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H: I need help for soldering. Is this good flux for use on PCB/Electronics?
this is a specific question to a product for soldering.
I am a newbie to soldering so be kind.
To be honest I was trying to learn soldering and watched videos and read some.
But in germany there seems to be an unwritten law to make finding the right soldering stuff as hard as possible (also confusing)
It is called soldering paste which someone else in IRC thought is something else but after he showed me what soldering paste is , some grey material for stencil , I can assure you it is not. This stuff is orange like flux should be I guess.
So I got me some tools and materials and one of them is these:
http://www.conrad.com/ce/en/product/588206/Solder-paste-Stannol-165018-Content-50-g-F-SW-26
If you go to the download sections you can see Datasheets.
Edit : It says "Activated resin mixture with petrolatum"
Now my question is can I use this on electronics? I want to solder on PCB, some pins to a microsd adapter for example.
This material is orange and I did not test it yet.
Here is a pic of it:
Also if you can tell me if that soldering tin (Sn60Pb39Cu1 with flux) is good (for my purposes)?
You can search on Conrad.com for it just enter this in search : 812803
I can not post anymore links thx to this restriction..
tl dr :
Is this good flux for electronics soldering?
Is this soldering tin good for electronics soldering?
Also if any tips, tell me.
Thanks for your read :)
AI: Both will work. In this particular case, the paste is actually essentially the same flux as the wire provides, so there is no penalty (other than a very high inconvenience level) in using it.
In general, though, solder pastes are acid-based, and much more aggressive than rosin-based. You can use such fluxes, but intense care in washing off the residue is required to avoid corrosion later. So be very careful about using any paste flux. Plus, of course, you wind up using a lot more flux when you use paste, since you have to spread it over the area to be soldered. Flux-cored solder provides the flux at exactly the point it's needed. |
H: Simple receiver unwanted selectivity
simulate this circuit – Schematic created using CircuitLab
In an earlier question I asked help building a simple RF receiver/amplifier. Thanks to help from Andy aka I was able to build the circuit above.
I'm using a function generator and a long piece of wire to transmit. To the input of this receiver I've connected another piece of straight wire. I connect my oscilloscope between the negative terminal of the battery and the transistor collector.
I turn on my function generator, touch the positive lead to the long wire and I see a nice amplified sine wave on my oscilloscope (the function generator and this receiver are a few meters apart). I also tried connecting the output to a bridge rectifier and I was able to blink an LED nicely.
So everything works nicely except one thing: It only works with a frequency of around 3.8Mhz, even with no kind of tuning circuit! It's quite selective: at 3.7 or 4 MHz the output is almost nothing.
Then I tried to use the function generator directly as an input: Now the problem is gone. Any frequency is amplified and with the rectifier connected I can blink the LED.
The transmitter antenna is a piece of wire around 4-5 meters long hanging from the ceiling. The receiver antenna is another piece of wire, around 1 meter long. I've tried both building the circuit on a breadboard and soldering it to a perfboard. I've checked my setup many times, I don't think I've made any mistake in setting up this experiment. I've also tried a loop antenna (around 7 turns of cable around 30cm in diameter). It also has the same selectivity at 3.8Mhz.
So my question is: How is the circuit showing such unwanted selectivity? I've also tried tuning it with an LC circuit, but it simply attenuates the 3.8Mhz frequency as well. Another thing to mention is that the circuit at 3.8Mhz also works only when the oscilloscope negative lead is connected to the ground of this circuit (regardless whether the positive lead is connected or not).
EDIT: Here are pictures of my setup. This quality is the best I could do. Please excuse my poor editing skills. This circuit on the perfboard has the amplifier as in the schematic connected to a rectifier whose output is connected to the small breadboard and the LED. This setup works very well on 3.8 MHz.
AI: 3.8 MHz has a wavelength of 79 metres and you have a distance between your antennas of maybe 1 or 2 metres. This means that they do not act as EM transmitters but they act as plates of a capacitor to each other. So this is answering one of your questions: -
Another thing to mention is that the circuit at 3.8Mhz also works only
when the oscilloscope negative lead is connected to the ground of this
circuit
You need a complete circuit to pass current capacitively and, earthing via your scope, connects your receiver to the earth on your oscillator through the AC wiring in your building.
And that brings in something else to consider - that AC wiring is a "system" in its own right and can offer all kinds of impedances at frequencies in the MHz range - who is to say that it isn't this wiring that is causing the series resonance at 3.8 MHz? If the wiring offered an inductance of 1 mH and the capacitance between your wires was 1 pF then this forms an effective resonant filter of 5.03 MHz. If your wiring offered 1.75 mH then your resonant frequency is 3.8 MHz!!!
So, in summary: -
You are not transmitting EM waves between the two wires
The gap being about 1 metre compared the wavelemgth being nearly 80 metres substantiates this.
This means you are coupling capacitively
Your coupling requires a return connection
That return connection is through your AC wiring
It could easily form a resonant filter with the capacitance between the two wires.
Try directly connecting the signal generator ground to your amplifier ground and see what happens. My money is on it being a gamechanger. |
H: Why won't you get electrocuted with a car battery unless you touch ground and positive?
First, if you have a better suggestion for title I'm really opened to suggestions, but it is the best I came up with.
If I touch an electric line and the earth, I will get electrocuted.
However, if I touch the positive of a car battery, people told me the same thing won't happen...why is it?
It seems I need to touch the battery positive and the battery ground.
AI: For current to flow through your body, two conditions must be met:
There must be a complete electrical loop. Static electricity excluded, electricity only goes somewhere if it has a path to come back. Electric lines are grounded in multiple locations. Your car battery isn't.
The voltage must be high enough to go through your skin, which is a weak electrical barrier. For a dry skin, a ballpark figure is 20V.
Touching only the positive post of a car battery, none of the two conditions are met. |
H: how powerful a stepper motor do i need
I had some nema 17 style stepper motors from an old prusia i3 and tried to use them to automate my blinds but noticed that they weren't powerful enough for my blinds. I used a basic newton meter to check the amount of weight i need to apply to move the blinds and it's roughly 25-30newtons. My stepper motor looks like it can only pull around 15-20 newtons. Is there anything i can do to increase the torque of the stepper motor(i'm using a L293D motor shield on an uno)? And i was thinking about adding gears but wasn't sure about the best type or the needed ratio so it doesn't reliably skip any steps, speed isn't really important? Any advice you have would be really helpful, thanks!
AI: Note that the amount of force (Newtons) available from a motor depends on how far away from the center of rotation you are, because torque is the product of force and distance. (We measure torque in Newton-meters or pound-feet or somesuch unit.)
A gearbox or pulley will give you close to the exact ratio of torque increase; a pinion with 12 teeth driving a sprocket with 60 teeth will give you about 5:1 torque increase, with the corresponding 1:5 decrease in rotational speed. Same thing for a belt or any other kind of rigid ratio.
Regarding "not skipping steps," you will want to add limit switches no matter what. Even with brushless motors, as suggested in the comments, you need to know how far to move, and unless the BLDC has an absolute encoder attached, you can't guarantee the exact stop position without limit switches.
Stepper motors are reasonably tolerant to light over-volting, or even over-current-ing, because they have very few moving parts, and no brushes to wear out. Also, if you only run them for a short amount of time, there won't be as much heat build-up, and thus you can run them at higher current for a shorter duty cycle. If you can crank up the voltage, and turn the motor driver up to maximum current, that might be enough. Or not -- the L293D is not a high-performance driver. |
H: Extending an HDMI port or Display cable?
I am working on building this project for my Raspberry Pi well I need to extend the HDMI port of this monitor http://store.hp.com/us/en/pdp/hp-22er-215-inch-display
I need to place the display in an enclosure, so the horizontal HDMI connector with a regular cable would cause the case to be too deep. So I need to basically relocate the port onto a breadboard and then connect to my raspberry pi. Any ideas? I was considering desoldering it and attempting to get wires to lineup and connect to it. Or somehow extending the ribbon cable coming from the board to the monitor so that I could put it in a separate box or something.
Any suggestions on if it would be a better option to extend the HDMI port or somehow extend the display ribbon cable?
AI: Since the issue is a horizontal port which would cause issues fitting in a case, there are only a few solutions.
Replace the connector with a right angle one with the same foot print. Highly unlikely.
Extend the connector with some wire. This may leave it fragile. Delicate soldering skills required. Be prepared for solder bridges. May result in errors and lower bandwidth.
Solder a cut cable directly to it, removing the connector completely. May be fragile, May also result in errors and lower bandwidth. Same soldering issues as 2.
If the video input board is separate from the display controller board, extend that cable instead.
Simplest solution, use a right angle adapter or cable with a right angle connector. This is a plug and play solution, and may only add 0.5 to 1 inch, not 3 inches. If you need more, then you can crack the plastic open. Or use a FPC version like https://www.ebay.com/i/262114109807 can't vouch for the quality but it has no depth beyond the connector and pcb.
Alternatively, use a display directly from the RPi's LVDS interface, bypassing HDMI.
You mention a display cable, likely the flat FPC that goes to the display internally. That is even harder to modify properly. |
H: How do dual RC filters interact?
A high-pass RC filter with a corner frequency of 340 Hz can be built with a 47 Ohm resistor and a 10 uF capacitor.
A low-pass RC filter with a corner frequency of 3400 Hz can be built with a 47 Ohm resistor and a 1 uF capacitor.
But what if I combine the filters, using only a single resistor? Will the corner frequencies of the low- and high-pass elements change?
The input is on the left, the output is on the right, the segment at the bottom is signal ground. The output feeds into a 10K input impedance amplifier input, and the DC blocking of the series capacitor is intentional.
AI: With the 10k load resistor, the high-pass time constant is 10k * 10uF = 100 ms. The cutoff frequency is 1/ (100 ms * 2 * pi) = 1.6 Hz. Assuming the DAC output impedance is very low, the low-pass corner would be as you expect. ( 1 / (2 * pi * 47 us)).
The two filters will not interact very much because their corner frequencies are far apart. The DAC output impedance will shift the low-pass filter cutoff frequency downward.
In addition, the series resistance (RDAC and R902) combined with the 10k load impedance will act as a voltage divider in the passband of the high-pass filter. Assuming the series resistances are small, this should be a minor effect. But if you were to alter R902, or if you discover that the DAC output impedance is substantial, this could become significant. Also, if you reduce the 10k load impedance, then the effect could become more significant.
For this reason, if you wish to shift the highpass frequency upward, I suggest you do so by reducing C902.
I would also like to point out that if this circuit will drive an external amplifier, and you cannot control the input impedance of the amplifier, then you may find that the highpass frequency shifts, depending on which amplifier is connected. Hopefully that is OK. |
H: Chip antenna PCB layouts and matching network
I am designing a PCB layout for NXP JN516x Wireless Microcontroller (http://www.nxp.com/docs/en/data-sheet/JN516X.pdf). Part of the design is the antenna, where I have to choose a 50ohm antenna. I decided to go with a chip antenna (https://www.digikey.com/product-detail/en/ethertronics-inc/1001312/939-1040-1-ND/4754415), since it has a small footprint. According to the NXP JN516X datasheet, There is a matching network for the antenna, as shown in the figure below (Page 89):
But I can see from the datasheet of the antenna itself, there is a recommended PCB Layout, where it is clear in the below figure that there is only two matching components, and the datasheet didn't even mention anything about the values or whether they are inductors or capacitors.
I don't have much experience with antenna design.
Which matching network should I use, the one in NXP datasheet or the one in the antenna datasheet?
Update:
I found the below statement on page 27 of the chip datasheet, it said that we need the three external components L1, L2 and C1, as they form a 50ohm resistive network. Does that mean I don't have to remove L1?
Thanks in advance.
What should I follow
AI: You need some of both.
L1 from the NXP document is for matching the antenna to the chip.
L2 is to provide DC to the RF amplifier. The inductor blocks the RF from getting back to the battery.
C1 is there to prevent the DC for the amplifier from getting to the antenna. If the DC were on the antenna, you could short the power supply by (accidentally) shorting the antenna cable.
You will need to combine the circuits from both documents.
You must have C1 and L2 from the NXP document. These are needed to provide power to the amplifier and to keep the DC from getting to the antenna.
The antenna documentation says you need a pi network to match the antenna impedance to the rest of the circuit. I'm not much help with that, but note that a pi network has two capacitors to ground, and the layout has two spaces for small capacitors.
The pi network would be used in place of L1 from the NXP datasheet.
You must follow the rest of the layout notes for the antenna. The clearances are needed for the antenna to work properly.
If I had to do it, I'd look and see if I could find an antenna with a more informative datasheet. The one you have might be fine for someone with more experience. It doesn't have enough information for me to be sure of getting it right.
From the appearance, I would guess that the trace between the two capacitors takes the place of the inductor in a pi network. I'd prefer that the document give an example of a complete network so that I could compare my own understanding with their calculations.
This is an example of an antenna I would be more likely to go with. Not recommending that part, just providing an example of a part with a datasheet I'd be more comfortable working with. It includes a layout and part values for the matching network as well as how to place it.
Matching network from my example antenna datasheet:
To be more clear:
You must have L2 and C1 from the NXP circuit. Instead of L1 from the NXP document, you use the matching network from the antenna document. Make sure to follow all of the layout specifications - these are important for the proper function of the antenna. |
H: What are the compulsory tasks of a charge controller for a lead acid battery?
Solar charge voltage regulators should coordinate the charging process of a battery. Suppose that I would like to design a very basic charging regulator controller with the following specs:
The battery charged is a 12V 12Ah lead acid battery.
The power supply is a 12V 10W solar panel.
During the day the battery is not used.
What are the minimum requirements and specs that are required? At the top of my mind I thought of the following:
Over voltage protection (the charging should stop when Vbattery > Vmax)
Over current protection (the current should be less than the maximum allowable by the battery) but this seems not to be an issue since the PV module is just 10W 12V.
If you need a battery for reference, then just take this one.
AI: The bare minimum requirements I would set at:
Undervoltage protection. Do not charge is battery open cell voltage is too low.
Overvoltage protection. To protect the charger.
Timeout. When during charging the voltage does not rise as expected.
But you can do many more things:
You can follow the proper lead-acid battery charging profile.
Source
You can measure battery temperature, and adjust charging rate respectively.
You can give a battery end-of-life warning.
You can measure the energy going in/out of the battery. Giving a pre-warning: "please charge", and disconnect if it gets very low. In order to preserve the battery. Batteries last longer when not depleted often. |
H: strange behavior for atmel 16
here is the code
#define F_CPU 1000000UL
#include <avr/io.h>
#include <avr/io.h>
#include <util/delay.h>
#include <avr/interrupt.h>
int main(void)
{
DDRA |= (1 << PA0);
DDRB |= (1 << PB0);
DDRD |= (1 << PD2);
MCUCR |= (1 << ISC01);
GICR |= (1 << INT0);
sei();
while(1)
{
if (PINB == 0b00000001) //when i trigger pbo it will light pa0
{PORTA = 0b00000001;}
}
//idont want any thing to happen i want pa0 is on for ever
}
ISR(INT0_vect)
{
_delay_ms(10000);
PORTA ^=(1<<PA0); //toggle pa0
}
and the circuit
the purpose is when i push switch PB0 led connected to PA0 goes on
then when i trigger the interrupt in PD2 this should toggle PA0 to turn it off
the strange thing when i toggle PA0 to turn it off it the led turn on by itself automatically again without triggering the interrupt why??
although this behavior dont happen in Proteus simulation ???
AI: Real world switches do not contact cleanly - they bounce and open and close their contacts many times over a few milliseconds before settling to their close position.
Almost certainly what is happening is that when you press the right hand button it closes, then opens again before closing finally.
The initial closure will cause an interrupt but the second closing will cause a pending interrupt to be latched into the processor.
When the ISR exits after toggling the LED there will be the second interrupt pending that will cause the ISR to be entered again.
(From The Lab Book Pages)
This is a very common problem and one of the reasons that it is not good practice to connect a switch directly to an interrupt.
The problem can be solved in hardware or software. This site shows some ways Switch debouncing.
Also as a matter of software design it is considered bad to put a delay loop within an ISR. Interrupt service routines should do as little as reasonable and exit quickly. All the time the ISR is executing the processor cannot process other interrupts. In your example it doesn't matter but in a real system it would. |
H: Set a reference voltage: using a zener diode vs using a buffer
Given the following hysteresis comparator circuit:
simulate this circuit – Schematic created using CircuitLab
R1, R2 and OA1 are used to set a reference voltage for the comparator with hysteresis OA2. The circuit works fine but I was wondering whether there could be a better (or more clever) way to set Voffset without using an "expensive" (or precious, if you wish) opamp. Would using a 6V Zener diode in parallel with R2 be a better choice for instance? What problems would it possibly cause? If you have other options please do point out advantages and disadvantages compared to the proposed circuit.
Sorry, I've just realized a few more specs could help:
Voffest must be around 6V with a +-0.5V tolerance.
The power supply is a battery, as far as my tests went, a +-1V on the supply rail should cover most scenarios.
I made some tests with a few spare LM741CN I had laying around. Speed is not a requisite. The V+ terminal is wired to 12V and the V- is wired directly to ground.
The hysterisis window should be of about 5V +-1V which is about what I got with my brief tests.
AI: I've used this circuit myself for everything from detecting moisture around toilets (there, using resistors in the area of around \$2.2-4.5\:\textrm{M}\Omega\$) to pretty much anything I need.
There are three parameters that vary meaningfully with a BJT: \$\beta\$, \$I_{SAT}\$ (which affects \$V_{BE}\$), and temperature (which affects both.) I've tested the circuit with random parts out of a box with good success, so long as the requirements aren't too precise. Yours aren't. So there shouldn't be much problem here.
I've arranged things to provide about \$5\:\textrm{V}\$ of hysteresis band width and I've centered it in the middle of your supply rail. (I'm assuming about \$12\:\textrm{V}\$ for the supply rail, but again it doesn't have to be a precision value.)
Here's the circuit:
simulate this circuit – Schematic created using CircuitLab
Simulate that thing before building it and see what you get. See if I guessed what you wanted to achieve. Play with \$R_6\$, up or down to the next nearby value, to see how it moves the band.
It'll do fine over a sufficiently wide supply voltage range. Given the values I used earlier for \$R_3\$ and \$R_4\$, the output wouldn't pull completely to the supply rail. So to make it close, I reduced their values for you.
Here's a Spice run using multiple source voltages (\$12\:\textrm{V}\$, \$13\:\textrm{V}\$, and \$14\:\textrm{V}\$), and doing a range of factor of three range on \$\beta\$ and a factor of three on the saturation current (which affects \$V_{BE}\$.) The two BJTs are not held to be the same, either. They are varied independently. You can see how the hysteresis is affected and see what the output voltage looks like, as well:
Adding temperature variations caused LTspice to complain about too many dimensions of variation, so I didn't add that. But I did do some runs to make sure that the behavior shown in the image above remains similar and it does.
It's cheap, works reasonably well, and qualifies for your \$\pm 1\:\textrm{V}\$ spec, even considering substantial variations. But it won't give you the predictable behavior of an opamp. |
H: How do you design a switch input to a MCU for ultra low power consumption?
I'm trying to build a low power door monitoring switch. I'm using an MSP430FR2311 with the below circuit. The switch is a SPST reed switch and there is an internal pull-down resistor on pin P1.0 of the MCU.
When the switch is open power consumption is 1 uA but when the switch is in the closed state consumption is 100 uA. I want to be able to run this on a small battery for years at a time. Is there a circuit that will get me low power consumption in both the open and closed states of the switch?
simulate this circuit – Schematic created using CircuitLab
AI: The internal pulldown has 20–50 kΩ, typically 30kΩ, which limits your choice of resistor to pull the input up.
