Measuring the internal resistance of a multimeter

When using a multimeter, most of us take for granted that the measuring device does not influence the D.U.T. (Device Under Test).

Most household multimeters have a specified internal resistance of 10 mega ohms. Some older meters might only be 1 mega ohm

But is this really the case?

Using a 1.5MΩ resistor and a 9V battery it is easy to determine the real internal resistance of your multimeter. Here’s what you do:

First measure the resistance of the 1.5MΩ resistor. We’ll call that R.
Second, measure the 9V battery, well call that Vbatt.
Now, connect R to your battery and measure the voltage between R and battery. This is Vm.

With my meter I got these values:

  • R = 1.475MΩ
  • Vbatt = 9.86V
  • Vm = 8.6V

Using Ohm’s law we can work out that calculating the meter’s resistance, we can use this equasion:

Rm = (R × Vm) / (Vbatt – Vm)

In my case this results in Rm = 10.07MΩ. This multimeter is right on the mark.

With a second, very cheap meter, I got these result:

  • R = 1.47MΩ
  • Vbatt = 9.52V
  • Vm = 8.02V

Rm = (1.47 × 8.02) / (9.52 – 8.02) = 11.79 / 1.5 = 7.86MΩ. Not so good this time.

Published in: on July 8, 2009 at 9:31 am  Leave a Comment  
Tags: , ,

The URI to TrackBack this entry is:

RSS feed for comments on this post.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: