equation(Is) has many dimensional terms. Does any of them changes when the Vce increases? I want to relate the effect of Vce to the above curves through the above equations..
AI: The Early effect causes the effective base width to reduce as the voltage on the collector increases.
This in turn causes the collector current to increase for a given Vbe. |
H: What is the name of this connector
Does anybody know the name of the cable that should be inserted in this connector? I lost it so I cannot make a picture.
I know it is a power cable where the other side is inserted in a 200-240 V (non grounded) Dutch (European) wall socket/out.
These types come in two sizes, the most generic is slightly bigger, this one is smaller (but for both I don't know the name).
The distance between the pins is about 6 mm, the horizontal distance 11 mm, and the height around 6 mm.
Brand: Philips
Type : Hair (not beard) shaver
Model: (I will add this later, not near the device)
Type : Power connector (other end of cable to ungrounded Dutch/Europen 220 V wall outlet/socket)
Marks: (I will add this later)
Update:
The model is Philips, HC3410. I could find an adapter but those are over 30 euro including shipping (more than the entire product). However, I found via AliExpress and a Dutch site a compatibility list so I hope it will work. I will not in 2 to 8 weeks.
For those who are interested:
Compatibility list: Compability list
Adapter: Adapter at AliExpress
AI: Your device is likely a Philips Norelco shaver. It is a DC low-voltage polarized connector which was used after it was decided that unconnected line-voltage cords and sinks don't mix. I believe it is proprietary, I haven't seen it anywhere else. If you want a replacement, just look for the proper charger for your model of razor. |
H: 1-dB Gain Compression Point in Amplifier
An amplifier has small-signal gain 15 dB and 1-dB gain compression at an input power of 13.2 dBm. The amplifier is used to amplify four modulated signals of equal power. To avoid problems with intermodulation products, the amplifier will be operated at least 5 dB below the 1-dB gain compression point. What is the maximum output power of each signal?
My workings:
1-dB compression point: 15 dB - 1 dB = 14 dB
5 dB below this value: 14 dB - 5 dB = 9 dB
5 dB below the input power: 13.2 dBm - 5 dB = 8.2 dBm
Divide by 4 input powersL 8.2 dBm - 6 dB = 2.2 dBm
Max output power: Gain + input power = 9 dB + 2.2 dB = 11.2 dB
Is this correct?
AI: I'd work it like this:
The 1 dB compression input power is 13.2 dBm
5 dB below this is 8.2 dBm
You have 4 equal-power inputs, so each must have 1/4 of this much power to avoid compression,
8.2 dBm - 6 dB = 2.2 dBm
This is the input power you'll be operating at. The output power for each signal will be
2.2 dBm + 15 dB = 17.2 dBm |
H: Transistor Current Source Biasing
I'm reading Art of Electronics, and in the section Transistor Current Source, they mention "The base voltage can be provided a number of ways. A voltage divider is OK as long as it is stiff enough. As before, the criterion is that its impedance should be much less than the DC impedance looking into the base (Beta*R_emitter)"
Why is this ?
simulate this circuit – Schematic created using CircuitLab
AI: The current sink will work regardless of their rule. It just won't provide the easily predicted value of current for the load, if you don't follow the rule closely. Let's see why.
After converting the base pair to its Thevenin equivalent:
simulate this circuit – Schematic created using CircuitLab
You can apply KVL and get:
$$I_B=\frac{V_{TH}-V_{BE}}{R_{TH}+\left(\beta+1\right)R_E}$$
Now you can figure the following:
$$V_B=V_{TH}-I_B\cdot R_{TH}$$
Given that the collector current (aka the load current) is \$\frac{\beta}{\beta+1}I_E\$, it must be the case that the load current is:
$$\begin{align*}
I_{LOAD}&=\frac{\beta}{\beta+1}\cdot\frac{V_E}{R_E}=\frac{\beta}{\beta+1}\cdot\frac{V_B-V_{BE}}{R_E}\\\\
&=\frac{\beta}{\beta+1}\cdot\frac{V_{TH}-I_B\cdot R_{TH}-V_{BE}}{R_E}
\end{align*}$$
Substituting in \$I_B\$ you get something like this:
$$I_{LOAD}=\left[\frac{\beta}{\beta+1}\right]\cdot\left[\frac{V_{TH}-V_{BE}}{R_E}\right]\cdot\left[1-\frac{R_{TH}}{R_{TH}+\left(\beta+1\right)R_E}\right]$$
Which can also be written out as (to make the ratio stand out and to emphasize that it is the ratio of two certain resistance values that is important in the following discussion):
$$I_{LOAD}=\left[\frac{\beta}{\beta+1}\right]\cdot\left[\frac{V_{TH}-V_{BE}}{R_E}\right]\cdot\left[1-\frac{1}{1+\frac{\left(\beta+1\right)R_E}{R_{TH}}}\right]$$
Note that the first factor is almost always very close to 1. So it can be ignored. The second factor is the current we'd expect when we designed the resistor divider at the base, in the first place. As you would expect that the emitter would be \$V_{BE}\$ less than the Thevenin voltage and of course this voltage across \$R_E\$ would produce the expected current there. That is, if you use the unloaded divider voltage!
Now, the third factor is the issue here. You want this to be 1, since that means your unloaded divider voltage is the right one to use in predicting your current sink value. But if it isn't 1, then the actual value will be different than the expected one (given no load on the divider.)
If you look at the third factor, I think you can see that if \$R_{TH}\$ is small compared to the value of \$\left(\beta+1\right) R_E\$, then the second term of that factor is close to zero and so the third factor will be close to 1. But if \$R_{TH}\$ isn't small in comparison, then that fraction (the second term of the third factor) will significantly reduce the third factor's value from 1 to something smaller. And so the predicted value won't be nearly as close to the actual value as hoped.
You can also see this as: "If the base current is small compared to the available current flowing through the base pair of divider resistors, then the predicted voltage at the divider will be close to the actual voltage present there and therefore the base will obtain that nearby value and reality will be closer to prediction." That's the qualitative hand-waving that also gets you to the same place.
But it all becomes quantitatively clear in the math, itself. The math not only tells you the same thing as the hand-waving does, it also tells you by exactly how much you might be off if you don't follow the rule by some amount. So it provides both the insight as well as quantities you can use if you choose not to follow the rules. |
H: Higher KV means lower torque for DC motors?
I'm in the process of trying to build an electric skateboard. On the websites I visited I read that torque and KV are inversely proportional which does seem counter intuitive. So I'm wondering why that is the case.
I think some beginning of answer can be found in this paragraph:
In summary, a low KV motor has more winds of thinner wire - it will carry more volts at less amps, produce higher torque and swing a bigger prop.
So yeah, lower kv motors have thinner wires. Thinner wires mean more resistance and less current for the same voltage. I don't really understand the rest though. I don't see the relation with more torque..
AI: Motors can be difficult to reason about, as there are so many variables. So let's analyse a motor and keep the speed, torque and power output constant.
Let's say it has two seperate identical windings. Each winding, for the sake of argument, carries 1A and 10v. Now if we connect these in series, the electrical supply to the motor will be 20v 1A. If we connect them in parallel, the supply will be 10v 2A.
Both windings are receiving exactly the same power, and the motor is providing exactly the same speed and torque, yet the voltage and current at the terminals are different. In both cases the electrical power to the motor is the same.
We could regard the two parallel windings as a single winding with twice the area. In fact, some motors are wound with multiple parallel strands, as it's easier to do than using a single thick wire.
We can see that the speed constant and the torque constant change in exactly the same way, as we fill the winding space available in the motor with either many turns of fine wire or a few turns of thick wire.
Notice that the heat lost in our two windings is exactly the same, whether we connect them in series or parallel. So the efficiency of the motor is also the same whether wound with many thin or few thick turns. All that matters is the size of the motor to receive the wire, and the power you want to run it at.
Most people will match the speed, power and weight of the motor to their vehicle, and then choose a KV to get a terminal voltage at that speed to match a suitable controller and battery, not the other way round. |
H: Is it possible to connect a phase control relay to an inverter (VFD)?
Is it possible to check a motor ON/OFF state by connecting a phase control relay to the inverter and the motor 3 phase terminals?
I have tested a phase monitoring relay to a main 3 phase supply from distribution and the relay works fine but when I connect it to the inverter/motor, it doesn't work. also in one of 6 different possible connection ways, relay get too hot and smoke came out of it! I want to know why? where is my mistake?
*inverter model is: Schneider ATV12H018M2
*phase control relay is: Schneider RM17TG20
AI: No, this is not recommended. The output of a VFD is not suitable of any electronics except the motor, you can't even measure it with your multimeter without putting it in a special low-z (low pass filter) mode.
The output looks like this:
If you'd want to do this, you'd need a harmonics filter in front of the three phase monitoring relay. And then there is no guarantee the device would survive the other features of the drive. (eg: braking and dc hold)
The best way to detect if a VFD is active is to use it's IO functionality. Almost all VFD's will have a one or more relays, analog (0-10V/0-20mA) outputs or digital (24V) transistor outputs that are freely programmable in the software. |
H: What Happens if a Device Is In a Device Tree but Not Physically Present
Probably a very newbie question:
I have a Zynq 7100, zc706 board. I am playing around with the ARM Processing System with Xilinx's PetaLinux distro. What would happen if I added a device to the device tree, say and EEPROM on the I2C bus, but in reality I never physically attached the EEPROM to the bus. I then try to use the I2C Linux driver to read and write to the address in the device tree where the EEPROM is. Will the attempt to open the file just fail? If that is the case what is the point of the device tree, just to tell the kernel to look for the devices/include the driver and build a certain way?
if((file = open(I2C_ADAPTER, O_RDWR)) < 0) {
printf("Failed to open the bus\n");
return -1;
}
buf[0] = addr;
buf[1] = reg_addr;
buf[2] = 0x10;
if(write(file, buf, 3) != 3) {
printf("Failed to write to bus %s.\n\n", strerror(errno));
}
else {
printf("Successful write\n");
printf(buf);
printf("\n\n");
}
if(read(file, buf, 3) != 3) {
printf("Failed to read from the i2c bus.\n %s\n\n", strerror(errno));
}
else {
printf("Successful read\n");
printf("Buf = [%02X,%02X,%02X]\n", buf[0], buf[1], buf[2]);
printf("\n\n");
}
Let's say I feed the code an I2C bus address (addr) of 0x5d and write to the base register reg_addr is 0x00, which is where my EEPROM is in the device tree. How does the system know the device location if I were to connect an EEPROM?
