The Meter is Part of the Circuit

In my last post, I drew a simple voltage divider circuit.

Two 6.8 Megohm resistors in series, powered by a 20 volt DC (20 VDC) power supply.
Vout was defined as the voltage across R1, or the resistor closest to the negative side of the battery.
Using the voltage divider equation, the output voltage was calculated to be 10 VDC.
The guys in the shop built this little circuit;
-and found that the actual voltage displayed was 7.45 VDC.



...and I asked, "Why?"

Well it turns out that a typical digital volt meter has about 10 Megohms of impedance.

Impedance is just resistance to current flow, but a term used when describing both AC and DC circuits for completeness - more on that later.  For now we can think of meter impedance as just the same thing as resistance.

When the meter leads are attached to the circuit above, you can see that we now have 10MΩ of resistance in parallel with R1.

So now we need to determine the equivalent resistance (Req) of R1 in parallel with the meter.

For two resistors, that's easy:

Use the product over sum rule, which is for two resistors in parallel -

Req = (R1xR2)/(R1+R2)

So Req = (6.8MΩx10MΩ)/(6.8MΩ+10MΩ)

Req = 68/16.8 MΩ  = 4.048 MΩ

So now we re-visit the voltage divider equation:

Vout = (Vin x R1)/(R1+R2)

So Vout now is:   (20V x 4.048MΩ)/(6.8MΩ+4.048MΩ)  =  7.46 volts

We got 7.45 Volts - so this was spot-on.  The difference between 7.46 and 7.45 can be attributed to the 5% tolerance resistors we were using, and really, .01/7.46 = .0013 which gives a 0.13% error.

Not bad!

Cheers for now