Measuring pA - cascading µCurrent units - Page 1
Forum section: electronics / beginners

author: SBS
author_rank: Newbie
timestamp: April 07, 2023, 03:06:16 pm
content: I am new to posting here so forgive if the format is not correct.We are trying to design a capability to measure pA within a glovebox, that can be used with our excisting potentiostat (for electrochemistry measurements).In short our (cost effective) idea was to cascade two µCurrent units, or modify the design so that we get a potential that is sufficiently high that we measure it outside the glovebox without a lot of noice.Does anyone have experience with a similar setup, or know of any pitfalls that we need to look out for.Thanks. (and agin sorry if the question format is incorrect)

author: Kleinstein
author_rank: Super Contributor
timestamp: April 07, 2023, 03:19:56 pm
content: To measure small current it needs a large series resistor (or a capacitor and integrate, but that is a different more advanced method and needs additional complications). The large resistor naturally has quite some noise. The µ current is a low dirft amplifier to measure moderate currents with a low drop at the shunt, but the principle is not really suited for small currents, like much below 1 nA.For the pA to nA range the normal method of choice is a transimpedance amplifier. So a kind of inverting OP-amp amplifier with the relevant resistor in the feedback. This way the voltage at the resistor can be quite large and the amplifier noise becomes less critical. So one can use low input current FET based OP-amps and does not need an AZ-OP amp like with the µC current.

author: SBS
author_rank: Newbie
timestamp: April 07, 2023, 04:45:25 pm
content: Thank you for the swift reply.I looked at the transimpedance amplifier (TI - pdf and wikipedia). If I was designing from scratch, I think I understand why this is the preferred route.However, our problem is a bit more complex, because we would like to integrate it into our excising setup, since the software, potential-source etc. is already in place.Presentation1.pdf(14.15 kB - downloaded 1131 times.)In short we have a variable current/voltage source, that can be controlled using the measured current (A) or the the potential from one of the two potential measurements (V1 or V2). The idea was to insert the new current measurement at the light blue box, and break the lower V2 connection and use it to log the response from the blue box current device. [The round symbol in the center is the electrochemical cell where current passes from the top ball (working electrode) to the flat plate on the bottom (counter electrode), the arrow is the reference electrode that "measures" the potential within the electrolyte of the cell (used as the potential reference for the rest of the measurements)].

author: ejeffrey
author_rank: Super Contributor
timestamp: April 07, 2023, 05:53:35 pm
content: How big are the voltages involved and how much of a measurement voltage drop can you tolerate? What is the dynamic range needed? How fast does the response time need to be? The uCurrent has a 10 kOhm shunt resistor which will be struggling with noise and resolution at the picoamp level. For picoamp measurements I would prefer to use at least a 1 M or 10 M resistor. However, this will increase the voltage drop in the circuit and limit the maximum current you can use. If your maximum current is a few nanoamps that's no big deal, if you need to go up to microamps that's going to be more challenging.A transimpedance amplifier eliminates the voltage drop across the resistor using a virtual ground. There is no reason you couldn't insert a transimpedance amplifier into the circuit at the location indicated.

author: metebalci
author_rank: Frequent Contributor
timestamp: April 07, 2023, 06:39:53 pm
content: What are the cons of a transimpedance amp ? I also had a need to measure ~1mA range AC current recently (with ~10kHz bandwidth) (at ~8 bit resolution so ~a few uA resolution) and the best and the simplest method I found is also using a transimpedance amp. I wonder why it is not common also in this range, or maybe it is common and I am mistaken.Edit: For the above example, I have a constraint that I could use max.1R current sense resistor.

author: alm
author_rank: Super Contributor
timestamp: April 07, 2023, 07:03:34 pm
content: A feedback ammeter (as the use of a TIA to measure current is more commonly called) is more complicated than a simple current shunt, and it draws more current: whatever current the DUT is drawing needs to be sourced from the TIA's power supply. For high currents there's the problem that the opamp needs to source the current, so for a 100 mA range you'd need a pretty beefy opamp. I would imagine stability could be problematic with difficult loads, like very inductive loads. Plus frequency response might be worse at higher frequencies where the TIA will appear inductive (run out of GBW). I don't see a reason either why you couldn't use it it the schematic you showed. For some more information about feedback ammeters see these two links:https://knowledge.ni.com/KnowledgeArticleDetails?id=kA03q000000x1AZCAYhttp://www.hep.fsu.edu/~wahl/phy3802/expinfo/labmanual/intlabdoc/instruments/keithley/Low_Cur_Meas_AN.pdf

author: metebalci
author_rank: Frequent Contributor
timestamp: April 07, 2023, 07:11:55 pm
content: Quote from: alm on April 07, 2023, 07:03:34 pmIt seems I was mistaken, many thanks for the info and the links.

author: Kleinstein
author_rank: Super Contributor
timestamp: April 07, 2023, 07:41:52 pm
content: A transimpedance amplifier usually needs a signal source that has as least some series resistance. So one often add some additional series resistance. For a TIA made to read down to the pA level one would normally have more than 1 Ohm.The TIA is an active circuit and thus needs an extra supply. For a low current battery supply may be OK.In the plan the ampmeter is shown in series with the supply at the low side. There is alread a amp-meter at the high side. Ideally the current would be the same for both, but common mode noise / hum from the supply can make a difference. So not sure the position as shown is a good choice.

author: MathWizard
author_rank: Super Contributor
timestamp: April 07, 2023, 08:20:26 pm
content: Back when they were doing the 1st radioactivity experiments, and atomic level experiments, what sort of charges and currents were they dealing with in their machines ? I thought I heard something just lately about Marie Curie's experiments, and measuring femtoAmps, or maybe femto-Coulumb's from quartz. Even if it was pico or nanoA or C, thats still beyond what most people can build.

author: SBS
author_rank: Newbie
timestamp: April 08, 2023, 02:57:02 pm
content: I do not know how to answer multiple post in one, so I put it here. (again sorry for the poor housekeeping)I looked more into the transimpedeance idea, and I hit a snag.Please see diagramPresentation1.pdf(124.91 kB - downloaded 55 times.)The electrochemical cell is here modeled naively as two resisters in series.Basically, the resistance ? is unknown, variable and intrinsic to the electrochemical cell. So I would think that I am no longer probing current but rather the "potential drop" across ? . Alternatively, I could given up the reference electrode and measure V1 across the electrochemical cell, but then I do not know where to put the current return path for the current sourced by the potentiostat i.e. where to put the negative lead.The compliance voltage of the potentiostat is +-15V. The dynamic range that I would need should be arround 1pA-10nA. Above this the noise is sufficiency low that we can just use the amp meter of the potentiostat.