AI: A device can fail anytime, and drivers have to handle such problems gracefully. So, don't oops and and don't run into endless loops. Instead, report an error code to userspace or upstream device driver. (This usually isn't a problem unless a device has VM-mapped RAM or doing DMA transfers.)
The purpose of the device tree is to take the knowledge of individual board designs out of the drivers, so one driver can handle any board which has a certain kind of device on it.
If you never plan to attach a certain I²C chip to the board, it's best to remove it from the device tree definition and recompile. If it's optional, you better place it into an overlay file so the user can decide if he had that hardware or not.
Look for the Raspberry Pi device tree stuff, it has numerous overlays for all kind of stuff to attach. |
H: How many capacitors in single RAM?
Just a simple question,
In https://en.wikipedia.org/wiki/Random-access_memory says that each bit of data in RAM/DRAM stored in a single pair of transistor and capacitor in memory cell.
So let's says the RAM has 4GB memory that mean it can store about 4billionx8bit = 32billion bit of data.
So that i want to ask is, is that true that there are at least 32 billion capacitor in a single 4GB DRAM ? and is that possible to put 32 billion capacitor that can fit on a single board of RAM size? or am I just wrong?
AI: Yes, they really have that many capacitors in that small of an area.
There are two dominant technologies to do this: stacked capacitor DRAMs and trench capacitor DRAMs.
Stacked capacitors basically use a number of layers of metal and insulator to build a capacitor of reasonable capacity in a small surface area.
Trench capacitor DRAMs basically etch a "trench" (a deep, V-shaped one) in the silicon, the deposit a layer of metal, another of insulator, and another of metal.
Either way, you end up with a relatively large capacitance for the surface area. The capacitance is still quite small by most normal standards though. For example as of 2017, a Samsung DRAM has a capacitance around 7.4 fF per cell.
To get meaningful results from such a small capacitance, most DRAMs actually have some extra capacitors in addition to those used for the storage.
To read a cell, you charge one of these spare cells (one that's physically close to the cell you want to read) with approximately half the charge you'd use to store a 1 in a normal memory cell. One easy way to do that is to use two capacitor cells together, so feeding the same voltage and duration of charge pulse into them results in half the charge in the capacitor.
Then you read back the values from the spare cell and the memory cell and feed them both into a differential amplifier (the "sense amp"). This helps cancel most common mode noise on the bit lines, so the signal coming out of the sense amp is a fairly clean low or high value, with substantially better noise immunity (and from it, improved reliability) compared to just reading the voltage from the capacitor by itself.
In addition, a typical DRAM will have some extra banks of memory. When the chip is being tested at the factory, they may find that one of the normal banks of memory has a defect. If so, the chip will typically include some fuses (or anti-fuses) that can be blown to substitute a spare bank for the defective one, so a chip can still usually meet spec, despite a defect or two.
Thus, a DRAM chip will typically have even more capacitors than you get from computing its size based on what it's rated to hold (though, in all honesty, the increase is fairly small, at least as a percentage--though with something like a 32 GB memory, even a small percentage works out to an absolute number that's fairly large).
As a final note: a DRAM chip has to have a fair amount of circuitry (decoders, sense amps, etc.) in addition to the DRAM cells themselves. As a really simple rule of thumb, figure that the actual cells occupy about half the chip area, and the associated circuitry the other half. |
H: How to remove folded battery from phone without starting a fire?
I was trying to replace the battery in my phone, but since it was very well glued to the phone, I accidentally folded it in half. This immediately caused some dark brown smoke to come out from it, so I took it outside.
A few hours later, it seems to have stopped making smoke, and the bottom folded part looks very swollen.
I'd like to take it out, but I'm scared it will do something like in this video. Is there anything I can do to avoid it from catching fire? Also, are the fumes it releases dangerous?
AI: a bloated Li-ion/Li-Po battery is very dangerous and should be handled with lot of care.Usually people would discard if the battery is in a dangerous position to be handled. I'm going to assume that your phone is expensive and you wouldn't want to discard it.
Since, you have mentioned that after keeping it out for a while it has stopped making smoke you could do the following:
1.Wear full sleeve cloths, thick fire safety gloves(most important) and some kind of face protection.
2.Always use some kind of prying tool/spudger that is used in mobile repair to remove batteries.
Something like in video could occur if the battery has some charge left in it.
Overall, its a real risk trying to repair the phone. Proceed with caution before you attempt to. Also keep it in a safe place where it wouldn't cause fire.
http://ehs.whoi.edu/ehs/occsafety/LithiumBatterySafetyGuideSG10.pdf |
H: What is the purpose of this NMOS transistor on an I/O pin?
I have the circuit structure that looks as shown below. An I/O signal, sigA, (active low) is driving the P1 port on unit U1. An NMOS transistor's drain is connected to this signal with the gate/source/bulk connected to ground. How does this NMOS transistor work and what is its purpose?
The gate/source are connected to ground, so I am not sure how current will flow through (i.e., what condition will make M1 turn on). It looks like a pull down device, but since sigA is active low, it doesn't make sense to me.
simulate this circuit – Schematic created using CircuitLab
AI: If this inside an IC it is for ESD protection and referred to as a snap diode or ggNMOS device.
It is intended to breakdown (without suffering damage itself) before the gate oxide of the IC when it is subjected to ESD due to static electricity.
It relies upon a parasitic bipolar transistor being formed during the normal CMOS fabrication process. This bipolar transistor avalanches when the voltage exceeds a safe level and dissipates the input ESD strike. |
H: Commutation of BLDC motors with respect to number of poles
I have a BLDC motor with HALL sensors. Datasheet reads the motor is 4 pole technology. Somewhere else it says "Number of pole pairs = 2". I can only assume that they both mean the same thing. It also features 3 HALL sensors at 120°
My question is, what would be fundamentally different in commutation of a BLDC motor with respect to the number of poles, like sequence or number of switching on/off for driver transistors?
For example, If I have a fully working code for a 6 pole or 8 pole motor, what would I need to do to make it work with 4 pole as well?
AI: 'Pole pairs' is the number of North and South magnet pole pairs. 'Poles' is the number of individual North and South poles (always an even number).
Unless it has a separate rotation sensor, The controller cannot tell how many poles the motor has or what mechanical speed it is doing. From the controller's point of view all motors have 2 poles and 3 phases, which requires 6 commutation steps per 'electrical' revolution. So you could have a 4 pole motor that needs 12 steps per revolution, and the controller will think it has done a full revolution when the rotor has only gone half way around.
All else being equal, a controller which works with 6 or 8 poles should also work with 4 or 2 poles. Secondary factors might have an effect, eg. 2 pole inrunners often have very low inductance and are designed for high rpm, which might not suit a controller designed for high torque outrunners. |
H: How to handle 2nd USB and Audio ground in a project powered from isolated 5V Wall Wart
I've read a dozen questions that are similar to mine, but have not found any responses that provides the specific guidance I am seeking. Here are some that I've read:
Ground Connection When Using AC Wall Wart Supply
When using USB and external power, how should I isolate the grounds?
How to safely connect the Arduino 5V Pin with a 5V wall wart and USB TTL converter for data transfer?
USB port and bench power supply ground difference
Common ground : Computer A and USB powered Arduino ( MCU ) from Computer B
Here is my situation: I've built an embedded device project that is powered from an FDA-compliant 5.4V floating wall wart. It connects to an earth-ground referenced laptop through a primary (isolated) USB serial emulator. In that scenario, all is well.
The problem, as expected, occurs when I introduce an earth-ground reference into the circuit. This can happen in several ways.
In the first case, I need to connect an FTDI serial emulation cable for out-of-band debugging output. The FTDI cable may go to the same laptop or to a different laptop, but the FTDI ground will always be connected to USB ground somewhere near the level of earth ground. The FTDI power pin is left disconnected since the device is self-powered, but the ground is required for D+ and D- signals to be transmitted.
In the second case an unbalanced audio connection may sometimes get plugged into a grounded mixer.
Finally, I need to be able to put an oscilloscope on the board, and the scope ground is at wall-ground.
By design, power and ground from the wall wart float WRT ground, and my scope shows that both positive and negative carry a 60 Hz AC potential of ~150 volts P-P relative to earth ground. My Fluke multimeter reads this as 48V (I assume this is RMS of the alternating voltage, which isn't really very sinusoidal). At first I thought there was something wrong with my wall wart, but I've tested a number of different models of 5.4V 2A wall warts from different manufacturers, and they all look about the same on the scope.
Within my project, the (floating) ground of the wall wart and the ground reference from the FTDI cable and/or the audio mixer are all connected via the ground plane of the PCB, but I am seeing some real problems with this.
When I connect the FTDI serial emulator ground, the I2C bus on my board runs inconsistently and eventually locks up.
When I connect the audio output of my floating PCB to one channel of a mixer, even a channel that's turned all the way down, 60Hz hum affects every channel of the mixer.
When I connect scope ground, the PC sometimes drops and reconnects its USB device and the embedded system may reset
These problems can be made to go away by tying the floating wall wart power supply ground directly to another physical ground (e.g. my scope ground). However, this introduces a huge amount of digital noise on the ground lines. This (and ground loop issues) originally led me to an isolated power supply and isolated USB signal path, and a lot of things got better after I installed isolation, so I'm not sure I really want to undo that if there's an alternative.
Can anybody point me towards an optimal approach to this problem? It would be possible (though expensive) to require that the FTDI connection also go through a USB signal isolator, and presumably something similar can be done for the audio signal. But is there a better way?