Fortunately, you can turn the internal pulldown resistor off, which is also the configuration used to measure the minimal leakage current reported in the datasheet.
simulate this circuit – Schematic created using CircuitLab
The datasheet suggests a leakage current of 30nA at 3V, so the internal resistance is on the order of 100MΩ. Whether you put an extra resistor in-line here does not affect the leakage current by much, so it can in principle be left out entirely if the inductance of the connection from the switch to the IC is low enough that the voltage won't overshoot. |
H: What is the function of these extra capacitors around the bridge rectifier?
I'm looking at different bridge rectifier circuits and found this one. Capacitors C5-C8 are smoothing capacitors. Both positive DC and negative DC have capacitors C1-C4 linking to both AC lines.
What is the function of C1-C4?
simulate this circuit – Schematic created using CircuitLab
AI: Those extra small value capacitors are to avoid the rectifier generating RF interference.
When a diode has been conducting and then current reverses it can stop conducting very suddenly when the charge carriers are depleted. This can occur extremely rapidly and generate interference to many MHz.
These sudden changes in current can cause interference to other parts of the circuit or even other equipment. It is modulated at 60Hz or 120Hz (US power frequency) so would be heard as a buzz in audio equipment.
The usual avoidance technique is as shown in the circuit. Ferrite beads may also be required in stubborn cases.
The technique can be used intentionally to generate high frequencies or narrow pulses, especially when high frequency devices are not easily available.
Step Recovery Diode |
H: Battery Terminals Connected to Earth
The circuit below contains a battery light-bulb circuit connected to Earth.
If the distance between points A and B where relatively small, then one could see that the earth would be acting as a sort of resistor (practically speaking, as soil has some resistance) and the lightbulb could be potentially lit.
If we however put A and B at large seperation, is current flow still possible? I say this because if the earth is considered an infinite electron source/sink, there would then be reason for current to flow with a battery as a voltage potential, thus lighting our light bulb
Reasons I don't think it will flow.
The potential over the battery is 12 volts, but as far as I know there isn't a relationship between any terminal and the earth, as voltage is relative.There doesn't appear to be a voltage difference, yet I still can't reconcile this thought with earth as an electron source/sink
AI: In the circuit you've drawn, you've accepted that there's a resistance between A and B through the soil, so current will flow. We have a complete circuit.
So the question is, how does the resistance between A and B change as we increase their separation?
It turns out that as we increase the distance between A and B, the resistance increases (no surprises there) but very, very slowly.
In fact, the resistance between A and B is dominated by the ground very close to A, and very close to B. When electricity codes tell you to set a ground connection, they'll tell you what length of rod, what minimum diameter, and to set it in soil rather than rock. The assumption is that the conductivity to any remote earth is dominated by the conditions around that electrode.
The reason is the scale, and the way the path changes as we increase the distance. Let's say we use a 1m long rod for A, 20mm diameter, so its surface area is about 0.06m2. Let's follow the current out 10mm, to a cylinder of soil 40mm diameter. To a crude approximation (I'm not going to do integral calculus here) we have 0.01m length of soil, 0.06m2 area of soil, so the resistance, length/area, is proportional to 0.01/0.06 = 0.16.
When we take the next step, we notice the surface area we are interested in has increased. So the extra resistance of the next 10mm out will be less than for the first 10mm. As we go further away from the rod, we have more surface area, more parallel paths to carry the current.
Once we are as far away from the rod as its length, we notice that instead of the current just spreading out sideways, we need to consider the down spread as well. The surface we are interested in is now getting to be a hemisphere, centred on our ground rod. Now the surface area is growing as distance squared. The resistance is increasing even more slowly than it was before.
Eventually, these surfaces will intercept subterranean watercourses, and the resistance will increase yet more slowly. |
H: How do I desolder these specific points?
I'm currently attempting to replace some bloated caps on an old PC motherboard but am currently having problems with some specific points.
Here's a pic of the motherboard:
.
The area outlined in fuchsia are the series of caps I am trying to desolder. Notice they are in pairs of square and round solder points. The round solder points are no problem to desolder. It's the square ones that won't desolder. No matter how much I heat them up, the solder on the points just won't melt and so I can't pull the caps off!
Is there a special technique or tool I need to perform this task?
Thanks in advance for any help you can provide.
AI: You need a bigger soldering iron, as in "more power."
The square connections are in the ground plane of the board. That is the large yellow area they are embedded in. That is a large piece of copper, and there are probably also large copper surfaces on the internal layers of the board.
Copper conducts heat very well, and it also radiates it away.
The large copper areas are basically sucking up all the heat your iron can provide and radiating it away fast enough that it can't get hot enough to melt solder.
The solution is an iron that can put in heat faster than the board can dissipate it.
So, you need an iron with more power.
Many irons are only around 30 watts. You'll need much more than that.
When I've had to do that kind of thing, I borrowed a huge 150 watt iron from my father in law. It isn't intended for electronics, but it has the raw power needed for large copper surfaces.
As for technique, high wattage irons often have wide tips.
I apply some extra solder to the heavy joint with the iron heating just the ground connection.
When that finally melts, I rotate the tip of the iron to heat both pads for that part.
The solder melts pretty quickly, then I can pull the part out. Afterwards (if you need to to replace the part) you can clean the holes with a solder sucker or solder wick.
While you are removing the part, you actually want as much solder as possible on the connection. Removing solder makes it harder to get the part out, not easier. |
H: Basic electronics: input currents of multi-port sub-circuits
I am simulating, with PSpice, the transient behavior of an analog circuit that contains several multiport sub-circuits.
When I probe the current on all inputs of a sub-circuit, I see that the sum of all input currents is different from zero.
Is it possible or is it just a modeling issue of the sub-circuit I am analyzing? Should't the sum of all input currents be equal to zero?
AI: It's a rather common model issue. Output comes from a controlled voltage or current generator which is referenced to global "0" ground node.
Global means it doesn't have to be explicitly "taken out" of the model, "0"s are just connected each other any hierarchy level they are.
Here's an excerpt from TI model file where (5) is the external output node
EGND 99 0 POLY(2) (3,0) (4,0) 0 .5 .5
...
RO1 8 5 63
RO2 7 99 62
...
VLIM 7 8 DC 0
This can be turn into the following schematics
simulate this circuit – Schematic created using CircuitLab
Now, neglecting VLIM which is just a current probe to have I(VLIM) somewherelse, it is quite clear that a part (most actually) of the output current sources/sinks from/to (0) global ground node.
That's why KCL seem to fail in your simulation. If you whish to take model's ground current into account you can just add a pin and rename all internal "0" nodes to this new connection.
All the above is a good lesson teaching "models are just models" they are all incomplete to some aspect. Awareness is key point in effectively using them |
H: Analyze Power Amp 2N3055 Wit Multi Diff amp
hi guys i have a problem on analyze on this circuit
and i don't know how i resolve these
1- There Is a CE One Stage After Of Diff Amp But I Don't Know Why used This And Why Used Capacitor?
2-After Of Last Stage There are Two Diode Junction On Transistors and
(variable resistor 100 ohm) and i don't know why used this on this stage ?
3- On Next Stage That There Is A Power Amp I Don't know How It's Work?
AI: 1) that CE stage provides the voltage gain for the "servo loop" to force the output voltage to be 21X the input voltage
2) the two diode-connected bipolars (perhaps Q9 and Q10), and 100 ohm pot, implement an adjustable biasing for the output stage; BEWARE these transistors need to be THERMALLY_CLOSE to the output 2N3055
3) output devices are emitter-followers, with bases driven by opposite-polarity bipolars capable of pulling close to the rails
4) overcurrent protection, by shunting base drive [D2/3/4/5]
Beware of the 100 Ohm resistors across base-emitter of 2N3055. These Rs are the only path to remove charge, and or handle Miller_Effect capacitance for the two 2N3055s. Thus high-frequency transient response will suffer. |
H: Multistage OA, DC or AC coupled?
I have a multi stage OA amplifier (pulse applications) and I am not clear if better use AC or DC couple stages (C8 in this example). I removed caps for better response, charging and discharging through high resistors.
The REFH is VDD/2 and simulation looks nice but I question there would be a real (life) benefit by keeping caps or not.
Thanks for suggestions,
AI: Overall you have a gain of 100 so if the DC value on your signal entering to the left of R10 is close to the value of REFH then there might be no advantage keeing C8. However, if the DC value at R10 is say 100 mV away from REFH then the final output will have an offset of 10 volts. Can you live with this offset on the output is what you should ask your self. |
H: How to dynamically regulate voltage from a DC power supply to motors
I'm currently trying to take an approximately 12V, 2 to 3 Amp, LiPO battery and dynamically (using either a microcontroller or a good old fashioned dial/slider) adjust the voltage it is supplying to a pair of hobby motors, probably bottoming out at around 5 or 6 volts, but being able to go lower wouldn't be a problem.
I've read a bit about voltage dividers, but it seems that it might not be the best way to go?
This question might be a bit obvious and for this I apologize, my knowledge of circuitry is pretty barebones. Anyone have any ideas?
AI: A simple way is to use a voltage regulator and if your motor current is 1.5 amps max then I would recommend using a buck voltage regulator like this: -
You can vary the output voltage with the resistor potential divider that feeds the FB pin. Power efficiency remains about 90% for a good range of loads and you will never get anything like this from a potential divider and, a potential divider would never give you a regulated output voltage.
If you need more current then this might be attractive to you: -
There is a version of the LT8610 that has an FB pin so you can set the output voltage. You pay for what you get and if you want good performance you need to pay a few dollars more. |
H: RC Submarine Charging Port
I bought a cheap RC Submarine. Similar to the cheap RC Helicopters it's charged via a cable from the Remote Control. The Remote Control is powered via 6 AA Batteries. --> Here's a picture of the Charging Setup
Question: I lost the charging cable and since it's a cheap device I'm not able/willing to reorder one :-). But since I'v got the correct type of connector for the Remote Control Site, I was wondering if I can just solder one with some jumper cables that fits the port on the submarine site.
If one of you could tell me of those 4 pins on the submarine is + and -.
I think there are 4 Pins to submit the charging state to the remote control (light goes of if fully charged). Is it feasible to just connect two Pins (+ and -) to the charger and ignore the other two?
On the Submarine Site, the charging port consists of 4 pins.
Here's a picture:
By closing the hold the pins connect as showing in the picture.
Additionally, here's a close up of the black framed part in the upper picture:
AI: There's no telling the polarity of things from the connector itself, so you'll have to figure this out by e.g. measuring whatever comes out of the charger and following the lines/traces inside the submarine to the internal electrical storage.
The "black framed part" really is just a jumper: it shorts toghether the two pins it connects. My bold guess is that it's only the submarine's way of knowing you've closed the lid. (especially since the charging cable is only two wires, these two pins will probably simply be unconnected in the charging connector, but will mechanically make sure you don't connect the charging cable the wrong way around) |
H: In a shared wireless channel, what are the factors that impact channel gain for each user?
Assume a wireless channel with bandwidth B is shared among N mobile device users. Let's say the size of data transmitted by each device is in the range [100, 1500] Kbits.
How does the transmission power of each device and the size of the data packets transmitted influence the channel gain for each user?
Can you please explain the channel gain calculation with a simple numerical example (say, with 5 users)?
AI: The channel gain that is required for a radio system is determined by the distance between the sender and receiver, the operating frequency and the types of antenna used. Other factors include fade margin and the potential background interferences of other non-associated devices. The weather can sometimes play a part especially at higher frequencies.
The data rate also determines the bandwidth of the receiver and higher bandwidths means potentially more incoming noise and therefore, to successfully receive good data, a higher transmit level is needed.
In a multi-user system, assuming that each user is allocated a time slot then the power required to transmit and successfully receive IS NOT determined by the number of other associated users.
Can you please explain the channel gain calculation with a simple
numerical example (say, with 5 users)?
No, because it does not depend on other associated users. |
H: How do I increase 3A fixed output current of a buck converter?
Basically I am making a solar inverter without any battery or charge controller that will directly convert the dc output coming from solar panels (6 connected in parallel) into 220V AC.I am using solar panels 50W each, having an open circuit voltage of 20V and the voltage varies between 15-20V during the entire day provided a minimum amount of sunlight is there.
Next, I am using a 12V buck converter circuit using an LM2576 and few more components to get a stable output voltage of 12V out of the panels.
Now this 12V DC is fed to an inverter circuit which converts it into a square/modified sine wave 220V AC at approximately 50Hz.
But, I am not getting desirable power output. From 6 panels, all I am able to power is a 45W LED bulb along with a small 3W LED bulb.
Probably, one problem is with LM2576 buck converter IC. This IC although providing a constant 12V output but it is rated at a 3A fixed output current. And I think probably this is the reason why we are unable to drive more loads. Is there a way to amplify current in this case? Or something else should be done which I am missing here ?
AI: You are correct that your selection of the buck converter is the limiting factor. Since you would have up to 300 watts available from the panels, ideally you would need a converter that can handle at least 25 amps for a 12 volt output.
You may already be aware that without an MPPT (maximum power point tracking) mechanism you will typically realize less power from your panels than optimum. You should also consider that the buck converter has inefficiencies as well. An estimate of 90% efficiency is a starting point.
Then consider that your 220 volt inverter has a 90% or so efficiency. Under optimum conditions, you could power a 220 volt, 240 watt (~1 amp) load .
Do a web search for high power or high current buck converters. The LT1339 is one example you can study. Be prepared that at this power level the design will require external FETs and that the inductor design becomes more critical (and probably not off the shelf). The overall circuit is not a simple one like your current design. Much more engineering will be needed to obtain the desired results. |
H: Headphones volume controls do not work after 4 pole jack repair
I have been repairing the jack for my pair of Sennheisers for Android (I have done 3 attempts so far) but I can't seem to make the volume controls work.
To clarify, after the repair, the sound is alright, the microphone also works but none of the buttons work.
I am not sure what I have done wrong. The soldered connections are all ok, at least it seems like they are ok.
There are 6 wires in total, 3 grounds and 1 for right sound, 1 for left and 1 for mic, something like this:
Then what I soldered the mic, right, left wires to the their corresponding positions, and then I soldered all 3 ground wires to the ground position (not sure if the grounding of all 3 ground wires together is correct), the positions look like this:
What have I done wrong?
Many thanks for your help,
Edit 1:
As Chris pointed in one of the anwers, the reason why the buttons were not working was because android uses CTIA specification and I wired things based on OMTP - a comparison is shown below.
After a rewired things based on CTIA as below, everything worked as expected.
Edit 2: Marking this question as answered although there is still one minor problem with the stereo where the right audio sounds very "echo-y" and not normal - for another thread.
AI: You have wired your connector according to the OMTP standard, with microphone on the second ring and ground on the final sleeve.
However, according to the general Android microphone specification Android normally expects CTIA wiring "Except in regions with legal requirements for OMTP pinout".
CTIA wiring places the common ground on the 2nd ring, and the microphone on the final sleeve.
Most likely, if this is your wiring problem, you would find that you don't have any microphone input (or you get by odd means), and that you have no stereo separation, since your speaker drivers may be effectively in series with one another and without any actual ground.
Of note, the Android document linked above also includes the expected wiring of headset function buttons to ground through resistors of specified value. |
H: What is the truth table for a logic NOT gate with two inversion bubbles?
I am currently reversw engineering a PCB and it contains a D-type register chip. The chips internal circuit diagram shows a NOT gate with two bubbles on it both in the front and the back. My question is if this is the same as a buffer or if this case some other meaning? And it would be great if a truth table could be included in the answer.
Datasheet: http://www.ti.com/lit/ds/sdls090/sdls090.pdf
AI: It is the same as a buffer. The bubble at the input is used to indicate that this input is active low. It drives the input of the reset, which also has a bubble at the input and is therefore also active low. |
H: SSR fires from nearby relay
I am using a SPF240D25R Solid State Relays - PCB Mount PCB SIP 240VAC, 25A load current and 4-15VDC trigger voltage. This is being triggered from a ULN2003A and used to drive a motor (pump).
On the same board I am driving a 5V coil relay which has 12VAC across its contacts with no load, this two is driven from another transistor with the same ULN2003A device.
When I fire the relay the SSR is firing. When I turn off the relay the SSR turns off. This ONLY happens when the load (motor) is connected to the SSR. For example, if I fire the relay and the measure the O/C load on the SSR it does not fire.
I cannot figure out the problem. I have a 100 ohm || 1uF capacitor snubber on the input to the SSR. I have a 1N4001 flywheel diode across the relay coil (even though the ULN2003 has them).
The main thing which is confusing is what does it not fire in error when no load is connected to it?
EDIT:
This is part of a large system and the schematics are over multiple pages. I have provided the various snippets below.
This is the SSR circuit. Please note that the SSR is a SPF240D25R (ignore the name on the schematics). Also note that the "pump relay" is not the offending relay. In fact when this relay is fired the same problem is NOT witnessed.
This is the lighting relay (recently added flywheel diode not shown)
AI: SOLVED - it turns out that the GND (pin 8) was floating. So even though it triggered other SSR and relay fine, it suffered from this interference. It would have been easier to find had it not worked at all! |
H: I need explanation on some digital multimeter (DMM) specifications
What is digital multimeter (DMM) ranges reading; resistance : 400 ohm - 40 Mohm supposed to mean?
would it mean the DMM cannot measure a material resistive property of less than 400 ohm ?
And what does tolerance 1% rdg. indicate which followed by: 5 dgt mean ?
Thanks in advance
AI: What is digital multimeter (DMM) ranges reading; resistance : 400 ohm - 40 Mohm supposed to mean?
It means that the meter has a number of different ranges for measuring resistance, anywhere from 400Ω full-scale to 40MΩ full-scale.
would it mean the DMM cannot measure a material resistive property of less than 400 ohm ?
No. The minimum value of resistance that can be displayed depends on the full-scale range and the number of digits in the display. For example, a "4-3/4 digit" meter on the 400Ω range would have a resolution of 0.01Ω.
And what does tolerance 1% rdg. indicate which followed by: 5 dgt mean ?
This describes the accuracy of the meter, which is a combination of a couple of factors.
The analog components limit the accuracy to ±1% of the actual value being measured.
The digital circuitry limits the accuracy to ±5 counts (i.e., in the least-significant digit of the display) of the actual value.
The overall accuracy will be determined by whichever of these two limits is larger.
For example, if you are measuring a 100Ω resistor, the first rule says that the actual value will be between 99Ω and 101Ω, while the second rule says it is between 99.95Ω and 100.05Ω — the first rule applies.
But if you measure a 1Ω resistor, the first rule gives 0.99Ω - 1.01Ω while the second rule gives 0.95Ω - 1.05Ω — in this case, the second rule applies. |
H: LTSpice, add real noise to opamp
If I will build this circuit in real life, I will probably see a mV noise with the scope due to high cascaded gain and probably some oscillations.
I use LPC358 which is specified to (Input-referred voltage noise) 178nV/SQRT(HZ). I am trying to figure out how small can be the input signal (now is GND at C3) to be masked by noise generated by circuit itself.
My simulation is clearly wrong, showing a clean output with some nV flicker.
How can I add some noise in LTSpice, based on the fact that any opamp has specified noise, that will emulate real worlds noisy output so I can select from several opamps, gain, etc?
AI: It looks like you have 30 kHz bandwidth so take the square root of 30,000 and multiply that by the first opamp's noise voltage then you have the equivalent noise at the input due to the first opamp voltage noise. Multiply that by the total gain and you have, near enough, the output noise as an RMS figure. That should be near enough to get a picture in your mind's eye. I wouldn't use any sim for estimating noise by the way. What I have described is the normal approach. |
H: 12 to 3V DC/DC conversion at 1A
I need to convert 12V input power to 3V (not 3.3). The output current needs to be 1A or better.
My gut-reaction was to get an off-the-shelf DC/DC, but it doesn't look like there are any commercially available that provide 3V at anywhere close to the current output I need.
I also need short circuit (over current) protection on the output side.
Design space of this PCB is limited. It will also be hand-assembled, so SMDs with pads under the part are not an option, through-hole would be preferred.
