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Suppose I have a capacitor and I want to observe its charge decay over time. How can I do that without affecting its discharge rate through measurement?

AFAIK a typical voltmeter runs current through a known resistance to determine voltage, but in the process this would discharge the capacitor being measured. With increasing complexity one could reduce the current required to make an accurate measurement, and then reduce the frequency of measurements, but in the limit the measurements will still drain some voltage.

In the hydraulic analogy it's possible to measure the pressure (voltage) by putting a spring gauge on a piston impinged by the two sides of the reservoir. No water flows from one side to the other, but we get a constant reading of the pressure.

So is there a meter, mechanism, or circuit that can do that for voltage on a capacitor or other power supply?

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    \$\begingroup\$ Do you have a gold-leaf electroscope available? Using an electroscope. \$\endgroup\$ – Andrew Morton Dec 13 '16 at 19:40
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    \$\begingroup\$ @AndrewMorton - jonk's answer seems to be proposing that. Still trying to get a sense of what sensitivity and precision one can achieve. (Also, curious as to whether these are just instructional toys or whether there is such a thing as a modern workbench electroscope designed for accurate measurements rather than just illustrating/estimating field effects.) \$\endgroup\$ – feetwet Dec 13 '16 at 19:55
  • \$\begingroup\$ @Optionparty - AFAIK that's not true of capacitor self-discharge: That occurs through the insulator, not between the electrodes. \$\endgroup\$ – feetwet Dec 14 '16 at 21:49
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Neat physics solutions aside, the practical way to do this is with a very low input bias current op-amp running in a buffer configuration. One of these op-amps with a properly designed layout can draw down to single-digit femtoamps of current from your cap, making disturbances pretty much negligible, particularly if you only connect the amplifier to the cap when you're taking a measurement.

Analog legend Bob Pease describes measuring leakage of a polypropylene cap using this method:

Now I will charge up some of my favorite low-leakage capacitors (such as Panasonic polypropylene 1 µF) up to 9.021 V dc (a random voltage) for an hour. I will read the VOUT with my favorite high-input-impedance unity-gain follower (LMC662, Ib about 0.003 pA) and buffer that into my favorite six-digit digital voltmeter (DVM) (Agilent/HP34401A) and monitor the VOUT once a day for several days.

[...]

Day 0: 9.0214 V
Day 1: 9.01870 V
Day 2: 9.01756 V
Day 6: 9.0135 V
Day 7: 9.0123 V
Day 8: 9.01018 V
Day 9: 9.00941 V
Day 11: 9.00788 V
Day 12: 9.00544 V
Day 13: 9.00422 V

The first day after soaking for an hour, their leak rate was as good as 2.7 mV per day. Not bad.

If you need to automate such a setup, a good old-fashioned reed relay has basically negligible leakage (better than even modern solid state analog switches) and can be used to briefly connect your amplifier to the capacitor under test in order to take a reading.

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  • \$\begingroup\$ Wow ... femtoAmp-seconds. When you put it in those terms I'm inclined to agree that this question is only interesting from a theoretical perspective. \$\endgroup\$ – feetwet Dec 13 '16 at 20:07
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    \$\begingroup\$ Drat, you mentioned Bob Pease while I was writing my answer :) \$\endgroup\$ – pjc50 Dec 13 '16 at 20:19
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    \$\begingroup\$ If you could swap in a low capacitance 'perfect' capacitor (maybe 20pF from plates in a vacuum) you could calibrate out the leakage from the op-amp/fixture and get even lower. 3fA /20pF would change about +/-150uV/second, easily measured. \$\endgroup\$ – Spehro Pefhany Dec 13 '16 at 21:13
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    \$\begingroup\$ "particularly if you only connect the amplifier to the cap when you're taking a measurement." note that the input behaviour of such op-amps is capaciance-dominated. So disconnecting the amplifier between readings is not likely to result in much improvement. \$\endgroup\$ – Peter Green Dec 14 '16 at 4:20
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Generally what you need to measure an electric field is an electrometer. The older gold-leaf electroscopes operate by the static repulsion between like charges, and if made of ideal materials would not leak any charge.

However, when you get really interested in the difference between a tiny current and no current flow, a large number of problems appear. All of your experimental apparatus has a finite (but very large) resistance. Electrons will happily tunnel a short way through solid objects. Alpha-decay in the materials generates a charge. Stray charge drifts in on the winds, or voltage is induced by passing fields.

The legendary Bob Pease has some good articles on the subject: What's All This Teflon Stuff, Anyhow? and What's All This Femtoampere Stuff, Anyhow?

