Op Amp buffer for high voltage divider  - Page 1
Forum section: electronics / projects

author: hrsx
author_rank: Newbie
timestamp: December 17, 2020, 08:51:14 pm
content: HiI built a high voltage power supply rated for -50 kV at 10 W and it works pretty well. By now I have spent far more time on a seemingly easy task than on the power supply itself: Measuring the output voltage. The voltage divider I'm using is made from 3x 500 MOhm resistors and one 500 kOhm potentiometer set to 75 kOhm. When I measure the voltage using a multimeter it works as expected. But I want to read the voltage using an ADC (ADS1115). Since the effective impedance of the ADC is pretty low I need a buffer. A simple enough task for an op amp, I figured (TL072). Just to make sure I added a low pass filter to the feedback loop. I attached a schematic of my circuit below.When I run the simulation it works fine and even when I test the real circuit with a (low impedance) lab power supply as the input it works. As soon as I connect the voltage divider to the op amp, things go wrong and it's driving me insane: Without the op amp the input is at -1V where it should be. Once the op amp is connected, the INPUT (and with that also the output) goes to +2V! HOW? The ratio of "old input" to "new input" is not constant over the range of the applied voltages. For about the first -500 mV everything works almost as expected but then the input suddenly starts to go positiive.I've triple checked all connections and done a lot of research on Google in the forum, but I am absolutely clueless what causes this behavior. Any help is much appreciated.Thanks!HS

author: Slh
author_rank: Regular Contributor
timestamp: December 17, 2020, 09:29:08 pm
content: Try moving the low pass to the Vout (I.e. not in the feedback loop). I expect some people who know the ins and outs of opamps could correct me but that's an unusual circuit.What's the voltage at the opamp output and does anything strange happen on startup?

author: magic
author_rank: Super Contributor
timestamp: December 17, 2020, 09:36:32 pm
content: Yep, take feedback frombeforethe resistor.Otherwise two things happen:1. the opamp tries to make the voltage after the resistor follow the input exactly, nullifying the lowpass2. it may end up oscillating at high frequency, something you won't see without a scope that can also result in the average output being wrongeditThen use the second half of the TL072 to buffer the filter's output to the ADC.

author: Zero999
author_rank: Super Contributor
timestamp: December 17, 2020, 09:56:09 pm
content: If you want a filter with a low output impedance, look up Salen-Key filter. The downside with this type of filter is the attenuation is lower at high frequencies, when the gain of the op-amp is lower, so a capacitor will still be required on the input side, which can just be a capacitor between the input and 0V.

author: David Hess
author_rank: Super Contributor
timestamp: December 17, 2020, 11:51:34 pm
content: There last example here shows how to drive an output capacitor without oscillation which might be desirable for a low AC output impedance:https://www.analog.com/en/analog-dialogue/articles/ask-the-applications-engineer-25.htmlAlternatively in the extreme case, there is a good chance it will work if the series resistance is reduced to zero ohms and a capacitor with high ESR is used at the output like a 10 microfarad solid tantalum or 100 microfarad aluminum electrolytic. The reason this can work is that the high capacitance implements dominant compensation and reduces the gain enough at high frequency to prevent oscillation. The high ESR adds a little bit of phase lead for better stability.

author: S. Petrukhin
author_rank: Super Contributor
timestamp: December 20, 2020, 02:05:27 am
content: Sorry, 1.5 GOhm and 75 kOhm is an interesting divisor. Wouldn't its accuracy be too weak with thermal drift? After all, the change in the resistance of one resistor by 1 degree will be gigantic, commensurate with the change in the resistance of another resistor at the same temperature.

author: magic
author_rank: Super Contributor
timestamp: December 20, 2020, 10:31:31 am
content: A divider is a divider. Either resistor changing its value by 1% changes the output by about 1% too.

author: Kleinstein
author_rank: Super Contributor
timestamp: December 20, 2020, 11:02:16 am
content: There are 3 effects with a very high resistance divider:1) the input impedance to the amplifier gets more important. With only 75 K at the low end this would not so bad here.2) Leakage currents around the large resistor get more important. It does not much extra leakage to add 1% to 1.5 G.3) The TC of very high value resistors is often relatively poor. This is because they are more like thick film, metal oxide or carbon film.One still has to balance with the heat at the divider. So it is natural to use higher resistors, with 50 kV. At this voltage 1.5 G Ohms is not even that high.