I don't need a complete solution, just an expert to steer me in the right direction.
Thanks!
AI: In general, you are looking for a buck converter. While many buck converters show applications for 3.3 volts due to the popularity of this voltage, many are in fact adjustable.
As an example take a look at the LTC3824. You will see the app notes show 3.3 volts but take note of the resistive divider on the output connected to VFB. This is what sets the regulated output voltage. Simply adjust the divide ratio and you have a 3 volt regulator capable of sourcing 2 amps.
Take note that this chip, as do others, has current monitoring that can foldback the output voltage if excess current is sensed. This is accomplished via a shunt resistor connected to the SENSE input. |
H: Finding percentage accuracy of instruction pipeline branch predictor
I need help in understanding the solution from solution manual. The question is from the exercise 4.24.4 and 4.24.5 of chapter 4 in the book Computer Organization and Design by Patterson and Hannessey (4th edition). The question is about designing branch predictor.
Branch Outcome Pattern
T,NT,T,T,NT
(T for branch taken, NT for branch not taken)
Question A
Design a predictor that would achieve a perfect accuracy if this pattern is repeated forever. You predictor should be a sequential circuit with one output that provides a prediction (1 for taken, 0 for not taken) and no inputs other than the clock and the control signal that indicates that the instruction is a
conditional branch.
Question B
What is the accuracy of your predictor from question A if it is given a repeating pattern that is the exact opposite of this one?
Solutions given are as follows:
Solution A
The predictor should be an N-bit shift register, where N is the number of branch outcomes in the target pattern. The shift register should be initialized with the pattern itself (0 for NT, 1 for T), and the prediction is always the value in the leftmost bit of the shift register. The register should be shifted after each predicted branch.
Question B
Since the predictor’s output is always the opposite of the actual outcome
of the branch instruction, the accuracy is zero.
Doubt
Now I have doubt with solution to problem B. If we have 5 bit left shift register initialized with the pattern T,NT,T,T,NT (T=1,NT=0) and if the branch outcomes are reverse: NT,T,T,NT,T, then the prediction will be correct for the middle (3rd prediction/branch instruction result) T, right? So the accuracy would be at least 20% (1/5). Then why the book says it will be zero?
I know this is a very simple question, but still am not getting it.... :\
AI: I guess the question meant the exact opposite with respect to the values, not the order. So the opposite of T,NT,T,T,NT in this sense is NT,T,NT,NT,T. Note that this is different from what you thought the opposite. |
H: Change of transconductance for MOSFET
This plot represents drain current versus gate-source voltage (with different ranges of temperature):
Often the transconductance of MOSFET is defined as .
What is the difference between upper equation and gm = Id/Ugs (without changes)?
AI: The transconductance is the ratio of the relative change of current to the relative change of gate-source voltage.
Looking at the black line of your graph, the current at VGS=4.25 is 0.6A and at VGS=4.5V it is about 1.1A.
The transconductance is given by gm=(1.1A-0.6A)/(4.5V-4.25V)=2A/V.
So, it can be used to have a linear expression to calculate the change of current for a change of voltage around the point where the transconductance was calculated. |
H: What do 3-digit reference designators mean?
While looking at many circuit boards and schematics, I often notice that 3-digit designators are used, for example R101 instead of R1.
What do these digits actually mean? The numbers are not in sequence, for example only a single relay, K101 may be present, still numbered 101.
AI: You should be able to figure out it looking at the device as whole. You can find such numbering in the devices with multiple boards, multiple physical or logical blocks in it. First digit may designate block #, other two (usually) designate component # in the block. |
H: Problems understanding a MOSFET Switch
I'm currently dealing with a circuit for communication between a Microcontroller and an e-ink Display. It uses an SPI-Bus. The display is from PervasiveDisplays, and at the bottom of this website http://www.pervasivedisplays.com/kits/ext_kit there is a download containing gerber files for the kit including the circuit for the extension board. The part I'm struggling with is the following:
To me this is a mosfet switch that conducts the microprocessor signal when the gate-source voltage is positive. The problem I'm having is that this input is connected to the drain. If it was connected to source (Drain being output), I could see the MOSFET conduct and put the drain pin on a low-voltage if the source is low, whereas if it was high, it would be pulled to 5V by the resistor. However, it's the other way around and I can't really wrap my head around how this part of the circuit works. Any help is greatly appreciated.
AI: It's a logic level translator. When drain is high Vgs is 0V. The gate is high (3.3V) and the source is pulled high via R25 to 3.3V. When drain is low, the source will also be pulled low by the internal body diode of the MOSFET. As the diode pulls the source low, the MOSFET will turn on (Vgs increases) and ground the source even better. MOSFETs doesn't really care which direction the current flow goes as long as the Vgs threshold is reached.
If needed, the effect of the body diode always conducting in one direction, can be circumvented by placing two MOSFETs "back to back" with their diodes in opposite directions. |
H: device to make a power supply sparks safe
I have an original HP 19V power supply for my laptop. From The first time I bought it I always saw a spark on plug in the wall plug. I notice the same in many ATX power supply.
The problem is that sometimes this sparkles are so huge that switch off the general ower switch of my apartment. Worst problem is the plug damaging. they burn and become black.
I know that this is caused due to the great inductive load of the power supplies.
I always wonder if exist a device or a way to stop this. I thought the easiest way should be install a plug with a switch like this.
but I'm thinking that this only change the victim of the lightening and do not solve the problem.
Maybe using some "step-resistors" that slow down the electric rush by charging the capacitors of the power supply and then disengage from the power line?
I don't even know if what I'm saying has a meaning, sorry but I'm not a pro with power electronics.
Thanks for help.
AI: Electrical plugs and sockets make poor switches, as you have observed. The problem is that they make and break contact very slowly.
A proper switch, such as built into the device you pictured, uses a toggle-action so that at a certain point the switch snaps and quickly closes or opens the contacts. There may still be some arcing but the contacts should be rated for the load being switched.
In Europe switched sockets are common. The socket can be switched off before plug insertion or removal. This will eliminate the arcing you observed. |
H: Are Adafruit RGB 30 LED/m flexi-strips suitable to wear in large numbers?
I am working on a project that involves using a large number of LEDs on a single garment. According to my initial design, I would need as much as 1800 RGB LEDs. I have seen these LED strips from Adafruit https://www.adafruit.com/product/285. Since the strips draws 60mA 600mA per meter, and I would need roughly 60 meters, this amounts to a total of 36 amps. Using them would imply (I guess) either using dangerously high amps or distributing the current by using unacceptable numbers of battery packs. My questions are therefore:
I am correct in my assertion that using the LED strips linked to above is not realistic (i.e. 1800 ish LEDs on a single garment)? Or can I overcome this problem in any other way, as for instance by setting software limits to the PWN controlling the LEDs, thus limiting the current flowing through them?
Are there any other types of LEDs that would be more suitable? They need to be easily controlled by a microcontroller without additional circuitry, and ideally RGB.
(Updated text to show the maximum "current per meter" value, 600mA per meter, quoted on that Adafruit page, instead of the originally written value of 60mA per meter; this change has affected answers.)
AI: Edit: Sam caught an error by the OP which changes my answer.
You will need to supply a total of 36 amps if you turn everything on a once. But this might be a bit boring because you will just look like a large white light bulb. Consider adding a small programmable controller that cycles through various colors and turns sections on and off to make it more interesting. This will also greatly reduce the energy that your battery will need to supply.
To help you with your battery calculations, if you had 1/3 of the LEDs turned on at any one time on the average, you would then only need ~12 amps. Battery life is expressed in amp-hours so a 12 amp-hour battery would power your LEDs for 1 hour. If you used a commonly available 7 Ah battery, this would power them for less than 1/2 hour.
The app notes indicate that the strips will work with 9 volts. If you used a 12 volt battery with a 12 volt to 9 volt buck converter, you could extend your battery life by at least 15%. The LEDs would be slightly dimmer but it may be worth it for the extra life. You could also consider Li Ion cells that add up to a voltage lower than 12 volts.
Do break your LED strings into strips that draw no more than ~4 amps each and put a 5 amp fuse in series with each such strip from the battery to head off a potential clothing fire. Monitor your battery temperature when in use to make sure it is not getting hot. Always charge your battery away from the clothing in a safe fashion.
Have fun with the project. |
H: What happens if you drive the output of a LM78CT05 when the input is not connected
I have a Fairchild LM78M05CT (datasheet) as a 5V regulator for a µC circuit that also has an ICSP port.
The actual circuit is the one in this question: Review of my first ever PCB design for a watering control robot.
I've seen some versions of the 7805 regulator data sheets where the application notes recommend a diode from Vout to Vin in order to protect the device from burning out if the output voltage goes higher than the input. However the datasheet to the Fairchild LM78M05CT doesn't show such a diode in the application notes. Hence I assumed it wasn't necessary with their implementation.
I went to program my device through the ICSP port so I disconnected it from the DC power (Vin to regulator unconnected) and applied +5V and GND to the ICSP header to power the target µC (which drove +5V to Vout of the regulator).
The programming failed, and the target device has a "electronic" smell not present before. Also a LED on the Vin side of the regulator shines indicating a current from Vout to Vin. When I next plugged in the DC cord I got a spark from the ground terminal (and there is a black residue from it on the solder mask) of the DC plug and immediately unplugged the DC cord.
Further I measure (in situ) only 500R or so between the Vout and GND pins on the regulator both ways which is much less than what I would have expected.
So my questions are:
Is my regulator shot?
Did I actually need the diode from Vout to Vin (would that have protected my device) in order to be able to program with ICSP?
Should I have had the DC power connected to the target? I would have to have had 5V and GND between the target and programmer (arduino working as ISP) to ensure matching signal levels so I would have had two +5V drivers.
Any other measurements I should make?
Status Update: Appears that I also bricked the DC adapter, LED doesn't come on and no voltage on terminals. Desoldered the regulator, all pin combinations with polarity measure within tolerance of a new one and it doesn't smell. Applied 5V over ICSP again with DC plug removed and no regulator mounted, smell came back and power LED lit up indicating RL1 is providing 5V on pin 11 in the circuit linked up top. Also the relay smells funny.
AI: It seems rather unlikely to me that this is the cause. The output voltage has to be greater than the reverse breakdown of the B-E junctions in the regulator to cause damage, and 5V is not. There also has to be significant current flowing back (both conditions are necessary for damage to occur).
For example, the TI datasheet specifically says:
I suggest proceeding with caution and looking for something else that may have caused what appears to be significant damage. For example, reversed power connections. |
H: Do I need to replace the transformer when switching from halogen to LED
Total noob question.
I have MR16 halogen downlights and want them replaced with LEDs. Someone told me that, if I simply replace the bulbs but not the transformer, then the power draw will still be the same (or perhaps not as improved as if I get new transformers or replace with newer GU10s with an integrated transformer).
Is this correct? Or nonsense? Am I over simplifying it and more information required?
Thanks.
AI: The answer is, "maybe".
The issue is that MR16 halogens can come in one of 3 voltages - 12, 24, and 120 volts. You need to check to determine what your system uses. Then you must check to make sure your LED bulbs use the same voltage. If not, you have a choice: change your choice of LED bulb, or change the voltage. The latter may involve changing transformer, or possibly eliminating it altogether. It all depends on what you've got. |
H: Buttons and encoder debouncing
I'm trying to figure out the way of debouncing encoders and tact buttons with Schmidt triggers, but I'm confused with 2 existing schematics:
simulate this circuit – Schematic created using CircuitLab
Which of those two options are right?
Values are shown for 5.0V inputs level, would the be the same for 3.3V?
Or for 3.3V I should use another Schmidt trigger IC (not 7414)?
Thanks in advance!
AI: The first is the better option. Notice that the second circuit does not provide the output gate with the intended 0 V ground when the switch is closed. It is supplied a low level voltage through a voltage divider.
However, when the switch is closed in the first circuit, the gate's input is pulled down to ground, but slowly- as the cap discharges. This is what you want.
Edit also read all the comments. I agree with mkeith that, if this is for a microcontroller project, consider denouncing in firmware.
And jonk is right (as he often is) that there may be better, prepackaged, options depending on your needs.
Edit: 3.3 V will work fine with a 7414, provided you supply the IC with 5 V nominal on the VCC pin to power the chip. The resistors are up to you. The values of the resistors and capacitor affects the capacitor's charge/discharge time. Higher values provide more denouncing, but increase the wait time allowed between subsequent triggers. You might also want to add a voltage follower as an output stage.
For Example:
and...
Regarding your 7414 question in the comments: |
H: Does half stepping produce less torque than full stepping in a stepper motor?
Would running a bipolar stepper motor at half step produce less torque but smoother movement than running it at full step?
AI: Properly done (sqrt(2)/2 driven half steps), half stepping gives the same static torque as full step, and slightly higher acceleration without skipping steps, so more effective torque with varying speed. Going all the way to micro-stepping increases the acceleration slightly more. Using 0.5 driven half steps results in less torque. |
H: Can someone help me figure out instinctively whats going on here?
V4 is intentionally left disconnected. I understand that this is an open circuit and current cannot flow. What I'm trying to understand is why does Vout = 5V and not 8.2 V. What key assumption am I missing here?
AI: There is no current through \$R_1\$. Therefore there is no voltage drop across it. Therefore, both leads of \$R_1\$ are at the same potential. Therefore, \$V_{OUT}=5\:\textrm{V}\$ since \$V_1\$ sets the voltage on the left side of \$R_1\$. Note that when you measure \$V_{OUT}\$, it is taken with reference to the "0" node (or ground.)
\$V_3\$ "hangs off" of that point. So the (-) end of it will be \$-3.2\:\textrm{V}\$. Etc. |
H: Why include a small resistance between op-amp stages
I have been reading through TI's reference material on constructing a current source.
http://www.ti.com/lit/an/sboa046/sboa046.pdf
In figure 52 there is a 150ohm resistor included between the output of the instrumentation amplifier and the input of the OPA633 op-amp.
I have seen this also done in some other circuits but don't understand why. What is the purpose of this resistor and how is its value decided.
AI: The OPA633 has a peak in its frequency response at about 200 MHz of nearly 5 dB when the signal source impedance is 50 ohm. If the source impedance is about 300 ohm that peak is about 1 dB. If you look at graphs in the data sheet you will see this on page 3: -
Using 150 ohm appears to be some form of compromise to avoid too much peaking whilst avoiding too much phase shift. Peaking of a circuit within a closed-loop can cause oscillation and too much added phase shift can do the same so, it looks to me like some compromise has been made with the 150 ohm value to optimize the possibility of the circuit NOT turning oscillatory.
The OPA633 also defaults to using 150 ohm when recommending it to be used inside the closed-loop of other circuits. See this in the data sheet: - |
H: Strange drop in voltage on Eneloop AAA-Batteries
I have a device in my flowerbox where I monitor soil moisture.
The device requires about 3.3 v (it's an Arduino pro mini with an ESP8266).
The device is running on 3 x AAA-batteries (should be 3.6 v)
The device runs a cycle of:
Powering on ESP and Sensor
Collecting sensor data
Transmitting it via WiFi
Powering off ESP and Sensor
Sleeping 20 minutes
It's been running fine for almost 75 hours, then it started continuously restarting.
I measured the voltage of the batteries. They were 2.86 v.
Then I waited 5 minutes, measured the voltage. Now they were 3.1 v.
I thought this problem might be caused by low outside temperature. I stuck the batteries in the fridge for 20 minutes, then tested them.
They were 3.1 v.
So question is: How come the batteries voltage change that much?
Best regards
Frederik
AI: They were 2.86 v. Then I waited 5 minutes, measured the voltage. Now they were 3.1 v.
You may be seeing the recovery effect as the energy in the batteries is depleted to a low level. When the load is removed, chemical or physical processes inside the battery (e.g. diffusion) allow it to recover partly.
I would use an analog input to monitor voltage and prevent restart looping.
I would not expect 3 x AAA cells to provide long term energy. At the least I would try to measure power consumption accurately for planning purposes. |
H: Grounding solar controller
I am wiring up the solar system in my van and I need help with claryfing some things up.
I have a Phocos CA08-2.2 controller and in the manual one can read:
Be aware that the positive terminals of the controller are connected internally and therefore have the same electrical potential. If any grounding is required, always do this on the positive wire.
Right, but there is a remark later on that actually corresponds to my situation:
If the device is used in a vehicle which has the battery negative on the chassis, loads connected to the regulator must not have an electric connection t the car body. Otherwise the Low Voltage Disconnect function and the electronic fuse function of the controller are short circuited.
So clearly I am missing something because it sounds contradictory to me. In the very first paragraph it says that "any grounding should be done on the positive wire" but later on it says "if you have your battery grounded on the negative side, then...". How should I understand this?
I figured the following circuit and after reading the manual about grounding the positive, I got my concern.
(I skipped the controller loads that are not grounded, and solar connectors in the solar controller for the clarity.)
So all in all - I have my 12V battery connected to my van chassis. Is it safe to connect the solar controller such that the controller negative connector will be grounded because the battery is grounded on the negative?
AI: I agree with you, there seems to be a bit of confusion.
But I think what they mean is, that in the case where the batter "-" terminal is connected to the chassis, thus grounded, then this should be the only point of grounding in the system.
That means the solar panel and the load(s) you have on the controller are not allowed to have electrical contact with the chassis, with neither of their two terminals.
So the conclusion is: I think you can indeed use the circuit you have in mind, with the battery negative connector grounded, but with the condition that the neither the panel nor the loads are grounded (have electrical contact with the car chassis). |
H: Why does the voltage double in a doubler-circuit?
I’m having a hard time understanding how the final voltage is achieved.
I understand that in the first half of the sine wave the first capacitor gets charged to 1V.
In the 2nd half the first one is already charged so the 2nd capacitor gets charged to 1V as well.
But how does this account to 2V?
Looking at it, it seems like these capacitors are in parallel, so I’d expect them to only provide 1V.
Where’s my mistake?
AI: The output of the transformer (A) is a waveform that rises up to some peak positive value and down to some peak negative value relative to 0 volts. Let's say it rises to +100 volts and drops to -100 volts (200 volts peak to peak). Typically, the frequency would be 1 kHz to hundreds of kHz and this means that the capacitor closest to the transformer will pass this peak-to-peak voltage virtually without attenuation.
However, passing this AC voltage without loss doesn't mean that it will still be + and - 100 volts at point (B) - the capacitor blocks DC so point (B) could be raised or lowered by some DC amount with the same AC content superimposed.
However, at point (B) the negative excursions are trapped and restricted (by the diode) to being no more negative than -0.7 volts. This means that the voltage at point (B) now only moves between -0.7 volts and +199.3 volts i.e. there is still the same peak-to-peak voltage but virtually the whole range is now positive.
This is the hard bit done because what feeds (C) is a diode peak detector and at (C) is smoothing capacitor thus, you will get a DC peak voltage of about 199.3 volts minus another 0.7 volts making it 198.6 volts. If you take current from (C) there will be a ripple voltage superimposed on the DC output just like any conventional half wave power supply rectifier circuit. |
H: Reducing voltage in power supply with diodes
Hello there i am building a variable power supply as a summer project. I have ordered some parts, and want to begin building it. I start with the transformer power supply:
I've ordered a transformer (230 V - 36 V) and i expected it to give a output voltage of 36 V peak (silly me), and then found out the output will be in RMS, i. e. an peak outout of 50,92 V. I can work with this, but the capacitors i've orderes is only rated at 50 V, so this will give me some smoking problems i guess. The rectifying diodes will give a voltage drop of minimum 0,6 V, which is not quite enough to go under 50 V.
Is it possible to just use two diodes extra? As seen in the simulation:
The diodes added is of course D5 and D6. Their only purpose is to lower the voltage, to under 50 V. This will theoretical be enough.
But will i need to go lower in voltage, as the overhead on the WV of the capacitor is too small?
And is the "extra diodes" way even a viable way of doing this?
