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H: Buck, Boost..etc Is there always an IC for that?
First of all I am still in school so perhaps I have not been exposed to material that may answer this question for me.
I have recently become interested in power electronics. I have purchased a book on the subject and I am taking an online course.
Like usual they start with basic models where everything is ideal and as the course goes on you begin to model losses.
I have breadboard a basic buck and it works great.
My question is that I feel a bit disillusioned because almost everything I learn "there's an IC for that"
I get the feeling that the fellas that design those IC are REAL engineers and everyone is just pluging and playing parts.
I understand the benefits of using an IC because it will reach an unparalleled level of efficiency versus discretes... but out there in the industry is there anyone making power converters not using a one size fits all IC? if so can anyone give some examples?
I have not taken a formal course on power electronics at school, so that fact contributes to my ignorance on the subject.
AI: It's always good to think about this kind of thing. You can build a circuit like this without using an IC — and in fact if you look at the datasheets of the ICs in question they usually have at least a block diagram (if not a complete circuit diagram) that would give you a good start on replicating what they do — but in practice nobody uses such a design commercially because it would be more expensive than using the IC yet provide no practical benefit. But for educational reasons, doing it yourself is a very good idea. I haven't tried building a switching power supply using this kind of approach, but certainly simpler projects like op amps, logic gates and so on, building it myself from discrete components has often been very educational. |
H: What is the current through this parallel circuit?
I'm trying to solve the attached circuit for I1 and I2. I think I1 is just 5A but I'm not so sure about I2, I think it might be just 10/5=2A but I'm not sure if the voltage in the top right part is still equal to 10V or not.
I'm new to electronics, thanks in advance!
AI: Look at both sides of \$R_2\$ and compare the voltage difference, and then make a conclusion based on Ohm's Law about the current.
Also, this is not a parallel circuit. Parallel components MUST share BOTH nodes. |
H: Control dual color LED without microcontroller
I have a latching, SPDT switch with dual color LEDs (the first page of the datasheet), that can either be Red or Blue depending on the polarity of the voltage. I would like them to be one color when the button is in position one, and the other color when the button is in position two.
The purpose of the button is as a manual backup switch for a relay board that is normally controlled over CAT5. The relays are tripped by shorting the board's gnd to the pins of the relays.
My voltage source is 110V wall outlet being stepped down to the 12V that both the relay board and LEDs in the button want.
My pinout for the button is:
C: going to gnd of the relay board
NC: nothing connected
NO: going to the relay pins
I found this guide which I believe demonstrates the principal of what I'm trying to accomplish, but with a momentary SPST button and no extra hardware connected, and the intention to use connected to a microcontroller. I can't use a microcontroller in this project as a) no room in the box for one b) no money in the project budget for one.
My main concern, and reason for not just tinkering around, is that the boards are super sensitive to voltage (I already fried one by accident), and I'm worried about sending extra voltage back down the line to the relay board.
Any and all help/tips/suggestions are greatly appreciated!
AI: I don't know what your specific restrictions are regarding space, but you should have something active. You can use a couple of gates from an inverter chip like the CD40106B, which can easily accommodate 12V.
simulate this circuit – Schematic created using CircuitLab
The inverter chosen really depends on the current of the LEDs. If you measure the current to be high, you can use an H-bridge chip or discrete transistors. |
H: What are the possible boot modes in the S32K146 (NXP S32K Series MCU)
I have been investigating S32K146 MCU boot modes and possible mode selection options.
I am interested to know the possible boot modes and how it can be set up.
From datasheet (Page -92): It can be seen that boot can happen from the flash memory.
After googling and few research: Came to know that booting is also possible from RAM using programmable VTOR (Vector Table Offset Register).
What are the possible boot modes in S32K146 and how this modes can be selected?
AI: It sort of depends on what you mean by "boot". At the lowest level, the boot process is what the MCU does when first powered on, before it begins actually executing user code. At a higher level, "booting" may refer to what you, as the developer, instruct the device to do before launching the actual application program.
Before the MCU can execute any instructions, it must know where to find those instructions. On ARM parts, the processor expects to first read the vector table. This is a list of at least four pointers:
The initial stack pointer
The reset routine location
The NMI ISR location
The Hard Fault ISR location
Additional ISR locations, for whatever interrupts the device happens to implement (technically these are optional)
This information is given in the ARM documentation
The vector table will typically be at the very beginning of your executable binary, so the processor will load the initial stack address into its stack pointer, and then jump to the location of the reset routine (the reset vector).
Normally, the processor expects to find the vector table at the beginning of flash memory. Some MCUs have dedicated configuration memory that can be used to instruct the MCU to look for the vector table at a different location. Some of these parts have built-in bootloaders in dedicated ROM, separate from the normal flash, and can be configured to run these bootloaders before starting the application to allow for updating the firmware without using a debug probe. However, it doesn't look like the S32K146 has these options, from a quick look. THis means that the MCU will always look for the vector table at the beginning of flash whenever it is powered on or otherwise reset. You would need to dig into the documentation, particularly the Reference Manual, to verify this. For an example of a processor that does have this feature, you might look at the STM32F4 series from ST.
At a higher level, you are free to write your own boot code, that runs when the device first starts up, and then "boots" into the application. This is a conventional bootloader architecture, where you place your bootloader at the beginning of flash, and when you're ready to run the actual application, the bootloader can do that by:
Updating the Vector Table Offset Register to the address of the new vector table (usually at the beginning of your application, wherever that's stored)
Updating the stack pointer to the first value in the new vector table
Jumping to the reset vector
At that point, the processor will begin executing your application, almost as if it had booted from that location in the first place. I say almost because it is important to realize that this sort of boot process will NOT reset the state of the MCU (in particular the state of peripherals) to their power-on values. This is one reason why a defensive programmer will be very thorough in their initialization routines.
One additional thing to note, on platforms like ARM, it is possible to execute code stored just about anywhere. So it is possible, for instance, to have a bootloader that loads an application from an external memory into the MCU's RAM, and then execute the application from RAM. This capability is sometimes exploited by debug tools, which can write a small application into RAM to do things like load data into an external flash memory. |
H: Opamp output sinewave is distorted
I'm using an opamp to amplify a 3.5 Vpp (peak to peak) 10 kHz sinewave which seems fine apart from a tiny noise mostly visible at its' peaks which does not particularly concern me much. The blue sinewave is the input and the output is the red sinewave amplified at 6.7 Vpp. There is a distortion and that is at both no-load and loaded conditions. This opamp is meant to be used as a resolver primary coil driver which is the case of my application. The distortion is something that worries me because it could affect the accuracy of the resolver-to-digital converter. My question is what can I do to minimize the distortion as I do not see any reason for it. The datasheet refers to zero crossover distortion and secondly the opamp is "obliged", to a certain extend, to provide an output signal based on the gain factor which is not so great that could drop the BW so low to affect amplification ratio. The BW is 600 kHz and the signal being amplified is 10 kHz.
Can anybody make any suggestions?
Regards
Images:
AI: I haven't gone through the whole circuit to figure out if there are other problems but you appear to be slew rate limited.
Slew Rate = 2πfV = 2*3.14*10k*6.7 = 0.42V/us
Datasheet shows 0.17V/us (typical)
Redo the test with 1kHz input signal and see if the output cleans up. If it does - that was your problem. |
H: NFC spectrum analysis, weird frequencies showing
I apologize for the picture quality, I did not have an opportunity to export the screenshots.
I have a problem properly interpreting this signal. The subcarrier peak is visible but the data seems to be noisy? I assume using a pre amplifier wouldn't get me a better result. There are also two peaks visible on the leftmost and rightmost side of the screen and I have no idea where they come from. This is the response of an ISO14443-A Tag:
Are the settings of the analyzer wrong or is this the proper spectrum?
AI: Firstly, info for non proffesionals:
ISO14443-A: https://www.msoos.org/largefiles/rfid_soos_pres.pdf
The subcarrier is at fc/16, so your subcarrier is at 13.56 * 15 / 16 = 12.7125
Glossary: https://www.nfc-research.at/index.php@id=40.html
PCD
Proximity coupling device: a transmitter that can read tags based on ISO14443 (PICC). The reader emits an electromagnetic field that powers a tag/transmitter by inductivity. Communicates with PICC using load modulation scheme.
PICC
Proximity inductive coupling card: a transponder that can be read or written by a proximity reader. Theses tags are based on the ISO14443 standard. Such tags do not have a power supply like a battery, but are powered by the electromagnetic field of the reader (PCD)
The info you provide
This is the response of an ISO14443-A Tag
So, this is how the card communicates with the base. Using the pdf previously provided,
PICC→PCD:
Load modulation
Subcarrier: fc/16 → fc/128
Coding: OOK, Manchester
Bandwidth: 106kbit/s→847kbit/s
The communication the card transfers to you is using OOK, on-off-key: https://en.wikipedia.org/wiki/On-off_keying.
One of the most visual caracteristic OOK using an oscilloscope is there is a lot of noise around the frequency of transmission. That's why you see lot's of, apparently, noise.
On-off-keying sends pulses (deformed pulses), so... where are the harmonics?
The peaks
The peaks your oscilloscope snitches are, indeed, the harmonics of the on-off-keying.
Did you perceive that the frequency between the center of transmission, 13.56, and one of the peak, is exactly the frequency between the center of transmission and your subcarrier?
That's exactly what happens in your image. |
H: Audio jack detection to turn on device without power loss
I'm trying to make simple headphones amp and I wonder - is there any way to use stereo jack connector with N-C switches like this:
to turn on the amp so the battery is mechanically disconnected?
I mean that, if my connectors had N-O switches, it would be easy, but with N-C I have no idea, whether it's even possible to make such "smart" power on, without any power loss (i.e. powering some transistors/logic gates that detect open circuit etc.) or maybe should I seek for female connector with N-O switches?
Best regards!
AI: Not really for a stereo headphone, but you can do it for a mono jack by hooking the battery negative lead to the ring contact on a stereo socket... Classic guitar pedal trick.
About the best you can do is a mosfet with a 22M resistor to bias the gate and use one of the headphone switches to pull the gate down when nothing is inserted.
Regards, Dan. |
H: Effects of Neodymium magnets on PCB
I'm currently doing a project that consists of:
Arduino Nano.
Battery.
Buck Boost converter.
GPS module.
TSL2561 light sensor.
BH1750FVI light sensors.
LSM6DS3 Accelerometer.
HDC1080.
MicroSD shield with microSD
All of the sensors are I2C based, GPS is UART and SD card is SPI.
My question is:
I'm planning to place the components inside a wooden box and mount the box to the top of the car roof with neodymium magnets so it is movable around the vehicle roof top for the measurement.
Will the wood box be sufficient to stop the magnetic fields produced by the neodymium magnets (for mounting) from interfering (if there is going to be any) with the components that are connected to the PCB board?
AI: None of the listed parts are sensitive to magnetic fields, so the mere presence of the magnet shouldn't bother them at all.
The sensors and magnets will be physically attached to the box, and so must move together. To induce current, the conductor and the magnet must move at different speeds. Since they are all attached to the box, all the parts will be subject to the same motion - no induced current, no problem.
The LSM6DS3 has provision to connect an external magnetometer, but you don't mention having one in your project. If you do have a magnetometer attached, things get more complicated. Probably better to use suction cups for mounting the box in that case.
Vibration could cause some low currents to be induced in your traces and wiring. Vibration can do this because the parts aren't perfectly rigidly attached to the box and will therefore move a tiny bit in relation to each other.
That shouldn't be enough to bother any of the circuits you mentioned.
The wooden box will have pretty much no noticeable effect on the magnetic field. About all it gives you is a little more distance from the magnet - which by itself reduces the strength of the field when it reaches your parts. But, the wood itself doesn't really matter - it might as well be air for all the difference it would make. |
H: Why does USB only use 2 lines for RX, TX instead of multiple data lines?
Wouldn't it be faster if there were multiple data lines (say 8) to transmit/receive data (say sequential bytes) instead of using a single line to transmit sequential bits?
AI: It would be faster indeed if instead of one line you'd feed multiple lines at the same symbol clock.
But, USB's primary and foremost goal is to provide easy, serial (hence the S in USB) interfacing between low-cost devices (hence the U in USB) with low-cost, lightweight cabling.
So, that's why USB doesn't do parallel data lines: It's simply not the niche it's supposed to fill.
Also, don't neglect that having multiple high-speed parallel lanes requires the transceiver system to introduce a relative high amount of effort to compensate different skews on different lines, which at high rates are inevitable.
It's often become cheaper to make something work twice as fast than building two of the slower variant, unless you're really directly talking to hardware that is in its raw principle bit-parallel (e.g. DDR memory chips). |
H: How to plot response of a non-linear state space model in Simulink?
I am trying to plot a non-linear model in order to compare it with it's linearized counterpart.
I am following this paper Nonlinear Model & Controller Design for Magnetic Levitation System and trying to reproduce the results the authors obtained. In particular I am trying to plot:
The above equations can be represented in vector format as follows:
I found no references on how to plot non linear state space model representations on MathWorks.
The Simulink state-space block is used to implement linear state-space systems and not non linear ones.
So, how can I plot the response of a non-linear state space model in Simulink ? Any suggestions would be appreciated.
AI: One method is to use a Matlab Function Block to implement the nonlinear equations. You can define the inputs and outputs yourself in this block.
The body of the function block will look something like this:
function [xdot, y] = nonlinearss(x,u)
% define constants
g = 9.81
% etc...
% nonlinear set of equations
xdot = [x(2); g-C/m*(x(3)/x(1))^2; etc...] + [0;0;1/L]*u;
y = x.'; |
H: How do I drive 60 back-to-back LEDs (Christmas Lights)?
I would like to reuse some old LED strings. I hope to replace the (failed) original controller with an AVR chip, but I do not know how to construct the differential MOSFET driver stage.
The sketch below is not complete (as the string is in fact 5 blocks of 12 parallel LEDs) but my problem is finding a way to provide the alternating current necessary to light firstly the odd LEDs and then the even LEDs?
simulate this circuit – Schematic created using CircuitLab
AI: What you are searching is a H bridge driver.
It's typically used in motor control application.
Here is perhaps something that can get you started with some good example There.
It is meant to drive motor, but will work the same with your LED. CW direction would be one LED type and CCW would be the other LED lighting up.
You also need to limit the current. Check whether those LED integrate the resistor (which is often the case in those led string). Otherwise you will also need current regulation. |
H: DC converters for solar cell
As we know, the voltage level of a solar cell is lower than what a rechargable battery needs, and thus a DC converter is placed in between.( With feedback control for optimum V-I curve)
This just confused me, so what the solar cell is delivering?? DC converter provides power from outlet not the cell. Wasn't the whole point of solar cell to provide power? We could just charge the battery from outlet without the solar cell.
AI: In most cases, in order to charge a battery, we need to control the input voltage/current and achieve the maximum power point (via a MPPT method).
Next figure shows a simplified model.
source: https://outbackjoe.com/macho-divertissement/macho-articles/design-guide-for-12v-systems-dual-batteries-solar-panels-and-inverters/
There are plenty of textbooks and sites that explains how DC/DC converters work and how to utilize them in a battery charging system.
For example, a Boost Converter (Step-up DC/DC Converter) increases the output voltage of a solar panel and tries to find the point where the power transfer from the panel to the battery becomes maximum (MPPT).
Charging a battery from a wall outlet is also possible - of course by using a proper AC/DC converter - but then you are using the energy that comes from the grid (eg. coal) and not a renewable source, like the sun. |
H: PoS-SoP convertion of function bellow (NOT using K-map)
Function F as: $$ F=\overline{C} \space \space \overline{D} +A\overline{D} + A\overline{B}$$
is a SoP (sum of products) which is given by grouping K-map's 1's.
I tried to complement the function \$F\$ to get PoS (product of sums), like: $$ \overline{F}= (C+ D) \cdot (\overline{A}+D) \cdot (\overline{A}+B) $$ but I checked the K-map's 0's to see if it's a correct PoS, but it is not even a PoS because it contains max terms that the correct PoS shouldn't have.
Does any one know where I'm wrong?
AI: $$ F=\overline{C} \space \overline{D} +A\overline{D} + A\overline{B}$$
DeMorgan's
$$ \overline F=\overline{\overline{C} \space \overline{D} +A\overline{D} + A\overline{B}}$$
$$ \overline F= (C + D) (\overline A + D) (\overline A + B)$$
First two terms:
$$ \overline F= (C\overline A + D\overline A + CD + DD) (\overline A + B)$$
Simplify:
$$ \overline F= (C\overline A + D) (\overline A + B)$$
Finish multiplying out last term, reduce and take DeMorgan's.
That should get you to your answer. |
H: Chimp climbing wires not electrocuted?
Reference video: https://www.youtube.com/watch?v=U5-Fy4ULdPg
The chimp is climbing both wires out of phase (i would have to assume, it is in Japan), do the power lines have to be closed for the chimp to do that? How is it possible that the chimp is not electrocuted? It was stated that power was out for 1000+ people so i would have to assume the lines had to be closed for the chimp to survive climbing this?
AI: It looks to me like the power cables are insulated. That's not common in my neck of the woods. If the power was turned off, it was for additional safety. |
H: 3 phase voltage problem when connecting to servo
I have 380V 3 phase with the following line to neutral voltages: L1: 238V, L2: 234V, L3: 224V. The system has a single phase servo driver that takes in L3 and N, however when hooked up L1, L2, and L3, change to the following: 394V, 400V, 0V. I'm not sure why this discrepancy is happening.
Thanks in advance!