AI: I would recommend isolating the FTDI chip by using opto-isolators on the UART lines. It is much cheaper and simpler than isolating USB.
The FTDI ground can then be referenced to external ground without affecting the rest of the circuit. |
H: Battery pack a little bit damaged. Should I throw it out or repair?
Today my 48v 15Ah LiFePO4 ebike battery fell off from my bicycle and got a little bit of road rash on the corner:
It still works. Does not seem to be heating or discharging on its own.
However, between battery minus and a tiny bit of the exposed area (not all exposed area in the picture is metal though) there seem to be +2V, which is making me somewhat nervous.
What would happen if I would simply apply liquid electrical tape on that area and call it a day?
Thoughts going on in my head:
Could internal short happen and battery catch on uncontrolled fire?
Does LiFePO4 chemistry play any role in this as well and add to the safety?
I don't know what form factor cells are inside the battery pack and if that plays any role as well. Obviously dont seem to be anything durable like 18650.
Should I simply throw away this battery and get a new one?
P.S. since its Halloween I am up to hear all horror stories on what may happen...
AI: LiFePO4 is a fairly safe chemistry and can take more abuse than most. The electrode material is far more stable than that used in LiPo/Li-ion cells, and most if not all incidents are electrical (i.e. short circuits leading to fire) rather than chemical in nature.
Nonetheless exposed metal surfaces should be insulated using a generous layer of silicone caulking or similar material (certainly not "liquid wire", since such a liquid is actually conductive). For additional protection the entire battery pack should then be placed in a silicone or rubber sleeve in order to provide further protection. |
H: 5A connector that fits through 1/4" hole?
I'm powering a load that draws 5A at 18V from a brick power supply. The power supply and load are separated by a thin metal wall that has 1/4" (6.35mm) holes in it. So I'm trying to find connectors that can fit through 0.25" diameter holes.
The best I've found is small JST plugs, but they're more like 0.29" so we'd have to file them down to fit through the holes.
I'd appreciate any suggestions, either for a 2-circuit connector or two 1-circuit connectors.
AI: A bit old fashioned by modern standards but 2mm plugs and sockets would do the job. This example is a line socket that is rated at 10A and measures 5.7mm in diameter, plugs would probably be thinner.
The only snag is lack of polarisation so you'd have to rely on colour coding. |
H: Switching Voltage Regulator Problems
I'm new here, so forgive me if I breach any forum protocols in the asking of this question.
I have designed a voltage regulator scheme for a guidance system on an aerial drone. All components require either ~4V (they have an input range) or 3.3 V exactly. My solution is pictured here in a schematic and a board layout. I used the TPS53318 switcher to regulate my battery voltage down to 4 volts with roughly 3 mVpp noise, then used an LP38502 LDO to convert that into nice clean 3.3 V with as little wasted power as possible. Datasheets here: TI TPS53318 and TI LP38503.
Schematic:
And board layout(image isn't very good, but I figured I'd include it just in case it helps):
So, my problem: It doesn't work. The output from the switching regulator is a very small oscillating negative voltage, which I assume is caused by some sort of LC oscillation in the absence of any real output.
I've isolated the LDO, and it does it's job fine when supplied with 4V from an external supply.
I have tried reflowing my solder, even replacing the switcher itself (I always order 3 times as many as I think I'll need) and I still see the same problem.
I've done my best digging around with a multimeter and a scope trying to find a resistor with the wrong value or something, but I'm coming up empty handed.
This is my first time trying to troubleshoot this type of circuit, or any electronics this small on a proper board, so I'm hoping that someone here might point out errors I might have made or possible bugs I can attempt to diagnose. The oscillating negative voltage output still seems odd to me, and I'm not entirely satisfied with my idea that it is caused by LC dynamics, so I'm also hoping that someone might be able to suggest a possible cause/solution for that effect.
AI: The first thing I noticed when looking through the datasheet and your schematic was your output components. First of all, your output caps look rather low in value. You have 4x 33uF where the 2 examples on the datasheet give you 4x 100uF and 2x 330uF. These are usually a good guide as to what your output capacitor values should be. I'm unsure of your exact requirements but double check your calculations for those. Do the same with your input capacitors too. Yours again seem quite a bit lower than the ones in the datasheet. I apologise if you have calculated these and they are correct, but it usually works for me to use the datasheet examples as a good guide. If they ARE too low, perhaps that could be a cause of your output not working correctly.
The other thing is your PCB layout. With these kinds of IC's (with a power pad underneath) it usually indicates they need ALOT of solid grounding work on the PCB. You should at the very least have yourself a nice big ground plane on the bottom side of your PCB and some vias underneath the IC to allow it to dissipate heat properly. When I first designed circuits using IC's like this, I had a number of failed units due to insufficient grounding, this is really critical in these designs. Your input and output capacitors should be located as CLOSE to the IC as possible which it doesn't quite look like they are.
I know it is a frustrating thing to redesign a PCB that you have already had made, but sometimes you need to.
Just make sure first that every component is in the right place, the values are correct and orientation of components is correct (on those that require it) before thinking about my suggestions.
I think it could most likely be the PCB design over anything else though.
If anyone would like to disagree and correct me on something I may have missed, please feel free as I know what I suggested is a pain in the a** to do! |
H: I painted myself in a corner. How to layout?
I'm a hobbyist working mostly in DC projects. For development and testing, I use breadboards, and later a perforated board. My projects are normally low count: an Arduino Nano/Pro Mini + a few ICs.
All my works are one-of-a-kind, and making a PCB is out of question for me.
I use the ICs and physical board itself to test the layout of components on board. I have Fritzing, but it's not up to the task for planning the physical layout; it's more like an ideal vision of the circuit where wires have no thickness and every thing fits neatly in the matrix.
My problem is that sometimes I paint myself in a corner: after soldering a few things, I found the next pin buried under other wires and components; no way to solder anything. I'd been searching for tips, found none. How is your flow work? How do you do these things?
Here some questions:
Solder all component first and wire them later?
Wire each component directly to Vcc, Gnd and shared signal or use rails?
How to layout for changes? Maybe I need to add something in the future.
AI: Your suggestions are spot on.
Solder your components first, but think about what you could use as a wire, e.g. a resistor can be used as a jumper. ICs are the main thing to put in first, as they are set and will have a lot of dense connections. Try and work in a matrix, with all wires/components running along rows and columns.
Rails are best - you'll have lots of things going up to V+ and down to ground so fit around that.
Leave yourself space in the prototyping phase. You can scale down later if need be, but once you've started small it's hard make it larger. A bit of planning will go a long way :)
You could also try some simple stripboard layout software. A number of them are paid (and surprisingly expensive), but PEBBLE is free and is a good starting point at least. |
H: What do these IAR IDE icons mean?
Quick and easy question - if you know the answer. I can't find it mentioned in the documentation. What do the icons circled in red mean?
AI: Found in the built-in help (Workspace window):
The column that contains status information about option overrides can have one of three icons for each level in the project:
Blank There are no settings/overrides for this file/group.
Black check mark There are local settings/overrides for this file/group.
Red check mark There are local settings/overrides for this file/group, but they are either identical to the inherited settings or they are ignored because you use multi-file compilation, which means that the overrides are not needed. |
H: Differential Amplifier Input Line-Level Signaling
This is the circuit I'm referring to:
This stage is often used as input stage of audio amplifier.
I am wondering, how can the input signal be normally amplified since there is no DC bias across the base of both transistors?
In my book, it says that we have to drive this stage with "line-level" amplitude signal (that is around 1V). But still, if the amplitude swings for 1V from its reference point which is probably 0V, then only a bit of a positive signal wave would be amplified, since transistor starts conducting at Vbe of approx. 0.7V!
Without base bias voltage, the output signal would look like this, right?
Instead of this, right?
AI: Since Io is a current source the emitters of the transistors will actually be at approximately -0.7V so the transistors will be in their active region.
If the emitter resistors are too low however the amplifier will saturate if the signal goes to high or too low though. Normally this type of stage is used where Vi2 is derived from the output of the following amplifier and provides negative feedback. |
H: Where is the best place to tap into this broken USB flash drive with a new cable?
Looking for some advice on attaching a USB cable to a USB drive of which the original USB connector has snapped off. It has a bunch of my girlfriend's coursework on it and of course she has no backup...
I have a reasonable foundation knowledge of electronics but am a bit hazy on PCBs.
From what I can make out from the image below, there are 4 solder pads for the 4 terminals of a USB connector. The connector simply solders onto these pads. Normally this would mean that I can simply reflow the solder to reattach the connector.
However in this case, I think that the solder pads themselves have broken away from the PCB as the material exposed seems non-conductive. With this in mind, the next logical thing is to try and expose some metal from the PCB wiring and solder to that. You can see in the image that I have started to do that with the leftmost terminal.
Both this and the rightmost terminal will be okay in this sense but I can't actually make out where the PCB wire is routed for the middle two terminals to scrape away at it.
Was wondering what people would suggest as a means of connecting into this? It's not got to last, just needs to survive long enough to pull the data off of it.
I have also included side and back images for reference.
Any and all help would be appreciated, thanks!
AI: Three of the four pads ripped off the board along with the connector. You will have to very carefully scrape the resist and connect to the vias for the D+ and D- lines as shown here:
The colors notated above are the wire colors you'll find in most USB cables if you cut off the "B" end and expose the wires.
EDIT: One way to make the resulting assembly a little more rugged would be to use another board as a strain relief for the cable. Hopefully, this sketch makes sense:
Get a piece of pad-per-hole board or even just perfboard, and tape the USB board to it, after putting something like a piece of cardboard or tape between the two boards to prevent shorts. Solder the wires from the USB cable to individual pads on the bigger board, and then use single strands of very fine wire to make the final connection to the USB board.
This way, any movement of the USB cable relative to the larger board won't put any strain at all on the delicate connections to the USB board. |
H: A capacitor stores half the energy when charged from a battery each and every time?
I understand that when a battery is connected to a capacitor that half the energy is stored in the capacitor while the other half is lost to heat or whatever other losses are involved once the capacitor is charged to the exact same voltage level of the battery.
