Measuring voltage across series capacitors... Impossible?

OK so this might seem like a dumb question, but I have been toying with it for hours.

Suppose you have this circuit...

simulate this circuit – Schematic created using CircuitLab

OK, hopefully we all know how to calculate that, given enough time:

$V_{c2} = V_{BAT1}/2$

So I want to measure that to check? How do I do that?

Before you snap-answer the question, remember that any current drawn from the join between $C_1$ and $C_2$ during that measurement will unbalance the charges on $C_1$ and $C_2$. When you remove the meter, or circuit, that unbalance will be locked in changing the voltages across $C_1$ and $C_2$.

Granted, given enough time, leakage current should settle things back to their balanced charge state. But let's assume leakage is very slow compared to your desired measure rate.

The question is, is it possible to repeatedly measure the voltage $V_{c2}$ with any real device or circuit? Or is this just one of those weird physics curiosities that has no solution.

EDIT: I realize BIG IMPEDANCES help, and are probable fine for scope/meter measurement, but this is a thought experiment.

EDIT2: I also realize there are electrostatic measuring devices, but I'm more looking for something you can build into a circuit board and monitor it with something.

• Let me make sure I understand your main concern. You mean by drawing current from the point between the capacitors, you are effectively removing some charges from both the upper plate of C2 and lower plate of C1, and saying this charge loss is unrecoverable? May 18 '17 at 18:22
• @EugeneSh. If the measuring device is not at the same potential as the measurement point, wither C1 will change and C2 will discharge or vica-versa. May 18 '17 at 18:25
• As a highly experienced engineer, you might know that there are some non-contact measurement devices but IIRC, they are for measuring high voltages. May 18 '17 at 18:27
• Basically you are asking, can a voltage be measured without taking any current. May 18 '17 at 18:36
• How are you going to guarantee that the leakage is even across capacitors, there aren't exactly specs on the leakage current\parasitic parallel resitance. How much does it vary? May 18 '17 at 19:22

Something like the TI LMP7721 Opamp is probably what you're looking for:

In the femto-Ampere range, we can be pretty certain that the measurement current-induced error is "rather low"; let's say a X7R ceramic cap has some 10 GΩ resistance. At a 1 V voltage across one of these, we'd have a 100 pA leakage. That's a solid 2 to 3 orders of magnitude above the input bias current of the opamp.

Bonus: having such an opamp allows to have a "voltage following" shielding around the nets of interest – further reducing leakage due to potential gradient in e.g. the PCB substrate :)

• +1 nice find. Yup, I'm starting to think that' probably the best one can do. Though it defeats the purpose of my original thoughts of using a capacitive divider in the first place... i.e. minimal DC current. May 18 '17 at 18:50

Since there's no specifics about the capacitors and/or budget:

• Step one: Get an op-amp with lower input leakage than the minimum likely leakage in the capacitors.
• Step two: Voltage follower.
• Step three: Done.

Practically this is not too hard when the capacitors are electrolytic. For ceramics it depends on... a lot. Silicium capacitors might leak too little for most integrated op-amps out there.

So, in the category "what to use if not a multimeter": The Electrometer would seem like the type of device you'd care about. If this is all about measuring DC voltages, the vibrating reed electrometer should effectively remove very little charge from the capacitors.

• YUp, I was familiar with that, but not exactly something you could connect to your arduino... May 18 '17 at 18:27
• @Trevor, nothing in the question asks about "something you could connect to your arduino". May 18 '17 at 18:30
• I frankly don't know! I could imagine one could actually just use a mechanical variable cap and a servo as well as sufficient DSP... May 18 '17 at 18:30
• @Trevor, how about an electrometer and a vision system? May 18 '17 at 18:30
• @ThePhoton I did say circuit, but maybe I should have been clearer. May 18 '17 at 18:31

Time is a solution. By quickly sampling the voltage across the capacitor, the change in its charge can be greatly minimized. For example, a sample probe with a 10 meg input impedance and a 200 uS sample time will keep the state of charge of your 1 uF capacitor at 99.998% of its original value. The 200 uS sample time is easily enforced with a reed relay, for example.

Assuming the battery can be viewed as an ideal source in this setup, just put two resistors of equal value in series across the battery. You could make one with a trimmer to get the voltage between them to exactly battery voltage over two. The measure from the center of the resistors to the center of the capacitors. If it reads 0, then no charge taken and the junction is battery over two.

The question is, is it possible to repeatedly measure the voltage Vc2 with any real device or circuit?

YES

Next question.

ticky tacky details

We can also estimate the decay rate from the load impedance and the step voltage drop from the uncharged load capacitance.

We can choose any instrument to perform the task and they will always have some leakage resistance and load capacitance.

Since this was hypothetical without real values, there is no explicit answer possible without an accuracy spec.

But the real answer is yes and we know what to expect with that known load impedance no matter how high it is.

This is essentially the way battery cell voltages are measured since all batteries are just really huge charged capacitors. ( with some imperfections)

This is also routine for HVAC measurements and is called a "Capacitance Voltage Transformer" with HV across the pF insulator part (e.g. vitrious enamel bushing) and LV across the external large fixed capacitor.

It is important to remember the limits on materials for RC time constants for these types of applications.