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  • \$\begingroup\$ Various non-op-amp electrometers: vibrating reed electrometer for micro-scale, field mill electrometer, wobbulator (vibrating plate) electrometer, quadrant electrometer (a panel meter with moving capacitor blades, good for 200V to 30KV full scale.) Lots of "Sensitive Research Inc." quadrant electrometers are commonly on eBay for ~$100 ea. The weak point in these is the surface cleanness and humidity re. their insulating posts (teflon, ceramic, phenolic etc.) Long thin insulators are best, heh, teflon spider-webs as physical supports? \$\endgroup\$ – wbeaty Dec 15 '16 at 2:18
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The better methods will depend upon the voltage difference you are attempting to measure. Same would be true for your hydraulic analogy.

But your hydraulic analogy fails entirely in another regard. The accelerating forces acting on electrons in a conductor are caused by very few charges. I don't think you have a feel for just how few electrons are needed at the surface of a conductor to accelerate significant mean-velocities for charges in a wire. If you bend a wire into a U-shape, it might only take one or two extra electrons at the bend to completely re-direct amps of current.

You can measure high voltage differences because the amount of charge difference reaches the point where sensitive (pith balls on a hair-like thread, for example) can be applied successfully. In this case, the impact on current is just as negligible as your hydraulic example's momentary impact due to very slight piston flexures.

For small voltages, this doesn't work because the charge difference is so absolutely tiny and any finite distance away from the bare conductor surface greatly reduces the tiny force.

The electronic equivalent to hydraulic pressure is \$\frac{\textrm{volts}}{\textrm{meter}}\$ or \$\frac{\textrm{Newton}}{\textrm{Coulomb}}\$. Copper's conduction electron density at room temperature is about \$1.346\times 10^{10}\:\frac{\textrm{Coulomb}}{\textrm{m}^3}\$ and their mobility is about \$4.5\times 10^{-3}\:\frac{\textrm{m}^2}{\textrm{V-s}}\$. Assume a wire with a cross-section of \$1\:\textrm{mm}^2\$ and carrying \$300\:\textrm{mA}\$ of current. The electric field required is about \$5\:\frac{\mu\textrm{V}}{\textrm{mm}}\$.

The charge difference over reasonable distances needed to impel that current is negligible (which resides entirely on the bare surface of the conductor) and you wouldn't be able to set up an instrument to measure it at any finite distance away. The only way to make this work is to add a conductor to the surface of that other conductor at some point and allow these tiny charge differences to act on their atomic scales so that their incredible forces can impel electrons in your measurement instrument as well. In short, you need to allow a current to flow, because this IS the most sensitive way available to you (at non-military budget levels) to make those pressure measurements in electronics.

It's nice to think about analogies, of course. But as you already know, the scale also matters. There's a huge difference between the distances separating galaxies and the forces that meaningfully act at that level and the distances separating atoms and the forces that meaningfully act at that level. Put in a more tactile level we humans can think in terms of, there's a huge difference between the forces that are important to us for walking and getting traction and the forces that act on fruit flies, who can easily land on the surfaces of walls and the ceiling because gravity is far less important at their scale compared to static charge and roughness for them.

Scale matters, too.

So the analogy fails here. In electronics, the very best way to measure these extremely delicate and tiny forces, which are all that is needed to impel practical currents in circuits, is to set up a measurement system that can respond to them. This means allowing a current to be affected. There's nothing more sensitive than that.

That said, I'll return to the fact that you can still make measurements without a current if and only if the voltage differences are large enough to set up enough charge difference to measure.

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  • \$\begingroup\$ Good explanation and background. Can you add an estimate of the magnitude of voltage differences between capacitor pins that should be measurable through field effects? \$\endgroup\$ – feetwet Dec 13 '16 at 19:26
  • \$\begingroup\$ @feetwet See youtube.com/watch?v=8BQM_xw2Rfo for an idea about the voltages needed. \$\endgroup\$ – jonk Dec 13 '16 at 19:34
  • \$\begingroup\$ @feetwet By the way, when watching that video, be aware that their test does actually transfer a very few electrons, which must be replaced in the wire itself to keep working. So it does have a momentary impact on the current -- just not one you could measure. About like that hydraulic pressure sensor you are talking about, which also only has momentary and very tiny impacts when changes occur. \$\endgroup\$ – jonk Dec 13 '16 at 19:41
  • \$\begingroup\$ Yes, that's a helpful video. In fact, you wouldn't have to "steal" charge from the capacitor if you precharged the foil from another source. Suffice it to note that kV differences are sufficient to see static, mechanical effects. Now, if you can do it with a piece of foil on a string at those voltages, it seems plausible (to me) that a carefully designed meter (which could charge its own "sensor plate" to an arbitrary voltage) could be 1-3 orders of magnitude more sensitive/precise, which would bring this into the realm of workbench utility. Does that sound right? Do such meters exist? \$\endgroup\$ – feetwet Dec 13 '16 at 19:46
  • \$\begingroup\$ @feetwet These tiny charges that set themselves up on the surface can be reasonably considered as a fraction of an electron charge. Nothing you can imagine will be anywhere NEAR as sensitive as placing a conductor AT those charges where distances are measured in Angstroms and the forces therefore can operate meaningfully. The moment you stand back and try and use a field effect at human measurable distances, those forces are pretty much zero and difficult to measure. \$\endgroup\$ – jonk Dec 13 '16 at 19:53
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There are a couple of ways to measure voltage without a current flow.