Only a one time direct connection with a simple switch from open to closed that charges the capacitor up to the voltage level of the battery is what is being referred to. My question is does this occur each and every time?
Once the capacitor is charged to the voltage level of the battery then there should be 7.2 milliJoules of energy in the capacitor.
If the answer is yes that half the energy is transferred into the capacitor each and every time then what would the capacitor voltage have to be in order to equal 100% of the energy if it were to come from the battery?
I understand that in practice this will never happen, which is why the statement equal to 100% of the energy is indicated. (Meaning by some other method of charging means)
simulate this circuit – Schematic created using CircuitLab
AI: Your statement that "half of the energy being stored in the capacitor and half being lost to dissipation" is not very specific to begin with. It is true that the energy lost to dissipation is equal to the energy stored in the capacitor once it is charged, but only if the capacitor is initially completely discharged. In this case, half of the energy removed from the battery went to the capacitor, and half was dissipated.
As the initial voltage on the capacitor approaches the battery voltage, the ratio between the energy transferred to the capacitor and that removed from the battery approaches 1. This explains why the efficiency of a charge-pump DC-DC converter theoretically approaches 1 as the load current decreases. |
H: Polar plot of s/(1+(s*T))
The magnitude at w=0 comes out to be 0 (=1/infinity) , which is different from previous problems (they didn't have any zero) I've calculated, is this right? I guess it should be right (because G(s)should be zero at w=0, that's what comes to my mind if u ask me meaning of having zero at s = 0) but I'm not sure how to do Polar plot if this is right.. pls help me
PS: Did I calculate magnitude for w = 0 right?
AI: As requested in the comments, here is the polar plot for various \$T\$ values as \$\omega\$ goes from \$0\$ to \$\infty\$. |
H: Trouble with voltage BJT circuit in active mode
For this circuit i need to find the Vi voltage to get Q2 in active mode.
The result i got is not close to the values in the simulator, probably theres something wrong with the circuit analisys.
Where´s the mistake ?.
Easy to see that:
Vce4 = 0.7v
from 20 - 1kI - Vce4 = 0 and vce4 - Vcb4 - 0.7 = 0 if Vcb4 = 0
Then:
20 - 0.7 = 1kI having 19.3v in R4 with 0.0193 A.
Also:
The Vce3 = 19.3 from 20 - Vce3 - 0.7 = 0.
Thats the easy part, now i know the Q3/Q4 is a current mirror. So, the I current I i just calculated is aprox to Ic3.
If i suppose Q1 OFF:
Ie2 = 0.0193A; that means Ib2 + Ic2 = 0.0193A.
If Q2 is in active mode:
Ie2 = Ib2(B + 1) , Ib2 = Ie2/(B+1) if B=100
then Ib2 = Ie2/101 and Ic2 = Ie2/(1+ 1/B):
Ic2 = 0.0191 A.
That sounds good but when Q2 is saturated. Vce2 = 0.2v and Vcb2 = -0.5v:
20 + 10 - 19.3 - 1kIc2 - 0.2 = 0
Ic2 = 0.0105 A.
Q2 is in saturation mode.
From Ie2 = Ib2 + Ic2, Ib2 = Ie2 - Ic2 = 0.0088 A.
Now:
If i suppose Q1 is in active mode having in mind that -0.7v + Vce1 = 0
Vce1 = 0.7v
With
Vi - 100kIb1 - 0.7 - 19.3 + 20 = 0
Vi = 100kIb1. This sounds weird.
Is Q2 is in active mode then: Ic2 = Ib2; 10.7 = Vce2 + 1kIc2; Ie2 = Ic2(1+1/B)
From Ic2(sat) = 0.0105 A that would mean: Ib2 = 0.000105A
Using:
Ie2 + Ie1 = 0.0193A --> Ie2 = 0.0193 - Ie1 and Ie2 = Ic2 + Ib2
Ie2 = 0.010605 then Ie2 < 0.010605 for 0.010605 < 0.0193 - Ie1
Ie1 > 0.008695 A.
Ib1 = Ie1/(B+1) and Ib1 (B+1) > 0.008695
Ib1 = 8.6x10^-5 A
Then:
Vi/100k = Ib1
Vi > 8.6 V for Q2 in active mode.
I simulated the circuit and for Vi = 8v the Vce2 = 0.4V.
Also the current Ice3 is always 24.2mA, and The Vce3 is not 19.3V, its aprox to 18.4V for low Vi voltages. But that would mean the Vbe2 is not 0.7V.
Thanks for your help and time.
AI: I agree with your work regarding estimating the current in \$R_4\$ as \$I_{R_4}\approx 19.3\:\textrm{mA}\$. So your analysis there is fine. Yes, \$Q_3\$ and \$Q_4\$ make up a current mirror, so you'd expect a similar sinking current in the collector of \$Q_3\$. (I'm also assuming here that we do not need to worry about the Early Effect, which in this case would otherwise probably be an issue.) \$Q_3\$'s collector should have a compliance voltage down to perhaps \$-19.3\:\textrm{V}\$. (That won't really matter because the emitter of \$Q_2\$ will keep it from getting anywhere below \$-1\:\textrm{V}\$.)
Now, you need to assume that \$Q_2\$ is active. This means \$V_{B_2}=V_{C_2}=0\:\textrm{V}\$. (Easily seen in the circuit.) Until this point, \$Q_2\$ will be saturated and a lot of the current mirror's sinking current will also come from the base of \$Q_2\$. But right at the point where \$Q_2\$ becomes active we can assume \$\beta=100\$ and its base current will be "normal." This also means that the current in \$R_3\$ must be \$I_{R_3}=\frac{10\:\textrm{V}-0\:\textrm{V}}{R_3=1\:\textrm{k}\Omega}=10\:\textrm{mA}\$. So the emitter current will be the collector current plus 1% for the base current, or \$I_{E_2}=10.1\:\textrm{mA}\$.
\$Q_1\$'s collector has the remaining difference of \$I_{C_1}=19.3\:\textrm{mA}-10.1\:\textrm{mA}=9.2\:\textrm{mA}\$. With \$\beta=100\$, the base current for \$Q_1\$ must be \$I_{B_1}=\frac{9.2\:\textrm{mA}}{100}=92\:\mu\textrm{A}\$. That has to be supplied through \$R_1\$. So the voltage across it must be \$\mid\:V_{R_1}\mid\:=92\:\mu\textrm{A}\cdot 100\:\textrm{k}\Omega=9.2\:\textrm{V}\$.
The only remaining question is, "what is the voltage at the base of \$Q_1\$?" Well, we don't actually know exactly. But it must be very close to \$0\:\textrm{V}\$. But I think your problem assumes all the \$V_{BE}=700\:\textrm{mV}\$; and in any case we know that \$Q_1\$ itself would just be barely active if its base were at \$0\:\textrm{V}\$. So I think it's reasonable here to argue that the problem wants you to take \$V_{B_1}=0\:\textrm{V}\$. And there isn't anything much here to argue strongly against that assumption. So there it is.
So this means the answer is, \$Q_2\$ becomes active when \$V_i\ge 9.2\:\textrm{V}\$. |
H: How can i find the current divider in the right resistor valued as "2 * R1"?
simulate this circuit – Schematic created using CircuitLab
My intention here is to find the current in the right resistor valued as "2*R1" using the current-divider formula. I don't know how to interpret the connections between these resistors.
AI: Since this is a homework question you just get a hint:
simulate this circuit – Schematic created using CircuitLab
Figure 1. (a) Original schematic. (b) Rearranged schematic.
You should be able to work it out from Figure 1b. |
H: Designing a 50Ω coplanar waveguide on 1.6mm FR4
I'm trying to design a 50Ω coplanar waveguide to connect the output of a LoRa module to a SMD metal antenna. The two images below show the top layer (red) and bottom layer (blue). Vias are green and both copper pours are connected to ground. C7, C9, C10 are tuning components - not all will be caps and not all will be fitted in the final design.
Board thickness is 1.6mm, dielectric is FR4 (which I know isn't ideal and doesn't specify exact performance - cost reduction is important to this design). Feed trace width is 1.5mm, gap between feed trace and copper pour is 0.3mm and copper thickness is 35um. Operational frequencies are nominally 868MHz and 915MHz.
Does this look like it has sensible dimensions for a 50Ω feed? I used a few different online calculators (each of which with varying degrees of variable control) and it seemed close - but I have no experience of any RF work so I wanted to run it past some more experienced eyes. Also any general comments on the RF section layout would be welcome.
EDIT: Following on from @ThePhoton's comments, I have shifted the antenna feed slightly - see updated top copper image below:
AI: Your trace width and spacing calculate out as a 50 ohm trace on a couple of calculators I checked, so that should be okay.
But I see a couple of other issues you should look out for:
In the original design, the return path is broken due to the cut-out around pad 4 of the antenna.
As shown on the app note from the vendor, the matching network should be as close as possible to the feed point of the antenna. The antenna is designed with plenty of space to fit the components beneath it. How to convince your CAD tool to allow this without a DRC error I don't know.
As shown in the app note from the vendor, the ground beneath the antenna component should be solid. This means the trace should be moved to an inner layer or the back side of the board.
With the matching network placed under the antenna (according to vendor recommendations), you may have to remove the antenna to adjust the matching network during tuning. Given the heavy thermal mass of the antenna and the un-connected pads that it mounts to, the pads are likely to lift when you try to do that. Be prepared to discard several boards if you have to tune that matching network. |
H: Application of superposition theorem
I find superposition to be a noob's approach towards the analysis of circuits containing multiple sources... An experienced person will probably not use it... How do we deal with various sources at a single time? Is there any trick which makes the analysis easier? How should we approach?
AI: There is nothing wrong with superposition. But I would go like this (EDIT - applies to the initially present but currently deleted example circuit):
simulate this circuit – Schematic created using CircuitLab
Two simple equations with two unknowns:
$$ I_1 = I_2 + I_3 $$
$$ V = I_2 \cdot R_2 = V_1 + I_3 \cdot R_1$$
Substituting the first one into the second one will give
$$(I_1-I_3)R_2 = V_1 + I_3\cdot R_1$$
The only unknown here is \$I_3\$ and I will let the reader to solve the rest.
