How to measure 50 pA current with high "noise" current? - Page 1
Forum section: electronics / projects

author: Kunibert
author_rank: Newbie
timestamp: April 25, 2020, 11:56:03 am
content: Hello,I am new to the forum and I am approaching you directly with a problem.I would like to measure a slowly varying (F<<1Hz) current from -50 to 50 pA. So far I have achieved this with a simple transimpedance amplifier using a resistor + capacitor in the feedback path and a guardring.However, in this case there is a parasitic capacitance of 3 pF between the measured object and a switched voltage source. A current of 150 mA couples in via this by voltage switching. Unfortunately, the coupling cannot be further reduced. These 3 pF are the only capacitance at the input.Basically, it should be sufficient if the transimpedance amplifier remains in the linear range?I have the following ideas:1) Two low-leakage diodes anti parallel to ground. But in simulations the controller becomes unstable.2) Input capacitor, which stores the introduced charge and limits the voltage rise. But the transimpedance amplifier can never lower the charge because of the high feedback resistance. In addition, the additional input capacitance causes instability at the controller3) The couplings are regular and predictable. It may be possible to decouple the measuring circuit for the points in time and then there would be about 50us left for settling and measuring.4) Making the feedback resistance variable also causes oscillations in the simulation.Does anyone have any ideas or literature/links? How to design switches with low capacity and leakage current? Reed relays and solid state relays will probably not work?Thanks a lot

author: ogden
author_rank: Super Contributor
timestamp: April 25, 2020, 12:09:06 pm
content: Usually you fight with such problems by eliminating them completely, in the source. Easiest solution: get rid of the switching supply. Depending it's one-off measurement or design of (mass market) product, just use battery + low noise LDO or mains transformer supply + low noise LDO.

author: Kunibert
author_rank: Newbie
timestamp: April 25, 2020, 12:37:16 pm
content: Unfortunately, this is not possible due to the principle. The voltage (~3750 V) generates an E-field, which is needed for the measurement. The coupling capacitance cannot be reduced because the measurement setup cannot be changed. The interference is not caused by the voltage source, but by switching the voltage. I didn't express myself very well. English is not my first language.I am aware that far more charge is transferred by the parasitic current than by the measuring current. I hope that the DC balanced interference can be compensated somehow.

author: dmendesf
author_rank: Frequent Contributor
timestamp: April 25, 2020, 02:07:31 pm
content: What about using a sample and hold in sync with the switching power supply to sample the amplifier output when the waveform is undisturbed?

author: Kleinstein
author_rank: Super Contributor
timestamp: April 25, 2020, 02:26:09 pm
content: One could try adding an electronic switch to connect the signal to ground and maybe also isolate the integrator.The usual CMOS switches have a charge injection in the 1-10 pC range, this is a lot less than some 3750 V*3 pF =11.25 nC. With something a an inductor / 100 K resistor towards the TIA one may not even need the switch to the TIA, just a connection to ground.Edit: using low leakage diode could be a problem, as the really low leakage diode show quite some reverse recovery. So much of the charge could flow back from the diode in the following 20-50 µs. If at all one could try a transistor junction (e.g. base+collector to emitter) : these are also relatively low leakage, but faster with a reverse recovery time of more like 0.1- 1 µs compared to some 20 µs for low leakage diodes.

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: April 25, 2020, 03:01:50 pm
content: Hi,To the original posterI just want to point out the obvious:I= C dv/dtto get 50pA with a 3pF coupling capacitancedv/dt = I/Cyou only need a dv/dt of16.6 V/secondto get 50pACan you post the schematic of your TIA?Regards,Jay_Diddy_B

author: ogden
author_rank: Super Contributor
timestamp: April 25, 2020, 03:58:21 pm
content: You would want to consider synchronous demodulator - means of measuring low-level signals in the presence of relatively high noise levels.https://www.analog.com/en/analog-dialogue/articles/low-power-synchronous-demodulator.html

author: David Hess
author_rank: Super Contributor
timestamp: April 25, 2020, 05:38:02 pm
content: A transimpedance amplifier cannot easily be clamped at its input because that will configure it to multiply its own offset voltage. See the attached example of how to fix this.What you can do instead though is clamp it from the output to the inverting input either with a low leakage FET or low leakage diodes. Transistor base-emitter junctions make good low leakage fast diodes.If the high voltage edge is consistent, then charge compensation with another edge and capacitor could be used.