The first thing that springs to mind is the piezoelectric effect. You would need to transfer enough charge from your capacitor in order to charge the crystal to the same voltage, but after that, there would be no current flow. This is the closest analogy to your hydraulic pressure gauge; you would read the voltage from the amount that the crystal flexes.

Think of something like a crystal phonograph cartridge. Movements of tens to hundreds of microns result in voltages on the order of millivolts, and this effect works in reverse. Obviously, you'd need a microscope of some sort to detect the movement — anything from an ordinary optical microscope to some sort of tunneling-current microscope, which would be very sensitive indeed.


For the second method, look up the original definition of potentiometer, which referred to a system that contained not only to the three-terminal variable resistor with which we're all familiar, but also an accurate voltage reference and a galvanometer to measure the current.

By definition, the current through the galvanometer is zero when the resistor is set to the unknown voltage.

Obviously, using a potentiometer to measure the self-discharge of a capacitor is problematic, because as soon as the capacitor voltage drops a bit, the potentiometer itself will start to supply current to recharge it. Therefore, you'll have to be constantly adjusting the resistor to keep the galvanometer nulled.

Of course, you could simply let the system come to equilibrium and read the capacitor's leakage current directly from the galvanometer, assuming it has a calibrated scale.

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  • \$\begingroup\$ I agree, the piezoelectric effect is the equivalent to the hydraulic pressure gauge. The walls of the crystal will deflect in proportion to the voltage applied. So, as the capacitor discharges, the walls will return to their "normal" state. With a calibrated microscope, you would be able to translate the walls movement to cap voltage, without the need of current flow! \$\endgroup\$ – Guill Dec 16 '16 at 9:08
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If your voltage is high enough, you can use a feild mill.

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    \$\begingroup\$ OK: I have a capacitor on my workbench. How do I use a field mill to measure the voltage across its terminals without running current between the terminals? \$\endgroup\$ – feetwet Dec 13 '16 at 18:01
  • \$\begingroup\$ "Extend" one of the poles to a largeish plate. Run your feild mill in proximity to it and you will have its voltage relative to ground. If you need the difference, use two plates, measure both and subtract one voltage from the other. It may be possible to "ground" it in one of the poles, but I have never experimented with it, only differential to ground. \$\endgroup\$ – winny Dec 13 '16 at 18:10
  • \$\begingroup\$ I've never heard of this being applied to non-ionizing voltages, and not sure how it could detect those. Can you elaborate, or give ballpark estimates of voltage sensitivity? \$\endgroup\$ – feetwet Dec 13 '16 at 18:23
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    \$\begingroup\$ Wait a minute: If the field mill is picking up a charge, then it has to be taking it from the capacitor, right? I.e., if a field mill can measure voltage on capacitor terminals then it will reduce the capacitor's voltage while running, which is the same problem with a conventional voltmeter I was wondering if one can avoid. \$\endgroup\$ – feetwet Dec 13 '16 at 19:31
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    \$\begingroup\$ Stealing charge? No, a field mill is like a nearby conductor, but wiggling. It can be yards distant from the measured object, or mm distant. mV resolution, or 100KV. Yes, it produces small AC charge-effects in the measured object. But no DC leakage. (Field mill is basically an electrostatic generator, where the measured object is the "field plate" of the generator, which never gets touched and so no DC femtoamps are drawn. Any energy in the measured voltage-signal is entirely coming from mechanical energy injected into the moving parts, NOT from the field plates of the generator.) \$\endgroup\$ – wbeaty Dec 15 '16 at 2:26
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Physicist here, probably about to be laughed out of the SE site for this theoretical answer, but here goes:

Why not measure the current non-pertubatively? Ideas:

  1. Put an ammeter on one leg of the capacitor. Integrate the current over time.
  2. Collect the lost charge onto a much bigger capacitor that's constantly monitored.
  3. Measure the electric field within the capacitor (assuming parallel plates or other accessible geometry).

Many low pressure gauges rely on the ionization of just a few atoms per second and measure the current caused by the now-free electrons hitting a cathode. Why not do the inverse and use the voltage over the charged capacitor to deflect ions in a high vacuum and measure their change in trajectory?

