Integrating Voltage over a Resistor - Page 1
Forum section: electronics / beginners

author: AQUAMAN
author_rank: Regular Contributor
timestamp: March 13, 2014, 02:45:56 pm
content: Hi allI am charging up a capacitor that has a very large internal resistance (its a gate of an IGBT/MOSFET) through an external resistor of about 1.5Ohms.You can see the current and voltage across this resistor (top graph) and the voltage on the capacitor (lower graph) in untitled.jpg.What I want to do is find a way of evaluating the charge on the capacitor after about 250-300ns. The internal resistance of the gate capacitor is high - about 2ohms and changes about 2mOhm/C. So using just the voltage across it isn't sufficient. I want to accurately measure the charge accumulated in the gate after 300ns at different temperatures, which will roughly be dependent on this change in internal resistance as all other parameters are stable in comparison.What I have thought of doing, is integrating the voltage seen across the gate resistor. Obviously the proper way to measure charge is to know the current and time, correct? But this voltage will be proportional to the current flowing into the gate.This is done with 2 opamps - 1 differential amplifier with gain 1, and then an integrator. I have attached an LTSpice pic. Obviously I do not get the charge on the capacitor, but I do get a voltage at the output of the integrator that appears to be proportional to the charge on the capacitor, at least until the opamp saturates.I attach the circuit and waveforms, does anyone have any comments for improvements and whether this is OK? My main questions are whether its OK to have an opamp swinging from +ve saturation to -ve saturation constantly, or whether I need another component to periodically discharge the capacitor in the integrator opamp?I am also concerned about when I put this output voltage into an ADC. Its rising almost 30V in 300-500ns. I am worried about the sample and hold time óf the ADC. If there are small fluctuations then the reading could be invalidated. If I increase the integration capacitance it slows the rise time but I get decreased sensitivity at the output - and my goal is measuring the variation of charge with temperature.Lastly, does anyone have any other ideas to do this? Other ways would be actually mirroring the current through the gate resistor into an external capacitor of similar size to the gate and measuring the voltage on this. But when I try to simulate this (putting transistor in the current path with a current mirror) it severly effects the switching of the MOSFET. Or perhaps you could mirror the current with a gain of 0.1x into a capacitor 1/10th of the size to reduce the power consumption of the measurement? I dont know how you would do this.I hope Ive explained everything well.Any help appreciated

author: uwezi
author_rank: Unknown
timestamp: March 13, 2014, 04:39:42 pm
content: What's your goal? What accuracy do you need?Are you trying to determine the temperature of the MOSFET by the temperature coefficient of the internal series resistance of the gate?Banging the opamp into saturation is not a problem for the opamp, but of course: it will slow you down when reversing and it will not give you any meaningful data. You should scale your amplifiers and the shunt resistor to avoid saturation - but then of course your integrator will also integrate the offset voltage and current of both amplifiers.The 100k resistor at the output of the first opamp doesn't have any function.Make R2 or C2 larger and you will avoid the saturation issue.

author: AQUAMAN
author_rank: Regular Contributor
timestamp: March 13, 2014, 05:40:24 pm
content: Yes, that is my goal.I can also measure the gate voltage directly from that potential divider in the gate path, I just put it into an ADC I think, but the sample and hold has to be triggered at the right point - right after the voltage has settled from the overshoot due to the inductance. Around here the gate current is around 5-7A and I can see 4mV/C. Ideally I want my integrator circuit to give something higher than 4mV/C. It should do as this will compensate for the slight decrease in gate current. IE if I knew the ACTUAL voltage seen on the capacitor at say, 300ns, the sensitivity would definitely be higher than 4mV/C.When I change the values so that the opamp doesnt saturate, the output doesnt return to the same point each time (see pic)

author: AQUAMAN
author_rank: Regular Contributor
timestamp: March 13, 2014, 05:43:13 pm
content: I have read there is supposed to be a large resistor in parallel with the capacitorBut when I put it in, it either does nothing or the opamp output starts at 0V or something. It changes the behaviour of the output anyway.This pic is with a 500k resistor in parallel. R2 5k and C2 30p

author: uwezi
author_rank: Unknown
timestamp: March 13, 2014, 06:20:09 pm
content: Your pulse train is asymmetric, the low pulses are shorter than the high pulses. This results in a slow built-up of charge on the integrating C2. Over time your integrator will run into either supply rail - again. Adding a bleeding resistor over the integrating capacitor actually gives the circuit a decay time constant, i.e. the integrated charge on C2 will leave the capacitor with a time constant of R times C. If your time constant is shorter than your pulses, the capacitor will never acquire a stable charge, if it is much longer then it will not help you, because you will still build up remaining charge as soon as your duty cycle is not exactly 50%. What you actually have then is a low-pass filter.What you really should consider would be a switched reset of the integrator when your gate is off:- section A.4.8 inhttp://www.ti.com/lit/ml/sloa091/sloa091.pdfGood luck...

author: AQUAMAN
author_rank: Regular Contributor
timestamp: March 13, 2014, 07:09:51 pm
content: But surely it doesnt matter how long my pulses are? The gate gets charged up and discharged the same amount no matter what

author: AQUAMAN
author_rank: Regular Contributor
timestamp: March 13, 2014, 07:37:43 pm
content: Actually I think it works.THis is with a 1Meg resistor in parallel. The upper voltage is what the integrator is integrating fyi.It doesnt matter that my duty cycle is changing I think - the gate gets charged up the same each time (or should do)But I am wondering whether I need to worry about making the circuit ready for continuous operation yet. At the moment I switch the MOSFET by applying a single pulse that is controlled by me when I click enter. So 1 pulse a second maximum. Surely I could just build the circuit I posted in my first post and the circuit should reset itself naturally, given that there are 10 seconds between measurements?

author: uwezi
author_rank: Unknown
timestamp: March 13, 2014, 10:18:54 pm
content: Quote from: AQUAMAN on March 13, 2014, 07:09:51 pmThe gate does, but I was talking about the capacitor C2 in your integrator!

author: AQUAMAN
author_rank: Regular Contributor
timestamp: March 13, 2014, 10:35:33 pm
content: This still doesnt matter I believe. The area underneath the area underneath that voltage across the gate resistor is the same in each pulse regardless of the switching frequency or duty cycle I use.I think that graph settles out if I leave it on longer

author: uwezi
author_rank: Unknown
timestamp: March 14, 2014, 08:00:14 am
content: Quote from: AQUAMAN on March 13, 2014, 10:35:33 pmYou are charging a (gate) capacitor through a resistor, which strictly means that you would have to wait infinitely long until the current really drops to zero for both pulse direction.For practical purposes you do not have this problem, because if you just wait 10 RC delays or so, all other imperfections in your circuit will have a bigger influence - for example the offset voltages and bias currents of both your operational amplifiers, which you will integrate as well.