Measuring the input impedance of a voltmeter

What is the input impedance?

An ideal voltmeter just measures the voltage without consuming any current from the observed circuit. But we live in the real world, our voltmeters have in internal resistance. For a modern multimeter this ist most times in the range of 10 MΩ or higher. In the equivalent circuit diagram this is a resistor parallel to the voltmeter.

Why do we need to know this?

Sometimes you have to measure circuits with very small currents involved. E.g. if you want to measure leakage currents of a diode, a fet or an integrated circuit like a CD4052. Or the leakage current of a capacitor!

For ‘higher’ currents you might simply use your Multimeter on the µA range – if you happen to have such a range. My Fluke87 can measure down to .1 µA, my Peaktech 3360 down to 10nA. That sounds like good enough, but in practice one can measure further down with little effort by using a high ohm shunt resistor.

In this example we use a 10 Meg Ω resistor. If we assume a leakage current of 1nA we get 10M * 1nA = 10mV. Most Multimeters can measure down to 1mV, some 10µV. My old HP desktop multimeter can measure down to 100nV.

But why do I need to know the input impedance of my voltmeter? Because it basically is forming a parallel resistance to R1. For a standard 10 Meg input impedance in the end we’d only get 5MΩ of effective resistance. Means our voltage drop will only be 5mV instead of 10mV.

How to easily measure the input impedance of a voltmeter?

All we need is a fixed voltage and a resistor. The voltage could come from a lab power supply or simply a battery. The absolute voltage doesn’t even matter. An even number just makes rough calculations easier. The resistance should be rather high. 10M or 20MΩ does work fine.

If we look at the circuit we see the external resistor is forming a simple voltage divider together with the input impedance of our voltmeter. All we need to do is to look at the voltage on our multimeter and do a bit of math. The measured voltage Um depends on the current times impedance.

Moving it around for R2 we get

Now let’s test out a few of my multimeters!

I’ve got a 20MΩ resistor

And I dial in 10V.

When adding the 20MΩ (measured 20.438M) resistor I get 3.313V in the 4.5 digit mode and 3.513V in the 3.5 digit mode. Which gives us 10.12 and 11.07 MΩ respectively.

The Peaktech scores similar values: 3.5219V, meaning 11.1 MΩ.

My Lab benchtop multimeter is another category though. While it has similar values (3.2876V) in the 10V range, seemingly weird things happen in the 3V range

The HP 3478A high-Z mode

For moving to the 3V range on my HP3478A I use my self built 3.000V reference (utilising a MAX6071AAUT30+T). It shows 2.99962V without the resistor.

But when adding the 20.4MΩ resistor I get weird reedings. The value is not really fixed but moves around from 2.99726 to 2.99742. This has to do with the internal multi-slope sampling on such a high impedance voltage source. Note that we now start to pick up noise in the cables. The change is in the microvolts range after all.

Doing the math again we get an internal resistance of 26GΩ – yes gigaohms!

When operating in the 3V range we basically get a direct connection to the internal FET stage. And note that – with every direct differential amplifier stage – we better don’t take it as linear resistance but like an OpAmp with a input current. And also note that this can be either positive OR negative – means maybe there is current coming OUT of the wires! Let’s check what we got here:

We simply take a 1µF capacitor with a low leakage current and measure it’s voltage. First we start with shorting out the capacitor. In this stage we measure a micro volt. When we open the short it quickly starts to rise..

Measuring the voltage of the 20.4MΩ resistor gives us values between 120 and 900 µV. Which means we have an input current of around -10 to -50 pA.

That’s pretty fine. But something to be aware when measuring leakage currents!