That said, it is not a very systematic approach, but it is easier sometimes for simple circuits like this one. For more complex circuits the superposition and other formal methods are better suited. |
H: Why the difference between the RSSI and Rx Level measurements is so big?
I'm new in Telecom. Today, I downloaded the Android APK called "Network Cell Info Lite".
In my house, I got this measures:
RSSI: -51
RXLEV: 60
I know this measurements represents the received signal. I mean... the greater the value, the higher is the power received by the phone.
I would like to understand which are the differences between two measurements and why they have so different values?
AI: The inverse square law attenuation of path loss or Friss Loss is the main factor followed by blocking effects of trees, or metal mesh or patio and then inverse phase reflection cancellation ( Ricean losses). If 1W is +30dBm then it quickly drops in level to some distance ? 500m ? to -80dBm to noise threshold -90~-100dBm. Depending on the RF band where speed and signal strength can affect performance tradeoffs, the % signal may be from 0 to 30% and excess signal strength from 30 to 100% , but this is just an estimate on my part.
Basically RSSi voltage is converted using log Amps with calibrated dBm and signal level in 0-100% is a linear conversion spanning more than 5 decades. (est.) |
H: How to connect two power sources to a switch?
I want to operate LED lights using either batteries or an AC/DC adapter wall plug. I have a SPDT switch but I’m not sure how to connect the two power sources to the switch. I might be using the wrong switch, if so what switch do you suggest? The main thing is that the AC power source is never connected to the battery and to have a choice by either operating the led lights by battery or AC/DC wall plug.
AI: simulate this circuit – Schematic created using CircuitLab
Figure 1. Wiring the components.
If your LED doesn't have built-in current limiting you will need to add a resistor to limit the current to a safe value. |
H: Is the 50% loss of energy when charging a cap from a battery a set rule in stone?
This question is similar to A capacitor stores half the energy when charged from a battery each and every time?
(And no it is not a duplicate it is different in that it has been made more clearer.)
What seemingly was originally thought as an extremely easy question was not well understood. So it has been reworded again to be even more specific. So not a duplicate question.
It is meant to be understood that a capacitor starts with 0J and 0V in it. Next the switch is closed in order to connect the battery to the capacitor in order to charge it up to the exact same voltage level of that of the battery.
There is now 7.2 milliJoules of energy stored in the capacitor.
I am only referring to the schematic presented, which works exactly like what the schematic portrays. This is a real world circuit done on the work bench. It has losses due to resistance in the wires, capacitor, battery, and switch.
It is not lossless and so it is not perfect, there are losses involved as mentioned immediately above.
Half the energy is lost in doing work to transfer the other half into the capacitor. In other words the amount of work it takes to transfer half the energy into the capacitor is exactly equal to the amount of energy stored in the capacitor. This is a well known fact of life and electronics.
In other words, if 2 Joules of energy came out of a battery, then only 1 of those Joules would make it into the capacitor.
What I am asking is: When charging a capacitor from a battery, is this 50% loss of energy and 50% stored energy in the capacitor a set rule in stone?
I am referring to this exact circuit only or any other similar circuit with different capacitance's or different battery voltages with the exact same process of simply closing the switch just one time only.
I am not referring to ramp charging, step charging, or anything other type of charging. No other components are involved or added into the circuit. Just a simple closing of a switch to connect the battery to the capacitor only.
I know that this is another question but it's directly related to the first question, I don't want to ask it in another question and risk having a duplicate question, which it shouldn't anyways.
Other question: If the 50% loss of energy in order to transfer the other 50% energy is a set rule in stone then what voltage would the capacitor have to be in order to double the energy in the capacitor to make it equal to 100%?
In other words, what voltage should the capacitor be when it has 14.4mJ of energy in it, which should be double or twice the amount of the 50%, if I am getting that correct at all.
It should not be inferred that I am implying that the capacitor is capable of receiving 100% of the energy transferred from the battery as this is completely impossible and preposterous with the given circuit, but only asking what voltage would the capacitor have to be in order to double the 50% of the energy transferred into it?
AI: So the voltage in such a capacitor is the integral of current over time (\$V=V_0 + 1/C * \int_0^TI dt\$).
Right away we see the problem. The perfect voltage source will force a voltage of 12V across the capacitor. Meanwhile, the ideal capacitor will force the voltage to be 0V, and start integrating the current that flows through it. The voltage cannot be 0V and 12V simultaneously.
Battery resistance to the rescue! The circuit has resistance. In fact, a good chunk of it is the resistance of the battery. So we really have a voltage source, a resistor, and a capacitor.
simulate this circuit – Schematic created using CircuitLab
This is more manageable. When you throw the switch, the voltage across the capacitor is 0, the voltage across the resistor is 12V, and everything is sane. At t=0, 100% of the power flows through the resistor. This means there's current flowing, and the capacitor can charge. This increases the voltage of the capacitor according to the following equation:
$$ V = V_s(1 + e^\frac{-t}{RC})$$
where V_s is the source voltage (12V), t is time, R is the resistance (of the parasitic elements like the battery), and C is the capacitance of the capacitor.
Now we can find the current. Since all the current flows through the resistor, and the resistor has \$(V_s - V)\$ volts across it, the current at any moment in time is \$\frac{V_s - V}{R}\$.
Now we can look at the power loss in the resistor: \$(V_s - V)^2/R\$. Integrating this over time, we get
$$\int_0^T \frac{V_s - V_s(1 + e^\frac{-t}{RC}))^2}{R}dt$$
$$= \int_0^T \frac{V_s^2}{R}e^\frac{-t}{RC}dt$$
$$\left[\frac{1}{2}V_s^2Cexp(-2t/RC)\right]_0^T$$
$$\frac{1}{2}V_s^2Cexp(-2T/RC) - \frac{1}{2}V_s^2C$$
Now we're interested in "completely charged" which is what we get as T approaches infinity. In this case, the first term drops out completely (because \$\lim_{x\to \infty}e^{-x} = 0\$). The final energy lost across the resistor is \$\frac{1}{2}V_s^2C\$.
Now, when the capacitor is fully charged, It's stored energy is defined just by the instantaneous voltage across the capacitor. \$\frac{1}{2}CV^2\$. Note that when \$V = V_s\$ (i.e. fully charged), this energy stored in the capacitor is exactly equal to the energy lost by the resistor.
This shows that if you charge a capacitor with nothing but a real voltage source (i.e. battery), you must lose 50% of the energy (to heat). |
H: Why is such amplifier configuration chosen for HB100 microwave sensor?
For HB100 microwave sensor module, I don't quite understand how this amplifier works and why it's been chosen instead of something simpler, such as two inverting or non-inverting op-amps. What can see for the diagram is that they are feeding integrator's output into an inverting differentiator's input. One friend suggested that a differentiator forms a band-pass filter and an integrator+differentiator combination cancels each other out, leaving you with a linear amplifier with a band-pass filter.
If that's that case, why would we want that?
How does that work, briefly described, and why has this circuit been chosen as an amplifier for this module (I am ignoring all the passive filters and coupling/bypass capacitors).
I and my co-worker are going to build it tomorrow and if anyone could shed some light I'd really appreciate it.
Datasheet (diagram at Annex 1): http://www.theorycircuit.com/wp-content/uploads/2016/09/HB100_Microwave_Sensor_datasheet.pdf
AI: Figure 1. The various filter componenents.
The 4.7u and 330k at (1) form a high-pass filter. Cut off is at \$ \frac {1}{2 \pi RC} = \frac {1}{2 \pi \cdot 4.7 \mu \cdot 330k} = 0.1 \ \mathrm {Hz} \$.
Figure 2. An active high-pass filter. Source: Electronics Tutorials.
The RC networks (2-3) and (4-5) form active band-pass filters.
The low frequency roll-off is at \$ \frac {1}{2 \pi \cdot 4.7\mu \cdot 1M} = 0.034 \ \mathrm{Hz} \$.
The high-frequency roll-off is \$ \frac {1}{2 \pi \cdot 2n2 \cdot 10k} = 723 \ \mathrm {Hz} \$ for the 10k and you can work out the 8k2 version.
One friend suggested that a differentiator forms a band-pass filter and an integrator + differentiator combination cancels each other out, leaving you with a linear amplifier with a band-pass filter.
The integrator and differentiator would only cancel out if they had the same cut-off frequencies. They don't so you get a band-pass filter.
If that's that case, why would we want that?
The datasheet explains on page 3: The magnitude of
the Doppler Shift is proportional to reflection of transmitted energy and is in the range of microvolts
(µV). A high gain low frequency amplifier is usually connected to the IF terminal in order to amplify
the Doppler shift to a processable level (see Annex 1). Frequency of Doppler shift is proportional to
velocity of motion. Typical human walking generates Doppler shift below 100 Hz. |
H: Coaxial DC power cable for switching AC adapter required?
Recently the DC cable from my AC/DC adapter to my laptop broke. I removed the broken section and it works again. Now the cable is too short for my taste and I would like to replace it. I heard from an electrician, that you should stick with the coaxial cable type (as it was used for the device, see picture below), saying something about the higher frequencies involved due to the switching-mode transformer. This is supported by the fact, that all major laptop producer seem to choose this kind of cabling.
So firstly I was wondering if that is true. After all, the output should be DC. Even if it contains spurious HF components, what would the coaxial cable help about it? Would using a "regular parallel cable" and a ferrite choke ring have the same effect?
Secondly I could not find any cable like this anywhere for sale¹. Do they have a special name under which they are sold?
¹ I was searching for "DC coaxial cable", "Laptop power coaxial cable", etc on DigiKey, ELV, eBay, Aliexpress, Reichelt and Conrad and found only either non-coaxial cables or RF-coaxial cables, but no high-current (4A) flexible DC coaxial cables.