AI: As a rule of thumb, it is always desirable to select capacitors which are 1.5 times (2times is even better) the desired voltage. So, in short, the answer is NO. Theoretically(ideally) it should, but practically it will be a bad design. An Even slight increase in input voltage or any small variation in the circuit than ideal and you might see the circuit smoked. |
H: Can I run a stepper motor rated 12V with a 24V or above power supply?
We are using this stepper motor (https://www.sparkfun.com/products/9238) in our project. In its datasheet, the motor rated voltage is mentioned as 12V.
How is motor rated voltage different from max Voltage?
How can we determine the max voltage of this motor since it is not specified?
Can we use a supply of 24V instead of 12V and make sure that the current remains the same in order to increase the speed of the motor?
AI: The 12 V rating probably refers to the voltage you can apply to a coil indefinitely. This limit is due to the heating it causes inside the motor.
You should be able to apply 24 for short periods of time without any problems. It would be fine, for example, to drive the coils with 24 V controlled by PWM. In that case, the PWM duty cycle should not exceed 50% on the long term.
Keep in mind that in steady state, the power dissipation goes with the square of the voltage. Applying 24 V to the motor will cause 4x the heating it can handle. It can be legitimate to do that for short periods of time, but then you have to ensure that the total average heating doesn't exceed what continuous 12 V would cause. The time for this average can be a few seconds to a minute or more, depending on the size of the motor.
I once did a motor controller where the processor modeled the motor temperature internally. When the motor was cold, it could be overdriven for short periods of time. The controller kept track of the energy heating the motor, and assumed some decay to ambient. It prevented driving the motor to the point of overheating. |
H: USB Descriptor: iInterface and iFunction - is there a point?
In a USB Configuration Descriptor you have the option of providing iFunction and iInterface string descriptor indexes to Function Descriptors and Interface Descriptors respectively.
Is there any point to implementing these descriptors? Adding them, I can see the strings I assign in lsusb -v -d vid:pid for the device, but that is the only place I seem to be able to see them at the moment.
Are these string descriptors used anywhere else? Maybe in another operating system?
Yes, I guess I could get at them with libusb in my own software, but that's besides the point - I'm looking at the OS level - is there anything (besides the odd diagnostic tool) that would make use of them?
BTW: this is for a composite device with a number of interfaces of different types.
AI: From: link
Permits distinguishing between two /dev/videoX entries from the same
physical UVC device (that naturally share the same iProduct name).
This change matches current Windows behavior by prioritizing iFunction
over iInterface, but unlike Windows it displays both iProduct and
iFunction/iInterface strings when both are available. |
H: Atmega32A and relay module cant turning water pump on
I have some problem with water pump.
So I use atmega32a and 5v relay module to turn water pump on for 5 second, but in reality the water pump can't last for 5 second. Just after water pump goes on, atmega32 has a strange behavior that makes another component like servo move randomly.
Components I use:
Atmega32A
Relay module 5v
Submersible water pump 5v
Micro Servo 4.8v
I already tried this:
Separating power supply
Change relay with transistor
Change microcontroller
Here is the schematic & relay:
I really need your help to solve this problem, and I'm sorry for my bad English. Thank you :)
AI: It looks like you are using a 5 volt submersible water pump but you haven't used a flyback diode so this is something to add - it will prevent a large back-emf when the pump turns off (the relay contact opens).
However, it is likely that your main problem might be a dip in the power supply voltage when the pump is activated. This might be a very small and transient dip but, it could cause t your micro to reset because it looks like it is sharing the same 5 volt power line.
Add decoupling capacitors and, in particular add one close to the relay pump circuit so that any sudden impulse of current is largely kept local to that part of the circuit.
Bread-boarding techniques and bad circuit layout can also contribute to this problem. |
H: How to change pad size in design rule in Eagle?
I’m very new to Eagle. When learned to create a library I followed these instructions from Eagle’s documentation tutorial:
“Default value for pad diameter is auto. It should not be changed. The final diameter in the layout results from the values given in the Design Rules.”
See the highlighted part in the picture!
After creating the library I’ve done a schematic and then its board, and I tried to change the pad size in board editor but nothing happend, and as far as I know there is no settings for pad size in design rule.
How do I change the pad size in board editor as Eagle says in its tutorial document ?
Please Help!
AI: Not clear from your question and from documentation - you created device in library editor, then to change this device's pad properties you need to go to library editor, do it there and then update device in the circuit/board; you will not be able to make change to pad which is a part of the device in board editor.
When in library editor, you can change diameter of the pad using info tool in the GUI.
The related design rules are located in the DRC -> Restring tab, where you can set pads' minimum and maximum diameter, and percentage of their drill diameter.
The recommendation It should not be changed prevents you making a lot of custom pads and vias which do not comply with settings on that DRC tab, and thus getting a lot of errors which you will have to manually correct in the library editor, or will need to change settings in DRC rules. For advanced EAGLE user it would be ok, but if you are a beginner, you will get lost and hate the tool at the end.
Let's look at what "pad diameter" is. It is a space to put solder into to fixate the component's pin. Default design rules are defined the way so that you would be more or less easily solder the pins, contact would be good and part would be fixed on the board well.
Why you may need to change the pad diameter from auto (in other words - from default rules' settings)?
To make its solderable area bigger (most probably not smaller!). I used to do it, it is normal, but as I said if there will be something wrong in the layout at the board level you will get an error.
To open some space on the solder mask - do not forget that pad is an area uncovered by the mask.
How to avoid making big "non-standard" pads? Just cover pads with polygon with the same signal name, for example, ground pad may be size of auto, but when you cover it with polygon they merge forming big conductive area - this way you have only small pad's copper area exposed for soldering, and remaining space covered by the solder mask. |
H: What is this white circle on this circuit?
I found this circuit inside a LED reflector that turns on and off at regular intervals, and there is this white circle in it. I am guessing that it somehow keeps time of the intervsle, but can someone explain it to me? Also, I found this same circle on another device with 3 LEDs that turn on and off in regular intervals.
AI: It appears to be a Glop Top or Chip On Board (COB) type of assembly. However a black material is usually used:
Instead of packaging the integrated circuit into a leaded case, the actual integrated circuit is placed on the Printed Circuit Board (PCB) and bonded out:
Go here for more about COB and the source of these images. |
H: Current limiter - charging battery from battery
I'm planning to charge auxiliary battery from car's alternator, by using voltage sensitive relay to connect positive poles of car's battery.
The only obstacle now is that auxiliary battery (12V) allows maximum charging current of 20A. Therefore I need current limiter in series with relay.
Which is the most simple circuit that I can use?
I've read that people use light bulbs due to their non-linear behaviour. In that case how can I calculate the needed wattage of (12V) lightbulb, to limit current to 20A?
EDIT: In other words I need 12V lead-acid battery charger that gets power from another 12V lead-acid battery with charging limit of 20A
EDIT:
System info:
Car battery: 100Ah 760A start current - regular lead-acid car battery
Auxiliary battery: 100Ah (C20), max charging current: 20A, 500A/ 5s start current - cyclical solar battery
Wires: 2x 10mm^2, cca. 2x 3m, copper
AI: This is not easy as a linear limiter and better with a PWM limiter and LC filter.
If the 12Vaux is fully discharged and requires (14.2-11.2)V*20A=60W series load dump . That is a lot of heat before the Imax reduces to a CV equal voltage to the alternator.
If one considers a light bulb a good solution , this is effectively a PTC almost constant current source when used in the 0 to 10% voltage range defined by DCR which is 10% of rated voltage R. e.g. 12V/20A where filament resistance drops to 10% of DCR at rated voltage at 3200'K Thus 3V*20A=60W , with a DCR of 0.15 ohms then at 10V and 10% DCR the bulb may be something like 30V 1.5Ohms or 45W which is really not a practical value. It might be something like to 12V 70W car headlamps in series/parallel unless you want an AUX light when charging.
So ok in lower currents, but not 20A. This is just a ballpark estimate, not a rigorous calc.
A better solution is PWM with a series choke rated for 20A such as those air coils found in ATX PSU's and a fast switch rate with a MOSFET switch rated for 50A ( hi side or low side. ) THe battery acts as a 10kfarad capacitor with some ESR from 5 mOhm to 1 Ohm when dead. but an RF cap will reduce EMI.
Another way is to just use a fixed 150 mOhm 60W heater wire (NiChrome) and use that to keep your coffee cup warm ;) Or use a Cap Pulse discharger rated for high RMS ripple current ( several large plastic caps in parallel) This is essentially an active SMPS current limiter with a 0.1V max drop at CV mode at 14.2V or 0.1V/20A = 5 milliohm (MOSFET + choke DCR)
I said to do this right is not easy and stay within limits.
Ultimately the simplest solution is get a bigger battery rated for the current of the alternator, but then you may end up blowing your Alternator diode bridge with both batteries are weak and drawing max current of alternator for long periods at 180'C junction temp.
You can look at NTC disc surge limiters, but they cannot protect a battery as they are designed to protect perhaps only 0.5Farad and not a battery with 10k~100K Farads. To understand read , https://www.digikey.com/en/ptm/a/amphenol-advanced-sensors/cl-series-inrush-current-limiters/tutorial. To dump 40 Watts of heat it either has to run extremely hot or be big like headlamp bulbs ( which will get hot and must be sealed from moisture). |
H: Current limit of switch mode power supply
I have a premade SMPS, it is 12V and maxes out at 8.5A. I have a TEC1-12706 on one pair of terminals and there will be a fan on the other terminals (I only have a total of 4 terminals for out). I would like to limit the current for the fan. How would this be done?
The PWM controller inside of the SMPS is the OB2269AP along with a coupler ic.
AI: As outlined in some of the comments, there are several ways to limit the current. One of them is to force a constant-current operation (CC) directly at the secondary side. Your SMPS is naturally designed to operate in constant-voltage (CV) mode and goes into short circuit protection if the feedback information is lost in the primary side (the opto collector is open loop). A simple way to limit the output current and force a CC operation is to break the ground path in the secondary side and insert a shunt resistance (\$R_3\$) in the below schematic:
Then the voltage developed across the shunt drives a bipolar transistor \$Q_1\$. When this voltage reaches the transistor \$V_{be}\$, it will start pulling the optocoupler LED to ground, forcing current to maintain a "constant" drop across the shunt. I say "constant" because the gain is very poor and it is a cheap system. If we consider a 0.65-V \$V_{be}\$ voltage at 25 °C, then for a 1-A constant current, \$R_3=\frac{0.65}{1}=0.65\;\Omega\$. \$R_2\$ can be dropped (unless you want to fine-tune the CC point) and \$R_1\$ typical value is \$100\;\Omega\$. In normal operation, when the output current is below the CC value, \$Q_1\$ is blocked and the added block is silent: the SMPS operates normally in CV. As the output current increases, the two loops being ORed, the CC takes the lead while the CV gives up and stop pulling LED current as \$V_{out}\$ goes down. As I said, it was a cheap way of doing CC in cell-phones chargers at the time Nokia was still number one, so quite some time ago : ) but it did the job well.
If I misunderstood the question and you actually want individual dc branches with their own current limit, then you have to go for a linear active series shunt solution but with the penalty of extra power dissipation when the current limit is reached.
Good luck with the mod and carefully test your changes with an isolated current-limited dc power source set at the lowest input voltage to limit damages in case something goes wrong. |
H: Variable DC Supply "Ignoring" Potentiometer
I printed a board meant to supply a variable 5–250VDC from this open-source schematic:
This was my board
with this schematic (excuse my silly device symbols)
but I can't get the output voltage to vary. With the 5k potentiometer (the box at the bottom center of the board), it measured ~320V, invariant of the pot's setting. For a 10k pot, it measured the same voltage, again invariant.
Here's the soldered print (excuse the poor quality, someone else has them currently):
Is something shorting? Is it somehow "skipping" the potentiometer? Is there anything obviously incorrect or any likely errors I could test for?
AI: Unfortunately, this PCB is unlikely to ever work. You will need to redesign it.
Switching power supply designs are very sensitive to layout. In particular, it's critical to minimize the area and inductance of the current loop. Unfortunately, your design places the inductor rather arbitrarily, which leaves the current loop all over the place. For more information on these aspects of design, you may want to read some application notes on switching PSU design, such as:
Linear Technology: PCB Layout Considerations for Non-Isolated Switching Power Supplies
TI: Five Steps to a Good PCB Layout of a Boost Converter
I also see a couple of generic issues with your PCB design, including:
Your layout is not making use of its ground pour. A good ground is critical for all circuits, but is particularly important for a power supply design.
One of the terminals of C1 appears to be unrouted.
As other users have mentioned, it looks like you may have chosen some parts which are not rated for the voltages you're trying to produce.
You are using rather arbitrary packages and footprints for a lot of parts. For instance, you're using a Kelvin resistor footprint for RSENSE1, but ignoring the kelvin terminals, and it almost looks like you've soldered an electrolytic capacitor to a SMD capacitor footprint on the back side. Switching power supplies are sensitive to lead inductance; these choices matter! |
H: What makes a good soldering station
I am a hobbyist looking to buy a soldering station. The market for soldering equipment ranges from $10 to $1000.
This begs the question:
* What is the reason behind these huge price differences?
* What attributes should I look for in a good soldering station?
I researched a bit and found out that a good soldering iron should be temperature controlled and should strive to pump enough wattage to maintain the desired temperature while the soldering iron is tranferring heat to the component it's heating.
Additionally it should be a brand that has easy to find replacement tips.
Surely there must be other things that I am unaware of. Please shed some light on these unknown unknowns.
AI: You don't need a $1000 iron unless you're going to be using it all day, every day, for a living, for the next 40 years. Same as any other tool. $100 will get you a good iron that will last you a long time (if you take care of it) and will do any hobby-level job.
Adjustable temperature is nice for versatility, but not necessary. You can overcome a fixed-temperature iron's limitations with technique and practice.
Personally I spent the first few years going through inexpensive (sub-$40) irons before finally setting aside $100 for a Hakko station. Things I noticed:
The control unit and soldering base were much sturdier/heavier, and didn't move around on me
The iron's handle didn't get hot after 30 minutes of use
Being able to adjust the temperature is nice when dealing with different heating needs
The iron heated up much faster
The iron sat with greater stability in its holder than previous ones
The included list of useable tips had like 80 varieties. I haven't used any of them yet, but I know they're available if I need them
You can, of course, get by with inexpensive tools depending on your level of use and the types of projects you're doing. Many projects are absolutely doable with a $20 35-watt iron. To me, $100 seems just about right for a tool I use reasonably often at a hobby level.
Edit per recommendations from comments:
Temperature control (not to be confused with adjustable temp) makes it much easier to get consistent solder joints. Temperature control means there is temperature feedback, so the iron tip will maintain its temperature when dumping heat into the joint (up to the power limit of the iron). This is especially useful when soldering to something that sheds heat quickly (like a ground plate).
Having enough wattage is key to good joints. If the iron doesn't have enough power to adequately heat the entire joint, you'll get "cold joints" which don't conduct well. At best they're annoying and give unreliable behavior. At worst they can be a fire hazard (they act like a resistor and can get very hot).
I submit that 35-40W is ample power for small projects that only involve small-gauge wire and component leads. I've used a 35W iron for things like swapping guitar pickups and little circuits you assemble on perfboard with 20AWG wire. More wattage is generally not a bad thing, as you'll generally end up with an adjustable-temp station above probably 60W or so (my Hakko is 70W). For soldering to a big hunk of metal (like a large grounding plate or block), you may eventually need a 100-150W gun.
I certainly haven't taken a survey of every available option, nor used them all, so as always YMMV. |
H: MOSFET Basic Biasing Problems
After a lot of theoretical studying of MOSFETs, I decided to try out at least the basics of it in practice.
Here is the first circuit I ever made using MOSFET:
simulate this circuit – Schematic created using CircuitLab
https://www.onsemi.com/pub/Collateral/BS170-D.PDF
I haven't used resistor in series with gate, because MOSFET is obviously voltage driven component (input gate current is negligible).
I chosen next parameters for the circuit:
Id(sat) = 10mA
Id(bias) = 5mA; where Vds = 5V
Vr1 = 5V
Rd = Vr1/Id(bias)
Then I didn't really knew which way to calculate Vgs (since the actual gate current can't be really calculated), so I tried next:
gm = Id/Vds, then Vgs = Id/gm = 5V,
After I made the circuit on breadboard, this are the values I got with measurement:
if Vgs = 1.8V or less, then Vds = Vcc
if Vgs = 2.4V, then Vds = Vcc/2 (I wasn't even close to this value...)
if Vgs = 3.8V or more, then Vds = 0
Where did I go wrong on this one?
AI: You should look more closely at the data sheet. Go to page 2, and about the 3rd item is gate threshold voltage. This is defined as the gate drive necessary to produce 1 ma drain current, and is specified to be in the range of 0.8 to 3 volts, with a typical 2.0.
In your breadboard circuit, in order to get Vcc/2 (5 volts) at the drain, you need 5 mA of drain current. Since a typical current at 2 volts is 1 mA, 5 mA at 2.4 volts seems perfectly reasonable. At voltages much below this the drain current will be right up next to 10 volts, since there is no current being drawn, and for much greater voltage the FET will be turned on hard and Vd will be near zero.
The only thing you really did wrong was computing gm. You seem to lost sight of the fact that this is your desired gm, not what you can actually expect. Go back to the data sheet, and you'll only find one gm value, specifically 0.2. However, notice the operating conditions - Vg is 10 volts and id is 250 mA, which is far, far away from the actual conditions you provided, so you can't trust it for what you were doing. |
H: Power output for Sager HVGC 65
I wish to get 12VDC @ 4 amps from Sager HVGC 65 Watt power supply. Datasheet: https://www.sager.com/_resources/pdfs/product/HVGC-65.pdf
I have attached a picture of the datasheet. I would like to use the HVGC-65-1050. I am confused because it says current adjustment is only from 650 - 1050 mA. But at 12 VDC I should be able to pull atleast 5 Amps right?
AI: It doesn't look like it works as you suppose.
At max voltage we get \$ P = VI = 62 \times 1050m = 65\; W \$ as specified.
It appears that the output is rated at 1050 mA maximum so the highest power you can get at 12 V is \$ P = VI = 12 \times 1050m = 12.6 \; W \$. (1/5 of the power at 1/5 of the output voltage.
This disappointing news can be explained somewhat by a look at the block diagram:
Figure 1. The SMPS block diagram.
Notice that the area in the green rectangle has to be designed to carry a certain current. In this case it is designed to carry a little over 1 A max on the low-voltage side. For it to work the way you expected it would have to carry 5 A in the transformer secondary, the diodes, etc.
The supply is rated as a constant current design and it's main market will be LEDs - with the possibility of dimming by adjusting the control input.
Note that the voltage is not adjustable - only the current. The voltage will rise and fall as required to maintain the set current.
These PSUs have an interesting dimmer control input.
Built-in 3 in 1 dimming function, IP67 rated. Output constant current level can be adjusted through output cable by connecting a resistance or
0 ~ 10Vdc or 10V PWM signal between DIM+ and DIM-.
To do this the controller input is probably something similar to the circuit below.
Figure 2. PWM controller input.
For DC voltage control we just apply the voltage, it gets to the controller with a slight lag depending on the R3/C1 delay and output power is set.
For PWM a pulse train would be used as shown in Figure 3. This time R3 and C1 filter the PWM to obtain the average DC value. Output power is set as before.
Figure 3. PWM signal transitioning from high pulse width (75%) to low (25%) and back again. Note amplitude remains constant.
I don’t know for sure, but if the PSU is able to sense a resistance connected to the input then it must be supplying a current to the input terminals as shown by the constant current source. We know that 100 kΩ gives full brightness so that means the voltage drop across the 100 kΩ is 10 V and I = V/R = 10/100k = 0.1 mA.
