Can a voltmeter itself disturb the exact measuring of voltage?

Question

I'm trying to understand the effect on a system when the voltage of the system is measured by connecting a voltmeter.

For example let's say we are trying to measure the voltage of a battery.

To measure the voltage of the battery, we must connect the voltmeter. Each probe will be connected to each side of the battery (positive and negative lead.)

Assume this voltmeter is a high-end one, there is almost zero current flowing through the voltmeter (infinite input impedance.)

Although it is, I'm concerned that this behavior itself could disturb the voltage of the battery because the Fermi level of each side of the battery will change.

When a probe of voltmeter is connected to one side of battery, their Fermi level would be in alignment for thermodynamic equilibrium because both the probe of voltmeter and the lead of battery are metal. If their material is different, the Fermi level of probe-lead connected metal would be somewhere between that of pristine probe and lead.

Anyhow, by connecting the probe to the lead for measurement, this changes the original Fermi level of the battery. The same thing may happen when we connect the other probe of voltmeter to the other side of the battery. So the voltage displayed in the voltmeter would be somewhat distorted because this is the result after some electrons from the voltmeter flow into or out of the lead of the battery.

Since the voltmeter measures Fermi level difference between two probes, the metal of the probe will probably have an effect on the result.

I've never seen or heard this kind of effect discussed. Is it because the contribution of the effect mentioned above is negligible or am I just wrong?

Answer 1

Yes, an analogue voltmeter, by itself, could disturb the precise measurement of a voltage by loading the source and changing its voltage.

The characteristic of an analogue voltmeter, which would decide how much it, by itself, could load the source and change the voltage being measured, is known as its 'sensitivity'.

The sensitivity of an analogue voltmeter is expressed in ohms/volt. The best analogue DC voltmeter would have a sensitivity of 20,000 ohms/volt. Such a voltmeter, with a full scale reading of say 1.0 V, would present a load resistance of 20,000 ohms to the source. Should the actual voltage be 0.2 V, the current drawn by the meter from the source would be 0.2/20000 or 10 μA only.

Answer 2

The disturbance should cancel out as far as the voltage reading goes because both terminals on the battery (the exposed terminals where you connect the meter) are the same metal, and likewise both of your probes are the same metal. On a car battery the terminals are lead or a lead alloy. Your meter probes are, let's say, nickel plated brass. As you follow the circuit around the voltage goes up and down across all these metal/metal junctions. But the net effect is zero because all the junction drops or rises cancel out. If you introduced an asymetric metal/metal junction on, let's say, the positive wire, by adding an extension wire made of aluminum, then you would have an uncompensated drop that would introduce an error.

Practically speaking, with a good voltmeter, you can trust the battery voltage on the display to 0.01V. With a better meter you can trust it to 0.001V maybe. After that you may have to worry about measurement setup things that most engineers never worry about.

Answer 3

No it can't. Hmm the voltmeter is connected in parallel to the branch we want to measure its voltage.

A perfect voltmeter has infinite resistance, and if you connect an infinite resistance in parallel to the load then the total resistance doesnt change at all and we have the same voltage drop on the load.

Answer 4

OK I'll consider an infinite input resistance voltmeter but it must have some input capacitance due to the insulation between + & - conductors must be non-zero capacitance.

So what do voltmeters measure?
Volts, of course with the potential to do work and conduct current but not until it is buffered by this zero-input current amplifier.

So there must be a charge transfer to the input of x pF. This equalization depends on the source having enough stored energy to transfer. It doesn't have to be a voltage source of 0 Ohm impedance. It just must not be affected by the energy dump to charge up this voltmeter $$ E=½CV²$$

let C be really small value in the voltmeter =\$1pF=10^{-12}F\$

All batteries are like UltraCapacitors with an intrinsic chemical cell voltage as the minimum. The Li-Ion cell is about \$10kF=10^4F\$

A drop of water between 1mm electrodes might be 100pF, as the dielectric constant =80 relative to air. Yet the battery will not notice the capacitance charge transfer to the voltmeter, but a charged drop of water might notice that load of 1pf. However, keeping the voltmeter attached might be sensible if the capacitance ratio was known and corrected in any resulting acquired voltage.

Solid State Electrometers have an input impedance of \$10^{14}\Omega\$ and can be modified with a reference to store a minimal energy of a few percent of a picocoulomb. [pC]

Answer 5

Charge is always taken from the voltage source.

In gold-leaf electrometers, that charge will be a transient event.

Thus the source can re_stabilize.

However, standard voltmeters demand a steady flow of charge, a current, to stimulate voltage_dividers (voltage_ratiometric measurments).

Some (old HP meters) may be AC_coupled, but that just means a AC signal must constantly provide charge.

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