AI: AC-DC adapters for laptop power supplies are of switching type. Therefore they do have some ripples, and the ripples can cause unwanted RF emissions. To mitigate this, manufacturers prefer to use "shielded" power cables, using coaxial cable construction. It is not a RF cable, but coaxial. I am not sure which particular cable they use, but as a good approximation you can look for "single-conductor cables" with "shielded" property, for example the Tensility cable (found via Digi-Key engine). |
H: If I have 2.4 amps on 5 Volts, how many amps am I using on 220v?
I know this sounds like a stupid question but I'm trying my best.
Anyways, I have a 2 meter RGB LED strip with 120 LEDs in total (60 leds/meter). I calculated the power consumption, which will be 2.4 Amps at 5 Volts but I can't figure out how many amps it draws at 220V.
AI: $$ P_{OUT} = P_{IN} \cdot Eff $$
Where P is power (W) and Eff is efficeincy.
Since \$ P = VI \$ this can be rewritten as
$$ V_{OUT}I_{OUT} = V_{IN}I_{IN} \cdot Eff $$
A decent PSU will have an efficiency of about 85% or so.
You should be able to work it out from here. |
H: Simulate isolation transformer in LTspice
Hi,
I am new to simulating transformers with LTspice, and I struggle getting the circuit to function as expected.
The source is a three phase TN-S (wye) connected. Then I wanted to add a 1:1 isolation transformer (wye/wye), but I cant get the output voltage to match 1:1 with the input voltage.
Played around with a lot of different values, but not getting any closer, so I dont know what I am missing...
Ignore "Rpe", the source was used in a previous model to simulate leakage current, and was used as a measuring point.
Help and pointers would be greatly appreciated.
AI: Increase the inductance of your transformer windings to something like 1H. At 1µH and 10mOhm, the voltage drop across the ESR of the inductors will be huge because the small inductance of the windings leads to a large current through the inductor. |
H: 4ِ-bit adder in Multisim
I want to design a 3-bit by 3-bit number adder circuit. The result must be in BCD. And I want to show the result by using 7-segment displays.
I built this circuit. I'm not sure from it, and when I run the circuit it doesn't show anything.
Where is my fault?!
AI: You actually got a lot of it right. But I don't use Multisim and can't help there, much. I think I see a wiring mess around your three gates -- you should look closely there. There might be "crossed wires" in there somewhere.
Also, you do NOT need a three-input OR gate. Just a two-input one. The logic should be something like:
simulate this circuit – Schematic created using CircuitLab
I'm not showing the 7447's there. But you should get the idea for those.
As far as the displays not showing anything? That's a different problem. They come with a lamp test pin. Use it to verify that your displays are wired up correctly. The outputs are open-collector, so you must be using a common-anode 7-segment display device and NOT a common-cathode device, for example. There are lots of possible problems there. Also, if you have a voltmeter use it to probe the voltages at each of the binary lines going in to make sure they are values you expect. Etc. But mostly, forget the rest of the circuit and just focus on making sure your digit displays actually work. If they don't, nothing else will. And they are easy to test. So do that FIRST. Before screwing around with the adder stuff. |
H: AC brushless motor as generator
How can i use an AC brushless motor to charge a battery? If I need to do it with a rectifier then what kind of rectifier should I use?? My motor is 1250kv and another one is 500kv. Can I use an Esc instead of a rectifier?
AI: You can not use an ESC. The best thing to use is a charge controller that rectifies the output of the generator and increases or decreases the DC output voltage to the level that is best for the battery. It might be possible to just connect a rectifier between the generator and battery, but generator voltage would need to be fairly close to the voltage required charge the battery. A transformer could be used to convert it to the proper range. The speed of the generator would then need to be controlled to some degree in order to avoid charging the battery with excessive current. If a three-phase motor is being used as a generator, a three-phase rectifier is required. If it is not a three-phase motor, some more research is required.
Note that I have assumed that by "brushless AC motor" you mean a permanent magnet motor that might be called a brushless DC motor or a permanent-magnet, synchronous motor. Induction motors are also brushless except for the rarely-seen, wound-rotor slipring-motor. It is best to state exactly what type of motor you intend to use.
With a permanent-magnet, synchronous motor, connected to an un-controlled rectifier, DC power is easily generated simply by turning the motor shaft. The generated voltage is unregulated and determined by the speed at which the generator is driven. However it should be relatively easy to get enough regulated power from that to control a voltage boosting charger.
With an ESC of any kind, considerable modification of the design would be required to convert it to a battery charge converter.
The only answer to a question of this type that is reasonable to provide here is to outline a general concept that is a reasonable approach to the problem. In order to proceed, you must first completely define the characteristics of the generator and the prime mover including the expected operating speed range and controllability of the prime mover. That dictates the requirements of the generator power processing system.
The second step is to define the power processing approach and determine the parameters of its basic elements. |
H: Does an ideal wire dissipate energy?
In this circuit, I calculate the current through R1 to be 10mA. What happens if I take into account the electromagnetic radiation in the ideal wire?
simulate this circuit – Schematic created using CircuitLab
edit: This is a pedagogical question, trying to establish that no, an ideal wire does not dissipate energy because that's not a useful definition of "ideal".
AI: If you accounted for the wires your circuit would look like the diagram below. It would have inductance from the magnetic field around the wire and resistance from the conductor.
The magnetic field effects are represented by the inductor and only apply if the current through the inductor is changing. Which in this circuit, once it is turned on the current is constant and the inductance does not matter.
If you had an ideal wire with electromagnetic effects, you could nix the resistors and just model the inductance.
simulate this circuit – Schematic created using CircuitLab |
H: Raspberry pi3 12v led circuit - broke down pi
I wanted to use 12v led, so I tried to switching 5v(gpio pin) to 12v.
This is my 12v switching circuit.
And this circuit broke down my raspberry pi.
When I wired this circuit, they worked well. But I switched on and off several times, the transistor and raspberry pi broke down.
Why my raspberry pi and transistor died??
AI: When those transistors are turned on, there's what amounts to a short circuit through the associated LED to ground. You need an appropriately sized resistor in series with those LEDs.
I suspect what happened (the failure mode you experienced) is that the transistors got damaged by overcurrent, and the failed transistor inadvertently fed 12V back into your GPIO pin and that baked your Pi. |
H: The output buffer of sample and hold
I saw some sample and hold circuits from the Internet,and i find there will be a buffer in the output,so i want to ask what does that buffer do for the sample and hold,can i use two stage amplifier as that buffer?
The two stage amp schematic
AI: Mostly the output buffer presents a really high impedance to the storage capacitor so it does not discharge significantly into the next stage. |
H: is it possible to protect workstation by removing VCC connection from workstation to circuit?
I am a newbie to electronics and currently working on my very first electronics project on Atmega16.I want to make sure that no matter how high current is supplied(by mistake), it shouldn't damage my laptop.So here's what I did till now,
I haven't soldered barrel jack near voltage regulator.
Here I connected MISO, MOSI, RESET and SCK from the connector to IC as shown in the image.
I was warned that if the voltage regulator doesn't work as intended and large current enters the circuit it will not only damage my circuit but also laptop's port and fry the motherboard.
So, in theory, I thought if I don't connect VCC coming from my laptop to USBASP then my laptop should be safe.
So my question is,
1)is it a good idea to not connect VCC from the computer and completely be dependent on an external source for power?
2)if yes then will I face some issues while programming the device as I am not supplying the VCC from a computer?
3)Also is it ok to disconnect ground from the laptop also?
I am not well with designing tools like eagle or ORCAD so here is the hand-drawn circuit diagram,
Also here is the back of the board.I want to remove the white and brown wire which are ground and VCC to ensure the safety of my laptop.
Is it a good idea?
And sorry for terrible soldering.It was my very first soldering experience.
I know this is not a discussion forum but still, if anyone finds a mistake in the circuit and if this platform permits such discussion then I will be pleased to hear it as I can correct them and avoid such mistakes in future.
If not then please ignore this part and let me know in comments I will remove this from question.
Thank you,
AI: No, it still will not protect against overvoltage on the serial ports. There are a few ways to do this here are two suggestions:
You still need to run the numbers on the current and size the resistor correctly. If the laptop port needs 1mA @ 5V to recognize a high or '1' condition, then you'd need 5V/0.001=5k max resistance, but you would also limit the current to the port and prevent a burn out. Diodes can also be used to clamp the voltage on the board to prevent the voltage getting higher or lower than VCC and ground.
You could also use a zener diode to keep the voltage below a certian value, in the other example, the zener will turn on if the voltage goes beyond 5V and shunt excessive current.
There are other ways these are just a few suggestions. Make sure your ground is connected to the laptops ground or you might have issues. If the VCC of your circuit is not of the potential of the port on the laptop (you haven't specified where you'll connect it with the laptop, then don't connect it. )
simulate this circuit – Schematic created using CircuitLab |
H: Will High Frequency explode an LED?
I have some basic LEDs and recently purchased some 4MHz Quartz Oscillators and since they haven't arrived yet, I was wondering what would happen if I hooked up the crystals straight to an LED in DC?
My guess is it may explode from being continuously pulsed 4 million times a second but then hooking it straight into the power is infinitely faster as there is no delay and they don't explode.
If it does explode, is there any kind of key or calculation you can perform as to obtain how many oscillations are required to fry specific components?
AI: High frequency will not explode an LED. it will simply turn it on and off very quickly. This is exactly what is happening in TV remote controls, albeit at 36kHz instead of 4MHz.
Normally, when you apply a frequency to an LED, when the signal goes high, the led turns on, and when the signal goes low, the LED turns off. Because of this, the perceived intensity corresponds to the duty cycle of the signal.
However, if one half of the period is shorter than the rise or fall time of the diode, then it will not raise to full power when the signal is high, and you can no longer compute the perceived intensity just from the duty cycle - the perceived intensity will be less than expected. |
H: Pull-up vs Pull-down for contact switch?
I am designing a simple contact switch(tactile, which conducts when you pushes, and springs back to not-conducting when released) interface with a microcontroller.
It's GPIO has option for both internal pull-up and pull-down.
1) Which one should I use? For 2 layered board, pull-up makes me easier to do PCB artwork, since one-end of switch connects to GND.