author: S. Petrukhin
author_rank: Super Contributor
timestamp: December 20, 2020, 05:29:37 pm
content: Quote from: magic on December 20, 2020, 10:31:31 amHow much will the resistance of the 1.5 MOhm resistor change and how much will the resistance of the 75kOhm resistor change with a drift of 1%? Don't you realize that the lower arm of the divider will remain unchanged, but the upper arm will float up? The drift is indicated as a percentage of the nominal value, not in units of resistance. I.e. large resistances drift more in absolute value. The proportions of the divider are strongly drifted.

author: fcb
author_rank: Super Contributor
timestamp: December 20, 2020, 05:46:54 pm
content: TL072 is a dual opamp part. Why not use one opamp as straight buffer (signal into +IN, OUT connected to -IN) - then any filtering - then the second opamp to buffer the filter and drive the ADC.As the ADS1115 is pretty slow (so your application has a low BW), you might want to add a capacitor across the lower part of your divider (perhaps a COG or Poly, something with very low leakage), a simple first order low pass filter for nearly free.

author: Benta
author_rank: Super Contributor
timestamp: December 20, 2020, 05:57:42 pm
content: That's no low pass filter, it's... something. At a first glance more like a high pass filter (don't want to waste my time doing the maths).Short R3, remove C1 and see what happens.

author: Kleinstein
author_rank: Super Contributor
timestamp: December 20, 2020, 06:02:56 pm
content: The absolute value of the resistor drift will be large for the large resistor - but who cares. It is about the change in divider ratio.With such a high voltage one essentially need such a large resistance (1 GOhms may be OK too, but 5 G could be an option too) to keep the power loss at reasonable level. 50 kV and 1 GOhms would still be 50 µA and 2.5 W , which is an awful lot for an accurate divider. So one would more like consider 10 G or even 50 G.The Sigma delta ADCs are integrating. So they have a kind of build in low pass filtering option.Some capacitance in parallel to the lower lag may be a good idea to compensate for parasitic capacitance parallel to the 1.5 GOhms. With a 1:20000 divider even 1 pF parasitic capacitance would want 20 nF to avoid peaking.

author: hrsx
author_rank: Newbie
timestamp: December 20, 2020, 06:33:52 pm
content: Thank you all very much for your contributions.Quote from: Benta on December 20, 2020, 05:57:42 pmThat is correct, I messed up there. Fixed it, but it had no effect on the circuit behavior.Quote from: magic on December 17, 2020, 09:36:32 pmThat really got me thinking about AC. I was very hesitant to use my scope in HV applications because as a kid I ruined a nice scope with high voltage. But I finally used it. Turns out the voltage divider is picking up several Volts of AC from the multiplier circuit that it is running next to. That causes the op amp input to hit the rails and cause some overvoltage protection to happen, which then changes the input and by that the output. The remaining signal has a wrong average, which is what I was measuring all this time. I could have actually found that out just by using a multimeter to measure the AC component of the signal. Also, the frequency measurement shows the 40 kHz of the driver for the multiplier. I have no idea why I didn't think of checking this before. I have now fixed the problem simply by using a (correctly implemented) low pass filter on the input signal for the op amp. It works perfectly, problem solved!The ambient temperature should not be a problem because that will stay pretty constant. However, the 3x 500 MOhm resistors will dissipate about 1,7 W in total. They are fully embedded in a generous ammount of silicone which acts as a heat spreader and also provides good electrical insulation, so leakage should be minimal. The 75 kOhm of the potentiometer sit in air but it doesn't even dissipate 0.1 mW. That is why I was not expecting a lot of drift. Looking back I probably should have used an even higher resistance. I'm only aiming for around 5% accuracy over all. I'll just have to see how it performs.