AI: simulate this circuit – Schematic created using CircuitLab
Figure 1. You have a broken neutral. |
H: Output voltage of op-amp in open loop configuration
What is the output voltage of op amp in open loop configuration when voltage in inverting input is greater than the one in the non-inverting input?
Is there a technical term for this output voltage?
Do datasheets refer to it?
AI: What is the output voltage of op amp in open loop configuration when voltage in inverting input is greater than the one in the non-inverting input?
The output will swing as far negative as it can go.
Is there a technical term for this output voltage?
Saturation voltage.
Do datasheets refer to it?
Yes. Maximum and minimum output voltage. With common op-amps this is 1 to 3 V from the supply rail. For "rail-to-rail" op-amps it can be close to zero at low currents. |
H: 74LVC and similar... can a logic 1 output sink current?
Say I am running my 74LVC125 at 2.5V. At the A input I have a logic 1 and at the nOE input, I have a logic 0, so the device should be driving a logic 1 onto the Y output.
With no load connected to the Y output, I should observe 2.5V at the output, obviously.
Now say I have a load along the lines of 1kohm to 5V. What happens? I choose 74LVC125 for my example because for tristate LVC logic, in the high-Z state (nOE high), applying higher voltage than Vcc on an output is allowed. But what about with the output enabled?
Will it sink something like (5.0V - 2.5V) / 1kOhm = 2.5mA?
Will very little current flow?
Will it sink more current than I would expect? The TVS diodes of the LVC device will not yet flow current, but will some kind of reverse bias damage happen to the rest of the LVC driver IC?
I want the first thing to happen, that is for the 'LVC125 to sink something like 2.5mA into 2.5V. Using a logic circuit like 74LVC125 is cheaper and than applying several MOSFETs for this purpose. Can it be done?
AI: It is not allowed, according to the datasheet, to apply any more than Vcc + 0.5V (absolute maximum) to the output.
However the built in clamp diodes can withstand a certain amount of current (not specified, but below GND -50mA is permitted.
So with the output high-Z and a pullup to +5 the output will be about at the Vcc rail + 0.7V (or ~5V, whichever is less). The Vcc rail may well not stay at 2.5V and may be pulled up to approaching 5V depending on what else is on the rail, since most voltage regulators will not sink current.
With the output enabled and high, the high-side MOSFET (p-channel) will be "on' and conducting (in reverse) and the above sentence also applies, however there will no longer be a diode drop involved. |
H: I have a college senior design question
I'm in my second to last year of mechanical engineering. So next year I'm going to be going into senior design. For those that have gone into any engineering course and have done senior design projects...
How are design projects chosen?
Do I chose one from a list of does a professor tell my group what he wants is to do?
If you can tell me anything about the beginning selection process thatd be really helpful to prepare me. Thanks
**
Edit: the responses are great and I appreciate them. I'm gonna take it one step further and also ask...
For those that went on to graduate school and received your Ph.D how did you choose your target research study?
There's a whole world of things you could have researched how did you choose?
**
AI: It depends on the school. At my school, the senior design capstone consisted of two semesters - the first semester was selecting an idea of your choosing to be your design project, usually by generating a list of ideas and vetting them with the help of the professor and/or your group members (if it will be a group design project). Part of that vetting process will be determining which of those ideas is feasible, and after that, which of those ideas is feasible in the time frame you are allotted. A justification is required about the marketability or value of your project idea as well.
Once you have your idea, you begin researching the topic in order to plan the project. In my case, I did this part quickly and began working on the project in the first semester to get ahead of things.
It sounds like you are trying to be proactive here, which is the best way to go with these things. A good way to find ideas is to use the university library databases and just read PhD journal papers that have been published. This will give you a broad picture of where advanced research is, and possibly spark an idea of how to incorporate some of the research into a project. I also talked a lot with my professors about things I was interested in, which helped too.
Oh, and something I did which was useful... I took a 2-credit hour self-guided "research" course. I think it served as a couple of technical electives for my degree program. I did this before the senior design semesters and basically used that course to get ahead in my senior design. I created a similar piece of hardware/software to what I wanted to do in senior design, and advertised it as something with a slightly different purpose. The research courses are very unstructured and really only requires showing that you learned something. I did about half of my senior design project's technical work during that course, so I was well-prepared when I got to the real thing.
What you choose should be relevant to your particular degree path (if you have "emphases" to add to your degree; I was a biomedical flavor of EE, so I did an ECG project). It should also be something that can either help society or be a marketable product; ideally both. Later, when you are judged, this is kind of important.
Typically, your ideas will be requested in the form of a proposal: some kind of multi-page report that summarizes your idea(s), feasibility, marketability, and a rough project plan that proves you have thought through the feasibility sufficiently.
This is based off my graduating in 2014, so somewhat recent. Hope that helps. |
H: Floating analog input using Arduino with 1m cable
I use an analog input from an Arduino to read the value from a piezo plugged in parallel with a 1M ohm resistor.
I followed this tutorial from Arduino: https://www.arduino.cc/en/Tutorial/Knock
Here is my schematic:
It worked perfectly until I decided to use a 1m cable between the piezo scheme (piezo || resistor) and the analog input.
With a short cable, I read 0 when there was no vibration nor activity. But when I change the length to 1m, I always read values higher than 0... This input is floating.
What would be the solution so I can read a precise value and stop this input from floating?
AI: Most of the designs I see in the comments seem overkill for what you are trying to do (BTW: you do realize you are applying negative voltages to your µC, right?). A very simple amplifier could do the trick.
Try this configuration, which is powered from the remote bias node on the µC side:
simulate this circuit – Schematic created using CircuitLab
If the gain is excessive for your application or you want to DC-couple to the piezo, you can add a resistor on the source of the FET and/or remove C2. |
H: Altium 18 PCB Filter Select X *except* if Y
I'm trying to create a PCB filter that will allow me to select everything on the board except for components that have an 0402 or 0603 package size. We have a parameter in our passive parts called "Package" which can be '0402' or '0603', so this would be a perfect way to isolate these components.
I can easily select these using this filter query:
HasParameterValue('Package','0402') And HasParameterValue('Package','0603')
However, if I try to invert this, it selects everything on the board, including the 0402 and 0603 components. I assume this is because the Not operator does not invert the results returned by HasParameterValue()...
Since we have a lot of footprints that are 0603 and 0402, and since these could change in the future, I don't want to specify a long list of Footprint = 'blah0402blah'. So I tried:
Not (Footprint LIKE '*0402*')
The Not has the same effect here. I attempted to do "Not LIKE" but that is not syntactically valid.
Does anyone know of a way to remove or except results of the query based on Parameter values or a similar method?
Edit: @LoganF
I have tried it with various paranthetical combinations. Just tried it like this to no avail:
(Not HasParameterValue('Package','0402')) And (Not HasParameterValue('Package','0603'))
AI: I will look for an answer using filter queries when I am in front of Altium 18. currently I have Altium 15 available, which doesn't have the function HasParameterValue(), but does have HasParameter(). I have found that you can use
IsPart and not (HasParameter('Value','10k') or HasParameter('Value','1k'))
to achieve a similar result, but I don't know if this will hold in Altium 18, since the Altium 15 equivalent of your example filter:
HasParameterValue('Value','10k') and HasParameterValue('Value','0R')
appears to be or instead of and:
HasParameter('Value','10k') or HasParameter('Value','0R')
However, in any recent version of Altium you can use "find similar objects" which is very powerful.
the quickest method since you already have a query which selects your unwanted parts:
run the query to select your 0603 and 0402 parts
right click on any selected part and choose "find similar objects"
change the "selected" attribute from "any" to "different" (last row in "graphical")
set the scope: in Altium 18, you get "current project/current document/open documents" before 18, you only get "current document/open documents"
make sure "clear existing" and "select matching" are both ticked.
when you run this it will invert the selection. you may want to narrow down the objects by setting component type from "any" to "same", object type=part should be set to "same" by default.
I will update this on Monday if I find a query based solution |
H: How to test a MOV with no visible damage
As far as i know cumulative degradation occurs in a MOV when multiple surges appears in a circuit, A way to minimize this degradation is having multiple MOVs in parallel.
This kind of damage can be internal and so not detectable by visual inspection.
In this situation i can use a digital multimeter to test the MOV for a short circuit between its terminals.
I'm aware that this test is not 100% precise since a MOV can be damaged and present a open circuit between its terminals.
Taking MOV above its breakdown with a high voltage, to see if it conducts and check the clamp voltage does not seems very safe or practical.
My question is:
Beyond looking for short circuits between the terminals with a multimeter is there any practical way of testing a MOV (with no physical damage visible) in a PCB?
Thanks
AI: A MOV can be tested with a small applied current (such as 1mA) and that voltage compared with the datasheet specification.
For example, a Bourns MOV-07D221K should drop 198V minimum, 220V nominal 242V maximum at 1mA. That test should be done quickly with appropriate equipment to limit the current to avoid heating the MOV too much, particularly with higher voltage MOVs. Obviously, you would do that out of circuit.
Chances are if the MOV looks okay (and tests open circuit) it is, they normally fail shorted and only go open when they are blown open by excessive current (which tends to be detectable visually or by odor). |
H: What step have I missed designing this differential amplifier?
First time posting on EE.SE and I hope this question isn't too broad. This will be a long post. I should also point out: I'm asking the process and not the solution the problem. I've did create a circuit that works, but I also found a circuit online that is simpler and probably performs a lot better in real life. The core question here is: what path/thought process could I have taken to arrive at the more concise solution? I understand my design well, and it works, at least in LT Spice, but it uses more components to get there. I understand the math behind both solutions and why they both work, so I don't need that to be explained either. It's only the way/path to achieve the latter solution that eludes me.
The task: Design a circuit that will amplify the difference between two input voltages by two.
My procedure: I cracked open LT Spice, and immediately started of with an op-amp. First I wanted to find a working topology (I can work out the exact values later). I hooked up both input voltages to the two inputs and ran a feedback resistor to give me a gain of two.
Pardon the cluster of a circuit (just a prototype). The difference between Vin1 and Vin2 is about 1 V, but Vout is 3 V. That's obviously wrong, and I quickly realized why. Sure, the difference is amplified by 2, but it is subtracted from Vin2, which gives not 2 V, but 3 V (Vin2 = 5 V, so 5 V - 2 V = 3 V at Vout).
The first attempt made me realize a couple of things. My desired result does appear in this solution, but it is 1) inverted and 2) with respect to Vin2, not ground like I expected. Having analyzed it, it's now obvious, but it wasn't before (beginner brain, oh well). Since the non-inverting input seems to be the reference here, I figured I just have to set it to ground and then feed the difference between Vin1 and Vin2 into one of the inputs.
Combining the voltages is easy, but one of them needs to be inverted so that they cancel out via superposition. The only thing that would be left would be the difference, which can be fed into the amplifier stage. I decided to feed Vin1 through an inverter. The other voltage gets superimposed directly, and the difference is fed into an amplifier with gain 2, giving the circuit like so:
The simulation verifies that the circuit works as expected. The task requirements are satisfied.
After having gone through the design process by myself, I looked up a differential op-amp amplifier online and found this guy (from electronics-tutorials.ws):
What the heck! How did I miss that? This hasn't even crossed my mind! After failing with feeding Vin2 into the non-inverting input, I reworked my design based on my observations. I observed that Vout will be a voltage offset with respect to the non-inverting input, and following that, I arrived at a working solution. But this is also a working solution! What could have I possibly observed about my initial attempt that would have led me to discover the simpler solution presented on electronics-tutorials? I feel like there is a mindset here that is completely lost upon me, and I'm trying to find it. I'm hoping you guys who would have come up with a design like this right away can help by pointing me towards that missing link. Ya'll are much smarter than I. :)
EDIT: I took a couple of days to ponder this problem, and I think I figured out how to get there. As per Reinderien's suggestion, I've taken the initial circuit I had and figured out the transfer function:
$$V_{out}(t) = \frac{R_2}{R_1}(V_{in2}(t)-V_{in1}(t)) + V_{in2}(t)$$
Which is basically what I already expected from the simulation I ran in LT Spice. I get the difference between the two inputs, multiplied by the gain, and offset by the input at the non-inverting op-amp pin.
I started to think about how I could get rid of the offset Vin2. I needed to somehow get both Vin2 terms to cancel out to get just the difference. The method dawned on me when I remembered that this is kind of what you get if you calculate a Thevenin's equivalent circuit. For example, in a simple voltage divider circuit, we end up with the source voltage multiplied by a factor determined by the resistors. The current here isn't of particular importance since the ideal op-amp doesn't draw any current itself, so it does not load the equivalent network (yes, I understand that in the real world, the input impedance would matter considering the voltage source would then be loaded, but for the sake of argument, and considering it is a basic electronics class, assume the planets align just right). So, I got on with the algebra as follows (to simplify latex, Vin1(t) -> V1 and Vin2(t) -> V2):
$$\begin{eqnarray*}
\frac{R_2}{R_1}(V_2-V_1)&=&\frac{R_2}{R_1}(\alpha V_2 - V_1) + \alpha V_2\\
&=&\frac{R_2}{R_1}\alpha V_2 + \alpha V_2 - \frac{R_2}{R_1}V_1\\
&=&V_2 (\alpha \frac{R_2}{R_1}+\alpha ) - \frac{R_2}{R_1}V_1\\
&=&\frac{R_2}{R_1}\left[ V_2(\alpha +\frac{\alpha R_1}{R_2}) - V_1\right]
\end{eqnarray*}$$
The equation on the left and right are the same, except for the (a*R2/R1 + a) factor. On the left, the matching coefficient is 1. There is only one possibility. Whatever a is, that whole thing needs to work out to 1, so
$$\begin{eqnarray*}
1&=&\alpha + \frac{\alpha R_1}{R_2}\\
R_2&=&R_2\alpha + R_1\alpha\\
R_2&=&(R_2 + R_1)\alpha\\
\frac{R_2}{R_2 + R_1}&=&\alpha
\end{eqnarray*}$$
I think that's fairly satisfactory. When we attenuate Vin2 by the complementary ratio (for lack of a better word) of the gain, the sums work out just right.
AI: You say you understand the math, so I won't (re)explain it to you. You say you want the process.
The process is -
Start with a basic understanding of the theory. One single op-amp can produce an output that's a linear combination of the two inputs. Any resistive divider prior to the op-amp will attenuate its input; you need this behaviour in this case.
Build a circuit with symbols instead of values, and calculate the expected output as a function of its inputs.
Insert "reasonable values" for some unknown resistors until the solution is not underdetermined. In this case, "reasonable" may be 1k or 10k, for instance, but this also requires knowledge of the application, including the input impedance of the op-amp and output impedance of the prior stage.
Solve for the remaining values.
Simulate to confirm your circuit.
Quite honestly, after doing work in this field professionally for any length of time, most people memorize the common circuits and skip the first few steps automatically. Whereas professionals understand the theory, they usually don't need to reiterate it once they understand and remember the common solutions. |
H: Is it OK that I solder inside my bedroom?
I'm a hobbyist solderer using tin lead solder to occasionally repair broken electronics. As I don't solder all day every day, I use my normal writing desk in my bedroom, which stands next to my bed, for soldering, cleaning it afterwards with soap and water to remove flux residue, etc. As solder pops and splashes, there may be some little solder leftovers lying around in the room.
Is this safe?
AI: It is a good idea to you wash your hands after using the leaded solder or else it can sometimes get stored inside the brain. Make sure you use a fume extractor (or if the area is well ventilated a PC fan will do the job).
Also lead-free solder can actually be worse than leaded solder due to the fact that they replace the lead with other chemicals which release toxic fumes. |
H: How can 400V & 690V on e-motors result in the same power
Forgive my ignorance, but how can two 3-phase ratings on the same motor result in the same output power?
The one I have is rated for 0.7kW in triangle for both 400V and 690V.
AI: simulate this circuit – Schematic created using CircuitLab
Figure 1. Connection possibilities.
On a 400 V phase-to-phase supply the motor windings are connected in delta (from the Greek 'D' or Δ).
On a 690 V phase-to-phase supply the motor windings are connected in star or wye (from the American letter 'Y').
From trigonometry you can show that the orange lines in Figure 1b are \$ \sqrt 3 \$ times the length of the black lines. \$ 400 \times \sqrt 3 = 692.8 \$.
simulate this circuit
Figure 2. Connection options.
How can two 3-phase ratings on the same motor result in the same output power?
In each case the voltage across each winding is 400 V so the same current will flow in each winding and the same power will be produced. |
H: Why do we use a CMOS for inverting a circuit when the PMOS already achieves that?
The output in a PMOS is as follows:
I/P O/P
0 1
1 0
Why can't I just use this instead of using a CMOS for inverting logic?
(Please explain in simple terms as I am a beginner in this topic and subject)
AI: In a word: Efficiency.
You can use a PMOS transistor to drive a logic output high (e.g. VDD) when the input is low (e.g. GND). However, you can't use that same PMOS transistor to drive a logic output low when the input is high.
When you drive the input high in your PMOS inverter, it turns off, leaving the output effectively high-impedance, which is not logic low.
Your actual truth table is:
I/P O/P
0 1
1 Z
You can overcome this inability to drive low, by using a resistor to pull the output low when the transistor is off. However to be able to strongly drive low, you need a low value resistor.
This resistor is always across the output, which means that when you turn the PMOS on to drive high, a large current will flow from the PMOS through the resistor to ground. This uses lots of energy. If you have billions of switches, you can see that the power consumption will be very high.
The better approach is to replace this resistor with an NMOS transistor. This is called CMOS. By using a NMOS device, you can think of it as being able turn off the resistor when the output is driven high (PMOS is on).
Using the NMOS you can also get a strong logic low because when switched on, the NMOS is effectively a short.
