# Can a voltmeter itself disturb the exact measuring of voltage?

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?

• Extend this to oscilloscopes and how they deal with a finite input resistance and, worst still, input capacitance. Commented Aug 2, 2020 at 14:49
• @Andy aka, Sorry, I didn't get it. I'm an electrochemist. You mean this problem is related to working principle of oscilloscope ? Commented Aug 2, 2020 at 14:55
• No, I'm just mentioning that all voltage measuring devices suffer from a non-infinite input impedance. Commented Aug 2, 2020 at 14:57
• @Andy aka, I see. But what I'm questioning is even if the input impedance is infinite, there is voltage change because Fermi level of probe wire connected to voltmeter and lead of the battery is equalized into some other value, which is different from the original Fermi level of the lead of the battery. Commented Aug 2, 2020 at 15:03
• @DonghoonLEE the effect you're referring to is the Seebeck effect: A difference in Fermi levels of different metals that causes an electrical potential difference at any temperature > 0K. Commented Aug 2, 2020 at 15:48

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.

• The concept of "voltmeter sensitivity" is not relevant today. It was introduced for the purposes of electrodynamic type measuring instruments, which were essentially ammeters. To convert them into voltmeters, they connected a resistor in series acting as a "voltage-to-current converter". By changing its resistance, they changed the range. In such a "rheostat arrangement", the voltmeter resistance depends on the range. Hence the weird concepts of "voltmeter sensitivity" and "ohms/volt". Today's voltmeters are real voltmeters using a "potentiometer arrangement" with constant input resistance. Commented Aug 4, 2020 at 19:03
• Thank you very much! Commented Aug 5, 2020 at 6:22
• You are welcome! Of course, I am aware that you know all this... but it would be interesting to comment on it. Few people know where this weird "ohms/volt" come from and can not understand why the input attenuator of the classic voltmeter is made in this way (as a switchable resistor instead as a switchable potentiometer). BTW I also never leave a comment unanswered because I think this is at least a sign of elementary culture and upbringing... Commented Aug 5, 2020 at 7:12
• Hi Circuit fantasist, Thanks to you, I reviewed my answer, found the word 'analogue' missing in 3 places and rectified that. I am in agreement with your views on this subject. Commented Aug 5, 2020 at 13:01

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.

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.

• Non-ideal voltmeters are perfectly in-scope as a part of applied electrical engineering. Most of the voltmeters used when discussing physically constructed circuits on this site are non-ideal. Commented Aug 2, 2020 at 16:30
• Yes it can. Perfect voltmeters with infinite impedance do not exist. Neither perfect wires. Nor batteries. Commented Aug 2, 2020 at 17:02
• I know that. It still affects the reading. Whether or not it makes any difference is another thing and it depends on the impedances of both the meter and source. In this case the OP is already concerned about extremely subtle electrochemical processes, they might have extremely high impedance. Commented Aug 2, 2020 at 17:46
• Sorry, didn't see the response to my comment at first. I would argue that electrochemical phenomena as part of a circuit being constructed and measured are reasonably within the topic defined in the help center, broadly construed, and secondly, the correct channel for dealing with an off-topic question would be voting to close, rather than an answer that dismisses/idealizes the elements that you feel are off-topic. In this case, such an answer doesn't really address the actual question being asked, in the form it is asked, and with the assumptions stated in the question. Commented Aug 2, 2020 at 23:04
• @Circuitfantasist since it is built from physical real world components, will it really be infinite, or just extremely large? Sure, specialized high impedance precision measurement devices can be built with using op-amps in NIC or INA configurations. You can already buy op-amps with below 1 femtoamp input currents and above 100 teraohm input resistance. Common handleld voltmeters with probes what you can buy have 10-20 megaohm input impedance. Benchtop multimeters can have 10 gigaohms. While not infinite, it may or may not be good enough for a given task. Commented Aug 5, 2020 at 10:50

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]

• Thank you for your comment ! Now I knew the input capacitance of voltmeter which is the amount of charge for the voltmeter to measure voltage at certain voltage of DUT is small enough not to bother the charge condition of DUT(In this case, battery). But does just connection of some metal wire to one of the battery electrodes also induce charge transfer for Fermi level alignment ? If it does, the longer the wire, the more deviation of the Fermi level from pristine one would be. Commented Aug 8, 2020 at 9:46
• A single wire on it's own has no capacitance. That requires 2 conductors with the dielectric between. A Fermi level is irrelevant compared to a battery level. Commented Aug 8, 2020 at 12:12
• Sorry, I don't get it. When two different metals are in contact, there must be charge transfer. This is the same situation as some random metal wire is in contact to a battery electrode. Is this right? Commented Aug 9, 2020 at 1:06
• a single random wire maybe, but depends on geometry relative to the return path conductors. But in reality a single wire draws no power from a baterry , so no "Fermi effect" but a volt meter on a drop of charged fluid might. Commented Aug 9, 2020 at 1:56
• I see. I'm trying to grasp the deviation of Fermi level caused by charge(electron)transfer. Away from this situation, If the same amount of different metal, say copper and lithium plates are in contact, then would the Fermi level of the combined metal be average of the two prisitne metal ? Commented Aug 9, 2020 at 6:11

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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• -1 for confusing everyone Commented Aug 2, 2020 at 16:02
• @Bimpelrekkie, I cannot agree with such a negative attitude toward more original answers that stimulate creative thinking; instead, they should be encouraged. They confuse those who do not want to think but only to know. Besides, such a demonstrative open downvote misleads others to vote negatively ... and here are 4 votes against. Commented Aug 4, 2020 at 8:56
• @analogsystemsrf, I share your thoughts. Voltmeters consume some, even small, energy (charge) from voltage sources. At first glance, the 19th gold-leaf electroscope is an "ideal" voltmeter that does not consume anything. But this is an illusion since, as you noted, it still consumes some energy but only in the first moment (the source is shorted). It behaves as a capacitor (differentiator) that should be charged... and after that, it really consumes nothing. Classic voltmeters contain input voltage dividers to expand the range... that continuosly consume some (even small) energy (current)... Commented Aug 4, 2020 at 10:04
• @analogsystemsrf, It strikes me that your answers are quite different from those of the others... and also that you do not respond adequately to comments... Sometimes I wonder if behind this name is a human being or a clever bot who manages to bypass the SE human verification? Commented Aug 5, 2020 at 7:27