Signal Distortion of Square Waveform for MOSFET Gate With High Current Pulses - Page 1
Forum section: electronics / projects

author: nikkoderic
author_rank: Contributor
timestamp: November 01, 2016, 07:47:23 am
content: Hello!I've been working on this project posted on the internet about testing power inductors (or any inductors?) by plotting the current that passes through the inductor over time because I want to be able to know how the inductance changes when an inductor reaches its saturation.Here is the project pagehttp://elm-chan.org/works/lchk/report.htmland here is the provided schematichttp://elm-chan.org/works/lchk/lchk_sch.pngThere are some changes that I made in the circuit.-I substituted IRF3710 mosfet in place of the mosfet used in the schematic-I changed the value of the current sense resistor to 0.1 Ohms using 10pcs 1 Ohm resistors in parallel-I used other general purpose power diode as the protection diode for the inductor. i think its a 1N4007 diode-I didn't connect the current limiter, it was messing with the output. probably because I used a different value for the sense resistorI was able to make the circuit working and I was able to actually determine the inductance of some of my inductors with fair accuracy. But I have some questions about the waveforms that I got because they are not what I expected them to be.Looking at the waveform circled in blue, the current exceeded the maximum current that can pass through the circuit when using a supply voltage of 10V. If for example I have a perfect 0 ohm resistor with no inductance placed across the test terminals, then I should get a maximum current of 100A using 0.1 ohm sense resistor. I tried placing a short between the test terminals and instead I got 43A. I know it's probably due to other resistances that are inherent in the circuit and to on resistance of the MOSFET maybe. All inductors that I tested maxes out (or decays?) to 43A (yellow line) so it implies that that is my maximum current that I can pass through the actual circuit if I place a perfect conductor across the test terminals.But the question is why is that some of my inductors specifically the low valued ones causes the current to overshoot. Is it because of kickback voltage? But I already have the diode across the inductor? Is it not enough? What can I do to prevent this?Another question related to this is that whenever I test inductors with low inductance, it causes the square signal that controls the gate to distort, shown in red. I tested an inductor with relatively high inductance (shown in the second picture), the gate signal looks better but it seems to be slightly affected too. What causes this? What do you call this phenomenon? What can I do to preserve the "squareness" of the signal? Is it somehow related to using a mosfet driver?To give you a background, I am an undergrad ECE student. This is project is not school related. I just want to create this circuit for personal use.I hope someone can answer. Thank you =)

author: BravoV
author_rank: Super Contributor
timestamp: November 01, 2016, 07:52:45 am
content: Quote from: nikkoderic on November 01, 2016, 07:47:23 amCan't answer you though, just sharing there is a similar circuit which is better imo, designed by one of the respected member here.Its proven and also succesfully built by few other members too ->Inductor saturation testeror just read the whole thread.

author: nikkoderic
author_rank: Contributor
timestamp: November 01, 2016, 08:06:24 am
content: Thank you. I might try building that circuit sometime. =D But I really want to know what actually causes these problems. I might learn something important. =)

author: _Wim_
author_rank: Super Contributor
timestamp: November 01, 2016, 08:19:05 am
content: Did you measure your 10V supply? This would be my suspect of causing the overshoot when it tries to react to the sudden (very high) current request. Connect channel 2 of your scope directly to the 10V supply and compare with channel 1. In the ideal cause, the 10V should not move at all...

author: xavier60
author_rank: Super Contributor
timestamp: November 01, 2016, 08:33:11 am
content: "I should get a maximum current of 100A using 0.1 ohm sense resistor"No, long before 100A is reached, the voltage drop across the sense resistor will cause the G to S voltage to drop too low causing the MOSFET to partly turn off. It will behave like a constant current sink.

author: bktemp
author_rank: Super Contributor
timestamp: November 01, 2016, 08:58:42 am
content: Quote from: nikkoderic on November 01, 2016, 07:47:23 amIt is most likely the parasitic inductance of your current shunt. The current does not overshoot, it is only the measured voltage across the shunt that is higher than the true current because of the inductance: u=L*di/dt. For small inductors di/dt gets large, therefore you will see a high voltage generated in the inductive part of the shunt resistor.You could either ignore it, compensate it or use a special ultra low inductance shunt or a current transformer.

author: nikkoderic
author_rank: Contributor
timestamp: November 01, 2016, 09:06:23 am
content: Quote from: _Wim_ on November 01, 2016, 08:19:05 amThe 10V from my bench power supply does drop a little more than a volt when the current overshoots. The moment the transistor turns off, the supply voltage spikes to 15V.Quote from: xavier60 on November 01, 2016, 08:33:11 amYes I know. That is assuming that the transistor is a relay or switch instead. I was just saying that because of the resistances that I have in my circuit and of course the transistor, the current should eventually settle to a certain current regardless of the inductance I place. And in fact, in the actual circuit, whatever value of inductor I place across the terminals, they all settle to a current of 43A. My question is why low value inductors overshoots that current first before settling down. Is it because the kickback diode does not discharge the inductor quickly?Quote from: bktemp on November 01, 2016, 08:58:42 amThat makes sense.Now, what really makes me scratch my head is that how the square waveform fed into the gate terminal is being affected by this.