CMOS therefore by using complementary transistors, has very low static power dissipation - when an output is being held either high or low, almost no power is consumed. |
H: Continuous, frequency-domain digital equalizer
I'm facing a real case of DSP application. I have a basic knowledge of digital filtering and telecommunication issues.
The input of my "black-box" is a digital audio signal (say a common WAV file, stereo 44100 Hz, 16-bit). I need to develop a digital equalizer and create the output audio samples.
The computation has to be done in real-time, i.e. with a live-stream.
I'm aware that it means I have to define a buffer large enough to feed the filters.
The basic approach is to create several pass-band digital filters to change the gain of the frequencies in each band.
Because the requested bands are a lot (> 16) I'm going to work in the frequency-domain.
My thought:
convert the signal in frequency-domain (DFT)
define a shape of the desired response
process the input samples with this shape
revert back to time-domain to retrieve the actual audio samples
Questions:
what is the technical name of this method?
do you think this approach is better(*) than use a lot of steep pass-band filters?
(*) better here is not just an opinion: I'm talking about code complexity, latency and accuracy
AI: Your question is better suited for the signal processing stack exchange, and this is perhaps the answer you are looking for. However, the basic issues are:
FFTs/DFTs are filters, however (1) their frequency response is not ideal as these overlap each other on the continuous domain, (2) are not spaced in a way that best matches our auditory response leading to wasteful processing, and (3) are circular convolutions which can lead to sample-to-sample discontinuities.
What you are looking for is an efficient orthogonal filter bank that is logarithmically spaced in the frequency domain. There exist several approaches to these.
An appropriate wavelet transform could be more efficient, as decimation is introduced on every step which bypasses the fixed bin resolution of FFTs.
In particular the Invertible Constant-Q Transform was designed with these problems in mind.
Although I am not very familiar with it, I can come-up with a methodology that might serve your purpose based on the wavelet approach.
Design a filter that extracts the top half of the frequency domain to the desired order.
Divide the domain in two orthogonal sections, by applying the filter and subtracting the filter output from the input.
Keep the top component for processing. This is the detail coefficient.
Decimate the lower frequency component by throwing away half the samples. This is the approximation coefficient.
Apply the filter again to the decimated half.
Repeat the process, successively dividing the spectrum in bands that are logarithmically spaced.
Apply the desired scaling.
Successively upsample (which includes low-pass filtering) each low band and add it to its corresponding high-band to reconstruct the filtered signal.
As the detail coefficients are band-limited as well, these can also be decimated and processed the same way to provide a higher frequency resolution.
Note that each lower-band takes twice the number of samples to process so this is an instance of variable-rate filtering characteristic of wavelet transforms. However, the total sampling rate remains constant due to the intervening sub-sampling.
There are some phase issues that would need to be addressed and would lead to practical limitations in the filters. Wikipedia goes at length on the Discrete Wavelet Transform. |
H: Rearranging transfer function for Bode diagram
Hi, I have been given this circuit and need to find its transfer function for then realizing its Bode diagram.
The problem is that I don't know how to arrange the terms to find the asymptotes of the diagram.
After some calculations I found : \$\frac{j\omega*R3*(R1+R2)Ca - \omega^2*R1R2R3CaCb}{j\omega*R1*(R2Cb+R3Ca) + R1 - \omega^2*R1R2R3CaCb}\$
I don't know if I am correct about this equation and even if I am I don't know how to find the cutoff frequencies of this transfer function can anybody help me please ?
AI: Welcome to the electronics stack-exchange!
I would go the approach of R.D middlebrook and do the analysis in the style of "design-oriented analysis" This means that I would first split up the circuit, and do algebra "on the circuit".
First separation:
We see that the RC filter can be separated out from the rest of the circuit because it is not loaded by the opamp:
simulate this circuit – Schematic created using CircuitLab
$$H(s)' =\frac{uo'}{ui'} = \frac{(C_a\cdot s)^{-1}}{R_3 + (C_a\cdot s)^{-1}} = \frac {1}{{R_3C_a\cdot s}+1}$$
first one down.
now to the opamp
simulate this circuit
We see that this is a common amplifier configuration with the formula
$$H = \frac{Z_2}{Z_1} + 1$$ where $$Z_1 = R_3$$ and Z_2 is R_1 in parallel with C_3.
So we see that
$$H(s)'' = \frac{uo''}{ui''} = \frac{R_2 || C_b}{R_1} +1 = \frac{(R_2^{-1} +(\frac{1}{C_b\cdot s})^{-1})^{-1}}{R_1} +1 = \left |\frac{1}{\frac{1}{R_2} + C_b\cdot s}\right|_{\cdot \frac{R_2}{R_1}} \cdot \frac{1}{R_1} +1 $$$$= \frac{R_2}{R_1}\cdot\frac{1}{1+R_2C_b\cdot s}+1$$
So the total transfer function is
$$\left( \frac {1}{{R_3C_a\cdot s}+1}\right)\cdot \left (\frac{R_2}{R_1}\cdot\frac{1}{1+R_2C_b\cdot s}+1\right)$$
from this you can see that we have 2 cutoff frequecies. the first one being $$\frac{1}{2\pi \cdot R_3 C_a}$$ and the second one: $$\frac{1}{2\pi \cdot R_2 C_b}$$ ( given that the input node on the opamp is virtual ground)
I hope that my answer gives you another perspective on how to analyze circuits, in an easier way. |
H: How to write data to end of line of existing file (STM32CubeMX FATFS SDIO issue)?
I have successfully coded write data into a CSV file is an SD card using STM32 CubeMS FATFS SDIO interface 1-bit mode.
But I can't find the file append function in HAL FATFS driver to write data to existing file. Is there another way to do this using STM32CubeMX and HAL drivers?
AI: Exactly as you would on other platforms.
Open the file for append
The f_open() function has a FA_OPEN_APPEND flag, which makes subsequent f_write() functions to append data at the end of the file.
Move the file pointer to the end of the file
If the file can't be opened in append mode (e.g. when existing data must be accessed first), then you can set the file pointer to the length of the file.
f_lseek(fp, f_size(fp)); |
H: How to drive p-channel MOSFET
I need a high side switch. I thought about using a p-channel MOSFET. The source will be connected to around 50V-60V. How can I drive this MOSFET? I can't find a driver which can switch between 60V and 45V in order to put the MOSFET on and off. They all switch between ground and Vcc, which is generally below 25V. How is switching a high-side p-MOSFET at 60V done? My control signal is at 3.3V.
EDIT:
I want to avoid a second power supply. The microcontroller used for the signal generates a signal for a low side n-MOSFET as well.
EDIT 2: The goal is to drive a chopper for a motor, so the switching frequency I was thinking about is around a few kHz.
AI: What you're talking about is generally called a level shifter. Assuming you don't want high-frequency switching, you can use a circuit like
simulate this circuit – Schematic created using CircuitLab
A few tips. The NPN must be rated for more than 50 volts. You can play with R3 and R4, but the total resistance should give a current at 50 volts of about 10 times the NPN base current or less. Larger values will dissipate less power in the resistors, but will also interact with the FET gate capacitance to slow switching and increase the power dissipated by the FET. Adjusting the ratio of R3 and R4 controls the gate drive voltage, so don't get carried away. A 1 to 3 ratio at 50 volts gives 12.5 volts and 15 volts at 60, which should be just fine. |
H: Is input current of a gate driver auto-regulated?
I have tried googling, but almost everything I can find on gate drivers discusses only the maximum current parameters.
I'm looking at a gate driver for a university project. The gate driver uses 3.3V logic, and on the datasheet lists the maximum 'Input Logic Current High" at 1uA.
I was planning on using a Raspberry Pi 3b, but the gpio pins can only be programmed to output 2mA-16mA. As it stands, is the Pi not compatible at all with this gate driver?
Part Number: MD1213 High-Speed Dual Mosfet Driver
Datasheet: http://ww1.microchip.com/downloads/en/DeviceDoc/20005713B.pdf
AI: The Pi outputs may be able to deliver up to 16 mA. The gate driver, or any other load you connect, will only draw the current it requires - same with any power supply and load. |
H: what is the German/Austrian nomenclature(Voltage and current direction) in electronic circuit?
AS I came from a country where we use the french system in defining the voltage current arrows in electronic circuit, I am confused about the German/Austrian system where I see the voltage arrow is directed from positive to negative and this makes me confused when it comes to analyse complex circuit especially with diodes and mosfet. any clarification is more than welcome
here one of the circuit if I use the German Terminology
AI: I think you stumbled over a convention called Verbraucherzählpfeilsystem. It's pretty common in German EE conventions.
All items are treated as loads, regardless if they are loads (Verbraucher) or generators.
simulate this circuit – Schematic created using CircuitLab
This idea is you don't have to put the arrows on the schematic. You simply write Verbraucherzählpfeilsystem somewhere prominent, then write
$$I_g = -100A$$
$$I_l = 100A$$
$$U_g = U_l = 100V$$
and everyone agrees \$I_g\$ and \$U_g\$ belong to a generator, because \$I_g\cdot U_g\$ is negative.
You could also write Generatorzählpfeilsystem somewhere and use the exactly opposite definition. But that's not so common in German EE.
Also note German convention is using U instead of V for voltage identifiers to avoid confusion with the V unit symbol. |
H: How fast do power plants need to respond to demand?
I've heard the layman's description of electrical grid response time before - it usually contains the word "instant" or "real-time". But I'm interested in the actual response times of the various types of power generation.
My limited ability to research has led me to articles on the three major "layers" of power generation: base load generation, load following generation, and peaking generation. I understand the high level difference between the three generation types, and roughly the kinds of plants that fulfill them. It seems that fuels like nuclear or coal serve the base load because they have slow response times. And on the other end of the spectrum, fuels like natural gas serve the peak load because the generation systems can be set up for fast response times.
That high level information has been easy to find, but I'm having trouble finding actual numbers. These are the only numbers I have so far:
Base load coal plants coal can take many hours to start up.
Peaking plants like hydro batteries can start up in tens of seconds, and simple cycle gas turbines can start up in tens of minutes.
But I feel like "start up time" is very different from response time, and I can't find any data on that. But maybe I'm working with bad assumptions - do these plants actually need to respond to instantaneous changes in load after they've started up?
If yes, which plants handle that kind of response, and how fast do they react?
Electrical engineering and power systems are quite far from my day job, so I'm not very good at constructing queries to answer my questions. Apologies if this has a trivial answer sitting somewhere else.
AI: The "fast" response you're thinking of is called frequency response. The grid consists mostly of generators and motors which "electrically" rotate in sync with the grid (the physical rotation being different by the slip). When the load on the grid exceeds the supply, they slow down and frequency decreases (slightly). When the power input to the generator is increased, or the power output of the load decreased, they will attempt to speed up the frequency will increase (slightly). The excitation current of the generator also affects this.
So the inertia of the grid provides the fastest response, where any sudden increase in load or decrease in generation is matched by a decrease in frequency. Then fast-responding plants will increase throttle/power input to the grid and frequency will stabilize. With thermal power plants opening the throttle like this does not change the thermal output of the plant, so if sustained would cause a decrease of pressure at the turbine inlet and loss of power. |
H: Why is the potentiometer setup in this configuration in this 555 circuit
I just started learning about chips and signals using the book Make:
Electronics, in one of the circuits, used to generate a pulse in Monostable mode
a potentiometer is used to vary the pulse length. What I'm having trouble understanding is why the right pin of the pot is connected back to the wiper and not to ground, I built the circuit on a breadboard and indeed when I connect it to ground the circuit does not function properly.
AI: The circuit needs a variable resistance there (AKA a rheostat). It does not need a potentiometer, but potentiometers are what you can buy in the store. You use a potentiometer as a rheostat by connecting it as shown, or by leaving one leg unconnected (it's usually better to connect it as shown, in case the wiper loses contact with the resistive element).
Note that potentiometers are not always connected between power and ground, or power and minus power, etc. -- they show up connected every which way; from power (or some reference, or input) to ground is the most common, but by no means the only way. |
H: Operating Voltage Range Question
Almost certainly a neophyte question. I'm trying to understand the difference between nominal voltage and operation voltage range. My intuition was that, for a given power supply, it would hold a voltage within a few volts (hence, it would have an operational voltage range within a few volts of it's nominal voltage).
So what I'm trying to understand, is why a device such as this has such a wide operational voltage range:
Image source: http://www.astronautix.com.
AI: It has a nominal power of 2200 W at the nominal voltage of 350 V which, I presume, is DC. This may be related to a standard power supply they manufacture.
My intuition was that, for a given power supply, it would hold a voltage within a few volts.
No, it would be up to the power-supply to hold the voltage, not the device.
So what I'm trying to understand, is why a device such as this has such a wide operational voltage range.
Because it can! An electric heating element may have a nominal voltage of 230 V. You can run it at 250 V and it will run a bit hotter (and may die sooner) or you could run it at a lower voltage but it would give out less power (but might last forever). |
H: Burst mode in synchronous buck converter with current mode control?
I read this previous question:
The meaning of Burst Mode
It says burst mode happens when the resistance of the load is light and the system stops switching the MOSFETs. However, I just cannot understand how the light load decreases the load current and therefore the inductor current, as mentioned in the link.
As I know \$Vload=Rload*Iload\$ and \$Vload=D*Vin\$. So if we have a low load and highest input voltage, this implies that the duty cycle is small and the current is going high - which means the inductor current must be high as well and never go to DCM!
What I am missing?
AI: However I just cannot understand how the light load decreases the load current
Light load is defined as low load current. High load resistance is just one way that may happen.
and therefore the inductor current, as mentioned in the link.
The inductor current is controlled by the converter circuitry; the low inductor current is not a direct effect of the light load; it is an effect of the controller reducing the duty cycle.
As I know Vload=Rload∗Iload and Vload=D∗Vin. So if we have low load and highest input voltage, this implies that duty cycle is small and the current is going high
Again, Rload is not necessarily an accurate model. Low load means that the load current is low, by whatever means that may happen. When the load current is low, the load voltage goes up, and the controller reduces the duty cycle.
What I am missing?
I'm not sure, but I suspect you're not taking the behavior of the controller into account. |
H: Diode on op-amp output to drive tube
I am working on a tube based guitar amplifier, and found an interesting schematic online. There is an op-amp that has diodes across a resistor in the feedback path.
I understand what it makes the op-amp do, instantly skip over any value less then abs(0.5V). What I'm asking is if anyone knows why you would add this to a guitar amplifier to drive a tube.
I have also included a simulation of Switch SW101a, basically how much resistance to ground on the negative input terminal.
The only answer I could come up with is "sound" or it somehow helps correct for non-linear responses in the tube.
AI: It's a form of deliberate distortion, which is popular in some guitar styles.
If you look at your blue simulation waveform, you can decompose it into two components:
the original sinewave (about 3.4 Vpp)
a square wave (about 0.8 Vpp) at the same frequency that has the same zero crossings.
This adds a series of odd harmonics to the signal, which gives it extra "buzz" or presence.
This is not the same as a clipping circuit, which only kicks in at high input signal levels. This effect applies at all signal levels. In fact, this effect is MORE pronounced at low signal levels. |
H: RF Antenna design for 2 different operating frequencies
I have noticed that my mobile phone sometimes shows it is using the 3G and then other times the 4G.
How is that possible if they are using the same antenna?
I assume it would use a microcontroller to switch the path going to the antenna to have the correct impedance (50 ohm). Or perhaps a Balun that can manage the different frequencies?
3G - 1.8 GHz
4G - 2.6 GHz
AI: Examining papers from Broadcom Corp, some years back, on all-silicon radios (no offchip discretes), they were using Varactors to TUNE the antenna interfaces. |
H: What's the source of this noise in my Arduino oscillator?
I've been experimenting with techniques to rapidly switch on and off the digital output pins in an Arduino. This is the code I use:
#define PIN3_ON PORTD |= 0x8
#define PIN3_OFF PORTD &= ~0x8
#define NOP __asm__ __volatile__ ("nop\n\tnop\n\tnop\n\tnop\n\tnop\n\t")
void setup() {
pinMode(3, OUTPUT);
}
void loop()
{
cli();
while (1) {
PIN3_ON;
NOP;
NOP;
PIN3_OFF;
NOP;
NOP;
}
}
When I observe the output under a scope, I see some sort of quickly decaying oscillation after the pin is switched off:
I have observed this on an Arduino Uno and Nano328. What is the reason for this oscillation?
AI: Try clipping your ground lead onto the ground pin that's physically closest to the pin you're looking at. If that doesn't change anything, and if you can, take up any slack in the ground lead of your scope by wrapping it around the probe.
I can't find a good article right off the bat (I'm sure that every scope maker out there has one) -- here's the few pointers I know:
Always ground the probe. If the signal is the least bit fast or important, ground each probe close to the signal.
You can do the ground-wrapping trick to make things a bit better.
If it's a really sharp-edged signal and you need to be sure, ground the probe right at the tip. If you pull the spring hook off the end of a typical O-scope probe, you'll see a metal sleeve around the tip. That's the probe ground. You can wrap a wire around that and ground it as close as possible to the signal you're probing (or you can look in the stuff that came with the probe and see if there's a little sprig-looking thing in there -- that's what it's for).
Make sure your probes are compensated, and calibrated if accuracy is important.
(If someone reading this knows a good article or YouTube presentation, please post a link -- this is one of those things that I've collected bits & pieces of over the years but have never had to articulate clearly). |
H: Troubleshooting TL071 to TA8207 100 Hertz hum
I worked out to build a pre-amplifier connected to a power driver in a single setups(a pcb). Problem arises when it's powered up. here are a few questions.