This theory is supported by the fact that if you use one pot to control multiple fittings that the required pot value is 100/n where n is the number of lamps. This makes sense as each PSU will drive 0.1 mA into the pot. So for five lamps in parallel on the one pot R = V/I = 10/0.5m = 2 kΩ.
• Finally, if nothing is connected the 0.1 mA will charge C1 to 10 V and give 100% brightness.
It’s simple and flexible. |
H: Choosing the right Schottky for the voltage/amperage
What do these ratings mean when it comes to reading datasheets for Schottky and how do you pick a diode based on these ratings:
Current, Forward, Continuous
Voltage, Repetitive Peak
Voltage, Forward
For example, I have a 12V / 3A line that i would like to put a Schottky diode on to prevent backflow but have no idea what any of the above ratings have to be so that i get the right one.
AI: Forward continuous current is what you might regard as the maximum current you are allowed to pass through the diode continuously given what the data sheet says (small print) about heat sinking requirements. This is a thermal specification as opposed to peak forward current (sometimes ten times higher than continuous) which is a "fusing" type of specification.
Repetitive peak inverse voltage is the blocking voltage that the diode can withstand when reverse biased.
Forward voltage is the volt drop across the diode when forward biased and is always specified at a forward current. For a Schottky diode this can be in the range 0.3 to 0.6 volts typically. |
H: Why does mechanical positioning of a cable has any influence on the function of electr(on)ic devices?
I think everyone has encountered this problem:
A device does not work anymore properly until you find out by coincidence a position in which it still works. For example my USB microphone has a faulty connection until I lift the cable slightly...then it works again.
I have now always assumed that it is a problem with solder joints and the repositioning reconnects the device. The problem is: After I bought a soldering kit some time ago (it is out of order, by the way :( ) I opened the microphone and was surprised that everything looked fine. The solder joints look solid and the cable itself did not show any marks or bends. Electrons should be kind of disinterested how the form of a cable or a connection is, so I am a bit confounded.
I suspect this question is a duplicate, but I did not find anything in this way.
What exactly are those problems which cause that devices/cables work only in specific positions (from most frequent to least frequent) and how can I fix them?
AI: This is a common occurrence in cables which are old and well used, as they go bad. The center conductor can become frayed or broken at a spot inside the jacket, and moving the cable around can make or break the connection.
This often happens at the connector, but doesn't have to. You may also find the foil, or ground of the cable intermittently making contact with the center conductor, creating a short. |
H: PCB surface mount alternative for JST-XH connector
I am having a very hard time finding an answer to this, most likely because it is rather difficult to search for items without knowing the acronyms or names for the connector types.
I have a PCB that contains what I believe to be a JST-XH connector with an accompanying female cable:
What I would like to do is make a new PCB that would contain the female connector so the boards would essentially plug into each other - stacking the boards. My problem is that I am having a very difficult time finding a female PCB surface mount JST (or whatever it is) connector.
Does anyone know if such a thing exists? If not, I see a few alternative options:
Remove the connector on the board and replace it with something like this
however removing existing components is typically a tricky business so
this is not ideal..
Connect the two PCBs with standoffs and use ribbon cables between the connectors; or
Strip away the wire housing and treat the wire like a pin to solder to the second board, which is less than ideal and will most likely be unreasonably messy.
Does anyone have any ideas for a scenario like this?
AI: YOu are not likely to find JST makes a female mezzanine connector in either SMT or THT that matches this Board-to-Wire male Jack with offset shroud and interlocks designed for cable for plugs.
just use a THT header socket that fits.
https://www.digikey.com/products/en?mpart=929974-01-08-RK&v=19 |
H: Recommended isolation paper for hobby circuitry
I'm having a hard time finding out what type of insulation is enough for my projects.
I will be building this: STM32 OLED Soldering Station Case Temperature Controller T12 KR BL ILS BC2 Handle Electronic Soldering Iron Tips 220v 70W
I'm wondering what type of insulation paper is needed for the metallic body, to protect the internal transformer and other bits.
This video shows what I'm going to be doing: https://youtu.be/25rww-pXqr0?t=5m39s
He states any insulating material is enough (0.3mm?). i'm wondering what is best bang for the buck so to speak. Looking at Aliexpress/Alibaba that is. I'm also planning to use the paper in a few other projects!
AI: As shown, that kit is a death trap waiting to kill an inexperienced buyer. Like you. You do not need paper. You will need insulated hookup wire to make connections. You will also need 4 longer screws in order to mount the power supply. Drill the bottom of the case to fit the mounting holes on the supply. Purchase 4 standoffs about 1/4 inch long which have a center hole (either threaded or not) which fits the screws, along with 4 nuts and lock washers which fit the screws. Push the screws through the bottom of the casing. Put a standoff on each (using threaded standoffs will make this easier, since you can screw them on and not have to worry about the screws falling out). Then put the supply board on the screws, then use the nuts/lock washers to hold the supply in place. Tighten everything down. The standoffs will hold the underside of the supply away from the metal case and no insulation will be needed. |
H: Help understanding current direction with transistor?
I'm an electronics hobbyist struggling with the concept of transistors. I made a simple circuit with a PNP transistor based off of a diagram, but don't understand how it works.
From the book containing the diagram:
Look carefully at schematic I. You see there are two paths for the current: one from base to emitter and another from the collector to emitter. The transistor is a PNP type. Notice the base to emitter path is open until you press the key.
What I dont get is the direction of current flow. I read that current has to flow against the direction of the arrow, but looking at the circuit, how could the LED light up if electrons can only pass the other direction? (opposite the blue arrow?)
Also I would really appreciate an explanation of the values for power sources (3V and 6V); On the board, V6 is 9 Volts while v4 is 6 volts so I'm unclear how they arrived at those values. Really appreciate your help; I've spent hours researching this and still am baffled.
Here's a picture of my circuit board:
AI: What I dont get is the direction of current flow. I read that current has to flow against the direction of the arrow, but looking at the circuit, how could the LED light up if electrons can only pass the other direction? (opposite the blue arrow?)
You are confusing conventional current flow and electron flow. In this circuit, when the switch is closed, electrons flow will be in the direction of your blue arrow, meaning that conventional current will be flowing opposite the direction of the arrow, exactly as required to light the LED.
Probably your book is trying to teach things in terms of electron flow rather than conventional current flow. In any case, they have drawn their schematic with entirely different conventions than the rest of the world uses. Throw that book away and get one that will teach electronics in a way that will allow you to communicate with other electronics hobbyists and electrical engineers. |
H: lower current than expected through simple circuit powering laser diode
I have a circuit I designed for a raspberry pi to turn a laser off and on. I designed it to supply approx 22mA but it's barely getting 11mA. I've tried numerous tests but I can't figure out what's wrong
Circuit:
Vcc = 5V, GPIO BCM 21 = 3.3V, voltage loss across laser (L1.V) = 4V
For that circuit, I tried to get R2 with the eq
$$Vcc−I2R2−L1.V−Q1.V_{CE}=0 \\
R2 = \frac{1}{22mA} (5V - 0.02V (negligible) - 4V)$$
which gave me $$R2 \approx 49\Omega$$
Using that, I've tested these values
$$
\begin{align}
I_{R2} &\approx 10.8mA (!) \\
I_{R1} &\approx 1.8mA \\
L1.V &\approx 4.6V \\
Q1.V_{CE} &\approx 0.02V \\
\end{align}
$$
edit updated L1.V from 4 to 4.6
update the laser isn't just a simple diode, it has integrated APC
Datasheets
2n4401 npn bjt
syd1230 laser (link is to HML1230 docs because SYD1230 docs do not appear to exist and laser closer resembles HML1230 in appearance, ie. has focusable
acrylic lens)
I've looked over all the data sheets (the whopping 2 of them), rechecked calcs, design... I can't figure out what I didn't account for. Why isn't \$I_{R2}\$ closer to the expected 22mA
AI: Did you check the power supply from which you power the LASER branch? Summing all the voltage drops in that branch gives
0.02V + 4V + 12mA × 49Ω = 4.6V, which are not the 5V you should have.
It seems that your power supply cannot handle so much current and is dropping out.
Check that your power source for that branch is not overloaded by other loads in your circuit, since 12mA seems a bit too low to overload a power supply, even small on-board regulators.
EDIT
After you corrected your question and added the LASER's datasheet it becomes apparent that you are outside of that LASER's specs. Just for a tiny 0.1V, now, but you could have damaged your laser during power-up when the whole 5V could have been fed to the LASER module.
That module expects 4V constant voltage supply (±0.5V max). Connect the module to a bench power supply set at exactly 4V and try if it's still working. If it is, then follow the advice in a comment of @sstobbe: get rid of R2 and put a diode in series with the module.
For extra safety, since at low currents diodes have a smaller voltage drop and so a 0.7V drop cannot be guaranteed (e.g., it could be 0.56V, depending on the diode actual model), it would be better to put two diodes in series, at least initially, just to be sure you work on the lower side of the specs.
After you measure the actual voltage drop across those diodes you might remove one and see if it works still in specs (but if it works with two diodes, maybe it's better to leave it that way, just in case your power rail has a small hiccup). |
H: An alternative to a regular glass fuse
I develop a home sprinkler controller, and would need a fuse. Traditional fuses (those glass cylinders with a wire inside) take too much space, and don't play well with SMD. My research found some poly fuses, including SMD poly fuses. However they seem to be for a small voltage (6-30V), and it is unclear whether they will burn if overvoltaged (think main connection or lighting strike). It is also unclear whether they can handle significant inductive load such as sprinkler solenoid connected via a long cable.
What are the honorable community members using nowadays in lieu of glass fuses? Is there a better alternative to the "traditional" glass fuse?
AI: Traditional fuses ... don't play well with SMD.
You're misinformed. There are plenty of SMD fuse holders available for that kind of fuse.
There are also smaller fuses with similar behavior (i.e. disposable rather than resettable like a polyfuse) available with SMD holders.
However they seem to be for a small voltage (6-30V),
Fundamentally, for a fuse to be able to withstand hundreds of volts in its fused (broken) state, there must be a certain distance between its terminals. There is a fundamental trade off between size and voltage rating here. So you need to decide your actual voltage requirement and then look for the smallest fuse that will support it.
it is unclear whether they will burn if overvoltaged (think main connection or lighting strike)
You should not be using a fuse to protect against over-voltage. Fuses protect against over-current, not over-voltage. And you should not expect a fuse to protect the fused device either --- the fuse is there to prevent over-heating from causing a fire. It is not intended to prevent damage to the fused device.
It is also unclear whether they can handle significant inductive load such as sprinkler solenoid connected via a long cable.
Typically inductive loads don't cause high currents, so they wouldn't be likely to overload a fuse. (You could, of course, contrive a circuit where an inductive load generates a high current. I'm speaking of typical scenarios)
They may cause high voltages when switched quickly. You should use an over-voltage protection device or a free-wheel device to protect against this. Not a fuse. |
H: Chip antenna PCB layouts
I am designing a layout for this Chip antenna "https://www.johansontechnology.com/datasheets/antennas/2450AT18B100.pdf". In the datasheet they have Test Board as an example layout. I have a couple of questions on their test board. As shown in the below layout:
I am not sure what no ground means here?
I know that the antenna pin with the brown mark is connected to the inductor, but what is the other antenna pin connected to?
Should the ground plane be exactly 46mm? What if I need it to be smaller?
The feed line should be 50ohm, can I change the line width, and decrease the 19mm length and keep the impedance the same?
Thanks in advance.
AI: No ground plane. Remove all copper from that area to the limits given in the datasheet. This is to allow the antenna to radiate properly.
Second antenna pin is connected to ... Nothing. It there just to hold the antenna in place.
You don't have to have a board the same size, but it is probably better to be larger than smaller - and keep other (large) components away from the antenna.
You can change the line width, but you need to use a line impedance calculator to make sure you get back to 50 Ohms impedance. The length nothing to do with the impedance. You probably do need to stay 19mm away from the antenna with the RF connector. It will interfere with the radiation pattern if you get it too close. |
H: Controlling power with Isolated DC-DC Converter remote on off vs Relay
I am designing a simple PCB with multiple DC to DC converters. One in particular is an isolated DC to DC converter with a remote on / off pin (to turn the converter on and off).
Currently i have in mind of connecting that converter to a general purpose or SS relay, controlled by arduino microcontroller chip, Atmega 328P. This will be used to open and close a mechanical/solenoid lock.
However, is it a good idea to skip the relay, and wire up the chip directly to the remote on/off of the converter? Or is it better to keep the relay? is the remote on/off pin of the converter isolated? Thanks.
Attached is the datasheet to the converter
http://docs-asia.electrocomponents.com/webdocs/0aca/0900766b80acafda.pdf
AI: However, is it a good idea to skip the relay, and wire up the chip
directly to the remote on/off of the converter?
I consider this as a contender - the on/off pin isn't isolated from the input end of the Traco circuit so if you need to control this with some isolation I'd consider using an opto-isolator. Please also make sure that when you deactivate the Traco, any back-emf from the solenoid doesn't try to push current back into the output pins to any great extent. You could use a zener diode across the output of the Traco to largely prevent this from happening. Choose a zener voltage that is a couple of volts higher than what the output voltage from the Traco is. |
H: Is connecting a 12V circuit to +12V and GND the same as connecting it to GND and -12V?
After reading What is negative voltage? and the answer given by kellenjb I wondered if I understood the topic sufficiently, and started theorizing. If the + or - sign is just a convention and the voltage is the potential difference to ground, and my PSU has a +12V, -12V and GND connector, would it make no difference wether I connected a circuit to +12V and GND or to GND and -12V? In other words, say I had a 12V LED, could I connect the anode to GND and the cathode to -12V and still have it work same as when I connect the anode to +12V and the cathode to GND?
AI: Yes, the LED would work.
Think of it like this: Voltage is basically the difference between the potential of two nodes.
So if you have the anode to +12V and the cathode to GND (which is basically our 0V reference point) then the LED will see a voltage of +12V-0V=+12V
But then, if you have the anode to GND and the cathode to -12V the LED will see a voltage of 0V-(-12V)=+12V. As you see, there is no difference in practice.
Also you say:
the voltage is the potential difference to ground
That is true for an absolute value of voltage, like the +12 or -12V of your PSU; they are referenced to its GND terminal. But as I said before, voltage is actually the potential difference between two points. In this sense, the absolute voltage of the anode and cathode of the LED is not important. |
H: Running high powered/current LEDs with high frequency
firstly i would want to thank all who are willing to help me in this problem that i am facing.
So to begin, i am a high school student doing a final year project. Basically i have to create a circuit where by a 1000Hz frequency is an input that i have to alternate it so that my output will be toggling with half of the input frequency.
Therefore i am using a single J-K Flip Flop which has a Q & Q(bar) output. That i do not have a problem....
However the problem comes in here. from the 2 outputs of my J-K Flip Flop, i have to use each output to power up 20 LEDs in series, which has to be 1.7V each LED and 700mA, making it 34V with 700mA.
But the output of a J-K Flip Flop does not produce enough voltage and current.
i have tried using a BuckPuck after the output, and realise the rise and fall time is too slow for the high frequency.
i found out that a NPN Transistir ( BD137 ) has a very fast Rise and Fall time, however after configuring the circuit, the output measure was 34v But only 200mA....
please help me to get a 34V and 700mA, thank you.
AI: The simple way is to use the signal to turn on a FET:
simulate this circuit – Schematic created using CircuitLab
The FET needs to have a threshold voltage (Vgs) well below the output of the J/K, ideally in the 2-3 V region if it's a 5 V flip flop. You also want the on resistance (Rds on) to be low enough that the power consumption at 700mA isn't an issue (power = I2*R). The maximum on current (Id MAX) should also be well over 700mA. Finally the maximum allowed drain-source voltage must be over 34V. |
H: How to show only Ground Plane in EAGLE
Is it possible to show only the ground planes like the one given in the figure?
Thanks
AI: To my knowledge EAGLE displays signals and layers, thus if you click show onto the ground signal, it will also highlight areas with all vias, pads and other tracks (depends on what you mean saying "planes").
One trick comes to mind is to make things you want to see separately in another layer, but I am more than sure then EAGLE will have issues connecting layers reporting unnconnected areas and trying to place vias in unwanted places.
Thus I believe you can highlight specific signal, but can not hide all other signals from the layout unless they are in another layer. |
H: What is a good alternative for the UA741 opamp in hobby projects?
Related to What's the uA741's appeal?, a lot of schematics use this rather dated opamp. My follow-up question would be: What is a good alternative?
Of course if you have specific requirements you just search for the correct part. What I'm talking about is, what should be the default opamp I stock in my parts bin?
Imagine the average hobby project with a breadboard and an Arduino, single-supply, probably 5V, frequencies in the audio spectrum, and accidental shorts.
It should just be a nice, single-ended, well behaved (no phase reversal) opamp with a large voltage swing (preferably rail-to-rail), and reasonable performance for audio.
Some comments on the linked question mention LM358 and LM324, but I've read the low slew-rate may be bad for audio, which is what most of my analogue signals are.
AI: In complement to Andy's list of criteria, I'll add that most of the time, one of the usual suspects will do just fine.
For example:
MCP6002 is a good one. Dirt cheap (bag of 10 is 3€). It's your crummy CMOS opamp. It is slow so it won't oscillate on your breadboard, has Rail to Rail In/Out, the input common mode even extends a little bit beyond the rails. Low power too. Performance is nothing to write home about of course, but it is good enough for many uses.
Microchip makes opamps? Well, yeah. They have a whole line of CMOS opamps which do absolutely nothing spectacular besides being cheap, RRIO, low-power, low-voltage. These work on 3.3V and 5V so a good match for Arduino, Pi, etc. If you want faster, try MCP6292 (10MHz).
Good thing with these cheap crummy CMOS opamps is they work well on battery voltages, and they don't have the LM358 gotchas, like the input common mode going to GND but not VCC, or the output being "kinda able to go to 0V but only if it never has to sink any current" and the like.
Yeah, there are other $0.50 RRIO opamps which would fit the bill. Do I want to spend 2 hours selecting a 50c opamp for a hobby project when I know this one will work? Probably not...
Also, cost. For amateur, hobby stuff, you gonna spend hours laying out that board, so when performance is needed, it's not worth it to skimp on the parts! Get a $3 opamp if it saves you a headache, it's well invested...
For example, if you want to filter and buffer a DAC or a PWM from an arduino and get a 0-5V output, there's no excuse for using an opamp which needs a negative supply voltage when you got 30c RRIO opamps which will do the job with only a 5V supply!
Audio, for example: if you need cheap, NE5532. If you need high quality, OPA1642, OPA1652, LME49720 and others in the same families are very good. |
H: Dual JK Flip-Flop Toogle Feature
I'm trying to make an 8-bit Display using 4 seven segment displays being driven by a single EEPROM (like in this video). From the video, I used the design of the 555 timer and a dual JK flip-flop to count 0 to 3. The clock works, however, when I wire the flip-flop in the same way as is done in the video (Both presets, clears, Js and Ks high and connect the clock to the clock1 pin and the clock2 to the output of the first flip-flop)
The two LEDs do not count up in binary as they do in the video instead, they both stay on.
I noticed that there are a few differences between the datasheets of the flip-flop I bought (CD4027BE) and the flip-flop he has.
On his datasheet, he has two preset pins and two clear pins whereas
I have two set pins and two reset pins (although I do not think this
is anything major)
On his datasheet, he has a toggle feature in the feature table where
as in the datasheet I have, there is not a feature table.
So my question is have I bought the wrong flip-flop or am I wiring it incorrectly and if I have bought the wrong flip-flop can I make it work with this flip flop?
Additionally, the desired output is:
0,0 - 0,1 - 1,0 - 1,1 and repeat.
AI: The TI data sheet for a CD4027B part shows this function table:
From this the immediate thing to notice is that the SET and RESET pins need to be connected to GND to keep them from having any action.
You can also observe that by connecting both the J and K inputs to a high level will cause the flip-flop to toggle its outputs at each clock rising edge.