author: Someone
author_rank: Super Contributor
timestamp: April 25, 2020, 10:42:06 pm
content: Quote from: David Hess on April 25, 2020, 05:38:02 pmThere are relatively common scientific instruments that have user tuned charge injection compensation (the measuring setup changes enough each use), its fiddly but works when the highest possible bandwidth measurement is needed.The OP hasn't mentioned what they are trying to extract from the signals, if its just the DC then there are many other options.Quote from: ogden on April 25, 2020, 03:58:21 pmThe problem here isn't uncorrelated noise (which lock-in or modulation will remove) but a large unwanted coupling from the exciting signal, its already modulated!

author: David Hess
author_rank: Super Contributor
timestamp: April 26, 2020, 03:39:52 pm
content: Quote from: Someone on April 25, 2020, 10:42:06 pmI am more familiar with charge compensation for sample and holds but the same principles apply. When implemented in curve tracers to remove fixture capacitance, it is called loop compensation.

author: dmendesf
author_rank: Frequent Contributor
timestamp: April 27, 2020, 02:09:28 am
content: You can treat the large signal as the modulating noise to me removed using a synchronous sampler...Quote from: ogden on April 25, 2020, 03:58:21 pmThe problem here isn't uncorrelated noise (which lock-in or modulation will remove) but a large unwanted coupling from the exciting signal, its already modulated!

author: Marco
author_rank: Super Contributor
timestamp: April 27, 2020, 02:40:29 am
content: Photorelay across the feedback resistor, trigger when output voltage exceeds positive/negative threshold?

author: Kunibert
author_rank: Newbie
timestamp: April 27, 2020, 04:25:32 pm
content: Quote from: dmendesf on April 25, 2020, 02:07:31 pmThe TIA doesnt reach a steady state due to the high current pulse and the high feedback resistanceQuote from: Kleinstein on April 25, 2020, 02:26:09 pmBut how to design a switch with a low off-state current and fast switching capability? The input offset of the OpAmp causes a voltage across the transistor, therefore several nA current flows. The body diode could also cause problems.Quote from: Jay_Diddy_B on April 25, 2020, 03:01:50 pmUntil now, its just a TIA like this (+Capacitor in parallel to the feedback resistor)https://upload.wikimedia.org/wikipedia/commons/thumb/f/f1/TIA_simple.svg/400px-TIA_simple.svg.pngIts a High-Voltage Source with good decoupling/ filtering. Voltage seems to be stable after switching, I will measure it again soon.Quote from: David Hess on April 25, 2020, 05:38:02 pmBut it needs to sink and source charge due to the positive and negative pulse. For this, the diodes would have to be connected in antiparallel. Wouldn't this lead to an undefined gain?Charge compansation is the right Keyword. In fact, the waveform does not have to be inverted exactly. With a calibrated compensation voltage this could work!Quote from: Someone on April 25, 2020, 10:42:06 pmIts "just" the DC-currentWhat other options do you have in mind?Quote from: Marco on April 27, 2020, 02:40:29 amI've been thinking about that, too. But most photocouplers are too slow (turn-on/off time up to some ms) and the off-state current is much higher than my measuring range. Could not make it work.Many thanks for the helpful answers!

author: Marco
author_rank: Super Contributor
timestamp: April 27, 2020, 04:45:29 pm
content: Quote from: Kunibert on April 27, 2020, 04:25:32 pmTLP3440S switches in 100 us and shows 1 pA or less leakage for a 5V signal at room temperature on its datasheet.I'm sure there are others with similar spec. What is 100 uS on a <<1 Hz signal? Make the trans impedance slightly higher bandwidth and do the rest of the filtering digitally BTW. It's much easier to do non-linear filtering to clean up the remaining glitch digitally.