Setup has a TL071CP as preamp with a unity gain. an input impedance of **10meg** through a 100n polyester cap. no mistery here. To notice: we used single star-gnd. Vcc is 12 volt not 17.5 as intended or 35 volts as preferred by the JFET.
Output stage is politely recycled 7W beautiful amplifier device from Toshiba Microelectronics division tagged TA8207K. Actually using 5% resistors at the gain configuration at 100 ohms. All the setup follows guidelines from the schematic.
These are close to original schematics next,
Tests:
I've tested *1meg* and *2meg* input impedance but decided later 10meg as Zin.
For the medium drive Zin tested several passive devices. in general the TA8207 has a 35kohm Ain Z. Ran it using 2.2k resistor to ground and seems to be nice but not completely noise-free at the speakers.
Thus replaced original 47kohms Ain Z at the driver for a 2.2k/2.2k which has a noticeable gain and frequency response along with noise again.
Then finally tried floating Ain to Vcc/2 with no success. In exchange used the voltage divider at TA78207's Ain.
Goals:
My idea is having 3/10 from the actual noise without rebuilding everything. Ideally replacing 5% to 1% bleeder hum noise filter.
Main question:
Do i need matching the output impedance from TL071 which is 120kohm at 10meg input Z to the 35kohm Z in JFET related equation i haven't read?. I've considered a bleeder notch-filter very likely solution to 100hz hum.
If i float using Vcc/2 or add an inductors or add resistors at the different grounds. still need a notch-filter and Vcc/2 at all Ains. ?
Thank you for all your comments. Diego
AI: This may be power line, or rectified power from transformer, coupling by electrostatic field into the (+) pin of TLO 71.
Or could be magnetic field of power line or of diode surge currents, coupling into a LOOP or into multiple LOOPS of the circuit. |
H: DC-DC step down 12V to 5V source versus straight 5V source
Here is the scenario: I have a project needing 3 supply powers:
3.3V to run the microcontroller (max 100mA)
5V to run a bunch of adressable LEDs (max 8A, the most current intensive)
12V to supply op amps (max 15mA x 6 = 90mA for all opamps supplies)
The question is: if I want to use only one power adapter brick to power this thing, is it more logic to use a 12V power supply and step down to 5V and 3.3V? Or is it better to use a 5V power supply since most of the current is used on this voltage and simply step-up to 12V for my opamp needs?
Note: All opamps are signal level, the 12V is to have maximized headroom to preprocess my analog signals.
AI: Assuming a generous 90 percent efficiency, that's still 45 Watts in for 40 Watts out, so a 12V supply just to bring it down to 5V would not be efficient compared to stepping up at 80 percent efficiency for 1.5W in 1.2W out
5 Watts vs 0.3W efficiency penalty. Even if you boost to say 14V and use a LDO for a cleaner signal for your OP amps, your still looking at 10 times worse power loss with a 12V supply.
Cost is negligible for supplies for 12V 5A or 5V 8 to 10A, but a 45 watt step down converter will cost more than a 1 watt step up converter.
So in my opinion, it makes more sense to go with your natural voltage need. |
H: RF Development board layout questions
I was looking at the development board layout for the MMZ09312B and I had a few questions:
datasheet: http://cache.freescale.com/files/rf_if/doc/data_sheet/MMZ09312B.pdf
Are large polygon pours away from the signal path a good practice? For example in the test circuit the top left and right corners are polygon pours of the signals at pins 3 and 7 respectively which are far away. Doesn't all this extra copper end up picking up a bunch of noise (even if there is via stitching)?
Since I want the traces of my input signal and output signal to have the same impedance as the SMA connectors (50 Ohms characteristic impedance) the trace width will depend on the distance between the signal layer and the ground layer among other parameters. With a standard 2 layer board (distance between layers approx. 60mils), my trace widths would be >100mil. With a standard 4 layer board (distance between layer 1 and layer 2 is approx. 10mils) if I set layer 1 as signal and layer 2 as ground I can use a trace width of ~15mils. But if I use the 4 layer board, what should I do with layer 3 and 4? Leave them empty?
Picture of dev board for reference:
AI: I think you made an error in reading that PCB layout. The two strings \$RF_{IN}\$ and \$RF_{OUT}\$ are not there to indicate that the polygons are connected to those pins, but that the connectors there are in and out.
Even if they were, keep in mind that at these frequencies, a DC short does not mean a short at high frequencies. This can lead to confusing results if you look at them in PCB software that is not built for high frequency design, as they don't take this into account.
The extra copper does not pick up extra noise (unless it happens to resonate and act as an antenna at a certain frequency). It will likely act as a shield, improving noise performance.
Since I want the traces of my input signal and output signal to have the same impedance as the SMA connectors (50 Ohms characteristic impedance) the trace width will depend on the distance between the signal layer and the ground layer among other parameters.
This depends a lot on the type of transmission line you are using. In this design they are using CB-CPW\$^1\$. This has a few advantages, such as potentially lower losses, more compact than microstrip and somewhat less sensitive to variations in thickness of the substrate. It combines the advantages of microstrip (big ground shield) and CPW (lower losses, less dispersion). Most important, however, is the fact that the ground plane gives you a very nice short between the two grounds on the top layer, suppressing the undesired odd mode. There are a few decent calculators online that should prove sufficient for designs such as this.
Layer 3 and 4 can usually be used for anything you want. It is however common to use at least one of them (generally layer 3) for your supplies, if you can't (or don't want to) do this on the top layer.
\$^1\$ CB-CPW stands for Conductor-backed CPW. Some literature will also call this G-CPW (grounded CPW) or CPWG (cpw with ground), most non-microwave/millimeter wave designers will just say 'CPW', but there are various types of CPW. 'True' CPW has one conductor layer with a ground-signal-ground mode, and the grounds extend to infinity. If you have CPW with a ground plane underneath, you have a somewhat different mode, and it impacts things. If you make it quite narrow (less than a wavelength or two of ground) it's really CB-FW-CPW (FW = finite width, so this would be conductor-backed-finite-width coplanar waveguide), as you start getting a significant error due to your grounds not actually being infinite.
The mode of CB-CPW is a hybrid between that of a CPW line and a microstrip line. |
H: USBL Hydrophone Array Spacing
I am trying to design a USBL acoustic pinger location system for a school project. I am struggling to determine if the frequency of the acoustic beacon ping places a constraint on the distances between hydrophones in the array. I have found in one source online that the hydrophones must be less than one half a wavelength apart to be able to implement the phase-comparison method for angle of arrival calculations. Is this the case?
AI: Yes and no.
Larger than that spacing introduces ambiguities in bearing. If the ambiguity can be resolved by other means (by knowledge of the source, or other hydrophones more closely spaced) then the extra receivers will still improve SNR and bearing accuracy. |
H: Why is there a grounding clip on this solder sucker instead of a third prong on the plug?
As most of these inexpensive tools tend to omit the ground altogether (such as in soldering irons), why would ground be added as a clip instead of a third prong?
Is there an advantage for it to be implemented this way?
AI: Some of reasons I can think of:
Some US plugs are still the original non-grounded outlets.
You don't necessarily want to ground directly to "earth". For example, you may an inline 1MOhm resistor to protect the user from electrocution (e.g. touching mains).
If you were using the earth terminal on the plug, you would not put a 1M resistor for avoiding confusion as to the purpose of the terminal - safety connection vs ESD protection connection.
You may not want to connect to mains earth at all. All you need is an equipotential working area - and that equipotential level doesn't necessarily need to be mains earth.
You could for example simply connect yourself (via wristband) and the solder sucker to an ESD mat. This would ensure you, your workspace, and your tool are at the same voltage as whatever device is in the workspace. |
H: Which is positive? Green or black?
I tried cutting my old Gameboy Advance SP Charger wires out since the adaptor doesn't work anymore. I am planning to attach it on a USB cord to replace the adaptor.
I have noticed that the wires aren't colored red and black which is the usual.
The wires are enclosed in a single black wire and inside there was a color green wire and a naked wire.
I want to know which is the positive and which is the negative among the two.
AI: Well, I bet on the green one to be the positive.
But don't trust me, use a multimeter to check the connection from the linkout port to your two wires.
Here's the pinout of the GBA side port, from this you can understand the pins on the charger's cable and check the connection to the wires: |
H: Unexpected behaviour of operational amplifier in an LDR light sensor
Here is the circuit I have set up in my breadboard
It is a light sensor using an LDR. I am using the op amp as comparator.
When indoors light is reduced, the LED lights up and I am able to set the lighting threshold with the help of the potentiometer.
But when indoors lighting is on, something strange (to me) happens. The LED is OFF except for a tiny region of potentiometer resistance. In this region, the voltmeter reads from -0.53 V to -0.29 V. This seems strange. When the voltmeter reads a value smaller than -0.53 V, the LED is OFF. This is logical. Additionaly, in a small region where the voltmeter reads positive values (from 0V up to 0.5 V) the LED is OFF. But if there is positive differential input, shouldn't the op amp go high?
Why the op amp behaves strangely in the aforementioned regions of the differential input voltage close to 0 V?
AI: When the lights are on, the LDR can have a very low resistance, say 10 k ohm or less. That means that the voltage at the - input of the LM358 will be around 9 V * 100k/(10k+100k) = 8.1 V
So the opamp then needs to compare at 8.1 V when it's supply is only 9 V. It cannot do that !
The input voltage range of an LM358 is 0 V to supply voltage - 1.5 V. See "Voltage Range" in table 6.5 of the datasheet
So at a 9 V supply voltage you cannot expect the opamp to work properly above input voltages higher than 7.5 V. To confirm, measure the voltages at the inputs of the opamp and see if these rise above 7.5 V
The simplest way to fix this is to make the opamp's input voltages low when there is a lot of light:
simulate this circuit – Schematic created using CircuitLab
This then "moves" the issue to the "dark side" (pun intended). If you want to make the circuit better in dark situations you could consider replacing R2 with a 1 Mohm resistor. |
H: Accumulate output from PID controller
I have come across a handful of examples of PID controllers where the process input is the accumulated PID output, i.e. the controller loop is u += pid(...) rather than u = pid(...) where u is the process input.
For the sake of example, say we are using PID to control the speed of a motor via PWM,
class PID:
def __init__(self, kp, ki, kd):
self.kp = kp
self.ki = ki
self.kd = kd
...
def pid(self, set_point, process_variable):
now = time()
dt = now - self.last_time
error = set_point - process_variable
p = self.kp * error
i = self.ki * error * dt + self.i_sum
d = self.kd * (error - self.last_error) / dt
output = p + i + d
self.i_sum = i
self.last_error = error
self.last_time = now
return output
My understanding of PID is that we should use the controller as
pid = PID(kp, ki, kd)
...
motor_pwm = pid.pid(target_speed, measured_speed)
But I often see it implemented as
pid = PID(kp, ki, kd)
...
motor_pwm += pid.pid(target_speed, measured_speed)
Now, for ki = kd = 0, the latter makes some intuitive sense to me; in fact, I think it actually gives you a kind of PI controller with ki = kp, and dt = 1 enforced. Once you introduce ki != 0 or kd != 0, however, I can't square this with any of the textbook explanations of PID I have read.
Is this 'accumulated' u += pid(...) controller simply an incorrect implementation of PID?
AI: It is not necessarily an incorrect implementation, if the feedback signal that your PID loop is tracking and controlling is the first derivative of the control output you need to emit. The continual accumulation of the PID output is equivalent to integration. |
H: Using less than 10 LEDs on LM3914
The LM3914 IC has a divider that works over 10 LEDs.
Does this IC enable using less than 10 leds and does it automatically adjust the scale based on the number of LEDs (eg 5)?
AI: Does this IC enable using less than 10 LEDs
Yes.
... and does it automatically adjust the scale based on the number of LEDs (eg 5)?
No. How would it know?
To do this you would set the reference voltage input to be double the input voltage required for five LEDs.
For example if you connect the internal 1.25 V reference voltage REF OUT to R HI then you need to attenuate your input signal to 1.25 V / 2 = 0.625 V to get five LEDs. Usually a simple resistor divider will do the trick. |
H: Opto-Isolater always on by higher voltage
I use an AQH3213 Opto-Isolater. When I use 3.3V on the 'dc motor side', everything works fine and it 'turns off and on in an interval of 1 second'. But when I use 5V, it's always on....
Do you have any suggestions what I do wrong?
The simple test code is correct, because it works with 3.3V; so must be a curcuit problem I guess.
AI: The AQH3213 is an optically isolated triac device for controlling AC loads. It should not be used on DC loads because the triac, once properly "fired" will remain on all the time until power is removed.
The reason it "appears" to work correctly at the lower DC voltage is that the triac will act as an on-off device if the current taken by load is below the threshold-latching level. Once above the threshold level it will remain permanently activated unless power is removed or cycled (as when used in an AC application). |
H: Does a plugged in 12v usb-charger use energy when no device is connected?
I hard-wired a 12v cigarette-plug into my a-pillar.
The whole point is to have an additional phone charger on the leftside where my phone mostly sits.
I connected it directly to the battery ( with a fuse ), so I can use it when the engine is off aswell.
When I connected the usb-charger to the cigarette-12v-plug an indicator-light came on. ( nothing else is connected ).
So obviously this is using power.
Will this drain my battery?
Do 12V usb-charger without an inidicator light use power when nothing else is connected?
if #2 is negative, how can i solve this?
AI: 1) yes, how long that will take depends on how much it draws and how long between car uses.
2) yes, they use power, more or less depends on the design & quality
3) Solve it by powering it from a fused accessory feed or ignition feed.
And, personally, I would do 3 whatever the power it draws... |
H: Ultrasonic Pulse Generator using Power Mosfet and Gate Driver. Confused about datasheet compatibility
I'm working on a project for University. The goal is to create an ultrasonic flaw detector that can measure flaws within a piece of metal. I have only taken introductory courses in the different branches of electrical engineering such as Circuits, Network Analysis, and Electrical Devices. Since my career path is currently in the software industry I haven't been focusing heavily on learning the topics I need to complete this project. Despite that, I've made some progress that I'm happy with.
Here is my current schematic:
The PT-12 has a resonant frequency of 5MHz, and operates best at a pulse amplitude of 220V. It is a dual-element transducer, meaning that the output and inputs of the transducer run on different cables, and both elements within the transducer are separated by an acoustic barrier.
Since the RF is 5MHz, that gives us a cycle time of 200ns. I've learned that the way these transducers typically operate is through a high voltage pulse of 1/2 cycle or less. This is referred to as pinging the device. So the goal of the pulse generation circuit is to provide a 100ns pulse at 220V amplitude.
My progress was going fairly well until I began thinking about how I'd be switching the pulse on and off again. I thought I could use a simple spst switch at first, and quickly realized that I was deluding myself. It seems that the only solution I could find was using a Power Mosfet.
I researched many different potential devices, as well as the parameters on the datasheets that would be the most important. I found that in order to get as close to 100ns pulse width as possible, I'd need to minimize both the Gate Charges (Qg,Qgs,Qgd) and Delay Times (tf,tr,td(on),td(off)). One potential device I found was SiHP6N40D, http://www.vishay.com/docs/91498/sihp6n40d.pdf.
Which had the following parameters:
It also specifies Rds(on) = 0.85 to 1 ohms. I learned while researching these power mosfets that Rds(on) typically decreases when more cells are added, which in turn increases capacitance and lowers response time. So a good Rds(on) value is near 1ohm for high switching purposes.
I found this equation on the following electronics.stackexchange post.
$$ f_{max} = \frac{1}{t_d + t_r + t_f + t_s} $$
And this seems to give 22.22MHz as the switching frequency, however the datasheet does not specify storage time. I'm hoping this parameter was just factored into the other delays. There was another time parameter, trr or Reverse Recovery time, but I'm unsure if this would affect the switching frequency. Assuming it can reach 22.22MHz, that seems to be quite good as it could turn on in 22.5ns and off in 22.5ns.
The next step was to interface the switch with my microcontroller so it could control the sampling rate. I realized the gpio output wouldn't be high enough to meet the Vgs of 10V, and additionally it would be too dangerous to directly connect my Raspberry Pi gpio pins to the power mosfet due to capacitance leakage through the gate. The solution for this was to use a gate driver, which in turn has its own delay times.
I researched some potential gate drivers and found 2 that had very low delay times:
ISL55110 --> https://www.mouser.com/datasheet/2/698/isl55110-11-1302115.pdf
MD1213DB1 --> http://ww1.microchip.com/downloads/en/DeviceDoc/20005713B.pdf
And it was at this point that I reached a threshold of confusion that I felt I needed to get help with. I don't know how to achieve my goal of a 100ns pulse width since there are so many factors to consider: mosfet delay, gate delay, microcontroller toggle speed. And ontop of that, I'm fairly certain now that in order to activate the mosfet gate, Vgs = 10V does not mean 10V is applied to the gate, it means that I would need to apply 230V. Applying only ~10V would put the mosfet into forward bias as according to this post.
In summary, I'd like help with figuring out how to go about approaching this to ensure I get as close to 100ns as I can. Also, if there are obvious problems that I haven't mentioned please inform me of those. Ultimately, if it isn't possible to get near 100ns, I'd be okay with simply being able to send A pulse to the transducer.
Component Calculations:
I've chosen to go with a sample rate of 2kHz, and used that to determine the components in my circuit schematic.
Since the capacitor should be almost fully discharged in 4 time constants (~98%), and I want the pulse to be 100ns
τ = 25ns.
C = 25ns/(50ohms + 1ohm) = ~490pF --> 470pF
*Where 50ohms is the impedance of the transducer and 1ohm is the Rds(on) resistance of the mosfet.
We'd want the capacitor to be charged in less than 1/2 of a cycle of the sample rate, or 250us.
