# Possible to measure voltage while charging?

Suppose we have a simple DC power source charging a battery or capacitor.

If we try to measure the capacitor's voltage in this simple circuit we will instead read the power source voltage:

simulate this circuit – Schematic created using CircuitLab

Is it possible to directly measure the voltage of the capacitor without disconnecting the power source and without knowing the capacitor's time constant and capacitance?

I.e., I'm wondering if there's a circuit that can accomplish this without exploiting the following solutions:

1. If we temporarily disconnect or turn off the power source, the voltmeter will read the voltage across the capacitor.
2. If we know the time constant and capacitance we could instead connect an ammeter and deduce the capacitor's voltage from that. (Or, if we knew it was fully discharged initially, we could integrate the amperage from the start of charging to calculate its voltage.)

I've been trying to picture some clever arrangement of diodes that might allow for measurement during charging, but that has left me thinking that this is theoretically impossible using a static circuit.

• If you think that you would only read the power source, then is the capacitor charging instantaneously? Dec 31, 2016 at 15:18
• @PeterSmith - No. The voltage across the power source is constant and always higher than the voltage across the capacitor. But, as connected, the voltmeter will "see" the higher voltage difference of the power source (ignoring the small delta associated with the fact that the voltmeter has a very large but finite resistance). Dec 31, 2016 at 15:22
• Your basic Physics is wrong. eg " ... The voltage across the power source is constant and always higher than the voltage across the capacitor. ..." is essentially not true. The cap and power supply voltages will differ only by the voltage drop in the wiring and connections. V=IR. Vdrop - Icharge x (Rwiring + Rconnections). As circuit resistance is low wiring drop will be low. Either current will be VERY high or the power supply will be "loaded down" to equal Vcap+ v_wiring-drop Dec 31, 2016 at 23:37
• @RussellMcMahon The context is charging a capacitor. I could have been more explicit, but to remove all doubt: The power supply can only supply some limited current. The capacitor is being charged, which means that, in isolation, the voltage across its terminals is less than the voltage of the supply. During charging the voltage difference across the capacitor increases (as a function of its time constant) up to the power supply's voltage in the limit. I added the "more correct" circuit diagram to the accepted answer. Jan 1, 2017 at 0:08
• @Feetwet My above comment is correct. The psu output terminal voltage and the capacitor voltage will be essentially the same at ALL times - the difference being due to V = I x R resistive voltage drop in the wire and any losses at the interconnections. You CAN measure the capacitor voltage while charging and nit will be close to the capacitor voltage you see when the psu is disconnected. The internal capacitor resistance (ESR) will also cause voltage drop inside the capacitor due to resistive voltage drop (v = I x R) as for the wire. ... Jan 2, 2017 at 7:10

## 2 Answers

As Peter says, in the ideal world with ideal components and circuits, you can't do this - the voltage measured on the capacitor is always the voltage source value (and the capacitor is instantaneously charged with an initial infinite surge of current - because these are ideal circuits).

Of course in the real world you can measure the charging voltage if your meter is much higher input impedance than either the voltage Thevenin and higher than the capacitor ESR. In the real world, these devices do not resemble the ideals so much.

You have to model reality differently: the voltage source becomes a Thevenin source with a series resistor, the capacitor has a series resistance and the voltage meter has an input resistance/impedance in parallel with the ideal infinite impedance meter. So your actual circuit should have 3 resistors, an ideal voltage source, an ideal meter and an ideal capacitor, like this:

simulate this circuit – Schematic created using CircuitLab

If you simulate this circuit, it will come closer to reality.

A key lesson here: circuit simulator results are always lies because they must assume a model of reality that is always has less fidelity than the real world (the only model that can have total fidelity is the reality itself - so build it and you'll have that): a model is a map, not the territory itself. It's a universal flaw in modern thinking to confuse the map with the territory! Of course models can tell some truths as well: the responsibility of the user of simulators is to know the point where the lies start.

• I just upvoted another of your answers (which deserved it) to offset the following: I very very seldom downvote any answer or question. I have downvoted this because, while it would be OK enough in isolation as a low scoring answer among many, in this case the OP has edited it and added an essentially incorrect circuit diagram and allowed himself to be mislead. It is not your fault but Claudio's answer is more correct and more useful to the OP. Hopefully we will sort through the pros and cons of this and com up with an OK result. |FWIW modern power supplies usually look more (not totally) ... Jan 2, 2017 at 5:07
• ... like CV CC supplies with CV applying until I rises to at or near CC and then V decreasing to maintain current near CC. This is similar to what you said but seeing it as CV + R allows the OP to confuse himself more. Jan 2, 2017 at 5:08

The circuit is wrong and doesn't represent a real world charger. A charger, even if modeled as a voltage source, will have a series resistance so when you measure the voltage across the capacitor you see its voltage and not that of the voltage source.

Most real world battery chargers have several operating modes, one typical mode of charging is that they operate at constant current until the battery reaches its nominal voltage and then at constant voltage until the battery reaches full capacity.