author: xavier60
author_rank: Super Contributor
timestamp: November 01, 2016, 09:21:11 am
content: Another possible reason is that because the MOSFET is working as a constant current sink, a drop in rail voltage after the MOSFET turns on will also cause a drop in Gate drive voltage.Edit, I just noticed that the control circuit is powered from a separate source.

author: nikkoderic
author_rank: Contributor
timestamp: November 01, 2016, 09:51:22 am
content: Quote from: xavier60 on November 01, 2016, 09:21:11 amYes. Additionally, the gate voltage actually rises more than 9V supply voltage

author: AlfBaz
author_rank: Super Contributor
timestamp: November 01, 2016, 09:58:08 am
content: Although I don't recall seeing this with the rise times (<1us) shown in your scope pictures, it may still be worth checking out, especially as it's an easy fix/checkExcessive lead length on your scope's ground lead and possibly return path from the shunt resistor back to the supply can cause this by placing a substantial amount of inductance in the return path.Try shortening the scope probes ground leadandensuring a low impedance/inductance return path from the shunt resistor, see if that changes anything

author: Andy Watson
author_rank: Super Contributor
timestamp: November 01, 2016, 10:12:09 am
content: Quote from: nikkoderic on November 01, 2016, 09:51:22 amConsider the gate capacitance - it is, in part, referenced to the source. When you have charged this capacitance and turned the fet on there is nothing to prevent the gate from tracking the source - so it rises with the source voltage. If the gate voltage rises sufficiently you risk reverse driving the 555 resulting in damage or odd timing behaviour.

author: 2N3055
author_rank: Super Contributor
timestamp: November 01, 2016, 10:21:46 am
content: Read the datasheet.. In order for your mosfet to reach 100A, it has to have Vgs of more than 6V.To reach reach 100A, that would be 10V on a 0.1shunt. Plus 6v for Vgs, it means you have to power 555 with 16V, at least...To reach 100A, you need to use smaller shunt, about 1-2V drop for max current..... And that's a lot, 100W + for resistor...

author: xavier60
author_rank: Super Contributor
timestamp: November 01, 2016, 10:33:57 am
content: Quote from: Andy Watson on November 01, 2016, 10:12:09 amWhen you see the internal schematic for the BJT type 555, when the output pin is high, it is able to rise well above rail voltage . Except R12.Edit, actually it is about 1.2V.

author: nikkoderic
author_rank: Contributor
timestamp: November 01, 2016, 11:15:40 am
content: Quote from: Andy Watson on November 01, 2016, 10:12:09 amI kind of getting it now though I'm not 100% sure if I'm thinking this right. So I can imagine it like having a capacitor tied to the gate and the source of the mosfet; that means that the other end of that capacitor is also tied to the current shunt. The voltage of the current shunt resistor plus the voltage across the gate-source capacitor is the voltage seen by the output of the 555 timer. Is that right?

author: Andy Watson
author_rank: Super Contributor
timestamp: November 01, 2016, 11:28:57 am
content: Quote from: nikkoderic on November 01, 2016, 11:15:40 amI think that is the mechanism. The details are slightly blurred because the actual capacitance is distributed across the fet device.

author: nikkoderic
author_rank: Contributor
timestamp: November 01, 2016, 11:41:30 am
content: I think that answer is sufficient for me to be able to sleep tonight hahahahaThank you

author: bktemp
author_rank: Super Contributor
timestamp: November 01, 2016, 01:58:11 pm
content: If you post a picture of circuit, maybe we can find some improvement.If you look at the original circuit, it is built very compact with almost no wires between the components (SMD current shunt connected almost directly to the mosfet). And the output to the scope is a coaxial cable connected directly to the shunt.By the way, at 100A your circuit is useless, because the voltage drop due to the resistance counteracts the current rise of the inductor. To get accurate results, the voltage drop should be less than 10% of the supply voltage, maybe 25% with degrated accuracy. So at 10V the maximum useable current will be less than 20A. That's why a low value shunt is very important. But on the other hand, a lower shunt resistance makes measurements of a fast rising current more difficult, because the voltage drop is lower, but the voltage generated by the parasitic inductance stays the same. To get a low voltage drop, you also need a low impedance power supply, large ultra low ESR capacitors are therefore mandatory.