R = 250us/470pF = 530kohm --> 560kohm
The current leaving the capacitor upon switching of the mosfet starts off high
Vc(1ns) = 220V x e^-(1ns/(470pF x 50.32ohm)) = 210.89V
dV/dT = (220V - 210.89V)/1ns = 9.11E9
ic(1ns) = 470pF x 9.11E9 = 4.28A
At the end of the desired time the mosfet should be open, ic would be
Vc(100ns) = 3.2V
dV/dT = (220V - 3.2V)/100ns = 2.168E9 V/s
ic(100ns) = 470pF x 2.168E9 = 1.02A
Extra info:
Why do I have both an Arduino and Raspberry Pi?
I began the semester with the intention of using the Arduino because it had analog input pins already on the board. I had already developed a GUI on an LCD screen I had, when I realized the 16MHz clockrate of the Arduino was not fast enough to control the mosfet switch speed of 100ns. I decided it would be too much work to rewrite the interface in Python. I simply added some code to communicate with the Pi serially which would then send the logic signals instead.
To determine distance, the Pi will simply count the number of clock cycles since the most recent HIGH output on its gate driver pin. Hopefully I'd know what the full delay of the circuitry would be by then, and then this could be factored out of the response time to get the true amount of time the wave took to travel across the measured medium. Specifically, the speed of sound in aluminum is 6320m/s (google). The dimensions of my measured object are fixed at 20cm x 20cm x 2.5cm. So if no flaw is present, the wave would travel the full distance and take 63.2us, otherwise a reflected wave would arrive back to the transducer quicker than that. It's a given that the defect upon testing will be near the middle somewhere, so roughly 31.6us travel time if a defect is present.
AI: Consider something like this
simulate this circuit – Schematic created using CircuitLab
And use a single sheet of copper foil for all your GROUNDS.
Make no slits in the sheet. |
H: Building circuit to amplify small sensor signals using an Instrumentation Amplifier and 24-Bit ADC
I'm a beginner when it comes to circuit design and I have spent the last few weeks doing research on this topic. I have now chosen some ICs (Instrumentation Amplifier and ADC) and would like to build the circuit. However I have some questions left, which I hope some experienced folks can help me with.
Sensors:
I have multiple of these CMP10 pyranometers: Datasheet of CMP10 Pyranometers
This passive sensors have two outputs: HI (+) and LO (-)
Here are some specs from the datasheet above:
This sensors measure the sun radiation and depending on the radiation they output a signal in the range of 0 mV up to 20 mV.
I would like to read it with a microcontroller, thus I need some good amount of amplification. In order to get a readout that is as accurate as the sensors, I would need to get to a resolution of 5uV per step. I know this is probably not trivial at all and requires a pretty good PCB design to get even close.
I calculated that in order to get down to 5uV steps, I need at least an ADC with 19-Bit operating at 2.048 V reference voltage (that would make about 3.906uV steps/resolution).
After my research, I learned that 19-Bit is pretty much what you can expect from a custom PCS design (if at all), that's why I have chosen an ADC with a ref of 2.048V (because going up to 5V ref, I would need at least 20-Bit to reach 4.768uV steps/resolution).
ADC:
So I have chosen this ADS1219 ADC (24-Bit): ADS1219 ADC Datasheet
I need an I2C-Interface which this IC offers. Also this ADC has an internal 2.048V ref and an internal low drift oscillator.
Instrumentation Amplifier:
To properly amplify the signals that I will feed into the ADC above, I chose this AD8227 instrumentation amplifier: Datasheet of ADS8227 Instrumentation Amplifier
To get a signal from the sensors (initially from 0mV - 20mV) amplified for the reference voltage of 2.048V of the ADC, I would use x 102 Gain (825 Ohm gain resistor) to amplify a 20mV signal up to 2.04V.
I want to connect multiple of these INAmps to the ADC (as the ADC is a 4-channel).
I can now connect the pyranometers to the INA like shown in the picture below (for differential measurement):
The complete circuit would look like this:
Some questions I have:
I'm using a rail-to-rail INA and supplying it with 2.048V only (using an LDO) to amplify the voltage to a proper value so the ADC (with a ref of 2.048V) can map the analog values accordingly. Although this is a rail-to-rail INA, do you think there could be problems missing values because it will not swing completely to rail on the output?
On page 21 of the datasheet of the INA it says a current return path must be provided. On the schematic I marked the return path I created between the sensors and the INA in blue color. Can I do something like this, when considering to measure the sensor signals differentially? If not, how would I provide a return path in this case?
Is it a good idea to work with 2.048V reference voltage in my case or would you go with 5V? As the ADC has an internal 2.048V ref and can be powered with up to 5.5V, it should not be a problem? Do you see any other problems regarding accuracy here?
The chosen ICs seem to be pretty accurate. Do you see huge issues in this case going for like 18-19-Bit accuracy/resolution here, with the provided system/idea?
On page 1 of the ADC datasheet, there is a schematic of the device. One can see that there are internal signal buffers right in front of the ADC, so I guess there should not be a problem going with the INA output directly to the Analog Input of the ADC?
Thank you!
It is a lot of text, but I still hope that some of you experienced people can give me some hints on my questions or if the direction this project is going is completely wrong. I highly appreciate any input from you!
I posted another question regarding amplification using only the ADC IC and its internal PGA here:
Question: How to use the internal PGA of an ADC IC
AI: The output will not swing completely to the rails - with a moderate load it will get to within 0.2 volts of 0 volts and positive supply rail and this sounds to me like a problem because it isn't guaranteed to do any better than this realistically (read page 6 of data sheet "output swing").
There isn't a problem here - your signals are DC coupled and they are referenced to 0 volts.
Is it a good idea to work with 2.048V reference voltage in my case or
would you go with 5V?
Go for a higher supply like 5 volts then at least you can guarantee the output can swing up to 2.048 volts.
The chosen ICs seem to be pretty accurate. Do you see huge issues in
this case going for like 18-19-Bit accuracy/resolution here, with the
provided system/idea?
Yes, I see issues - the input offset voltage is specified in the 100 to 200 uV range and this is an error added to your input signal that you appear to be expecting not to be there.
I don't see a problem here.
If accuracy is a problem I don't see any reason why you shouldn't use a standard rail-to-rail op-amp but one with excellent input offset voltage specifications like the ADA4528. It has an input offset voltage of 2.5 uV and can swing its output to within about 10 mV of positive rail and ground. You'll need two gain setting resistors because it would operate as a conventional non-inverting op-amp amplifier circuit. |
H: Single battery for different voltage components
So I have a 12V DC motor and a 6V stepper motor. Ideally, I would like both of them to draw from the same 12V battery pack (due to build constraints). The DC motor will be fine but the extra 6V running through the stepper could damage it, right?
Is there a way I could have one power supply (12V battery pack) and have both motors drawing from it? So DC motor drawing 12V and stepper motor drawing only 6V. I've read about voltage dividers, but would that mean wasting 6V in the circuit through resistance?
Can I use a buck converter in series between the DC and stepper motors to step down to 6V?
Or can I simply connect them in parallel so that 4 batteries to go the stepper motor then 8 batteries go to the DC motor, then they both connect to ground?
AI: The first is the best one - using a buck converter- which increases overall system power efficiency.
the second option will drain the cells in you pack with different characteristics- which will effect the overall battery pack performance - which is highly not recommended.
4 of your batteries will be overused and 4 of your batteries will be under used |
H: IC Identification - TT617-18-1
Does anyone have any ideas what this chip is please? Tried searching but to no avail.
(I think the chip Identifier is the TT617-18-1).
Thanks in advance.
AI: It is a "Call Progress Tone Generator"
please follow the link
https://people.ece.cornell.edu/land/courses/ece4760/FinalProjects/s2001/pr57/M-991.PDF |
H: Power NMOS used as voltage switch
I've designed a PCB schematic that uses a MOS device to switch one of the higher power rails. I've looked through the datasheets many times, and believe it should work, according to the Current-Voltage characteristics of the device. However, I'm very apprehensive about this, as I am worried that I missed something, and the drain-source voltage drop will degrade the available power on the rail.
I'm hoping someone could double check my design:
Here is the schematic:
The regulators shown here are fixed voltage, Medium current (<5A). The MOSFETS are planned to be IRLB8743PbF NMOS. Here's the datasheet.
According to this graph in the datasheet, I believe this device should operate without a significant voltage drop on the output rail, at a Vg of ~3.3V. It is important that the drop be no more than half a volt to prevent sensor brownouts. Also, I anticipate the rails using very low current (mostly powering microcontrollers).
I would really appreciate any personal experience tips, or any help finding stuff I may have missed in the datasheet.
Thanks in advance!
AI: You have gotten great feedback already. It's obvious that the best approach is to use a PMOS instead of the NMOS. You could use something like this, with a PMOS (and NMOS) to get what you are looking for:
simulate this circuit – Schematic created using CircuitLab
With this you can use a low voltage signal (e.g 3.3V, 5V) to turn on the PMOS. With no signal at the NMOS gate, both the NMOS and the PMOS are OFF. When you apply 3.3V at the gate of the NMOS, that will turn it ON, and will pull the gate of the PMOS down to 0V—this will turn ON the PMOS since now VG-VS (e.g -5V, -12V) is enough.
Just need to pick the two the MOSFETs. For the NMOS, something like the DMN3067LW works great (turns on with low voltage). For the PMOS, you need to pick one with the 5A current capability you want.
One caveat though, if you were to keep the same configuration as you have it, you need to pick a PMOS capable of turning on properly with 3.3V, and 5V. The 8V regulator should work fine because you have the PMOS before the regulator. So you have a high voltage signal (12V), and when you pull the gate down to 0V, VG-VS=-12, enough to turn on most high current transistors (and with lower RDS(on)).
In your 3.3V and 5V regulators, you have the PMOS, after regulation which puts the source either at 3.3V or 5V which means that pulling down the gate will have a VG-VS of either -3.3V or -5V, enough to turn on the PMOS but possibly with higher RDS(on) and at 5A that maybe a drop you would not want (plus heat). For those, I'd place the PMOS/NMOS control circuit before the regulators (just like for the 8V regulator) so that there is enough drive (approx. -12V). You could find transistors that will turn on for 3.3V or 5V but at the expense of higher RDS(on).
Hope this gives you more insight. |
H: what is simultaneous switching output noise?
Can someone explain to me what simultaneous switching output noise is (SSO) is it the same thing as simultaneous switching noise (SSN)? from what I understand when lots of pins are switching together it affects current draw from the source and this is a problem for parallel loads. this also may be crude way of putting it, but are these the Vcc version of ground bounce?
Thanks.
AI: When MORE THAN ONE logic output switches at the same time, the VDD and Ground rails of an MCU are upset.
This also applies to multiple-output logic devices such as buss drivers. |
H: Understanding ladder PLC logic
I'm trying to understand the flow logic of ladder language. Is it sequential like C/C++? "i don't think so"
Particular case: I've been trying to something very simple but cant make it happen as I want!
Figure 1. ladder logic for counting and reset
The timer never resets, I've tried with a lot of moving things around. Also, the coil is not blinking.
By the way (Q0.0 is 4 Hz coil).
I was trying with temp variable 'starting with #', but I read that it gets destroyed (I assume the networks running sequentially and in while(true) manner)?
ct0_reset _______________________________________________
__________________________________
Tag counter > 4 _____________|
_ _ _ _ _ _ _ _ _ _ _
Out 0 __| |_| |_| |_| |_| |_| |_| |_| |_| |_| |_| |_
Tag counter up __|___|___|___|___|___|___|___|___|___|___|____
______________________________________________
cnfg_port _|
Figure 2. Timing diagram variables/coils
comments asked about how the Q0.0 was done. here is the total code this is network 1 in (figure 1), and (figure 2) for blinking showing the total code.
Figure 3. ladder logic for blinking
How i solved this:
after many tries i did this: (although i feel its hard to read the logic here "i understand it" and "it works"). but i dont think its easy to read, so how would you do this?
i also edited blinking logic to the following (i feel this is more right):
AI: If you have a latter version of TIA that allows mixing LAD and SCL you can add a simple code, instead of LAD:
IF Tag_Counter1 >= 5 THEN
tmp_cnt0_reset := TRUE;
cnfg_port := NOT cnfg_port;
ELSE
tmp_cnt0_reset := FALSE;
END_IF;
Regarding your code is totally wrong: Q0.7 is writen multiple times, as well M1.0.
You can also replace the CTU with SCL:
Out0_PosPulse := Out0 AND NOT Out0_pos; // Out0_pos ...static variable
Out0_pos := Out0;
IF tmp_cnt0_reset THEN
Tag_Counter1 := 0;
ELSIF Out0_PosPulse THEN
Tag_Counter1 += 1;
END_IF;
OR you might like the compact version:
Out0_PosPulse := Out0 AND NOT Out0_pos; // Out0_pos ...static variable
Out0_pos := Out0;
IF Out0_PosPulse THEN
Tag_Counter1 += 1;
IF Tag_Counter1 >= 5 THEN
Tag_Counter1 := 0;
cnfg_port := NOT cnfg_port;
END_IF;
END_IF;
But there are still issues, Out0 and cnfg_port are PLC outputs, you shouldn't access them like any variable, you should only write into them. You have o use intermediate variables of type static or temp for that purpose.
You have a lot to learn about PLC programming. |
H: How to automatically assemble lithium polymer battery?
I'm redesigning a product for automatic assembly (DFA / DFX). The current product includes a lithium polymer battery with solderable leads, very similar to this:
Right now the leads are hand-soldered.
How do I make this into something that can be assembled automatically?
I don't think there is any battery like this that can survive reflow, so it has to be added after the main pcb is assembled. There are batteries with JST connectors (adds cost, and still wouldn't be easy to automatically plug in), and batteries with metal foil tabs (no protection circuit - it would have to be on the other pcb; and the tabs probably have to be hand soldered anyway).
I'm sure this is a very common problem in the manufacture of small wearable devices like mp3 players and bluetooth headsets, but how do people usually solve it?
P.S. It turns out everything else in the product can be assembled by (essentially) pick and place: picked up with a vacuum tip and dropped in from above. The parts drop onto alignment pins in the bottom half of the case, then the top half of the case snaps on.. the battery is the one headache.
AI: Ask the battery supplier to provide the battery protection PCB (the PCB at the end of the battery) with contacting areas, and use "pogo pins" (i.e. "spring-loaded contacts") on your PCB. You'll then only need to make sure that the enclosure of your device presses the board onto these pins without exerting pressure on the cell itself (don't mechanically stress Lithium batteries for explosive reasons).
But really, I'd presume that, yes, the cheapest way here is hand-soldering. I don't know how your PCB ends up integrated mechanically in the end product, to be honest, and that's the step where an additional attachment action probably could happen, anyway, so I like the connector idea – especially since batteries with connectors are produced in large quantities, so you could benefit from economics of scale and don't come out that much more expensive. |
H: How do I keep this VCCS stable?
I have designed a circuit to measure impedance of certain tissues. It injects a constant current from two electrodes and measures the voltage at different electrodes. The system mainly consists of a frequency generator, voltage controlled current source, some multiplexers connecting the electrodes and then the measurement circuitry.
When I connect different resistors between OUTA and OUTB I can verify the circuit is behaving as expected. However, when I connected the secondary board with multiplexers, the output suddenly changes to a distorted waveform centered around 5V. The same occurs if I bring my hand close enough to the OUTA terminal. This does not recover until I restart the frequency generator.
AI: There's a pinpoint comment which brings up the key concept: the stability. You have designed a feedback loop which tries to push AC as a constant current source. Capacitive load causes immediately 90 degrees phase lag and that can be enough to make your loop an oscillator when added to already existing phase lag.
Your options:
reduce the gain at high frequencies DIFFICULT! Causes easily more phase lag.
insert a permanent minimum load resistor and take it into the account in the calculations. That resistor hopefully reduces the phase lag caused by the capacitance.
redesign the regulation to work in DC mode ie. the current is forced to have amplitude which causes the right peak voltage to the current sense resistor
ADD: The case is extremely complex. Capacitive load affects more at the lowermost end of your current sense resistor, but if we assume the differential amplifier function still be valid at high frequencies, the current sense voltage is actually RC high pass filtered. There's a slight possiblity to compensate the case by adding a small capacitor in parallel with the current sense resistor. No quarantee, but easy to test.
More complexity is caused by the fact that wirings have also inductance. There can be parasitic resonance circuits at megaherzes or tens of megaherzes. DC mode regulation would solve HF phase problems. |
H: Ltspice ac anlysis with dc offset
I am doing ac analysis of a simple passive RC low pass filter. The input is a 2 V peak to peak sinusoidal signal. R=1.2k and c=12nF. I want to add a 2.5 V dc offset to the input signal. When I change the dc offset value of the signal generator I always see the same magnitude value. With a 2.5 V offset it must be 13.1 dB at low frequencies but I always see 6 dB. So, how can I see the effect of the offset voltage?
AI: AC analysis uses a linearized representation of the circuit around the bias point. DC analysis calculates the bias point.
Given that R and C are linear components, the element equations are the same. However, your offset is simply a bias point, which will be removed from the circuit during linearization, thus having no influence in an AC analysis. |
H: Conductivity formula for Metals
Is there any expression for the direct change in conductivity of metals with temperature ?
I am aware of change in resistivity with temperature.
AI: Since
$$\sigma = \frac1\rho $$
and, for a good approximation on typical temperatures where the metal is solid and and warmer than 180 K (==-90 °C):
$$\rho = \rho_0 + \alpha T$$
(with $\alpha$ being the temperature coefficient) it follows that
$$\sigma = \frac1{\rho_0 + \alpha T} $$ |
H: Help identify the discrete component in this battery pack
I am rebuilding a NiMH battery pack for a RF remote control device which can be recharged by plugging the device in.