Lastly if you want it to count binary in the manner stated you will want to connect the CLK2 pin to the Q1_NOT pin instead of the Q1 pin.
Another thing you need to put suitably sized resistors in series with the LEDs to limit the current to a safe level....both to protect the LEDs and the outputs of the flip-flop. |
H: How to use two of this DC-DC converter synchronously to get more current output?
This is the DC-DC converter: http://cds.linear.com/docs/en/datasheet/37911fb.pdf
There is a clockout pin in the chip and in the datasheet it says:
CLKOUT (Pin 33): Clock Output Pin. A 180° out-of-phase clock
is provided at theoscillator frequency to allow for paralleling two devices
for extending output power capability.
This is the block scheme of the chip:
I searched in application notes and example designs but there were no use of this feature in them. I need a clear explanation of the working principle and how to do this.
AI: My first reaction was that you should use two of the controllers in parallel and connect the CLKOUT pin of the one to the SYNC pin of the other. It believe it is clear enough:
SYNC (Pin 34): External Synchronization Input Pin. This
pin is internally terminated to GND with a 90k resistor.
The internal buck clock is synchronized to the rising edge
of the SYNC signal while the internal boost clock is 180°
phase shifted.
and
Frequency Synchronization
The LT3791-1 switching frequency can be synchronized
to an external clock using the SYNC pin. The falling edge of CLKOUT corresponds to the rising edge
of SYNC thus allowing 2-phase paralleling converters. The
rising edge of CLKOUT turns on switch M3 and the falling
edge of CLKOUT turns on switch M2.
You basically use the CLKOUT clock as the external clock to which the other controller is synchronized.
By searching a bit online, I also came across this in the LT journal, issue October 2012.
That should clear any doubt! |
H: BLDC motor emergency stop
I'm into a preliminary design of a eletrical vehicle that uses BLDC permanent magnet motors for motion (3 phases 36V 300W hub motors in wheel). For safety compliance this vehicle needs to provide a breaking mechanism that stops the vechicle when there's a power failure. My first choiche will be using reverse-action brakes/clutches that will triggers in case of power-loss unfortunately I have no room around motor and wheels.
I'm thinking about another solution that uses relays.
My idea is to place an SPDT relay (with coil bound to the primary power source) between motor driver phase output (relay NO) and BLDC motor phase (relay COMMON)
all relays NC will then be connected togheter (trough a resistor?).
In normal operations relays will be forced into NO position allowing drivers to handle motors.
In case of power loss all relays will return to their NC position and thus all phases will be shorted togheter. If the vehicle is then forced to move (because a slope or something pushing at it) BEMF will react in opposition to the motion.
I don't think this idea is something new I will only know if someone sucessfully used this or if there are hidden dangers/pitfall into my approach.
AI: This type of system is fairly common but you must remember that this is a reactive system. It will not HOLD the vehicle but rather, will retard it's speed.
That is, the amount of brake force generated will depend on the speed of the vehicle. When stopped, there is ZERO brake force.
A vehicle parked on a steep slope will continue to run down the hill, all be it at a much slower and fixed rate. As such this mechanism can in no way be called an "emergency STOP" system.
ADDITION:
Current limiting would also be prudent to prevent the motor from burning out. This would be especially a problem if the vehicle is on a really steep slope. At higher speeds this would of course cause braking to be reduced. At some point, maximum braking would be achieved as the vehicle slowed.
ADDITION 2
For emergency stop though it should be considered supplemental braking at best. Some sort of mechanical braking system capable of stopping the thing with or without the assist would still be a MUST-HAVE. In the end you need to decide if the extra cost and reliability reduction of the effort is justified. |
H: Controlling BLDC with IR 2130 - stuck at fault, unable to reset
I'm working right now with the following circuit, trying to drive the BLDC motor with it.
I don't exactly know how it works (especially the overcurrent sensing part) and the man that designed it is currently unreachable so I can't simply ask him.
I'm trying to drive the BLDC motor with it and Arduino but the IR2130 is stuck at FAULT state (so it's not sending any output to the transistors) and I can't reset it in any way.
I've tried resetting it by setting all LINs in high state for 10us as the datasheet says but it won't work. I'm monitoring the FAULT state before the main controling loop begins and I still get 0 at it. There shouldn't be any currents flowing through the motor right then so why would it detect any overcurrent? I don't think undervoltage is the problem because I'm getting nice and stable 12V at Vcc
AI: the /FAULT output is an open-drain output. You need an external pull-up resistor to your VCCIO logic supply voltage (3.3V or 5V).
The open-drain output has a low-on resistance in the 55-75 Ohm range. Any pull-up value comprised between 1k and 10k should do the job. |
H: Crosstalk through I2C
I'm using an Arduino to communicate with I2C magnetometer sensors. The parts list and a basic block diagram of the wiring is shown below:
HMC5883L Magnetometer:
https://learn.adafruit.com/adafruit-hmc5883l-breakout-triple-axis-magnetometer-compass-sensor/overview
TCA9548A I2C Expander:
https://learn.adafruit.com/adafruit-tca9548a-1-to-8-i2c-multiplexer-breakout/overview
Standard 4-pin Ribbon Cable:
https://www.digikey.com/product-detail/en/3m/8125-04-100/ML04G-100-ND/1107830
Standard 6-pin Rectangular Header:
https://www.digikey.com/product-detail/en/sullins-connector-solutions/PPTC061LFBN-RC/S7004-ND/810145
Standard 4-pin Header:
https://www.digikey.com/products/en?keywords=3M9449-ND
simulate this circuit – Schematic created using CircuitLab
I apologize for the schematic, but I had to simplify it to make it clearer. Essentially, we're using an Arduino Uno to read from six magnetometer sensors by using an I2C multiplexer. The multiplexer is attached to a PCB board made for the Arduino, so it's very close to it, and the sensors are at a distance 1.5 meters away via ribbon cable connected via headers to the PCB board. The schematic only shows one sensor since including all six would make it messy, but the other sensors are also connected the same way.
Now, from what I read about I2C, there are some issues with crosstalk, and according to another question posted, it is recommended that VDD and GND be placed between SDA and SCL in order to reduce and/or prevent crosstalk from occurring. The HMC5883L fails in that regard, so with my ribbon cable, I crossed the wires so VCC and GND are between SDA and SCL. It's not the most elegant solution, but it's the simplest solution. The crossing only occurs at the sensor's end of the cable (less than 1 inch), not in the middle of it, but I was wondering if this would be an issue with I2C communication. To be honest, I tested this with two sensors, so I'm pretty confident that this configuration will work with all 6 sensors, but I wanted to ask just in case that some issue might arise that I never considered.
Also, in case you were wondering, I had to add the 10k pull up resistors on the SDA and SCL lines so that the sensors could be read from 1.5 meters away. Initially when I had it set up on a breadboard, they were unresponsive, and it wasn't until I added the resistors that they could be read from a long distance. The sensors themselves have 10k pull up resistors, but I suppose that decreasing the resistance allowed for longer distance. The resistors are 10k SMD resisors that are on the PCB board itself.
AI: What is your I2C speed? Depending on the speed and distance, I2C can go as low as 1k for high-speed low-impedance bus. See http://www.ti.com/lit/an/slva689/slva689.pdf
What do you exactly mean by:
The crossing only occurs at the sensor's end of the cable (less than 1 inch), not in the middle of it
By "crossing" the wires, the good practice consists in not having the two I2C lines running side by side. For example, this is bad practice for a long ribbon cable:
VCC
GND
SDA
SDL
A "better" way of doing things:
VCC
SDA
GND
SDL
This way, GND shields SDA from SDL. However, GND itself can generate noise if the bypassing isn't sufficient on either side of the cable.
The best way would be to have additional wires, for the sole purpose of shielding the signals:
VCC
GND
SDA
GND
SDL
GND
This is used on all IDE ribbon cables for example. |
H: A Series-parallel circuit combination with current as input source
I am asked to find:
1.) the voltage drop |V|
2.) Current i1
3.) Current i2
So far I only managed to find the total resistance and total voltage which is 6 ohms and 72 volts respectively. I am really confused on where to start or what to find first. Can anyone help me with this? Thanks
AI: The voltage across the 4-ohm resistor, |V|, is the voltage across the left-side 12-ohm resistance (\$V_a\$) minus the voltage across the 6-ohm resistance, (\$V_b\$). Since you know the voltage across the 8-ohm resistance (72 V), you can determine \$V_a\$ by calculating the equivalent resistance loading the left-side 12-ohm resistance: \$R_{eq}=12||(4+6||12||4)\$. For the symbolic expression, do not develop the terms and keep the || as it is easier to read. Once you have that \$R_{eq}\$ value, you see a resistive divider with the 20-ohm resistance and \$R_{eq}\$. From the 72-V level you have calculated, apply that resistive divider and obtain \$V_a\$. Now that you have \$V_a\$, apply another resistive divider considering the 4-ohm resistance and the parallel combination of 6, 12 and 4 to obtain \$V_b\$. Calculate \$V_a-V_b\$ and you have what you want. With \$V_b\$, you have \$I_1\$ and \$I_2\$.
As a side note, below is a resistive divider involving \$R_1\$ and \$R_2\$ in the left side.
The voltage at node a is equal to \$V_a=V_1\frac{R_2}{R_2+R_1}\$. If you now load \$R_2\$ with \$R_3\$ as shown in the right side, you have an equivalent resistance equal to \$R_{eq}=R_2||R_3\$ and the new division ratio becomes \$V_a=V_1\frac{R_{eq}}{R_{eq}+R_1}\$. |
H: Lms43PD-05-CG Photodiode Troubleshooting
Datasheets:
Will post below since I can't post more than two links.
LED circuit schematic:
Photodiode circuit schematic:
Details:
The resistor value for the LED circuit (top right) is 24 ohms rated at 5W.
The resistor value for the photodiode circuit is 33 ohms.
The op amp is an LM358.
Problem:
I've hooked up these schematics to an Arduino uno and have been pulsing the LED at a frequency of 10 kHz. For some reason, the photodiode isn't reacting when I point the LED towards it. I've tried several different resistor values and can't get the photodiode to react, unfortunately. I've read several different threads and haven't found the information I need. I can post the code, if needed, but I doubt that's the culprit considering I've tried the same code on a different photodiode.
AI: Assuming you fix the obvious problems with the circuit mentioned in comments, you will still have some difficulties.
Let's say you are coupling 10% of the light from your LED into your photodiode. There's no quantum efficiency spec on the photodiode, so just for giggles we'll assume it's 100%. Then the expected photocurrent is
$$\frac{ P_{src} \eta \lambda e}{hc}$$
where \$P_{src}\$ is the source power, \$\eta\$ is the transfer efficiency, \$e\$ is the fundamental charge, and \$\frac{hc}{\lambda}\$ is the photon energy at your operating wavelength. Plugging in numbers, I get,
$$\frac {(10 {\rm \mu W})(0.1)(1.6\times 10^{-19} {\rm C})}{(0.3 {\rm eV})(1.6\times 10^{-19}{\rm J/eV})}$$
or about 30 uA.
The dark current of your photodiode is specified as 8 - 25 mA.
This means you have a signal to noise ratio of not much more than 0.001.
You will probably need to do some fairly careful design to make this system work. Consider these changes:
Increase your source power dramatically
Use a lock-in amplifier to separate your signal (10 kHz) from the background noise of your receiver
Use well-designed optics to optimize the optical coupling from the source to the receiver. Optics for 4 um wavelength will need to be chosen specifically for this wavelength.
Shield your receiver carefully from any background light
Cool your receiver to reduce dark current |
H: Conjugate, non-conjugate and single poles in an RLC circuit system
In all the texts I encountered so far, I find the following pole-zero diagram example for an RLC series circuit:
The transfer function for the above circuit can be found as:
H(s) = XC/(R + XL + XC) = (1/sC)/[R + sL + (1/sC)]
H(s) = 1/(LCs² + RCs + 1)
But when I plot the pole-zero diagram of the above circuit for different R,L and C values, I don't always obtain the poles as conjugate.
Below are three different pole-zero diagram of the above RLC circuit for different R,L and C values:
1-) Conjugate Poles
R = 10;
C = 0.00001;
L = 0.001;
2-) Non-conjugate poles
R = 10;
C = 0.001;
L = 0.001;
3-) Pole at the origin
R = 100;
C = 0.001;
L = 0.001;
Question:
What can we say about the real response of the circuit for each three case above by considering the locations of the poles? In other words, what does it mean poles being conjugate, being asymmetric/non-conjugate and being at the origin only for an RLC series circuit?
(Especially, in the second case there is two different non-conjugate poles which is the most confusing situation to say something about the system)
Edit:
Following the answer I was able to reveal the higher pole for the third case:
AI: This general form for this type of low pass filter is: -
\$H(s) = \dfrac{\omega_n^2}{s^2 + 2\zeta \omega_n s+\omega_n^2}\$
And if you solve the quadratic in the denominator (to reveal the poles) you get: -
\$s = \dfrac{-2\zeta\omega_n \pm 2\omega_n\sqrt{\zeta^2-1}}{2}\$
\$ = \omega_n(-\zeta \pm \sqrt{\zeta^2-1})\$
Then, if you analyse the square root, you can see that for low damping (low zeta) you get the square root of a negative number hence that part of the equation involves "j" and you get conjugate complex poles at some fraction of +/-\$\omega_n\$.
When the damping (zeta) reaches unity, there are no more complex poles and a single pole lies on the real axis at \$-\zeta\omega_n\$. This then splits into two poles (along the real axis) as zeta rises above 1.
A low value of zeta is under-damped hence you get a peaky response in the bode plot and you get conjugate poles. When zeta = 1 you get critical damping and when zeta is greater than 1 you get a rather sloppy 2nd order filter that starts to look like a 1st order filter as R dominates over \$X_L\$.
To get numbers we need to know how zeta and omega relate to R, L and C values: -
\$\zeta = \dfrac{R}{2}\sqrt{\dfrac{C}{L}}\$ and
\$\omega_n = \dfrac{1}{\sqrt{LC}}\$
For R = 10, C = 0.00001 and L = 0.001, zeta = 0.5 and Wn = 10,000 and this is as you display the conjugate poles on your first graph.
For R = 10 and C = L = 0.001, zeta = 5 and Wn = 1,000 so the poles are at: -
\$s=1000(-5\pm\sqrt{24}\$) = -9899 and -101 and I can't precisely say if this corresponds with your graph but it looks close.
For R = 100 and C - L = 0.001, zeta = 50 and Wn = 1,000 so the poles are at: -
\$s=1000(-50\pm\sqrt{2499}\$) = -99,990 and -0.01 so you are not able to see the higher pole on your graph but otherwise I would say I get about the same result.
To substantiate the theory a bit more, this picture may be useful: -
It's also noteworthy that when both poles lie on the real axis (i.e. the over-damped situation), pole positions are: -
\$ = \omega_n(-\zeta + \sqrt{\zeta^2-1})\$ and \$ = \omega_n(-\zeta - \sqrt{\zeta^2-1})\$
And, if you did the math you would find that one pole is the normal conjugate of the other with respect to \$\omega_n\$ i.e. if one is ten times \$\omega_n\$ then the other is one-tenth of \$\omega_n\$.
In other words \$ = \omega_n(-\zeta + \sqrt{\zeta^2-1})\$ is the inverse of \$ = \omega_n(-\zeta - \sqrt{\zeta^2-1})\$. |
H: Caesar cipher - digital circuit
How do I create a combinational circuit that encrypts a message using Caesar's cipher?
And another circuit that decrypts this message?
I thought about using modular arithmetic, but I did not find a simple way to do that either.
AI: I thought about using modular arithmetic
Yup, that's how you do it, alright.
but I did not find a simple way to do that either.
And, if you're going to build it up out of ICs you're not going to find a "simple" way.
Your problem is that (as you've noticed) the 94 characters you've arbitrarily picked (I assume you took the printable characters and excluded the space) are offset by 33 from zero.
So,
1) start by offsetting your characters. Use a pair of 4-bit adders to subtract 33.
2) Convert your key the same way.
3) Add the two using another pair of adders.
4) Use either a pair of 4-bit comparators, a pair of 4-bit adders or an 8-bit comparator to detect a result greater than 93.
5) If step 4 gives a positive result, use another pair of adders to subtract 93 from the results of step 3.
6) Now add 33 to get valid ASCII
That's a preliminary total of 12 16-bit DIPs, which is not simple, but not outrageous, either. Using wirewrap panels I've designed systems with multiple hundreds of chips, so I may be a bit blase about this. YMMV. |
H: AC vs. DC in Low Speed, High Torque application
My club is building a 1/4 scale utility tractor (around 800 lbs) and for the past couple of years we have been using an electric hybrid drivetrain. Our system has been a 32 hp gas motor powering a DC generator which we convert to 3 phase AC to power our independent wheel motors with an operating voltage of 72V AC. My question is would we be better off using a DC motor considering we don't require high speeds but need very high torque for our pulling competition.
The motors we are currently using are rated at around 40 ft*lbs at 5,000 RPM and go through a 31:1 gear reduction which we thought was plenty but when they start seeing a high load, the current spikes and faults out our controllers.
The first year we ran this system the controllers faulted out because we didn't limit the amount of current the motors could have, we changed the system this year to have a current limit so the controllers wouldn't fault out and it basically shut our motors off under a high load, I guess the question isn't necessarily which is better but is one more efficient under high load to make the most out of the current we can generate.
RPM: Our final drives have a 31:1 ratio so anywhere between 3000 to 5000 RPM, HP our gas motor is 32 hp so we sized our current motors to be 15 hp peak per motor so we could equal the power capability of the gas motor, Budget is hard to gauge, we have a decent amount of funding but sometimes companies sponser us or give discounts, Torque: we initially thought 1200 ft*lbs per wheel was plenty but now it is looking like we would need more around 2000 ft lbs per wheel to be able to compete with the other teams at our competition. The pull lasts maybe a minute or two maximum so we are pulling peak current that whole time but we have long rest cycles between pulls and events so the motor doesn't see prolonged current draw. We have a .66 F Capacitor in our system to smooth the DC current we provide to our controllers.
Links: Motor Controller
AI: What you have is a permanent-magnet synchronous motor. That is the same thing as a brushless DC motor. The motor may be too large for your controller or the controller may not be adjusted properly for the motor. If the motor and controller is suitable and adjusted properly, the problem could be that the engine-generator set is being loaded too heavily. Check to see if the DC voltage drops when the load is high.
The motor-controller combination that you have should perform as well as a brushed DC motor and controller or an induction motor and controller. One technology is not necessarily better than the others, but some models and designs of each may be better than the others. With any of them, the weakest link is the one that will be the limiting factor. A good design, with properly matched motor, controller and power source is essential. When the limit is reached, the controller should limit the current, not fault out.
Re comments and further consideration:
With the optimum design, a permanent-magnet synchronous motor (PMSM) should be capable of providing the most torque per amp of any motor technology. However, all designs are compromises involving various measures of performance and manufacturing cost. Selecting the best motor is a mater of evaluating the specifications of motors under consideration.
The performance of the motor must be evaluated based on a careful analysis of the requirements of the application. Tractor pulling is a contest in which a load is pulled as far as possible. As the load is pulled, the required force is increased by shifting the load weight from a wheel and axle to a skid thus increasing the friction between the load carrier and the ground. It should be defined whether the load increase is based on time or distance and whether or not there is a time limit.
It would probably be advantageous to operate the engine at maximum power for the entire pull. Since power is torque multiplied by speed, that would mean reducing speed and increasing torque during the pull. The most efficient means of doing that would likely be a continuously variable transmission (CVT) such as a hydrostatic transmission. A manual gearshift would be just as efficient, but probably not quite as effective. However the gearshift or both the gearshift and the CVT may be prohibited by contest rules.
A series connected, brushed DC motor inherently slows down and increases torque as its load increases. That characteristic may make it preferable to more efficient brushless types of motors for this application.