Other than that, is there any difference or reason for me to use one configuration over another?
2) Is there any possiblity that the noise from physical contact make MCU malfunction?
AI: 1) Which one should I use?
Whatever suits your overall design best
1) For 2 layered board, pull-up makes me easier to do PCB artwork, since one-end of switch connects to GND.
That is exactly the reason pull-ups are used more often than pull-downs (and some chips only have pull-ups): a ground line is often more conveniently available for the other side of the switch (and open-colledror/drain outputs are more common than their top-side counterparts).
1) Other than that, is there any difference or reason for me to use one configuration over another?
2) Is there any possiblity that the noise from physical contact make MCU malfunction?
I would guess that (in a reasonably well designed system) the power is more noisy that the ground, so an active-high switch would potenetially induce more noise. But even that should be no problem on a digital input with sufficient margin.
So by all means, go for pull-up and an active-low switch. |
H: WAGO PLC Datasheet I/O' module specs
In "WAGO" PLC datasheet of Analogue and digital I/O's modules what does the following measns :
1- Isolation: 500 V system/supply for analogue i/o module
2- Isolation: 500 V system/field for digital i/o module
And how can i know the type of isolation (channel to channel or channel to earth )
AI: Figure 1. A Wago modular PLC I/O system. Modules are assembled by sliding together from front. Blades on the left edge connect IO bus power from module to module while a set of gold-plated contacts near the top rear provide databus connection between the modules.
Figure 2. Module close-up. The 'system' side of the device is most likely confined to the area inside the red box. The remainder of the device will be opto-isolated from the system.
Definitions would help:
System: The CPU side of the parts including the data bus.
Supply: Usually the 24 V supply to the PLC / modules.
Field: The sensors, transducers, etc., connected to the input module.
Isolation: 500 V system/supply for analogue i/o module.
It means that the CPU side of the system is isolated from the analog module external supply. There is no ground connection. The isolation is rated to 500 V.
Isolation: 500 V system/field for digital i/o module.
It means that the CPU side of the system is isolated from the analog module inputs. There is no ground connection. The isolation is rated to 500 V.
And how can i know the type of isolation (channel to channel or channel to earth).
The type - whether opto-isolated or transformer isolated you could only tell by opening a module and having a look for opto-isolators or a transformer.
Figure 3. A well laid out opto-isolator board with good creepage clearance. Source: Soft Solder.
Usually the isolation zone - no-man's land - is clearly visible.
Channel to channel isolation: To have a breakdown between channels it must breakdown between the channel and the 24 V supply or channel and system. It's still 500 V.
Channel to earth isolation: This is up to the user and how it is wired. You have the option to have, for example, the 24 V supply to the analog modules remain floating with no ground reference. You need to determine the best solution for this yourself, depending on your application. |
H: What is the purpose of copper planes in a switching power supply?
I'm looking at including a buck converter to power a 3.3V microcontroller, and I used TI's Power Designer to generate a recommended layout for my parameters.
I noticed that the copper planes are quite large here compared to the footprints of the components involved. I understand the value in having a plane for the ground, since it's a common reference point, but why are there such large areas for the other connections? Is it for heat dissipation, or other reason(s)? (Or am I misunderstanding something about how to read the diagram?)
AI: Lower track impedance
In a switching regulator, the track impedance matters a lot. Not only resistance, but also inductance, and both are reduced when using wider tracks (or planes).
Heatsinking
A switching regulator produces heat, which has to be channeled out of the component. Copper is a very good heat conductor and is used as radiator in many switched mode power supply designs.
PCB Manufacturing issues
When producing PCBs, manufacturers often ask for a certain percentage of each layer to be copper. This is to ensure an even thickness on the whole PCB in the plating phase, as well as an uniform expanding and shrinking under temperature variations. |
H: What to use in finding frequency response : \$G(s) \cdot H(s) \$ or \$ \dfrac{G(s)}{1 + G(s) \cdot H(s)}\$?
While studying frequency response of control systems, all of the sources I'm referring to is considering unity feedback system i. e. H(s) = 1 & drawing Bode plots and polar plots on basis of that.
But what if feedback is not unity? Do I have to do analysis and Bode plot with
\$G(s) \cdot H(s)\$ or \$ \dfrac{G(s)}{1+G(s)\cdot H(s)} \$ ?
and why?
Note : G(s) is forward path transfer function.
I thought it should be
$$\dfrac{G(s)}{1+G(s)\cdot H(s)}$$
but one not-so-reliable source (tutorials point website) states that it has to be
$$G(s) \cdot H(s)$$
Also why \$G(s) \cdot H(s)\$ is called open loop transfer function?
AI: What is the "open loop transfer function"?
To avoid misunderstandings and to be conform with the relevant literature, I like to clarify:
1) The "open loop transfer function" is the forward transfer function G(s) of the active device only without any feedback (when the loop is "open").
2.) The transfer function of the whole loop (but OPEN) is the so called "Loop gain" L(s)=G(s) * H(s).
3.) The transfer function of the whole circuit with feedback applied is
A(s)=G(s)/[1+ L(s)]
Note that this formula applies for negative feedback only. Otherwise the "+" sign has to be replaced by a "-" sign.
4.) For stability analyses, we only need the loop gain response L(s) in the BODE plot (magnitude and phase separated) or in the Nyquist plot (magnitude and phase combined). |
H: How to calculate maximum gain that can be achieved by an op amp?
I am designing a differential amplifier for a Wheatstone bridge which have small voltage variations like 0uV to 20uV. so need to design an instrumentation amplifier. I do not know how to find the maximum gain that can be get from an opamp (I know how to calculate overall gain from resister values used). data acquisition frequency is not a huge matter in my application. 5Hz or 5 samples per second reading would be enough via an microelectronic.
UA741 how to find maximum gain that can be archive from this opamp(frequency at 5Hz)?
AI: Maximum gain is limited to the open loop gain of the op-amp but, for a linear amplifier you should not try and implement a gain value that gets too close to this value because distortion will be an issue and you will also find that the frequency response may be much less than you require. DC accuracy is also made worse with a larger gain so this should also be considered.
Here is the typical open loop gain of an op-amp: -
At a frequency that is somewhat less than 10 Hz the gain is flat down to DC and very high (10\$^5\$ = 100,000). At about 7 Hz (in the example above) the gain begins to fall at 6 dB per octave and unity gain is seen around 1 MHz.
So if you want a closed-loop gain of (say) 1000, your frequency response will be limited to about 1 kHz and, as your input frequency rises from about 100 Hz, you will notice the output signal gets more distorted because there is less open-loop gain headroom to perform the corrective action that negative feedback brings to the party. Above 1 kHz, the gain will follow the open-loop response. |
H: Can a short circuit damage nearby devices?
This happened yesterday and left me scratching my head.
We had a badly built power extension cord in our office. Inside it has 3 wires - live, zero and ground (hope I got the terminology right). When plugging in a phone charger, the live wire touched the ground wire and blew a fuse. However it appears that a notebook power supply that was already plugged in the same extension cord was also damaged. Now, when plugging in that notebook charger (anywhere), it too blows a fuse.
What I'm wondering is - how could this happen? As per my understanding of electricity, the live wire touching the ground wire should simply cut the power to the notebook power supply, since all the current would now flow from the live wire to the ground wire and away. There should be no high voltages produced in the nearby devices. Is there any other mechanism as to how this could have happened?
AI: The problem is you don't know what was going on when the power was shorted out, the charger could have been causing an over-voltage condition on the line and that could have blown out the notebook power supply.
Or the charger could have been injecting current into the ground and the preferred pathway could have been through an improperly built notebook power supply (of chinese manufacture without ETL testing).
There is no way to really tell what when on without schematics for both power supplies and how the cable was miswired.
Can a short circuit damage nearby devices?
Apparently, yes through your account of the damaged charger, it can |
H: Why are old digital ICs families still so "popular"?
Why is it so? The basic 74 series, the 4000, are such poor ICs compared to the newest families (by the 74: AUC, AUP, ALVC...), which only have advantages over the oldest one: lower power consumption, lower propagation delay...
AI: Why do you think TTL 74/54 series are popular? They are a relic of the past now (unless you have to maintain old hardware).
Metal gate CMOS 4000 series is also somewhat obsolete, but it has some advantage in some specialized digital application where its wide supply range (up to 15V at least) is convenient. Of course it's slow and power hungry, compared to newer CMOS families.
74HC/HCT is mainstream because it is mature and stable (but it's losing terrain).
The other, faster, families are used only if you really need their speed. A faster digital device is not necessarily an advantage in a circuit that doesn't need that speed.
If you are a student, I suspect that your intuition is skewed because many high-school labs (and even university labs) purchased tons of TTL (and CMOS 4000) chips and many teachers love beating a dead horse (sadly). |
H: What are wavelengths of RGB diodes?
Disclaimer: I am not an electrical engineer, and nowhere near studying it either. This is just a curious question which I got and couldn't find an answer to.
I am aware of how several pixels are used to make up a television screen. Each pixel consists of 3 LEDs; one red, one blue, and one green. By varying the intensity of either the red, green, or blue; it's possible to bring about any possible color that exists.
My question is, what are the wavelengths of the light that these blue, green and red LEDs emit? Does it depend on TV to TV or is it general?
AI: First off, leds typically transmit on a narrow band of wavelengths, shown below are the curves for a blue, green and red LED.
You can see how they match up to the response of the human eye, so with only three colors, you can cover a wider range of colors. Usually with a TV they use an LED that covers more wavelengths than a standard LED.
They also go through the effort to match brightness and calibrate the LED's to a known color standard so that the color you see from the matches that of the media played and the camera that recorded the media. |
H: How much Energy is Stored in this Capacitor?
This is a followup to my previous question: A capacitor stores half the energy when charged from a battery each and every time?
This is a real world circuit with losses. The losses are in the resistance of the battery, the switch, the capacitor, and the inductor.
The inductance value of the L1 inductor has been left out because it was not known or measured. It's not a value of 0 however as it has some inductance.