The battery pack is connected with three wires:
black to the negative
red to the positive
BLUE through a mysterious discrete component to the negative
The component in line with the blue wire looks a bit like a glass-body diode but my multimeter cannot measure a diode voltage. (It measures infinity, not zero, if this helps.)
However, I can measure about 11.5 kOhm across the component in both directions.
Can anybody enlighten me, what this component is likely to be? I would like to verify that the component is still functional because it is unclear whether or not the device still charges properly.
AI: Its a thermistor like this (photo from internet, not spam related):
This kind of resistor depends on the temperature of both batteries.
EDIT: In this photo, is a Negative Temperature Coefficient. There are also Positive Temperature Coefficient resistors. The main difference between them is how the resistor decreases or increases when the temperature grows, respectively.
That's a cheap component and a cheaper way to determine when both batteries are charged up.
When a NiMH is almost charged, its temperature starts to grow. If you measure the temperature based on a voltage divider into a DAC, you can measure the temperature and, therefore, activate the charge or deactivate it.
EDIT: The thermistor must be very close to the batteries to read a correct value, so that's why this discrete component is located close to the batteries.
The temperature based graph to read when stop charging the battery is like this:
Related: Voltage input for charging NiMH Batteries
If you can measure the temperature, you can check when it's charged up. |
H: Problem with interrupt-based timer on STM32F411
I was following this online tutorial and cannot get the interrupt-based timer to work. I know the timer is running because if I uncomment the five lines in the main loop, I get a nice square wave output. But with those lines commented out, the timer interrupt is not firing and I cannot figure out why. Can someone suggest what I'm doing wrong?
#include "stm32f4xx_hal.h"
void SysTick_Handler(void)
{
HAL_IncTick();
HAL_SYSTICK_IRQHandler();
}
static TIM_HandleTypeDef s_TimerInstance = { .Instance = TIM2 };
void InitializeTimer()
{
__TIM2_CLK_ENABLE();
s_TimerInstance.Init.Prescaler = 160;
s_TimerInstance.Init.CounterMode = TIM_COUNTERMODE_UP;
s_TimerInstance.Init.Period = 100;
s_TimerInstance.Init.ClockDivision = TIM_CLOCKDIVISION_DIV1;
s_TimerInstance.Init.RepetitionCounter = 0;
HAL_TIM_Base_Init(&s_TimerInstance);
HAL_TIM_Base_Start(&s_TimerInstance);
}
void InitializeLED()
{
__GPIOA_CLK_ENABLE();
GPIO_InitTypeDef GPIO_InitStructure;
GPIO_InitStructure.Pin = LD2_Pin;
GPIO_InitStructure.Mode = GPIO_MODE_OUTPUT_PP;
GPIO_InitStructure.Speed = GPIO_SPEED_HIGH;
GPIO_InitStructure.Pull = GPIO_NOPULL;
HAL_GPIO_Init(LD2_GPIO_Port, &GPIO_InitStructure);
}
void TIM2_IRQHandler()
{
HAL_TIM_IRQHandler(&s_TimerInstance);
}
void HAL_TIM_PeriodElapsedCallback(TIM_HandleTypeDef *htim)
{
HAL_GPIO_TogglePin(LD2_GPIO_Port, LD2_Pin);
}
int main(void)
{
HAL_Init();
InitializeLED();
InitializeTimer();
HAL_NVIC_SetPriority(TIM2_IRQn, 0, 0);
HAL_NVIC_EnableIRQ(TIM2_IRQn);
for (;;)
{
// int timerValue = __HAL_TIM_GET_COUNTER(&s_TimerInstance);
// if (timerValue == 50)
// HAL_GPIO_WritePin(LD2_GPIO_Port, LD2_Pin, GPIO_PIN_SET);
// if (timerValue == 100)
// HAL_GPIO_WritePin(LD2_GPIO_Port, LD2_Pin, GPIO_PIN_RESET);
}
}
AI: Looks like the UIE bit in TIM2->DIER is not enabled. There must be some HAL function to do it, I don't know. I would not recommend using HAL for timing stuff anyway, there are issues with its accuracy.
Starting a timer is just a few lines of code.
void InitializeTimer(void) {
__TIM2_CLK_ENABLE();
TIM2->PSC = 160; /* Set prescaler value */
TIM2->EGR = TIM_EGR_UG; /* Update prescaler */
TIM2->ARR = 100; /* Set timer period */
TIM2->DIER = TIM_DIER_UIE; /* Enable interrupt generation on timer overflow */
TIM2->CR1 = TIM_CR1_CEN; /* Start counting */
}
void TIM2_IRQHandler()
{
/* Clear interrupt flag first. It cannot be the last instruction in an interrupt handler. */
TIM2->SR = 0;
LD2_GPIO_Port->ODR ^= LD2_Pin;
} |
H: Does anyone know how the low tone control of the Boss Metalzone works?
https://electricdruid.net/boss-mt-2-metal-zone-pedal-analysis/
https://sound-au.com/articles/gyrator-filters.htm#s9
I have been using these websites to try and work out how the eq works, but I've had a bit of a problem when it comes to the gyrator used for the low tone control. I can't quite seem to work out the relationship between the frequency response and the components.
AI: According to the Wikipedia article, the simulated inductance has a value of
$$L = R_{012} R_{013} C_{009} = 10.34\text{ H}$$
This inductance is in series with a resistance equal to R012, giving an L/R time constant of 4.7 ms, which corresponds to a frequency of 33.86 Hz. This is the lower frequency limit for the bass control's effect.
Similarly, the series combination of C008 and R012 has a time constant that corresponds to a frequency of 328.8 Hz, and this defines the upper limit of the bass control's effect, and corresponds directly to the inflection point shown on your graph.
Note also that the series combination of C008, R012 and the simulated inductor is also in series with a minimum source resistance of 22 kΩ (either R014 or R015), which insures that the circuit is overdamped (no resonance). If this source impedance were reduced, you would start to see a resonant peak at around 105 Hz. |
H: Differential transimpedance amplifier
I have found an opamp schematic that I don't understand.
As far as I know a transimpedance amplifier "converts" my input current into an output voltage.
But in this case the positive and the negative inputs are driven by an input current.
Does that mean that my current difference will be amplified?
I tried to simulate this schematic in LTspice, but unfortunately I cannot explain this schematic to myself.
It would be very helpful if someone can name this schematic.
AI: A simple trans-impedance amplifier is this:
Here is the link to it.
This is a simple one, there can be different variations. You are correct when you say that an output voltage will be produced from the current source input you give.
But obviously no current goes into the Op-Amp terminals. |
H: Creating DNF with variable in output column
I am trying to model an Arbiter in VHDL. For this I've created a Moore automaton and am currently mapping the output.
The output should be when mapping from the state:
z1 z0 | y
0 0 | 0000
0 1 | inB
1 0 | inA
1 1 | inA
So basically depending on the state z either inB, inA or 0000 should be passed to the output y.
How can I model this in a DNF? I would usually process by finding the y rows with a 1 in it, but here I don't know how to proceed.
AI: Something like
y <= inA WHEN z1 = '1' ELSE inB WHEN z0 = '1' ELSE "0000"
Or did I missunderstand your question? |
H: Does anyone know how the Boss Metalzone sweepable mids affect the bandpass?
https://electricdruid.net/boss-mt-2-metal-zone-pedal-analysis/
https://sound-au.com/project150.htm
Like in the original question - I'm struggling to understand the relationship between the op-amp on the right to the op-amp on the left. As far as I know, a Wein Bridge network is being used here but otherwise, I'm stumped.
AI: The opamp on the right is just a voltage follower, and provides a low-impedance source to drive the bandpass filter (actually only half of a Wien bridge).
The opamp on the left is a simple inverting amplifier. VR02a picks off some amount of input signal (either in-phase or inverted) to feed to the filter. Since the source impedance at the wiper varies with position, the other opamp is used as a buffer so that this does not affect the bandpass filter response.
The output of the bandpass filter is then summed with the original signal to produce the midrange boost or cut. |
H: TP4056 to charge xS1P li-po li-ion
Very related to this question.
While it's already clear that several TP4056 can't be used to charge and discharge several cells, it's not clear to me if it's possible to only charge a battery pack with several cells in series.
The input and output GND pads are connected, but what about the battery terminals? From the TP chip datasheet I fear that B- pad is also connected to GND, and therefore i can't charge a battery pack from it's balance connector, but i'm not that good in diagram reading and understanding and i hope to be proven wrong (maybe thanks to the additional protection chip, dw01 if i'm not wrong).
Thanks for your answer.
AI: it's not clear to me if it's possible to only charge a battery pack with several cells in series.
It is clearly answered in the datasheet:
DESCRIPTION
The TP4056 is a complete constant-current/constant-voltage linear charger for single cell lithium-ion batteries.
At the end of page 3 there's a schematic which will be very similar to what is on most TP4056 modules:
Note how the - terminal of the battery is also grounded. On the module you can also easily see (and measure) that IN- / GND and B- are shorted. So indeed, the TP4056 is not suited for balanced charging of series connected cells.
If you need to charge series connected cells then you need a different IC. |
H: How make a dual +-12V supply from a 24V SMPS
I'm trying to power a home-made load cell transmitter using a 24V single SMPS. I need to make +12, 0 and -12 Volts that are capable of 50mA. I wish to power multiple channels of opamps and bridges.
I don't have much budget and availability of components in India.
I have an idea to use 1 LM7812 an 1 LM7912(negative) linear voltage regulators and a voltage divider setup to do this as per the circuit below.
simulate this circuit – Schematic created using CircuitLab
Would this work? I've modified it from the suggestions and articles elsewhere.
Somebody suggested me one other circuit but I am concerned about the current capabilities of the opamp.
simulate this circuit
Would this work? If yes, please suggest suitable op-amp.
Are there any other techniques that would do the job economically?
AI: You first idea will not work at all.
Your second idea will work, but many OP-Amps aren't going to deliver more than a few mA on their output, which limits the current your circuit may draw from the virtual ground. There are Power-OP-Amps available which may deliver up to a few ampere, but if you cannot get your hands on one, you can use a PNP/NPN transistor pair to increase the output current:
simulate this circuit – Schematic created using CircuitLab
The OP-Amp will take care of stabilizing the output so it matches the voltage set by the input voltage divider. Take care of capacitive loads, as Spehro noted in his answer, though. |
H: Is it better to control an amplitude of audio signal or gain?
I'm creating simple headphones amplifier with an opamp (basic non-inverting config with RC high-pass on NI) and I wonder, what's better - to control an amplitude of input signal, or a gain of an opamp?
What are benefits from gain control over signal control or vice-versa?
Best regards!
AI: Generally, you want to have the most gain at the beginning of your signal chain. The reason for this is that the noise added by the gain stages gets passed on and amplified by the next stage.
The downside to having too much amplitude before going into your opamp, is you may more easily saturate the inputs to the amplifier and cause clipping at a lower overall sound level.
Typically, however, you would design your circuit such that you control the amplitude before going into your opamp, so that you can make maximum use out of your total headroom. |
H: UART - Writing and reading AT commands
I am working with a B-L072Z-LRWAN1 board and a 8pya00-simcom-evb v1.02 board with SIM7000E. This SIM module operates through AT commands and I need to send these commands through an UART connected between mentioned boards.
I have already set up the UART in the B-L072Z-LRWAN1 and I am able to transmit bytes of information, but I am not sure how to send AT commands.
Which is the format I have to use to send the commands? Do I use char variables? Do I use uint8_t variables and add '0' at the end to make them chars? Do I need to add \n\r at the end of the command?
I am a little bit confused so I need your help.
Thank you!
AI: char and uint8_t should be equivalent on ARM, in other words, char should be unsigned, but there are toolchains out there with char by default signed. It does not matter in this case anyway, because AT commands would only use characters in the [0-127] range.
Adding a \r at the end of the command is mandatory, \n and \x00 would probably be ignored, but better not send them. Responses are terminated with \r\n (not \n\r) |
H: Can a magnet cause a microwave oven to unsafely radiate energy?
I've already successfully warped the colors on a large old CRT Television by setting unshielded speakers on top of it. If I understand correctly, this was due to the deviation of the electro-magnetic (EM) radiation beam from its intended target, as influenced by the speakers' magnets.
If I further understand correctly, the magnetron of a microwave to produces a beam of EM energy which is directed at a reflector or scatterer, and then into the cooking cavity which attenuates the energy through notch-filtering and corner-wave traps. (I'm probably using slightly incorrect terms, feel free to correct my terminology.)
Is the magnetron contained within the cavity/[farraday ]cage, and if not, can the beam of EM be deflected such that it could escape or deviate enough to emit radiation into the surrounding space? (If the magnetron should be situated within the containment structure, are there regulations stipulating this?)
This situation is more practical than hypothetical: We have a neat kitchen gadget with a strong (rare-earth, I think) magnet on its base, designed to hold itself to an oven door with substantial certainty. Attaching it to the oven isn't an option, so it's currently on the over-range microwave oven. In a moment of analysis, I realized this may be a poor choice...
If my consideration of risk due to magnetic deviation is unfounded, is there another concern, short of magnetically affecting the safety switches (which shouldn't be possible without intentional compromising) to having a stronger, rare-earth magnet attached to a microwave?
AI: If I understand correctly, this was due to the deviation of the electro-magnetic (EM) radiation beam from its intended target, as influenced by the speakers' magnets.
No. There's no "EM radiation beam" here. There's an electron beam. "EM radiation" is radio waves, light, x-ray, in short: all the electromagnetic waves.
Your CRT works by accelerating electrons, which very much act like particles and not like waves at the scales and energies involved.
So, what happens is that these charged particles are subject to Lorentz force. You should probably read the wikipedia article on that, if you want to understand how the deflection coil in a tube TV works (or why your speaker magnets have that effect).
If I further understand correctly, the magnetron of a microwave to produces a beam of EM energy
Nah, not really. The magnetron is a cavity in which a wave is kept resonant and amplified through a strong magnetic field. Not a beam, at all!
which is directed at a reflector or scatterer,
Since this is a wave, and doesn't behave like a beam at the sizes of microwave ovens: Be a bit careful with your wording; this is all wave phenomena, so while you can reflect the e-Field of the wave, this is not a "specular" reflection like a laser pointer on a mirror.
and then into the cooking cavity which attenuates the energy through notch-filtering and corner-wave traps.
No, the energy can't be attenuated. There's conservation of energy. There's just constructive and destructive interference, leading to higher field strengths in some places than others.
The energy in the interchanging electric and magnetic field is then absorbed by molecules of the food you're intending to heat and converted to exactly that: heat.
Is the magnetron contained within the cavity/[farraday ]cage, and if not, can the beam
not a beam!
of EM be deflected such that it could escape or deviate enough to emit radiation into the surrounding space?
sure. We'd call that opening then an antenna.
(If the magnetron should be situated within the containment structure, are there regulations stipulating this?)
No matter where the magnetron is located, the amount of energy that a device can emit is draconically regulated by national and international bodies; in the US, that'd be the FCC; in Germany, the Bundesnetzagentur, in UK the OFCOM, and so on.
Generally, microwave ovens aren't allowed to emit any significant amount of EM radiation, especially not an amount able to hurt bystanders.
This situation is more practical than hypothetical: We have a neat kitchen gadget with a strong (rare-earth, I think) magnet on its base, designed to hold itself to an oven door with substantial certainty. Attaching it to the oven isn't an option, so it's currently on the over-range microwave oven. In a moment of analysis, I realized this may be a poor choice...
A static magnetic field should have no adverse effect. You can verify that by adding a static magnetic field component \$\mathbf B\$ to Maxwell's equations and realize that for the static component, \$\frac{\mathrm d}{\mathrm dt}\equiv0\implies \mathrm{curl} (\mathbf E) = - \frac{\partial}{\partial t} \mathbf B = 0\$, which simply means that the E-field isn't bent by the static magnetic field, and hence, the wave can't take a different direction. |
H: Driving LED with 555 timer and BC639 transistor
I am trying to produce an LED dimmer circuit using a 555 timer circuit and a BC639 NPN transistor. I know there are better transistors that would suit the task better, but the BC639 is what I have to work with.
I've got the 555 timer configured so that it is outputting a ~12V 1.4kHz signal.
I am using a BC639 NPN transistor to drive a white LED, with a maximum forward current of 50mA. I have put a 330Ω resistor in series with the LED so that the current is 26.36mA.
I'm trying to figure out how to make it so that the LED turns all the way off when my 555 timer is outputting the lowest average DC voltage.
I have been told that this is down to the value of R3. I have calculated a value for R3 of 51.4kΩ:
IC=26.36mA
hFE=120
IB=IC/hFE
IB=0.00022A
R=(V-VBE)/IB
R=(12-0.7)/0.00022
R=51.4kΩ
This resistor value does not allow the LED to be turned off fully. Through trial and error I have found that the closest E12 resistor value that does allow the LED to turn off fully is is a 680kΩ resistor.
AI: The value of the base resistor has nothing to do with the range of the PWM.
R3 should be more like 10K to fully saturate the transistor (forced beta in the range of 20).
When the output is high (LED 'on'), the capacitor has to charge through R2, which means that the "on" time is limited by the time constant of R2 and C1.
You can try "flipping" the output (if you have an ancient design bipolar 555 you may need a pullup to turn the transistor off fully), so use a PNP transistor and the same value of base resistor. Then reverse the CCW and CW sides of your pot.
