How to find out what is the internal resistance of a voltmeter? I imagine the higher the better - it'll have less influence on the measured circuit.

What is the approximate internal resistance of cheap DMM on the voltmeter setting? Do more expensive multimeters have better (higher) internal resistance? Is there a significant difference in internal resistance of the voltmeter between a Fluke and a 5$ DMM?

What is the best method to measure the internal resistance of a voltmeter?


5 Answers 5


Apply a known voltage over a series resistor. This resistor in combination with the internal resistance will form a voltage divider. Say you apply 5V over a 1M series resistor, and the DMM shows it as 2.5V, then the internal resistance is 1M.

Now that I reread it, I guess it's not completely unambiguous. By "applying a voltage over a series resistor" I meant you connect the + to the resistor and the - to the ref. input of the DMM.

  • \$\begingroup\$ Just a note: The voltage should be applied in series with the resistor. So you will have GND-$V_\rm{source}$-1$\,\rm{G\Omega}$-DMM-GND. \$\endgroup\$
    – jpc
    Apr 14, 2011 at 15:49

This is quite easy, you only need a 1.5 volts battery and a 1 megaohms resistor.

  1. First measure the voltage battery with the DMM. Let say it reads 1.609 volts
  2. Second measure the resistor. Let say the DMM reads 1.008 megaohms
  3. Then measure the voltage again but this time use the 1 megaohms resitor in the positive pole of the battery so you take the measure in the resistor leg.
  4. If the DMM reads 0.801 volts the DMM has a internal resistance of 1 megaohms, if it reads 1.461 volts then the internal resistance is 10 megaohms

The formula for using different input voltages or resistors is:

DMM internal resistance in megaohms= ("DMM voltage measured " x "value of resistance used in megaohms") / ("input voltage" - "DMM voltage measured ")

You can use this simple Excel file for the calculations: DMM_IR calculator


To figure the input impedance of your voltmeter, do the following;

  • Measure a voltage with the meter and at the same time with ANOTHER meter, measure the current going through the VOLTMETER. (amp-meter)

  • Using Ohm's Law, calculate the following... Voltage measured by the meter, divided by the Current measured by the Amp-meter = Impedance

Example I measured 13.8 volts, and with another meter in series with the voltmeter, I measured the current through the voltmeter at 1.2 micro-amps.

13.8 V / 0.0000012 A = 11,500,000 Ohms or 11.5 Meg Ohms

Most DMM's today are 10 Meg Ohms input impedance minimum, (even the free one from Harbor Freight) so the current measured will be in the micro-amp range. Therefore you will need a meter that can measure micro-amps...

And yes, the higher the input impedance the better. However for most uses today, 10 Meg Ohm impedance is all you need.

  • Better multimeters /voltmeters in lower ranges (up to ...10..20V, using input buffers) have higher input resistance, usually >10 GOhms.
  • Multimeters with FET input buffers have 0.2....2TOhm input resistance (depending also on humidity, dirt, leads isolation). Typical Hi input bias current is about 20...50pA, Lo+Guard input bias current is about 0.1nA (to GND).

This value should be stated in the manual of the DMM. My UT-60 is specified to have \$> 10\,\rm{M\Omega}\$.

  • 3
    \$\begingroup\$ -1: question asks how to measure it \$\endgroup\$
    – DarenW
    Apr 14, 2011 at 17:18
  • \$\begingroup\$ @DarrenW But there were several sub-questions about the resistance of cheap voltmeters and the method to determine one. IMHO looking at the specs is the simples way to determine the resistance, especially since it is commonly included (even in the specs of the cheap models). Tomorrow I will add info about my Escort. I may even try the measuring method. \$\endgroup\$
    – jpc
    Apr 15, 2011 at 2:08

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