An induction motor can be controlled to operate with constant torque over the low end of its speed range and constant power with declining torque at higher speeds. With a current limiting controller, that would provide performance somewhat similar to the performance of a series DC motor.
The torque capability of a PMDC motor will be pretty much constant over its entire speed range with no capability to trade speed for torque. The torque capability is limited by the current available. In this application, the motor must decelerate to a stop pretty quickly when the current limit is reached. There may be a way to trade voltage for current in the supply and control systems. That could have the effect of trading speed for torque if the motor can operate at a higher current for the required time.
Before all else, it important to make sure that the selection of the fixed ratio of the drivetrain gearing provides as much torque as possible by reducing the speed as low as the application will tolerate. |
H: Why does this optocoupler's datasheet recommend splitting a current limiting resistor in two?
And not only that, they also want to keep a ratio of R1/R2=1.5
The opto in question is ACPL-M21L by Broadcom
Here's a link to the datasheet:
https://docs.broadcom.com/docs/AV02-3462EN
Thanks.
AI: What @Andy said is basically correct but the reason for the split R's is more complex.
It has to do with the insulation transient impedance ratios for the lightning simulation test specification.
The device capacitance in the off state = 77pF @ 1MHz @ 0Vdc and the Partial Discharge (PD) rating is in uC.
Highest Allowable Overvoltage (Transient Overvoltage)
V IOTM 6000 8000 V peak
In this case the impedance is not the ESR of the diode or the resistance of the Series resistors but the relative capacitance divider ratio.
The diode capacitance being larger than typical resistor capacitance (including dust, Pollution degree 2) (over a wide range of case types) permits the device to ensure it can handle over 7x the DC insulation value of 1140 V peak for isolation.
8kV is a standard impulse test spec and thus they found 1.5 anode/cathode R ratio to work best for immunity to 1us rise time pulses.
I won't attempt a mathematical answer but it requires model each part as a // RC equiv cct each in series.
simulate this circuit – Schematic created using CircuitLab |
H: How would I make a circuit that takes a voltage (V) as input and outputs min(V, Vref)?
Simple question, how do I turn the following block diagram to a real world working circuit?
I know I could use the following:
1) A comparator, but that will give me only either +Vcc or -Vcc.
2) I could use a rail-to-rail comparator and set V- at 0 Volts and the maximum allowed voltage at V+.
3) I could use a Zener diode... But I've zero experience with using them...
simulate this circuit – Schematic created using CircuitLab
AI: It depends on how accurate you need this to be. Here is a simple concept:
D1 and D2 provide the MIN function. The voltage at the top of the diodes is the minimum of Vin1 and Vin2, plus a diode drop. It should be obvious how to expand this to any number of inputs to take the minimum of.
D3 tries to compensate for the diode drop, so that Vout is the minimum of all the inputs with the diode drops cancelled out. If within a few 10s of mV is OK, then this might do.
Vhigh and Vlow are voltages you have to supply. Vhigh must be a bit higher than any input voltage of interest, and Vlow a bit lower.
The impedance of the input voltages need to low enough to overcome R1. The output impedance at Vout is higher due to R2 needing to be high to not interfere with the signal.
That's the basic concept.
The next step is to realize that BJTs can be thought of as diodes with gain. Here is the same concept carried out with BJTs using their gain to advantage:
Look carefully and you'll see its really the same thing, using the B-E junctions of the transistors in place of the diodes in the previous circuit. The advantage is that the input signals don't have to supply anywhere near as much current. The B-E diodes still do the MIN function, but most of the current comes from the negative supply.
The gain is used the other way around so that the B-E junction of Q3 loads the signal much less. Due to the gain of the transistors, this circuit has much lower output impedance while having higher input impedance.
One drawback of using the B-E diode of BJTs this way is that the max input range is narrower. This is because the reverse voltage characteristics of the B-E junction is usually fairly low. If you're only doing this over a 5 V range, then there should be no problem. Of course, always check the datasheet of whatever transistor you plan to use. Choose ones with high B-E reverse voltage capability if you want a wider input voltage range. |
H: Encoders debouncing with 7414 at 3.3V
I'm trying to make a small module that has 2 rotary encoders (both have build-in push buttons), hardware debouncing circuit and have to be compatible with both 5.0V and 3.3V microcontroller logic inputs.
What I've found, that most typical schematic involves the inverting Schmitt trigger (74HC14 or 74LS14) and should be looks like:
simulate this circuit – Schematic created using CircuitLab
The D1 could be omitted since it is just for decreasing charging time of capacitor C1.
Could you help me with questions:
1) Will 74HC14 (74LS14) work with 3.3V input?
2) Could 74HC14 (74LS14) use the 3.3V as power supply?
3) Could 74HC14 (74LS14) produce the 3.3V output to be connected with STM32F103**** (or other 3.3V logic level microcontoller)
Or should I use the SN74LV14A instead of 74HC14 (74LS14)? Looks like SN74LV14A is more suitable for lower voltages.
AI: You want the 74HC14. The HC series will operate at your 3.3 Vcc. The LS will not.
For your questions 1-3, yes to all for the HC series. |
H: Finding state equations for an RL circuit with inductors in T configuration?
I'm working on an exercise which asks to "Identify state variables and write apropriate state equations" for the following circuit.
simulate this circuit – Schematic created using CircuitLab
Trying to find the answer I get the following system
$$
\begin{cases}
v = v_S-R_1i_1-(L_1-M){i_1}',
\\v=(L_2-M){i_2}' + R_2i_2,
\\v=M{i_3}',
\\i_1=i_2+i_3.
\end{cases}
$$
from which I can't properly derive the requested state equations - I end up with equations like \$0=0\$
The difficulty seems to me to be purely mathematical but as the problem is related to circuit design I guess this forum is better place to ask this question than the mathematical one.
So, I need some help.
Thanks
AI: In comments you said that what you're really after is to "Identify state variables and write appropriate state equations", rather than solve the circuit. I'll answer how to write the state equations.
In an analog circuit, the state variables are inductor currents and capacitor voltages. So here, the state variables are \$i_1\$, \$i_2\$, and \$i_3\$.
To find the state equations, first, substitute your 3rd equation into your first two:
$$Mi'_3 = v_s -R_1 i_1 -(L_1-M)i'_1$$
$$Mi'_3 = (L_2-M)i'_2 + R_2 i_2$$
and re-write your fourth equation in terms of derivatives:
$$i'_1 = i'_2 + i'_3$$
Now move all the time-derivative terms to the left and other terms to the right:
$$ \begin{align}-(L_1-M)i'_1 + Mi'_3 &= -R_1 i_1 + v_s\\
(L_2-M)i'_2 + Mi'_3 &= R_2 i_2\\
i'_1-i'_2-i'_3 &= 0\end{align}$$
Now you have a system with the form
$${\bf M}\left(\begin{matrix}{i'_1\\i'_2\\i'_3}\end{matrix}\right)={\bf A}\left(\begin{matrix}{i_1\\i_2\\i_3}\end{matrix}\right)+{\bf B}v_s$$
Now you just have to pre-multiply each side of the equation by \${\bf M}^{-1}\$, or, equivalently, do algebra to eliminate all but one of the three derivative terms in each line, and you'll have the expected state equations:
$$\left(\begin{matrix}{i'_1\\i'_2\\i'_3}\end{matrix}\right)={\bf M}^{-1}{\bf A}\left(\begin{matrix}{i_1\\i_2\\i_3}\end{matrix}\right)+{\bf M}^{-1}{\bf B}v_s$$
Finding \${\bf M}^{-1}\$ will be tedious, so I'll leave that to you since this was your homework. |
H: NPN transistor wired correctly?
i'm experimenting with transistors, and wired one to my circuitboard. Though I got the LED to turn on with a NPN transistor, I'm not sure it's setup properly.
I read that current moves against the direction the arrow is pointing for the transistor symbol (emitter to collector), so I wired the emitter to a higher voltage (7.5V) than the base (4.5V). The problem is, when I remove the voltage source to the base, the LED remains lit. Shouldn't it turn off if the base truly acts like a switch for current flowing from emitter to collector?
I included two schematics below, one a crude, most likely incorrect representation of my circuit, and another from a book I'm learning from. My attempt at making a circuit follows my failure to get the schematic in the book to work - please let me know if you spot any errors in either schematic.
simulate this circuit – Schematic created using CircuitLab
AI: That is truly a very poor diagram in your book, throw it away. You have also made mistakes in transcribing it.
This is much better. It is drawn as a conventional circuit, with GND at the bottom, and increasing voltage generally up the page, with conventional current flowing downwards (hint, in the direction of the BJT emitter arrow), which makes it much easier to read and interpret.
simulate this circuit – Schematic created using CircuitLab |
H: Measure small voltages in a high voltage circuit
I'm troubleshooting an oscilloscope where the cathode voltage is ca. -2000V respect to ground and chassis. There are parts of the circuit (eg. the heater filament, or the blanking / un-blanking circuit) which generate low voltages but they are connected to HV supply lines. Eg. the heater secondary is generating 6.3V in AC but one of the wires is connected to the -2000V to keep the voltage difference of the cathode and the filament within spec for the CRT.
How can I safely measure these low voltages (eg the 6.3V across the filament) within the high voltage circuit?
I have a CAT II 1000V multimeter and CAT III 1000V probes. I assume that the isolation they provide is not sufficient to safely measure in such an environment. If I understand correctly the rated voltage is not just for the voltage I want to measure but it's also a rating of isolation towards the "outer world". Meaning that even though the multimeter and the probes are floating at -2000V the isolation they provide is still rated at 1000V making them not suitable for such a measurement. What options do I have?
AI: You are right in assuming that you cannot rely on the meter insulation between you (0v) and the -2000v circuit you want to measure.
You can make the measurement safely, if a little more slowly, by isolating the meter and its probes and wires from the ground. Get a suitable insulator, say a glass or plastic bowl or chopping board, and place the meter on it. This is obviously a battery powered portable meter, not a mains powered one. With the target powered off, set up the meter and connect it. Dress the probes and wires so there is an air-gap, which is resistant to the set-up being bumped or knocked, between meter probes and their wires, and any ground conductors. This is why a bowl may be a better solution than a board, you can allow the excess meter probe wire to sit in the bowl with the meter. Power on, read the meter without touching it, and power off. Wait for high voltage capacitors to lose their charge before unclipping the meter. |
H: Meaning of Sigma in Laplace transform
I found the question below in a forum and I also have same problem. Unfortunately, there is no answer at all so I will post it here hope someone could make it clear.
A gain relation in a circuit of RCL and dependent sources ends up in an H(s) which is a quotient of polynomials in s. Number of poles is the number of energy storing elements independent of each other (you can assign independent starting conditions) and zeroes depend on the behavior of H(s) when s tends to infinity and the number of poles. Some zeroes and poles can be found by inspection, this is done knowing the above and observing some conditions and values of s so that the gain becomes zero or infinity.
The questions:
s is supposed to be sigma + jw, and sigma arises so that the transform integral converges. However, when constructing bode plots, this is completely ignored. Why? Is there a physical meaning to sigma?
The effects on the bode plot of zeroes and poles are to change the slope in 20dB incrementals (bode magnitude), the gain isn't really infinite on the poles, since s is replaced by jw and if the poles are real, s being complex will never have those real values so that the bode plot goes infinite. What is the meaning of this?
A region of convergence (ROC) for s can be found, which is a range of values for s so that the integrals converge. What is the ROC useful for? What happens when the frequency is outisde the ROC? Textbooks are pretty good in showing how to find these ROCs, but not in making clear what effects they have in your circuit.
AI: As most folk know, \$s=\sigma+j\omega\$ (where \$j\omega\$ is the frequency along the x-axis in a bode plot or spectrum analysis). However, in a bode plot, \$\sigma\$ has no apparent meaning but it is actually the "unseen" z-axis (in and out of the screen/page).
If the natural resonant frequency of the RLC circuit were 1 radian per second then the z-axis is purely zeta, the damping ratio. That's the simple answer. The fuller pictures (hopefully) should be apparent with this image showing example bode plots along the top and the pole zero diagram in 3D at left bottom: -
As you should be able to see, the "z-axis" (or \$\sigma\$ axis) has values corresponding to \$\omega_n\$ (the natural resonant frequency aka \$\frac{1}{\sqrt{LC}}\$) and the damping ratio zeta (\$\zeta\$).
s is supposed to be sigma + jw, and sigma arises so that the transform
integral converges. However, when constructing bode plots, this is
completely ignored. Why?
Hopefully you can see that now.
Is there a physical meaning to sigma?
Basically it's damping ratio multiplied by natural resonant frequency for values of zeta between 0 and 1.
The effects on the bode plot of zeroes and poles are to change the
slope in 20dB incrementals (bode magnitude), the gain isn't really
infinite on the poles, since s is replaced by jw and if the poles are
real, s being complex will never have those real values so that the
bode plot goes infinite. What is the meaning of this?
The gain IS infinite on the poles, indisputably. The rest of that particular question is probably irrelevant due to that misconception.
A region of convergence (ROC) for s can be found, which is a range of
values for s so that the integrals converge. What is the ROC useful
for? What happens when the frequency is outisde the ROC? Textbooks are
pretty good in showing how to find these ROCs, but not in making clear
what effects they have in your circuit.
I have no idea what this means or how to answer it, sorry.
the transfer function 1/(s+1) has a pole at s = -1. This means that
sigma = -1 and omega is zero. However, when make Bode plot, s is set
to be equal to jomega that is pure imaginary number. So the transfer
function at s = jomega is not infinity at all. If it is not infinity
then what value it would be also why are we interested in s = j*omega
instead of s = -1 for the function above?
Think like this - you have a table and on that table you stand your pencil upright in the middle (this is the pole). You then place a really thin and pliable handkerchief over the end of the pencil. The contour made by the handkerchief produces a tent effect but more perfect: -
At any point from that pole position, the handkerchief has a numerical value that is perfectly defined. For instance, if you draw a circle around the pole, all points on the handkerchief would have the same amplitude. If the pole is at -1 then, the amplitude when s = 0+j0 (the origin) is 1. The amplitude at 0+j1 needs a little more thought - the distance from the pole to 0+j1 is 1.4142 therefore the amplitude is the reciprocal of 1.4142 i.e. 0.7071 (aka the 3 dB point of the simple RC filter your numbers numerical example describes).
At 0+j2, the distance from the pole is \$\sqrt{1^2+2^2}\$ = 2.236 and the amplitude is therefore 0.4472.
At 0+j10 (ten times the 3 dB frequency), the distance is \$\sqrt{1^2+10^2}\$ = 10.05 and the amplitude is 0.0995.
This can be extended to complex pole pairs and zeroes too. At any point on the jw axis (called X below) the amplitude is: - |
H: How to interface SIM800L to UART on STM32F103
I have problem to interface SIM800L with STM32F103. I just read SIM800L datasheet that UART logic level is on 2.8 V. But the STM32F103,can be operated in 3.3 V. I connected my TX RX pin directly to my MCU. I send AT command but nothing happen. Am I doing something wrong?
AI: You have to check the I/O and serial port characteristics in the datasheet of the devices to be sure that logic levels are matched. You need the VIL, VIH, VOL, VOH respectively for the SIM800L and the STM32F103. For the SIM800:
For the STM32F103 the input values that are tested in production:
VILmax = 0.35 * VDD = 0.35 * 3.3 V = 1.155 V
VIHmin = 0.65 * VDD = 0.65 * 3.3 V = 2.145 V
Theoretically VIL a bit higher and VIH a bit lower.
As you can see there is no problem here as the SIM800's VOLmax (0.1 V) is below the STM32's VILmax (1.155 V). And the SIM800's VOHmin (2.7 V) is above the STM32's VIHmin (2.145 V).
The other direction, STM32F103 output values:
There are several cases, you should calculate with the worst case values, so when the IO current is +20 mA. VOLmax = 1.3 V and VOHmin = 3.3 V - 1.3 V = 2 V.
Now, here come the problems as the SIM800's VILmax is 0.3 V but the STM32's VOLmax is at least 0.4 V. There is a 0.1 V gap in which the input to the SIM800 is undefined.
Also the STM32's VOHmin can be 2 V, 2.4 V or 2.9 V. The 2 V is far below the SIM800's VIHmin 2.5 V. The 2.9 V would probably be fine.
All in all logic level conversion is suggested, it is mentioned in the SIM800's datasheet as well along with a reference circuit.
If your problem persist after doing the level shifting, here are a couple of items that you should check:
You should check that you connected the SIM800's TX pin to the STM32's RX pin and the SIM800's RX pin to the STM32's TX pin.
SIM800 has autobauding enabled by default, and autobauding supports only the following baudrates: 1200, 2400, 4800, 9600, 19200, 38400 and 57600. So if your baudrate is 115200 then that should be changed.
Make sure that every AT command you send ends with \r\n sequence. |
H: Would an ATX PSUs internal fan work as the dummy load for a benchtop power supply conversion?
The PSU has 2 12V rails listed in its specs on the side, however both show continuity between them when I tested it with my multimeter. I then tested continuity between the positive from the fan and 12V1 and it also showed continuity. The fan is rated at 0.7A. This is a higher load than the 22 Ohm 10W resistor I was originally going to use (0.55A). Should this provide enough of a load? Would it be necessary to put a load on one rail or the other if they are both connected internally anyway?
When I started it up with no other load connected, the grey wire came to +5V which I believe means power is OK. Is this a good sign that it likely doesn't need an extra load?
Finally, should I also put a dummy load on the +5V rail?
Extra Info:
I was able to dig up these specs, although they aren't from the manufacturer so it's hard to tell how accurate they are. They do suggest that extra load might be required though.
Min Load:
+3.3 - 0.3 A
+5V - 0.5 A
+12V1 - 1A
+12V2 - 1A
Although given that both 12V rails seem to be connected I'm not sure if that means 1A each or just 1A for both.
AI: The problem is that most of the output voltages are generated directly from different windings of the same transformer, so it is not possible to regulate them independently.
If you only load the +12V output the controller will increase power to the transformer to compensate for voltage drop in the rectifiers etc., but this causes voltage on the other unloaded outputs to increase. If the +5V output exceeds its maximum permitted voltage the crowbar will trip and shut the PSU down.
Loading the +12V output down more will only make it worse. To prevent the +5V output from rising too high you need to draw a reasonable amount of current from it, typically 2-3A. |
H: Can a person suffer mild electrocution if he answers a charging phone with wet hands or sweaty palms?
This texas girl got electrocuted and died while using her cellphone in the bathtub. Can a person with wet hands or sweaty palms face mild electrocution if he answers the phone while it is being charged?
https://www.usatoday.com/story/news/nation-now/2017/07/11/texas-girl-electrocuted-while-using-cellphone-bathtub/467225001/
A 14-year-old girl from Lubbock, Texas, died Sunday after being
electrocuted in a bathtub while using her cell phone, according to
local reports.
Madison Coe was electrocuted after she either grabbed her phone that
was plugged in or plugged in her phone, her grandmother Donna O'Guinn
told KCBD-TV.
AI: It wasn't simply the wet hands that killed her, it was also the fact that her body was thoroughly grounded via the bath water and the metallic plumbing of the bathtub. This, coupled with a faulty charger, created a dangerous situation. Only a GFCI/RCD would have saved her.
This is why you don't use cheap no-name chargers. If there's any doubt, plug it into a circuit that's protected by a leakage current detector. This is also why all modern building codes require such protection for all circuits installed near plumbing, as in bathrooms and kitchens. |
H: LTSpice resonant circuit simulation non-idealities
I just simulated the following circuit in ltspice XVII:
A parallel resonant circuit is tuned to 125KHz with varying capacitor sizes. Then a voltage divider is formed with a resistor.
These are my settings for the inductor, capacitor and in the Tool Panel. In my understanding, this should now simulate an ideal inductor without losses and the same for the capacitor.