That will mean your LED output won't go as close to 100%, but that will have no visual effect-- the slight amount 'on' is much more noticeable than a slight reduction in full brightness.
The output will still not be completely off, but it will be greatly reduced in brightness. A small capacitor, at a guess around 1nF between base and emitter of the transistor can eliminate that short turn on, at the expense of the transistor running a bit warmer. |
H: Speaker crossover wiring
Came across some old Arcam speakers (cut open) and using the parts to build a bluetooth speaker box. Only problem is I can't figure out the wiring on the crossover/filter. The original speakers are pretty old so Google didn't help much.
Image of component side
Have a 2-channel audio amp board and trying to split the signal to one tweeter and one mid on each side.This is what I'm trying to achieve:
So question is: does anybody know what the cables are, and how can I cut/solder to achieve my goal?
AI: I'd say that connections 1 and 5 ("I") are inputs. There's a notation on the board that the input is "bi-wire" (presumably to support a bi-amp configuration).
All of the connections labeled "E" are common connections.
The connection labeled "T" goes to the tweeter. This half of the board is configured as an LC high-pass filter — capacitor in series, inductor in parallel with the load.
The connection labeled "B" goes to the bass driver. This half of the board is configured as an LC low-pass filter — inductor in series, capacitor(s) in parallel with the load.
So,
1 and 3 are the input from the amplifier for the tweeter
2 and 4 go to the tweeter itself.
5 and 6 are the input for the woofer (bass)
7 and 8 go to the woofer itself.
If you are single-amping this, then you can connect 1 and 5 together, as well as 3 and 6. |
H: What is a "reverse type" tsop?
I was looking through a datasheet, when I noticed this sentence:
TSOP package is suitable for cards, and reverse type TSOP is also
provided.
What is a "reverse type" TSOP? I am familiar with TSOP packages.
AI: The difference is shown on page 3 of the datasheet you linked to. It is not a difference in package, but a difference in pin arrangement. Compare the pins between the HM628128ALT and the HM628128ALR -- the pin functions are reflected. Page 2 indicates that the ...ALT version is "normal type" and the ...ALR version is "reverse type". |
H: L293D Enable pins can reduce chip heat?
I use a L293D to drive a 7.4V (280mA/phase) stepper motor.
I rotate the small motor 1 rev per DAY.
The L293D is hot all the time even though the motor turns only once per day.
I saw there are “Enable” pins, and I keep them high all the time.
If I set the Enable pins LOW most of the day, will this reduce the heat I feel on the chip?
AI: The enable pin can disable the output driver.
Turning off the output drivers will reduce the current consumption of the unit.
However, it will also remove all torque from the motor and put in free-running as described in Table 3 of TI's datasheet. |
H: Reversing P-channel mosfet for polarity protection component selection
Situation: I'm trying to implement a reverse polarity circuit using a p-channel mosfet and a voltage divider using a resistor + schottsky diode. More information here if you need a refresher.
The key point here is that the MOSFET is reversed. The current flows from Drain to Source, first through the internal diode but after turning on through the normal MOSFET path.
I've been reading various datasheets of various MOSFET's and they all seem to indicate that the maximum Vdrain-source voltage is negative. If I took this at face value it means that a MOSFET can't conduct in both directions, which I know to be false. Example.
Can I ignore this negative number and assume it's a maximum rating both ways or is there more at play here?
Edit: the actual circuit taken from Hackaday
AI: FETs ideally don't care about the direction of the current. You simply have to invert your conventions and switch Source and Drain in your equations. Everything works exactly the same way.
The only reason that power FETs do care is because there is a Source-Body connection put in place during device fabrication. This creates a parasitic diode that conducts when the voltage is reversed. However, if your drain-source voltage is below the diode voltage, the device will still behave as a FET.
If you have access to the fourth, body, terminal. You can bias it externally so that the device always operates as a FET. |
H: Blown oddball fuse
I have a device that has a blown Daito BD20 DC400V 2A fuse. Datasheet here: https://www.furutaka-netsel.co.jp/pdf/daito/bd.pdf . These are rather odd looking, solder-in fuses and they are apparently a bit hard to find. The fuse also has an interesting internal construction; the top just snaps on so I originally thought it was a fuse holder and popped the top off. The fuse element itself appears to be contained inside of a rubbery silicone compound, and the cap contains some padding that surrounds the silicone. Is there anything special about this particular type of fuse that I should look for in a replacement?
AI: As with many things, the answer is: Depends.
It's impossible to know what the designer's requirements are when they specced the part, so without asking them, we can make some educated guesses.
Within a certain range of products, the key specs that will matter will be current rating, and "slow blow" or "fast blow". These may or may not be written in the datasheets, but they will be shown as a time/current graph.
https://www.digikey.ca/product-detail/en/littelfuse-inc/80712000440/F4219-ND/3130994
This seems to meet most of your key specs. |
H: How do you measure quiescent current?
How do you measure quiescent current?
(For instance, a Battery connected device)
AI: You measure it like you would active current use. Nothing changes. Quiescent current is the amount of current a circuit, that is doing nothing, takes. You would connect it to a ammeter just the same. For very low quiescent currents, you may need a more precise or quality meter. Some devices or circuits have IQs in the micro and nano amp ranges, which a cheaper meter would not be good enough to register properly. |
H: Step down open collector circuit
I'm trying to use the LPD3806-400 incremental encoder with an Arduino DUE.
I have the problem that the datasheet says 5 V to 24 V and also the output is open collector. Therefore, I have to pull up the output.
I guess this can be done with the standard pull-up resistor method.
However, my Arduino DUE just receives up to 3.3 V, so how should I fix it if I can't add the voltage divider anymore because of the connection to ground conflicting with my pull-up resistor?
As you can see in the attached image, the fact that the 5v of pull up now are grounded too, makes the voltage drop dramatically.
The transistor collector is my encoder output, and the voltmeter is the Arduino input
AI: I took a look at the Japanese datasheet and could see enough to make sense of it. (My Japanese is limited to asking a few dumb questions like, "What is this called?")
The datasheet specifies the device as operating on anything from \$5\:\text{V}\$ to \$24\:\text{V}\$. This means that the red wire requires a voltage between those two. If you look over the datasheet, you can see where that is specified for that wire, too.
This, however, doesn't mean that the open collector outputs also must be run from the same voltage. While it is true that the sample schematic they provide suggests the idea, you actually do NOT usually have to comply with that idea for open-collector outputs. In fact, that's one of the advantages of having them be open-collector. You can use different rail voltages for the pull-up!!
The open-collector outputs are made either active or inactive, using the internal circuitry that is powered entirely by the \$5\:\text{V}\$ to \$24\:\text{V}\$ supply rail you provide on the red/black wire pair. So you do need to supply a voltage rail in that range or else the circuitry inside the unit can't actually work properly. However, so far as the green and white wires are concerned, that internal circuitry simply enables or disables the open-collector outputs. When active (or closed), these outputs on either the green or white wire are "shorted" to the black wire. When inactive (or open), these outputs are "floating" and aren't directly connected to anything. (This allows you to add a pull-up resistor to any reasonable voltage [which you can fairly take to mean anything \$\le 24\:\text{V}\$] which will take over when the green or white wire is inactive.)
The only limitation for the rail voltage used by the pull-up resistor is when the pull-up rail voltage exceeds the maximum \$V_\text{CE}\$ of their transistors used to create the open-collector output lead. The datasheet doesn't specify a limiting value here, but most small-signal BJTs can handle \$24\:\text{V}\$ (\$30\:\text{V}\$ to \$60\:\text{V}\$ isn't at all uncommon.) And since they specify a schematic showing a direct pull up to the red wire voltage, I'd have to assume the open-collector outputs can readily handle up to at least \$24\:\text{V}\$. And that is far more than you need. So there should be no worry about using \$3.3\:\text{V}\$ for your pull-ups.
In short, you need just two voltage rails: \$5\:\text{V}\$ (for the red wire, referenced to ground with the black wire) and \$3.3\:\text{V}\$ (for your Arduino as well as the resistor pull-up rail used with the green and white wires.)
Ignoring speed of rotation issues, the only possible remaining question is whether or not the active output voltage of the green and white wires (A/B open-collector outputs, when actively "shorting" their wires to ground) provide a low enough voltage output that will meet with your Arduino input level requirements. I would tend to guess, otherwise ignorant due to the datasheet limitations, that this output voltage will readily be \$\le 300\:\text{mV}\$ if you use reasonably valued pull-up resistors. This is well within the usual guidelines for digital inputs powered off of \$3.3\:\text{V}\$ rails. So I don't expect a problem here, either. (But you should verify this, once you select your pull-up values.) |
H: How do I create a working bandpass filter in LTSpice?
I am attempting to create a bandpass filter in LTSpice. I am fairly new to the program, which is probably contributing to my difficulties.
The filter I am creating needs to have a center frequency of 1200 Hz, with the cutoff frequencies at 1080 Hz and 1320 Hz. While the initial circuit design didn't seem so difficult, I cannot seem to get any working simulations. Here is my current schematic:
Note I initially started with a single stage that also produced similar results to this one. The second stage was added to narrow the bandwidth.
Here is the frequency curve this circuit gives me:
I honestly don't know what to make of this curve... It sure doesn't look like what a bandpass filter is supposed to be. I have triple checked all of my values and the general layout for the circuit, and I am at a loss for what is wrong. I have also a wide variety of other RC values and different OP amps with no success.
Just for testing, I also went ahead and built an active low pass filter. This one is also giving me a very strange frequency curve. Here is the schematic:
and here is the frequency curve:
Any help with this would be greatly appreciated. I am currently at a loss for other things to try. Thanks!
AI: There is no path from your power supply to ground. The opamps are not biased properly.
The only reference on the power supply is through the opamps and, given that these are basically single nodes and no DC current can circulate between them due to the AC coupling, the bias voltage will stabilize to whatever value makes their output current zero. Probably saturation to a rail.
You should always run a DC analysis and check your bias points before you attempt an AC one. |
H: Is there a relay that disconnects BOTH positive and ground wires?
My latest time-lapse camera rig, was failing to run due to the constant ground connection on the camera power input.
This is an off-grid, solar-powered rig, combining a Raspberry Pi and an Arduino, to share tasks using a minimum of energy.
As a novice in electronics I felt I was making progress, having used GPIO controlled relays to break the positive wire connection on several of my rigs.
However, in this rig, I use two different methods to take the pictures:
1/ An in-camera script (C.H.D.K. Canon Hackers Development Kit) which takes a picture automatically, on power up of the camera by the Arduino,
2/ A PTP connection to the camera's USB port when the Raspberry Pi is active and supplying the power to the camera as well.
Due to the ground wire of the Raspberry Pi remaining connected to the Camera power input, I found that my on-board script failed, due to it falsely detecting a USB connection. On disconnection of the ground wire, the script works as intended.
I imagined this need for 'double disconnection' ie. of both the positive AND negative wires, would be a common issue in electronics and that there would be a relay component ready made for the job but Google doesn't turn one up.
I suspect I simply just don't know the right name for the task, among the many versions of relays out there.
Please advise, or, if there's another approach in circumstances like mine, I'd love to hear it. it seems wasteful to be using a two channel relay to operate a single power switch. ie. one channel for the positive wire and one for the ground.
I hope I explained myself well enough. If not, I apologize and will correct as best I can
Thanks in advance
AI: Hi you really need a DPST ( double pole single throw) relay.
but DPDT (double pole double throw) will be mostly available with the hobby stores.
You can use any of the above
Pls follow the link
https://www.evelta.com/relays/power-relay/hongfa-hf115f-024-2zs4 |
H: Simple way to avoid current draw on resistor divider sense?
This is a simple, but complicated question.
I measure voltage on several lines, more than 3'000. The lines can be up to 100V.
I only need to measure at certain time, and would like to avoid current draw from the sense lines when it's not being measured.
I was thinking to use a mosfet and detach the whole VSS from the bottom side of the resistor divider, but the AN are connected to sampling logic, which has internal protection diodes to 3V3, so it will still draw current.
Given that I have a lot of those lines and the application is cost sensitive, I would like to avoid to have a P Mosfet on each line with the resistors and zener that would be needed.
Is there some trick possible in here ? Maybe some component made for it I don't know about ?
Update:
The answer of Edgar Brown is valid, but is a bit complicated to drive the gate, given the voltage of the mos will vary.
Also I do not have space on the PCB to add mos between the two resistor of the divider (or would need to do quite a lot of re-routing).
I have another possible solution:
Q53 would decouple the sense divider from the ground and the other Q would decouple from the sensing IC.
If the gate of Q54 is driven as a open drain, Q53 would be driven first, bringing Q54 to a few V to VSS and then Q54 can be driven.
Only one resistor from all the Q54 gates to the Q53 Vd sould be enough?
Update 2
I think I found a better solution, adding a bias voltage before the VSense. Simply biased at the same voltage of the off state.
AI: Just use NFETs as high-side switches under the 1M resistors. You might need an extra voltage to drive their gate, or to reduce your measurement range, but once you bias them in their triode region (which the 1M resistors make quite easy) their impedance will be negligible.
Just make sure to use a FET with low leakage across the reverse diode (or better yet, with a separate body connection) and to drive their Vgs to zero or negative when off, to stay out of the subthreshold region.
This is the basic idea. Assuming that each NFET is activated independently from each other.
simulate this circuit – Schematic created using CircuitLab
PFETs could also be made to work. But their biasing is not as clean as (1) you would have to use the sources themselves to provide some of your gate biases, (2) you would need level shifters to reach 100V, and (3) you increase the leakage paths due to the level shifters. And any gate drive whatsoever will put you in subthreshold and that means leakage. |
H: Can Flash controllers make a card readonly?
It is said in this answer here that
Some of them have a controller chip that will permanently lock them to
read only if they detect a write error, as a preservation measure.
Is it possible for a flash controller to permanently lock a SD card into readonly such that it cannot even be formatted?
AI: Yes! For SD Cards writing 1 to CSD Register's Bit 12 and 13 can enable write protection, permanent or temporary.
Regarding the Q&A you have referenced over here. I am not sure or I can confirm but while working with SD Card's I read on some material that when SD card starts to detect that it's life cycle is getting over,It puts itself into Read Only mode so you dont put up your valuables over there and can read all the stuff that you have stored on it.
I have just READ, I haven't confirmed it because I don't remember the source neither have I tested it practically. So take it with a pinch of salt. |
H: Output voltage of boost converter in shutdown mode (no switching)
Is it common that a boost DCDC converter has an output voltage which is the same as the input voltage, when the switching does not occur?
I'm reading the TPS55340EVM-017 datasheet. In the caption for Figure 14, it is described that the EN voltage goes low and the output voltage ramps from 24 V to Vin. I'm curious why the output voltage does not decrease to zero.
Also when the switching does not happen, isn't there limitation in the output current?
AI: Because the diode (D in the schematic above) in the boost converter allows current to flow from input to output. When you drop EN, the device stops switching and thus stops boosting but the voltage will not drop any lower than Vin (save for a diode Vf) because of said path is always available. |
H: A question about reverse polarity protection diode
Is the function of 1N4001 before the 5V regulator reverse polarity protection?
Is this a fine method? And I sometimes see 1N4148.
Can any diode(besides zener) be used for such function?
AI: Yes, this diode is probably for reverse polarity protection. It does its job but has the disadvantage that about 0.6 V is dropped across it. It needs to be rated for the current, both continuous and surge, as well as the reverse voltage it might encounter. 1N4148 is fine for lower current applications. A shunt diode with polyswitch/fuse is also an option. More elaborate schemes with MOSFETs are possible. |
H: Convert bit to bin Xilinx file
How can I convert a .bit (output from ISE Project Navigator) into a .bin file?
For what I understand, .bit files are similar to .bin but with a header that specifies things like the board and stuff.
AI: Better than convert, it is possible to directly generate the bin file with ISE Project Navigator.
Before double clicking on "Generate Programmng File", first do secondary button on it and go to: "Process Properties..."
In there, under "General Options" tab check the -g Binary option.
I know there are other ways for converting the .bit file to .bin files directly but I don't know how. |
H: Do poles enhance frequency or cause it to dip
I'm a bit confused about the idea of poles. In some of my courses such as digital signal processing, we are taught that in filter design, you place a pole at a frequency you want to enhance and a zero where you want to cause a dip.
In other courses in which we deal with Bode plots, I notice that a pole causes a -20dB/decade slope decrease?
Why is there an incosistency?
AI: Here's a bit of a crash course in poles: -
The upper pictures show a bode plot with frequency peaking in various amounts. The lower left picture attempts to show a 3D view of bode plot and poles and the lower right picture is the conventional pole zero diagram.
In other courses in which we deal with Bode plots, I notice that a
pole causes a -20dB/decade slope decrease? Why is there an incosistency?
There is no inconsistency; a single pole causes a peak in the spectrum and, due to the mathematical relationship (1/distance) between the pole centre and more distant positions on the jw axis there is a slope that results.
For a 2nd order low pass filter (example) there are two poles and the reciprocal-of-distance relationship leads directly to the ability calculate any point on the jw axis: -
Hope this helps. Related question and another related question and yet another related question. And another. |
H: Using a usb-c laptop charger to charge a phone
I don't really know anything about electronics and I'm not sure if it's the right place to ask, but still.
Is it OK to charge my phone with my laptop's charger?
Here's info on the laptop charger:
And here's info on the phone charger:
AI: Generally speaking, the fact that both devices do have the Type-C connector doesn't mean that both devices will be interchangeable to full functional capacity. The Type-C specifies connector and its use for USB connectivity, with basic provisions do discriminate few power supply options at 5-V level. The connector can use any other means to communicate with chargers on the top (or instead) of basic functionality.