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  • \$\begingroup\$ Very last idea sounds interesting, and actually sounds like it could be turned into a practical and sensitive bench meter. I wonder if there is a commercial incarnation of it. #3 is not possible with most practical capacitors, though you can see the leading idea in other answers and comments is field measurement off the capacitor's terminals . #1 and #2 aren't helpful in this case because the idea is to look at the isolated self-discharge rate of the capacitor. That won't produce the same data if we're "discharging it but keeping track of what discharge is attributable to the measurement." \$\endgroup\$ – feetwet Dec 14 '16 at 4:13
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You can use an AD549 (costs about 30 EUR) as unity gain follower. The input resistivity is larger than the resistivity of standard wire insulation or standard PCB material in a typical circuit.

Note: There is a typo in the AD549 datashet (2014) page 9 it should be pin 6 where pin 5 is printed.

You should look for the Keithley (now Tektronix) whitpapers on low current measurements. Unfortunately the website is so user unfriendly that I found no way to create a link.

If you need something more intelligent, one can apply a voltage to the capacitor and regulate it so that there is no current. But this is not trivial and makes only sense under laboratory conditions, with very expensive low noise wires, good shielding, stable temperatures...

Have a look in the manuals of

  • Keithley Nanovoltmeter Model 2182A
  • Keysight NanoVolt Micro-Ohm Meter 34420A
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Your understanding of voltage measurement is incorrect. A voltmeter has a high impedance input (>1M\$\Omega\$, typically around 10M\$\Omega\$). Almost no current flows into the meter while taking a voltage measurement. The same is true of an oscilloscope.

You may be confusing voltage measurement with current measurement. Multimeters contain low resistance 'shunts,' through which the current you are measuring flows. The shunts have a low, but precise and known, resistance. Current flow through the shunt creates a voltage across it. That voltage is measured. Since the shunt resistance is known, the meter calculates \$I = V_{Shunt}/R_{Shunt}\$.

Measuring the voltage capacitor with a high impedance meter will cause charge to flow out of the capacitor and into the meter. Whether or not this will skew your results depends on the rest of the circuit and exactly what you're trying to measure.

Note that real capacitors are not ideal, and will discharge naturally over time. Depending on the type of capacitor, this self-discharging is significant or not. High quality film capacitors are very stable and will hold charge for hours or days depending on the circumstances. Aluminum electrolytics, not so much.

You can improve the accuracy of your reading by connecting the capacitor voltage to a high input impedance buffer, and then reading the output of that buffer. This way, your meter will draw tiny current from the buffer output, rather than off of the capacitor. A JFET-input op amp can have input resistances in the 1G\$\Omega\$ to 1T\$\Omega\$. This might be too high, and can cause problems of its own.

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  • \$\begingroup\$ What you have described is the mechanism for voltage measurement I described in the question. I acknowledge that the current flow in a typical voltmeter is small in absolute terms, but so long as it is non-zero and continuous it will always be significant for some capacitor, voltage, and/or duration. \$\endgroup\$ – feetwet Dec 13 '16 at 17:56
  • \$\begingroup\$ In general, any measurement factor can be significant or insignificant. A very small amount of current flows through the input termination resistor (1-10Meg) in the meter, true. But is your capacitor completely isolated in your circuit? Are there paths in the circuit through which charge can bleed away from the capacitor much faster than they would through the meter? Some tiny current flow is unavoidable in physical reality. Whether or not it is significant cannot be answered in general. \$\endgroup\$ – vofa Dec 13 '16 at 18:01
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    \$\begingroup\$ This question is about nothing other than measuring the self-discharge rate of the capacitor. I am asking if in practice (or even in theory) it is possible to do this without creating current flow between the terminals of the capacitor (other than trivially at the moment the meter is connected). Your comment says current flow is unavoidable. That is true of voltage meters of the type we describe. But is there a law or proof that it is true of voltage measurement in principle? \$\endgroup\$ – feetwet Dec 13 '16 at 18:06
  • \$\begingroup\$ The voltage measurement device will have some input resistance. As you increase that resistance, less current will flow. Even at 100Teraohms and 1V, 10fA will flow. If that current flows for 1 second, over 600,000 electrons have flowed through the termination resistor. To my knowledge, you're never going to have zero current flow. You can have staggeringly low, totally irrelevant current flow, but not zero. This page might help: robotroom.com/Capacitor-Self-Discharge-1.html \$\endgroup\$ – vofa Dec 13 '16 at 18:13
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    \$\begingroup\$ I've not downvoted this, but I think opening by saying the question is wrong is not applicable here. \$\endgroup\$ – pjc50 Dec 13 '16 at 20:15
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Measure the instantaneous voltage across the cap with a high input impedance oscilloscope, this will be good enough for practical purposes.

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    \$\begingroup\$ The input impedance of a typical scope might be 10 MΩ or 100 MΩ. If you read the rest of the discussion on this page, you will find that such an impedance is still far too low. \$\endgroup\$ – uint128_t Dec 21 '16 at 6:17

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