Using ordinary AC-calculation, the voltage over the resonant circuit should be:
$$ V_{res} = V1 * \frac{\frac{1}{\frac{1}{j \omega L} + j \omega C}}{R + \frac{1}{\frac{1}{j \omega L} + j \omega C}} $$
$$ = V1 * \frac{1}{R*(j \omega C + \frac{1}{j \omega L}) + 1} = V1 * \frac{1}{j R*(\omega C - \frac{1}{\omega L}) + 1} $$
$$ \Rightarrow V_{res} = V1 \quad \forall \quad \omega C = \frac{1}{\omega L} \Rightarrow L = \frac{1}{C \omega^2} $$
So at resonance, when \$ \omega C = 1/(\omega L) \$ the resonant circuit poses an infinite resistance and \$ V_{res} = V1 \$.
However, LTSpice gives the following simulation results (where smaller C values yield larger amplitudes):
Now I'm wondering how this is calculated by LTSpice. Are there any non idealities that I didn't switch off? Or am I missing something in my calculation?
Any hints would be very appreciated,
Thanks in advance.
AI: ltspice (or any simulator really) is only an approximation to both, reality and ideal components. Reality because it can not model all the details reality depends upon, and ideal because it can not run with infinite precision in values and time.
Basically how any spice works is that for the next timestep it checks all involved complex formulas and matrices, and if they converge within the required number of iterations into values with error tolerances below those specified with the *tol options, then it will go on. If not, it will lower the timestep and try again, until it either reaches a limit and errors out, or the tolerance is met.
reltol is the paramter used to specify a certain accepted error relative to the next timestep. The error is estimated using a polynomial to "predict" the value at the next chosen timestep, then it is actually computed there, and the difference taken. If its too big, make the timestep smaller.
This also means that instead of those parameters, you can make the simulation more accurate by forcing a really small timestep like 1n but that makes things really really slow, the dynamic timestep feature is one of the things that make it much faster.
Together with trtol (which specifies a factor on overestimation of the actual error) these are the major knobs you want to play with to either make the simulation more accurate, or faster.
Additionally, ltspice internally uses floats, so sometimes .opt numdgt=7 (anything over 6) is needed to force it to use doubles instead, which may or may not make things more accurate. |
H: Can I read EEprom in circuit when connected to a SH7055 MCU?
I am trying to read a microwire 93C66 EEprom which is connected to a Hitachi SH7055 MCU. The device is a complete automotive ECU. I need to be able to do this in circuit.
Currently I get an error when I try to read so I assume I need to hold the MCU in either reset or hardware standby mode.
The Eeprom is connected to the MCU as follows:
Pin1 - PIN193 PJ15/TI9F
Pin2 - PIN192 PJ14/TI9E
Pin3 - PIN191 PJ13/TI9D
Pin4 - PIN190 PJ12/TI9C
Can I just tie the HSTBY pin to GND or would I have to power the whole board?
Datasheet for the MCU
Datasheet for the Eeprom
What else would I have to do in order to read this EEprom in circuit?
AI: Keep the entire board powered. You will need to pull the HSTBY and the RES pins low. This will put the external processor bus in a high impedance condition. You should then be able to exercise the EEPROM provided there are no other driver ICs on the EEPROM lines and that the EEPROM is not tied to either of the above pins. |
H: Power on configuration for MCU using CPLD and pin multiplexing
simulate this circuit – Schematic created using CircuitLab
Suppose we have simple same here. The logic behind this is very simple. During MCU boot it tests pins CONF1 and CONF2 for data on it (LOW or HIGH) and configures its internals based on this. Also during the startup sequence it holds BOOT_COMPLATE HIGH for example. After MCU goes into ready state it drives BOOT_COMPLATE in the opposite direction (LOW in this example). After that pin on the MCU are used for DATA1 and DATA2 functionality. Vendor of MCU advises to use CPLD for this kind of work. But provide no shame for free.
My thoughts about how this should work and looks like all together:
simulate this circuit
Power up CPLD first (assume we have the proper program already set up in it)
Drive IO2 and IO3 LOW or HIGH as needed by config
Test signal level on IO1 and wait for LOW in this example (indicates MCU boot complete)
Set IO2 and IO3 tri-stated (floating)
Done? Or am I missing something?
Also it would be nice to have VHDL or SystemVerilog example for this simple situation
AI: It looks fine to me. It fits with the description you gave. I think this is basically what the MCU vendor had in mind.
Only additional thing you should consider, is what are the "Other components" do during the boot period. More specifically, how do they influence the CONF1/CONF2 signals. It could be that they also pull the lines to one direction (not actively of course, but due to internal pull-up/down or through the ESD diodes).
If it is the direction that suits you (the same as the CPLD) then all is ok. If not, there could be problems, depending on which side pulls stronger. If this is the case, you could/should use "isolating" components between the "Other components" and your MCU/CPLD structure.
That is most commonly series resistors that limit the current and thus the influence that the "Other components" can have to the configuration signals during boot.
OFF-TOPIC: I'm just asking myself why do they recommend to use a CPLD for something that simple?! Why can't you just use fixed pull-up or pull-down resistors? |
H: Speaker vs. amplifier wattage matching
Is there any rule-of-thumb for choosing amplifier wattage for the given speakers wattage?
For example, I have speakers, 100 W (max. input power) / 6 Ohm each. Is it ok to match them with 44 W (min. continuous power) / 8 Ohm / channel?
In general, is it ok to hook up large speakers to weak amplifier? Is it ok to hook up small speakers to powerful amplifier?
I've read that speakers should be able to handle twice the wattage of amplifier output. True?
AI: Is there any rule-of-thumb for choosing amplifier wattage for the given speakers wattage?
Speaker wattage can be defined in many ways and the vendor will choose one to suit his intended market.
It could be the maximum continuous power that the coil will dissipate without burning up.
It could be the maximum power that the speaker can convert into sound with a certain level of distortion.
It may be the maximum power in a narrow band of optimum frequencies that the speaker can reproduce adequately accurately.
You can see from this that the power handling specification is not the full story.
For example, I have speakers, 100 W (max. input power) / 6 Ohm each. Is it ok to match them with 44 W (min. continuous power) / 8 Ohm / channel?
The power is OK but the speaker impedance is a bit low. At low volumes you'll be OK but at higher volumes you're asking the amplifier output transistors to deliver more current than they were designed for. Best not. Load the amplifier with greater than equal to the specified speaker impedance.
In general, is it ok to hook up large speakers to weak amplifier?
No problem.
Is it ok to hook up small speakers to powerful amplifier?
Yes, provided:
- The amplifier switch-on thump doesn't destroy them.
- You control the power (volume).
Use your ears to tell you when you're over-driving the speaker.
I've read that speakers should be able to handle twice the wattage of amplifier output. True?
It's generally a good idea. An amplifier will "clip" when it runs out of voltage at high signal levels. This happens at a very well defined point. The speakers, on the other hand, will gradually increase in distortion as the cone reaches the limits of travel or the suspension starts to behave non-linearly.
A high-fidelity speaker system will be specified in sound level output and will give distortion figures for the full spectrum.
From comments:
"The power is OK but the speaker impedance is a bit low". The amp data-sheet says: "Speaker impedance 4 - 16 Ohm" so I thought that those 44W/8Ohm rating could be "converted" to match the 6Ohm of speakers, adjusting the wattage a bit (say 55W/6Ohm). Isn't so that I can freely convert between various impedances, provided the amp (data-sheet!) supports it?
If the amplifier has an output transformer there may be a switch to convert or optimise for 4, 8 or 16 Ω speaker load. Since load should be >= amplifier output impedance you would set the amp to 4 Ω for a 6 Ω speaker. Transformer outputs were common on valve amplifiers but less common on transistor amps. If you wanted you could purchase one but I don't think it would be worth the trouble.
Bear in mind with all this that your ear has a logarithmic response to volume. To double the perceived volume you need ten times the power. A (right side) mismatch between amplifier and speaker impedance will not be very noticeable in most cases as you won't be running the system at full power. |
H: What is the failure mode of a solar panel?
I have understood that solar panels don't last forever. Warranties are typically a couple dozen years, and you can expect your panel to last for perhaps twice the warranty period.
But what does exactly happen when the panel is failing? Does it fail suddenly like computer hard disks, or does its output degrade like the capacity of batteries? What are the physical principles behind the failing solar panel? Is the failing in some way related to heat?
Do solar cells made by a reputable brand last longer than cheap Chinese cells?
Is it possible to make a solar panel that would last essentially forever given a high enough price is possible? Such a panel might prove useful if it turns out that the low interest rate environment will continue.
Of course, there are many types of solar cells, so the answer may be limited to the most common types, i.e. polycrystalline and monocrystalline silicon cells.
AI: Reading here and a couple other places makes it sound like solar panel degradation varies widely. Manufacturing origin doesn't appear to be correlated to longevity or if it is, it may be opposite what we expect (China appears to do well). The general gist is that you'll lose a fraction of a percent every year on average. It's likely due to high energy photons slightly changing the structure over time. Weathering is also a concern. Wind-blown sand scratching the surface and dust blocking light are two other ways cells degrade. Most solar installations appear to be able to handle 20-40 years of use without issue, but some don't appear to handle thermal cycling well. In that case, you can have catastrophic failure of one or many cells causing poor solder bonds to break, delamination to occur, or entire cells to crack. Corrosion of the cell and connectors could be another late game failure mode.
I think more of a concern than the cells degrading is the supporting electronics (inverter) dying. The cost of installing a solar installation these days is largely being determined by peripherals rather than the panels themselves. Power electronics to support the system and their failure mode is really what I would be researching if I were in your shoes as I believe the likelihood of catastrophic failure there is much more likely in a much shorter timeframe.
This looks to give a great rundown of many manufacturers and their lifetimes. I've included a diagram from there below: |
H: Why do inductors like these do not work
I was trying out a series RLC band stop filter on hardware but for some reason it was not working, I later figured out that the culprit was the inductor after testing both inductor and capacitor via the RC and RL filters, I tried different inductors of the same kind, even of different values of inductance but they yielded no success till I used the coil that is usually found in energy savers, and then the filter worked perfectly, so is it that the inductors like these that i used previously have some sort of limitation in the amount of current that can pass through them or are they not good for filters?
I had used Vpk =1V from a function generator and R=1Kohms
The inductor values that I had checked ranged from 20 micro Henry to 25 milli Henry, sorry I cannot tell you the model number or anything else because in my area, 98% of electronic shops just sell these in local cardboard boxes, the boxes just contain inductors of some specific value that they have bought and placed in the box
.
AI: Of the three basic passive components (Resistors, capacitors, inductors) found on your shop's shelves, inductors are least ideal. Where the box label says "200 uH", you should recognize that the shop owner has likely measured it with an instrument under very specific conditions - likely differing greatly from your intended use.
Resistors and capacitors have much more dominant primary properties (resistance for resistors, capacitance for capacitors) than inductors. Notice the queries from other responders, who have all identified things that influence the success of the filter using your inductor: frequency, current levels, wire resistance, capacitance between turns of wire.
For your specific application (band-stop filter), your inductor choice might consider:
Does the inductor have to deal with high power levels?
Is the inductor designed to be used at your filter's band-stop frequency?
How much attenuation is your band-stop filter expected to achieve?
What frequency accuracy is expected at the filter's suck-out frequency?
A high-performance band-stop filter, requiring great attenuation at a very precise frequency likely requires a high-quality inductor having an accurate inductance that doesn't vary with temperature. A tiny inductor might be wound on ferrite that can easily saturate at high power levels. Resistance of the inductor at your filter design frequency may be far higher than resistance measured with a DC ohmmeter. Inter-winding (parasitic) capacitance can cause its inductance to vary greatly at different frequencies. It is not worth your shop-owners time to characterize all these effects. |
H: Solutions to store GUI elements in an embedded systems
I have a project in mind to create a GUI using a STM32F4 microcontroller. I am currently working/playing with a STM32F469I-DISCO board that embeds a MIPI Display. I understood how to display images and others stuffs but I was wondering what are the solutions to store icons more "efficiently". What solutions are commonly used.
For now when I want, for example, display a specific icon, I create two files :
a source file icon.c that contains an array in which each element describes a pixel ;
a header file icon.h that simply contains the array prototype.
Then I include the header and use my icon when I need it. But I noticed that icons are not negligible in terms of space requirements (depends of course on the icon size, color format...).
Anyway, generally, in an embedded system (using a STM32 for example), what are the most "efficient" solutions to store everything that is related to the GUI itself ? Is an EEPROM correct ? SD Card or something else ?
Thank you !
EDIT : By efficient I mean a solution that takes the less space possible (physically). I don't have any requirements on the power consumptions side.
AI: If you have a number of graphics and they are taking up lots of space, then you'll need space to put them somewhere and more space to work with them.
If you are simply compiling them as c or h files, then they are likely falling into your internal Flash space. So the question is if they are taking up too much space and need to be kept elsewhere. Note that even if you are setting the compiler to keep them in RAM, then they will still have an image in Flash that is loaded to RAM at startup.
So then you have off-board options. This could be an EEPROM chip, which are relatively cheap and easy to work with. An SD card brings more space but also more overhead - you'll need to get a FAT system working. I am assuming you are not using an external memory bus with Flash on your processor, but if you were, then that would be a good place.
If you take this route, you have three obstacles to overcome:
The driver and API. Depending on what you choose, you may have to write your own. From wherever the driver came from, you'll have to get it working and validated in your setup.
Initialization of the memory. You'll need to get your images into the memory device somehow. If it is an I2C EEPROM or an SD card, you'll have to figure out how to get your images in there in a way that your processor will recognize.
Local copies in RAM. When you pull an image out of memory, you'll need enough RAM space to hold it while you work with it.
I don't know enough about your application and specs to make a recommendation. If the graphics are likely to be modified a lot for aesthetics before they are done, then SD card may be most convenient as you can update it easily with your PC. But this will have a significant scope impact on your software project. If you are under a cost target and the images are simple and not going to change, then maybe I2C EEPROM is the best way, but if the images are large, there may be some time lag. If you can suffer the internal Flash space, then that may be easiest in the long term. If you have control over processor selection and PCB design, then you could get an external memory bus with tons of Flash.
There is also the option of image compression, which can be done in embedded systems, and while I've seen it done I don't know enough of the pros and cons right now to comment intelligently. |
H: How can I temporarily short-circuit two adjacent poles on a circuit board cheaply?
On the image below I need to short circuit GP100 to enter flash mode on an esp-chip. I can't afford a soldering iron atm (and also don't know how to solder). I'm currently trying to do it by holding small screwdriver between the pins. I need to hold down a button and plugin a USB at the same time, so it's kind of tricky. Is there any other good trade-trick for doing this?
AI: Too poor to buy a $10 soldering iron?
You need something which can be held with one hand yet make contact on the two pads at once, and it also needs pointy bits to punch through the oxidation.
Check your medicine cabinet for a set of tweezers: |
H: How does the Nintendo 64 detect a controller is connected?
To give some context, I am trying to emulate controller input to the N64 via a GPIO pin on the Raspberry PI.
In an attempt to accomplish this, I have one of the PI's GPIO's in a pull-up configuration connected to the data line of the N64 controller input on the front of the system. Here's what that looks like:
In this configuration, I see the N64 system pull the line high when it's turned on, but there is no data request or any further change.
However, when I additionally attach a controller to all three inputs (data, ground, vcc) like this:
I am able to monitor data flowing between the system and the controller with the PI.
So, finally, the question: how is the N64 detecting the controller is connected, and how might I go about emulating that? Is it the resistance between the 3.3v and GND line? Something else entirely?
This is my first post, so let me know if I've betrayed any stack exchange policies, or if you need any additional information.
Thanks for your help!
AI: It turns out I wasn't sampling the GPIO long enough. I wrote a little Raspberry PI kernel module to sample the GPIO as fast as possible for 3 seconds, and record a semi-accurate trace. Here's what showed up on the wire after one second:
In summary, the N64 waits one second after boot to ask the controllers to identify themselves with 0x00 followed by a stop bit.
Thanks Chris Stratton for your help! |
H: Reverse current blocking on switching regulator
I was thinking what is the best solution to add a reverse current blocking on switching regulator. My first thought was to look for an ideal diode circuit but all of them were too much expensive. The cheapest solution was to put a simple diode but they've got significant amount of voltage drop. I thought to hide the diode's voltage drop inside the switching regulator. Connecting the feedback pin of regulator at the diode's cathode the Vo would be maintained to just at the desired set point.
What are the pros and cons of this circuit?
Is there a better solution?
AI: Try connecting a simple diode as shown in the circuit below
simulate this circuit – Schematic created using CircuitLab |
H: μC <-> USB <-> μC advice please!
I'm working on a radio receiver project based on using a Uniden BC125AT, which will be controlled by an external microcontroller.
The original plan was to tap into the BC125AT keypad to issue keypresses etc, an ugly solution, but it works.
But the BC125AT has a USB port for programming and control, and of course this is a USB-serial interface. When I pulled the BC125AT apart I was hoping to find an FTDI so I could tap into the serial and and use our microcontroller to issue commands. Unfortunately they're using a Renesas V850ES/JG3-L 32-bit micro with on-chip USB controller:
http://www.farnell.com/datasheets/1581873.pdf
So now I'm left wondering if there's anything I can realistically do to interface with it? Of course I could use a single board computer... but that's overkill just to achieve comms and I think we'd just go back to hacking the keypad. I really just want to see if I've missed any options?
Cheers,
AI: What you are looking for is USB Host, sometimes called USB OTG.
Not all microcontrollers can do this, but you should be able to find some.
That is only the hardware side though, you will also need software. For example, what device classes does the radio use? (it can have multiple) Does it enumerate as virtual serial port, or does it use custom drivers?
If you know this, you can get onto host software. Mbed has a library. |
H: How to charge this lipo back?
I have made a lipo battery pack with 2 identical 3.7V 1400mAh battery packs connected in series.
This pack is being used in a wearable project where I wear this pack on my wrist.
Now to charge it up, I have a TP4056 1A single cell lipo charger. Now my charger allows me to charge one cell at a time. I was thinking that if I wire individual cells to the charger in parallel, I could use the same charger for both the cells and could keep my battery pack compact and would have a simple plug-in solution for charging it up.
Will this method work?
AI: That will not work.
Not without extra care, attention and electronics.
If one cell when you start charging is 3.3V and the other is 3.4V, depending on the quality of the cell and thus its internal resistance, already you can have a situation where one cell forces several amps into the other the moment you connect them in parallel. Doing this regularly will severely reduce the life of your batteries. If the difference is larger the currents can be big enough to cause serious and immediate problems.
Apart from that, you should be aware that your device or pack should turn off the current drain once either one of the cells falls below a safe minimum voltage (3.2V for example, or 3.0V), so that they can't discharge to a point where damage occurs. This is why many 2cell packs have an internal BMS board that monitors both cells for over and under voltage and turns them off if either is at danger.
With home-made wearable tech I'd also strongly suggest you add current-limit protection inside the battery pack, so that a single mistake while wearing has a much lower risk of short-circuit based fire and/or injury. |
H: Doubling 150VDC
I'd like to double the 150–220VDC 50mA supply from this Nixie power circuit, for use with electrowetting, but I'm not sure of the best method. Most boost converters I've seen require a low input voltage, and I haven't seen any other circuits that do DC-DC voltage multiplication (though I am almost completely new to electronics).
This seems like a super-simple question, and I'm sure it has a simple answer. Are there just higher-rated components out there I should use? Anything that isn't a big box that requires a quote?
AI: This seems like a super-simple question, and I'm sure it has a simple
answer. Are there just higher-rated components out there I should use?
Anything that isn't a big box that requires a quote?
There are a few flyback boost designs on the internet such as this one from Linear technology: -
The output voltage can be chosen by setting the FB (feedback) resistor ratio from the output. Read the data sheet.
Offerings from TI are also likely to be similar. |
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