To begin, your laptop charger (Xiaomi?) supports some sort of "fast charging" capability and lists several "power profiles". However, it is not clear whether it does it over standard Power Delivery Specifications, or via Qualcomm QuickCharge proprietary specifications. Still looks more like a PD.
Your Sony charger seems to support either basic Type-C 5V1.5A profile, or it might use Sony proprietary signature over USB 2.0 lines. Again it is not clear.
Bottom line is that it might charge your Sony with full blast (5V1.5A), or the phone might resort to basic 5V 900 mA (or even 500 mA) charging if it uses Sony proprietary charger signature.
In both cases you can safely plug it in and see if the phone gets charged fast enough. |
H: How to 'fix' this in an enclosure without glueing?
I have an enclosure, and want to fix/put a STM32 inside, a relay and probably later a small circuit board.
However, the STM (STM32F103C8T6) and the relay do not have holes to fix them... and I do not want to glue them because maybe later I want to change/add components.
What is the best way to do this?
Note the USB cable will be get a separate hole to attach (or maybe even a separate connector in the enclosure. The enclosure itself will be attached to a box that will be moved regularly (even during working conditions).
Also, I wonder if holes would be needed to prevent too much heating. The STM32 will run on a very low frequency initially, but later maybe on 72 MHz.
(I know putting this is not entirely an electronics question, however, I think it's better appropriate here because more people doing electronics have experience how to handle my problems. It's the first time I use an enclosure).
AI: Cut a piece of pcb stock (or any sheet plastic) to fit properly in the housing, including screw holes to match those bosses.
Then attach your components to this base plate either with double sided tape, velcro, or cable ties going through holes in the plate.
Run the cable through a notch in the edge of the housing rather than a hole so it can be easily removed, but won't pull out by accident (unless you want it to be removable in the field). |
H: How to find value of unknown resistance when only net voltage and current across unknown resistor is given (in parallel)
I have just started learning thus as it was not my core subject. I searched on google and tried to search for a relevant youtube video. But I was not able to find a proper answer.
AI: Actually some quidance is already given. Here's another: Assume unknown U from node B to F. At node B you can write Kirchoff's current law equation (100V -U)/10Ohm = U/10Ohm+2A. Then calculate your R from Ohm's law. |
H: How much current can a 0.1" pitch pin header pass?
I'm curious how much current it is safe to pass through a 0.1" pitch pin header. I'm sure it will depend on the gauge and stranding of the wire attached to it as well, but assuming that is not the limiting factor how much could be safely passed?
AI: Samtecs TSW standard 0.1" header has a variable voltage and current rating based on the matching connector it plugs into. This is due to how much surface area makes contact with the header. That says, it averages 4 to 6 amps per pin.
http://suddendocs.samtec.com/catalog_english/tsw_th.pdf
Based on the common design and material, most headers will have similar properties.
When used for stacking, as in two boards soldered to the same header, the max amperage is 3 amps @ 80 degrees C (I'm assuming per pin). Except for some of their dedicated power strips which have 8+ amp. |
H: Raspberry PI controlled LED strip flickering
I have the following circuit, the gate is driven by a pwm pin on a raspberry pi zero.
The LED, successfully light when my program commands it to.
However, the LEDs flicker occasionally when the duty cycle is set to anything less than 100.
What can I do to stop the LEDs from flickering?
simulate this circuit – Schematic created using CircuitLab
AI: I'll turn my comment then into an answer:
If your PWM is done in software it will flicker as the PI is not a real-time OS so regularly the CPU will have something else to do and your lights will flicker.
You can TRY to give our program a higher priority. The 'nice' command does that. Beware that a lower nice number is a higher priority.
Better is to use the hardware PWM in the Pi. That comes out of specific GPIO pins. |
H: Understanding Zener diode datasheet
I am a beginner in electronics, and if you have a look at my profile, you'll understand that I'm completely confused in nomenclature of components.
I recently downloaded a datasheet of a Zener diode series, but I need a bit of help in studying it. I can't understand most of the column headings.
The above one is a picture of the diode characteristics table.
Please correct me (or help me) in identification of the column headings:
Zener voltage - this is the voltage that can be applied across the diode in reverse bias state.
\$\mathbf I_Z\$ - the current in the zener diode in reverse bias state.
\$Z_Z\$ @ \$I_Z\$ - as far as I can guess, it's the resistance offered by the diode in reverse bias mode.
Leakage current - I haven't come across this term. Does it have something to do with the efficiency?
\$T_C\$ - the temperature conditions in which the diode works the best.
\$C\$ - the capacitance of the diode.
AI: I'll keep this short. (I assume you are just asking generally and not about a specific application. Writing a book on all uses of the zener is beyond the scope here.)
The zener voltage is probably better seen as the "break down voltage" for the reverse-biased zener diode. If you impress a higher voltage on it, it will collapse from the applied voltage and allow huge currents to flow if there isn't something else to limit those currents.
So if you take the 6V2 and hook up a \$10\:\text{V}\$ power supply to it, you will pretty much destroy the zener.
However, if you put a \$680\:\Omega\$ resistor in series with the zener and that same \$10\:\text{V}\$ power supply, then at first there will be no current (for all intents) and the voltage drop across the resistor starts out at \$0\:\text{V}\$ (for the tiniest moment.) This impresses the entire \$10\:\text{V}\$ across the zener, which immediately begins to collapse and start a current flowing. The current rapidly rises and as it does the voltage drop across the \$680\:\Omega\$ resistor increases, thereby reducing the voltage difference across the zener (good thing.) Eventually, the whole process stabilizes when the voltage drop across this resistor is about \$3.8\:\text{V}\$, leaving the desired \$6.2\:\text{V}\$ across the zener itself. At this point, the zener stops increasing the current and just allows the impressed voltage across it to remain stable at this value.
Different zeners will be designed to reach this stable point at different voltages. It is your job as a designer to make sure that the current that results in the zener is the rated value (approximately.) In your datasheet example, this current is \$5\:\text{mA}\$. So, with the \$680\:\Omega\$ resistor I mentioned, we can expect about \$\frac{3.8\:\text{V}}{680\:\Omega}\approx 5.6\:\text{mA}\$. And this is close enough to the spec that you can expect about the right voltage across the 6V2 zener.
From this discussion you have your answer about \$I_Z\$.
Note also that this datasheet includes maximums and minimums for the zener voltage. This means that you cannot actually expect a precise \$6.2\:\text{V}\$ from the 6V2, but instead \$6.2\pm 0.4\:\text{V}\$. This is over the range of parts you might find in a box, or in a bunch of different boxes of them bought at different times. They are telling you that you cannot expect too much accuracy from these devices.
The value of \$Z_Z\$ can be used to estimate the worst case variation of the voltage across the zener, if you know the current variations. So let's continue with the 6V2 with \$Z_Z=10\:\Omega\$. We just computed an estimate of \$5.6\:\text{mA}\$ using a \$680\:\Omega\$ resistor and assuming an exact zener voltage (that we now know we can't be entirely sure of.)
Let's see where that takes us. The zener voltage for the 6V2 should be \$6.2\pm 0.4\:\text{V}\$. Assuming a 1% resistor of \$680\:\Omega\$, we may have a current ranging from \$\frac{10\:\text{V}-6.6\:\text{V}}{680\:\Omega+1\%}\approx 4.95\:\text{mA}\$ to \$\frac{10\:\text{V}-5.8\:\text{V}}{680\:\Omega-1\%}\approx 6.24\:\text{mA}\$. A difference of about \$1.25\:\text{mA}\$. While we don't know the exact voltage for some specific zener here, we can still estimate that there will be an additional variation of about \$1.25\:\text{mA}\cdot 10\:\Omega\approx 12.5\:\text{mV}\$ due to \$Z_Z\$.
This is actually not so important here, though. It's just mathy number twisting, really. Where it becomes important is instead when you add a circuit that uses the zener voltage. Often, this is an emitter follower BJT. (See this question: Explain the logic of a 12 V to 9 V conversion.) The base of the BJT will require some base current and this base current will vary depending on the load requirements.
So the point here is that a designer can estimate the load current variation for some larger circuit that uses the zener. And from this load current variation estimate a base current variation. And from this base current variation and \$Z_Z\$ estimate how much the zener voltage will vary due to the load current variation.
This may be important (or not.) But it gives you a starting point to estimate how bad it might be once you calibrate your circuit and start applying a realistic load, now.
The value of \$I_R\$ includes a referenced voltage in the table. This basically helps you to understand how much prior leakage you can expect from the zener diode if the impressed reverse voltage is LESS than the face value. So if you have the 6V2, you can see that they specify it for \$4\:\text{V}\$, which is well below their minimum of \$5.8\:\text{V}\$. (But it is also as big as possible, short of that, so that the leakage current will be a "worst case" scenario.) So if you didn't use a \$10\:\text{V}\$ power supply but instead applied \$4\:\text{V}\$, then this value of \$I_R\$ is the worst you would expect to see (about \$3\:\mu\text{A}\$.) This would produce only about \$2\:\text{mV}\$ across the \$680\:\Omega\$ resistor, in the example case I've been discussing here. But there are other circumstances where this leakage might be more important to know. |
H: Single push button relay with reset
I have an automotive application where I would need a 12V relay/circuit that can be toggled with single button between on/off. But it should reset itself back to when input voltage drops to zero.
I found this: https://www.the12volt.com/installbay/uploads/E33_Latching_Relay_2.gif
but it seems to require two buttons.
Edit: I am controlling 12V 6A max.
Edit 2: I found promising looking one button two relay circuit: http://www.homofaciens.de/technics-base-circuits-relay_en.htm and there figures 32-41. Any comments on that?
AI: if you're not averse to using speciality relays like ratchet (alternate acting) and delay you can do it this way:
simulate this circuit – Schematic created using CircuitLab
or this way:
simulate this circuit |
H: What does it mean to "gate the clock"?
In my lecture notes I keep reading "do not gate the clock". I tried searching on the Internet, but I'm unable to find the exact meaning of this phrase.
AI: To "gate the clock" means put a logic gate in the clock line to switch it on or off.
simulate this circuit – Schematic created using CircuitLab
The diagrams above show and AND and OR used to gate the clock. One forces the clock low the other high.
To prevent clock pulses which are 'too short' either high or low ("runt pulses"), we must make sure that:
The control signal for the AND gate should change only when the clock is low.
The control signal for the OR gate should change only when the clock is high.
Gated clocks are very useful for reducing power in CMOS as the logic stays 'quiet' while the clock is stopped. You will find that modern synthesis tools have special option to insert clock gating automatically.
simulate this circuit
Above are two circuits which safely generate a gated clock. The circuits rely on the fact that there is as a small delay (clock to Q) for the control signal to come out of the register. Thus the control signal changes at the gate when the clock has a known polarity. |
H: SPST Power Switch with LED for Guitar Amplifier
I would like to have a power switch for the solid state guitar amp I am building, but I would like an LED indicator to know when it is on. I know how to do this using a dpdt on-off, but would it be possible to use an spst and still have the LED?
AI: Yes. Put the LED in parallel with your amp:
simulate this circuit – Schematic created using CircuitLab |
H: Is this an encoder? Correct naming of circuitry
Newby to the digital circuitry world please be nice.
For a rock paper scissors game, I've created the following to determine a winner, however I don't know what this circuit would be called in the realm of correct terminology.
I'm tempted to simply call it an encoder but am both unsure on how loosely that term can be used and also as to how the 2n IN = n OUT could be interpreted to validate this.
Thanks for all help and suggestions]1
AI: https://en.wikipedia.org/wiki/Encoder
"An encoder is a device, circuit, transducer, software program, algorithm or person that converts information from one format or code to another, for the purposes of standardization, speed or compression."
You are converting information from one format to another. But it's the score of the game. In this case your entire game is the entire system. In my opinion an encoder is a part that can be put in between two systems. So I'd say that... it's in a really gray area but it leans towards not being an encoder.
I'd call these things for encoder:
Rotary encoder, it's bridging information between something that rotates and some microcontroller. This is important.
Forward error correction codes, it's about converting some bits of data to some other bits that you are then sending through a medium, like wireless through the air. And then the receiver change the bits that was sent back to the data that was encoded on the sender side. Without this encoding it would take you minutes to load this page. This is important.
UTF-8, it's one of many text encodings. This one allows me to write the Swedish "åäö" and some Japanese "きみ わ". This is important.
It's like a human translator allowing someone who's deaf to communicate with someone who's not deaf but doesn't know sign language.
In your case there's... gahh.. this is so gray. There are three systems, you, your opponent, and the game. But no, my list above that shows what I'd call an encoder are vital parts of the system. Or things that matter really much, without those encoders the systems have to use other much more difficult alternatives. Your game can be played without the hardware, which might be even simpler. Your hardware version of the game is not important.
An encoder shouldn't be the main focus of a system, it should be something that another system can use for some positive effect. Your game is the main focus.
So no, I stand my point as classifying your game as not being an encoder.
I'd call it what I've done throughout my answer most of the time. A game. But if you want something more nerdier then I'd call it a "decider" circuitry, since neither you nor your opponent will decide who's the winner, but the circuitry will, and both of you will take it at face value. Or I'd call it "game circuit" since that is a pretty correct description of what it actually is. |
H: Headphone Audio Noise
I built an audio mixer, and tested it with two different pairs of headphones. When I test it with one pair of headphones (I'll call Headphones #1), there is no noise (except for a few "pop" sounds when it first connects). When I test it with the second pair of headphones, there is very audible noise/crackling.
The only difference in specs that I'm aware of between the two pairs of headphones is the impedance. Headphones #1 is about 30 ohms, and Headphones #2 is lower impedance at around 8 ohms.
Given the below circuit, in which V1 and V2 are input audio signals, and the output is connected to a headphone channel, can you see a reason why noise would occur with Headphones #2? Why would a lower impedance headphone be subjected to audible noise with this circuit, or could it be due to a different characteristic?
The 1k resistor at the output is a discharge resistor I put in to prevent popping.
AI: This is a preamp IC. Its output current capablity is too low for 8 Ohm load and 100mV output voltage. I guess that strong bass peaks cut off everything else momentarily. The headphones are complex loads, 8 Ohms is the nominal value, the actual impedance is heavily frequency dependent and can be at some frequencies much lower.
You should use a speaker amp IC. |
H: Is it necessary to integrate the output voltage of Rogowski Coils, but not the output voltage of current transformers?
I have researched how Rogowski Coils works, and I have seen the integrator circuit at the output of the coil, I have seen the waveforms, done the math, I know that it's necessary to integrate the signal. However, current transformers are magnetics core with the coil around them, and I haven't seen any circuits after the coil or anything like it, they work on the same principle, so am I missing something? Is the integral of the output voltage necessary when I'm using a current transformer to get the current waveform of a conductor? Thanks in advance
AI: A current transformer is a transformer. The primary conductor threads a magnetic core, as does the secondary. Both primary and secondary currents add vectorially to create the core H-field. As the output is ideally shorted, the winding voltage is very low, hence the core B-field is very low. With a low B-field, and a very high (ideally infinite) core permeability, the H-field is close to zero, so the secondary current is a good facsimilie (scaled by turns ratio of course) of the primary current. Both the very low load and the high permeability are needed for high accuracy. You say you haven't seen any circuits following a current transformer - the short circuit, usually a very low value resistor, is a vital component.
A Rogowski coil is a field sampler. The primary conductor sits in free space, and generates a magnetic field around it, defined by the free space permeability. The turns of the secondary define a toroidal volume. The changing magnetic field in this volume generates a voltage in the turns around it. This voltage represents the rate of change of the field, Faraday's Law, so integrating it gives the primary field, and hence primary current. No current (ideally) flows in the Rogowski coil, so its effect on the primary magnetic field is ideally zero.
In photographs of Rogowski coils, it often appears as if the secondary coil is wound round a core. This is not a magnetic core, it is there to ensure the secondary coil is of fixed size and is easy to handle. It also provides a dielectric tube for the return wire of the coil. For symmetry, and so rejection of unwanted currents, this return wire has to run back down the centre of the windings. To this end, the core of the winding is invariably a length of coaxial cable with the outer removed.
The correct geometry for a Rogowski coil is achieved when the pickup toroid wraps the primary an integer number of times, and coils are usually calibrated for one turn. With any gap or overlap in the pickup coil, the gain becomes sensitive to the position of the primary cable within the loop, and the coil starts to couple to fields outside the loop.
You might like to consider this. A transformer is controlled by topology. As long as the primary threads the core, and the secondary threads the core, they are coupled, and it doesn't matter what shape the turns take. A Rogowski coil is controlled by geometry. If we double the diameter of the coaxial cable forming the coil, then the enclosed volume quadruples, the total flux monitored quadruples, and the gain increases by 4. We calibrate the gain of a Rogowski coil. When we calibrate a current transformer, we only check how close to ideal it is. |
H: Solenoid circuit current draw doesn't match calculations
I have a circuit of 6 solenoids connected to Darlington Array (http://www.ti.com/lit/ds/symlink/uln2803a.pdf) and driven by 18V (2x9V batteries) as can be seen below:
From what I understand, each load should draw 200mA (18V/88Ohm) and when all of them are switched on at the same time, the total draw should be 1.2A. They are all in the active state for a brief moment (20ms).
I wanted to confirm my calculations by measuring the total current draw with a multimeter connecting it in series with the current source.
The measurements showed a maximum of 47mA draw instead of 1.2A.
What could be the reason why the actual measurement does not match the theoretical?
AI: Your meter is not fast enough to capture the 20ms current pulse it is averaging it over a period of time to give you the reading you see.
If you have a scope monitor the voltage across a small resistor connected in series with the voltage source to see the real current. |
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