The capacitor starts out with an initial energy of 0 Joules as well as an initial voltage of 0 Volts.
The switch is closed only one time and left closed. The capacitor charges to the same voltage level as the battery of 12.33 V therefore it now has 76.014uJ of energy in it.
The question is:
Does the 50% loss of energy that normally occurs when charging a capacitor from a battery (without the inductor) still apply to this circuit, even with the L1 inductor in place?
In other words, the capacitor will charge up to the same voltage level as the battery of 12.33 V as confirmed with an actual physical circuit. Has only 50% of the energy been transferred into the capacitor than what came from the battery?
The actual physical circuit has actual inductance in the inductor, but it is unknown, but it's not 0.
Summarized and to the point:
In this circuit, how much energy is transferred from the battery and into the capacitor when everything comes to rest and all processes are completely done? 50%?, 99.8%?, how much energy was transferred from the battery to the capacitor?
simulate this circuit – Schematic created using CircuitLab
AI: The LC tank, that you've drawn, will oscillate / ring, and as you've stated in your question, "This is a real world circuit with losses", it will come to a still. During the oscillations the same current will go back and forth several times through the same losses, aka resistances.
So having an inductor will drain more power than having no inductor. The total power lost (aka energy) will be greater with the inductor than without the inductor. I'm implying the same thing with different words.
If the LC tank is critical damped or overdamped then the current will only pass through the resistances once. If it's underdamped however.. Then you get oscillation / ringing.
Hmm, this is my gut feeling though, so I may be wrong. I don't feel that I have time to put up the math to prove myself wrong or right. It's not worth the effort.
When it has stopped oscillating it will reach a steady state and no more power will be lost. And then it will reach your strange 12.33 V.
And to reach the climax of your "question". The capacitance hasn't changed, the voltage hasn't changed, the equation for the energy stored in a capacitor hasn't changed. It doesn't matter how it was put in there. After the steady state, when everything has come to a still, the energy stored in the capacitor can be calculated, which you've done beforehand. So I'll take your word on it being "76.014uJ".
Your question is weird and I feel that it doesn't matter what I write, will the correct answer give you anything meaningful?
When I feel that I'm becoming misunderstood, I take a step back and ask myself "Who is my audience?"
"Do they know what I know?"
"Do they need to know something else to grasp the information?"
In other words, The problem is 99% of the time with me. Perhaps.. you need to reword your question so you don't have to make 3 nearly identical questions within 24h.
Try #2
LC tank
At T=0 when the switch closes, the following things will happen:
The capacitor, together with the inductor, is like a loaded spring. But it will "decompress" and surge current and start oscillate. I'm explaining it weirdly, but it's my view of it. And according to the link, the energy stored in an LC tank which is what you got in that setup, is equal to \$\frac{1}{2}CV^2 + \frac{1}{2}LI^2\$. At T=0 then the current will be 0. If we know the voltage and the capacitance, which we do, then we can calculate it. In other words, the total energy in the LC tank will be what you had in your question. "76.014uJ". However, this energy will be moving around in the circuit. At some point in time the majority of it will be in the capacitor, some other point in time in the inductor.
Since the oscillations are going in both directions, then... at some point in time the capacitor will be fully charged in one direction, such as having 12.33V across it's terminals. Fully loaded. Then in some other point in time it's voltage will be entirely inverted and it will have the same amount of energy. Such as having -12.33V across its terminals. This is probably how you are seeing a value above 50%. The equation is \$\frac{1}{2}CV^2\$, if V is -12.33 or 12.33 then the energy stored will be the same, right? But if you calculate some weird energy difference then you might see 99.8% because it's swinging to the top of some cycle and then to the bottom of some other cycle. It's hard to explain but I can see that it's possible that you've come to terms that it's 99.8%. While in reality it's 49.9%.
After trying it out in a simulator I've come to terms that at T=0, both the AC and DC energy are put into the LC tank. After the AC part has decayed, then the DC part is left which is equivalent to \$\frac{1}{2}CV^2\$ J. In other words, at T=0, \$CV^2\$ is inside of the LC tank in the form of AC + DC, the AC will have an amplitude that is half of the DC, so the \$V_{PP}\$ of the AC is the same value as the DC level.
Solving it for DC first, you get what you had in your question, "76.014uJ", solving it for AC after a tiny amount of time has elapsed so everything is stored inside of the capacitor and none is inside of the inductor, you get "76.014uJ" as well. Then you put the AC component + the DC component together and you get an answer that is close to that of laptop2d, his has decayed a little bit which is why it's slightly less than 100%. I used superposition in order to solve this problem. I'm not a clever man. It took me 3 tries.
So the 50% part I spoke about earlier that was "swooshing" around, that's not a lie, but it was also a 50% DC component laying on top of everything which confused me and most likely you.
TLDR; At T=0 the system contains 2×76.014uJ≃152µJ. After the AC component is gone, the DC component, 76.014uJ is left inside of the capacitor. |
H: Why doesnt this motor activate? edit
Its 3v motor and my source is about 5.14v.
AI: You can't wire an LED and motor in series like that.
When the motor is stopped the LED gets the full start current of the motor and the voltage dropped across the LED will be large. The LED will burn out quite quickly and it will also drop enough volts while it does that for the motor to not start.
If you want the LED on when the motor is on, wire it in parallel as shown below.
simulate this circuit – Schematic created using CircuitLab |
H: Question about MOSFET ground connection
In the circuit above, the sources of the two MOSFETS should be grounded to the same ground as the FET driver as shown in the schematic ?
AI: Yes, the MOSFET source pins, FET driver ground and microcontroller ground are all connected. Otherwise, they wouldn't be able to drive the FETs.
Also, please be careful with that circuit. If anything goes wrong with the thing, it might catch fire or explode when it shorts out the mains. Your microcontroller will be at live voltage, so don't measure anything when it's connected. And don't leave the programming cable plugged in when you try out the circuit, it'll blow up your computer otherwise. |
H: power at voltage variation
220V/5W means consumption of 5W at 220 volts. That implies 0.0227A current.How will this device operate at 110 volts? What will be the power consumption and how much will be the current?Sorry for bad english
AI: Power, voltage and current are related by the formula $$ P = VI \tag 1$$.
For a resistive load we can make some substitutions from Ohm's Law, \$ V = IR \$. This results in $$ P = \frac {V^2}{R} \tag 2$$ and $$ P = I^2 R \tag 3$$.
From (2) we can see that if R remains constant and we halve the voltage then the power will be 1/4. So if your 220 V, 5 W load example is a resistor then on 110 V it will dissipate 5/4 = 1.25 W.
Figure 1. A bulb running on reduced voltage.
Your bulb example is a resistor load but the resistance is temperature dependent. At high temperatures the resistance increases. This means that at 110 V a 220 V, 5 W bulb might consume somewhat more than 1/4 - maybe 1/3 of its rated power, 5/3 = 1.6 W. The bulb will glow a nice warm shade of orange. |
H: LTSPICE Logic Buffer: What does the extra pin do?
I give up... what is the extra pin for?
LTSpice Documentation is horrible....
AI: As described in the help under LTspice → Circuit Elements → Special Functions (near the bottom), it is the "common" pin, a.k.a. "node 8".
If you leave it unconnected, it is automatically tied to the global node 0 (ground). |
H: Improved Differential Amplifier
I made this circuit as shown below:
simulate this circuit – Schematic created using CircuitLab
After those improvements with current mirror and constant current source were considered, this is what I got.
The emitter currents through both of Q3 and Q4 were calculated to be 1mA, but when measured approx. 900uA each. Voltages were split equally between each of sub-stage of amplifier - voltage across current mirror = 10V, voltage across differential transistor pair = 10V and voltage across constant current source = 10V (measurements actually showed up a bit different results - Vmirr = 0.8V, Vdiff = 19.2V, Vccs = 10V).
As seen measured values aren't really as expected.
Why is there only 0.8V across the current mirror (Vmirr + Vre1)? Why are both Q1 and Q2 in saturation?
AI: Let's start out by looking at your \$Q_5\$ current sink.
You are using a voltage divider using \$R_5\$ and \$R_6\$ to set the base voltage. \$Q_5\$ is being treated as an emitter follower, using \$R_{e_2}\$ to set the collector's sinking current. I just wrote a page on this, here. There, you'll find the following equation:
$$I_{C}=\left[\frac{\beta}{\beta+1}\right]\cdot\left[\frac{V_{TH}-V_{BE}}{R_E}\right]\cdot\left[1-\frac{1}{1+\frac{\left(\beta+1\right)R_E}{R_{TH}}}\right]$$
In your case, \$R_{TH}\approx 44.329\:\textrm{k}\Omega\$ and \$V_{TH}\approx 1.705\:\textrm{V}\$. I can guess from this that you assumed \$V_{BE}=700\:\textrm{mV}\$ so you were expecting about \$1\:\textrm{V}\$ across your resistor, \$R_{e_2}\$, making for a current of \$\approx 2.13\:\textrm{mA}\$. However, the \$3^{rd}\$ factor above says about 65% of that would be more realistic; or about \$1.4\:\textrm{mA}\$. This is because your resistor divider network is not stiff. It's floppy.
So how might you improve it? Well, you could stiffen up the divider. That's the obvious path. Another might be to convert this to a current mirror. Either way, you should provide a means by which you can set the value to a preferred value (or, at the very least, sit down and actually measure what you got.) But being able to make adjustments to get a preferred design value would really be better.
Better still would be to work out a method that is independent of the voltage rail for at least one order of magnitude change. (It can be done with discrete parts!!) But it involves the use of three transistors in an odd arrangement (Wyatt) and only really works at a design current that is perhaps too low for your desire. But it is stable over temperature by taking advantage of the temperature coefficient of metal film resistors and of copper wire itself. (On the order of 3300 ppm per K.) But I think this would be for a different day.
I'd recommend working out a current sink that doesn't move much if you heat it up or cool it down a bit and also stays close to a nominal value if you swap out transistors. Want to start there? Or do you feel that you do not need any help on this point and want to take a different step? |
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