author: Kleinstein
author_rank: Super Contributor
timestamp: April 27, 2020, 04:54:53 pm
content: Campling to ground can work, if there is enough resistance between the switch and the TIA. With some 100 µV offset and 100 K ohms the current would be in the 1 nA range. This is not good but could be still manageable for the TIA without going to saturation.I think the clamping to the output should be the better solution. With the regular clamping to the output one could use a charge amplifier and see the phase with the high current pulse as integrator reset. So one would measure the current from the rate the charge accumulates after the reset.The input of the charge amplifier is reasonable close to ground, so that switching should not be so difficult and with a well defined charge injection. This could be a JFET switch with some compensation of the gate charge. A modern CMOS switch chip is probably easier to use. Suitable switches can be below 1 pC charge injection at a suitable voltage. Switching speed can be below 100 ns, which should be fast enough. It more like needs several µs for the integrator reset.One may have to do a separate zero measurement to compensate for the leakage currents from the CMOS switch and the OP. Leakage current can be below 10 pA, at least for a typical value.

author: David Hess
author_rank: Super Contributor
timestamp: April 27, 2020, 09:17:01 pm
content: Quote from: Kunibert on April 27, 2020, 04:25:32 pmOr anti-series so they clamp at a higher voltage. Diode bridges can be used to clamp at a precision voltage.Quote from: Marco on April 27, 2020, 04:45:29 pm2N4117 JFETs would have been used in the past and are much faster. 3N series MOSFETs would be good but are basically unavailable now.

author: Someone
author_rank: Super Contributor
timestamp: April 27, 2020, 11:00:52 pm
content: Quote from: Kunibert on April 27, 2020, 04:25:32 pmAs always the optimal solution depends on the specifics of your measurements, their environment, constraints etc.You've already the simplest answer in your first post:Quote from: Kunibert on April 25, 2020, 11:56:03 amNow all you need to do is calculate some settling times and find out if there is low enough error left in the measurement.

author: Kunibert
author_rank: Newbie
timestamp: April 28, 2020, 10:38:52 am
content: I have not found any photocouplers that switch below 1µs. The interference signal has 13 kHz and 75 ns pulse width. Anyway, I have found suitable CMOS analog multiplexers/ switches. The IC has better specifications than I could achieve with a discrete JFET solution.Quote from: David Hess on April 27, 2020, 09:17:01 pmanti-series = -----|>|---|<|----- or -----|<|---|>|----- ?Can you explain this a bit more detailed? I do not know what that means.I will try all suggested approaches and will report as soon as I have first results.

author: graybeard
author_rank: Frequent Contributor
timestamp: April 28, 2020, 08:49:42 pm
content: Find a way to chop (turn off-and-on) the current you want to measure. Then run the noisy measurement into a lock-in amplifier synced with your chopping signal.

author: Kunibert
author_rank: Newbie
timestamp: May 05, 2020, 09:59:07 am
content: I found an analog switch with a charge injection of 0.35 pC.Do one simply connect a switched 1 pF capacitor to an adjustable power supply (~0.35 V)? Or do one use a dummy switch of the same type?Does anyone have experience how to set up a capacitance compensation?

author: Kleinstein
author_rank: Super Contributor
timestamp: May 05, 2020, 10:10:47 am
content: I would not worry much about 0.35 pC of charge injection this would be no big problem, more like corrected mathematically if needed.If really needed it could be some small capacitance (possibly adjustable) and a square wave signal with some 3-5 V. However the quoted charge injection depends on the input capacitance and DC voltage level. So the compensation may not work perfect in all cases.A dummy switch maybe used on the same chip, but hardly externally, as the charge injection is different for on and off (for the off -> on transition one gets both sides, off only one. So if at all one would need a switch of halve the size.Chances are a very low charge injection switch mal already use some compensation inside the 0.35 pC may be just the typical left over.One may than be able to adjust the DC level for the switch to get zero charge injection at that one voltage that is needed.The compensation idea was about the 3.7 kV times some 5 pF causing the large spikes.

author: Kunibert
author_rank: Newbie
timestamp: May 05, 2020, 06:41:48 pm
content: Quote from: Kleinstein on May 05, 2020, 10:10:47 amThe charge injection of the analog switch is large compared to the charge introduced during the measurement period. However, the measuring range can be adjusted as required.I will save further attempts for the V2, if the charge amplifier is promising.Quote from: Kleinstein on May 05, 2020, 10:10:47 amI will give it a try!Thank you!