author: magic
author_rank: Super Contributor
timestamp: December 20, 2020, 09:09:07 pm
content: Quote from: S. Petrukhin on December 20, 2020, 05:29:37 pm1% drift on 1.5G is 1.515G.1.5G/75k = 200001.515G/75k = 202001% error in the division ratio.BTW, I cheat because it should be (1.5G+75k)/75k but it doesn't matter. Do the math.Quote from: Benta on December 20, 2020, 05:57:42 pmAlways look at feedback.It's a unity gain buffer which makes the TL072 work needlessly hardIf you took the output from the OUT pin,thenyou would see unity gain at low frequencies and rising slope above the RC cutoff frequency, then falling slope as the TL072 runs out of open loop gain and oscillation as it runs out of phase margin at 2MHz or whatever.My guess.

author: David Hess
author_rank: Super Contributor
timestamp: December 20, 2020, 10:52:02 pm
content: Quote from: Kleinstein on December 20, 2020, 11:02:16 am4) When high voltage is present which is often the application for high resistance dividers, the voltage coefficient of resistance becomes a significant source of error if the divider resistors are not of the same type and value. Stacking the same resistor to make a divider should compensate for this to some extent.Low voltage circuits also suffer from voltage coefficient of resistance at the part-per-million level.

author: shridmaster
author_rank: Contributor
timestamp: December 21, 2020, 08:29:22 am
content: Ironically, going with such a simple solution is needlessly complicating this.First of all, Gigaohm resistors are....not fun to use. You run into non-idealities. Avoid when possible. While it's important to conserve current in voltage detection, I'm sure you can spare more than 30uA on a 50kV system. Reduce that 1.5G if you can.Secondly, you're better off measuring current and converting that to a voltage. You can use a shunt resistor and feed that into an op-amp, or use an IC like a Hall sensor. There's lots of options for current sensing, use whatever fits your needs.

author: S. Petrukhin
author_rank: Super Contributor
timestamp: December 21, 2020, 08:53:26 am
content: Quote from: shridmaster on December 21, 2020, 08:29:22 amHow to measure current at 50kV?

author: fcb
author_rank: Super Contributor
timestamp: December 21, 2020, 09:26:09 am
content: Quote from: S. Petrukhin on December 21, 2020, 08:53:26 amI'm assuming they meant measure current at the 'bottom' of the HV dropper resistor?The TC of the dropper resistor will be the biggest PITA, also I suspect it's cheaper to pay for a higher accuracy resistor than calibration of the 50kV. Nicrom in Switzerland make nice pieces.

author: shridmaster
author_rank: Contributor
timestamp: December 21, 2020, 11:58:45 am
content: Quote from: S. Petrukhin on December 21, 2020, 08:53:26 amCurrent sensing resistor in series with something less than 1.5GOhm, as explained.

author: Harm314
author_rank: Newbie
timestamp: December 24, 2020, 11:39:48 am
content: Having no experience with high voltage systems, I tried to estimate an appropriate value for a current sense resistor for this voltage as an exercise for myself. In short, 75 kOhm seems pretty close (but could be lower). As pointed out above, the current through R1+R2 is completely dominated by R1 (with R2 only changing it by 50ppm). At 50 kV and 10W, the power source can only deliver 200 micro-amps, of which already 33 micro-ampere are used for the measurement. In theory R1 could be lowered to 250 MOhm, but would then need to dissipate the full 10W.R2 could be changed slightly (lowered) to optimize the input range for the opamp: For a 50 kV source the voltage over the sense resistor should be 2.5 V, not the measured 1V. With R1 potted there is possibly some leakage current.To suppress AC overload, you could add a low-pass filter at the input of the opamp; or use the two-stage approach suggested above.

author: bson
author_rank: Unknown
timestamp: December 24, 2020, 09:07:29 pm
content: A divider has Vout/Vin = R2/(R1+R2), and including TCs, you get (TC2*R2)/(TC1*R1+TC2*R2).If the two resistors have similar material properties such that TC1 = TC2 then they cancel.In reality they will differ by some epsilon e, TC2 = e * TC1 and you get Vout/Vin = R2/(R1+e*R2) where e = TC2/TC1.For example, TC2=1.00% and TC1=1.05%, then e = 1.0100/1.0105 = 0.9999. That's four nines!This is then further scaled as R2 << R1 and basically disappears. Or, a requirement on e could be calculated based on the resistor ratio to derive acceptable TC specs.