Basic: Voltage Divider question

[Auke Haarsma]

This site explains a voltage divider quite clearly: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/voldiv.html#c1

If I have two 10k's dividing the voltage (9Vcc) I get 4.5V in between the two R's, right? Now, theory says that if I provide a load of 1M in between the two R's, I'd get something like 4.47 V. So far for theory. Now, *my* practice.

I put it on the board like this:


9V+-----10k
|         |
|         y ----1M---x
|         |
-------10k

Now, if I measure with my DMM at point y, I get indeed the mentioned 4.5 V. But if I measure at point x, then I get 2.37V, where theory predicts 4.47. I'm sure this is an error on my side, but I don't see/understand it yet.

Anyone to shed some light?

[brett]

Hi
your multimeter has a resistance of about 1M.
So while it "nominally" reads 4.5V, it actually is reading only half of the 4.5V (ie 4.5V x 1M/(1M+1M)), or about 2.25V.

Better quality DMMs have an input impedance of 10M or more, so that the problem is only encountered when measuring very high impedances.
cheers

[Auke Haarsma]

thanks for your reply. If I understand correctly, my DMM is not suited for measuring circuits under a load of ~1M or more.  Could I simply increase the resistance of my DMM by placing a 10M resistance (eg.) between the probe and the circuit I'm measuring?

[Paul Perry (Frostwave)]

thanks for your reply. If I understand correctly, my DMM is not suited for measuring circuits under a load of ~1M or more.  Could I simply increase the resistance of my DMM by placing a 10M resistance (eg.) between the probe and the circuit I'm measuring?

That won't work, because the resistance of your DMM would form part of a voltage divider with the 10M resistor attached to it.
What *MIGHT* help, is to measure using a higher voltage range, because that might have a higher input impedance. Try and see if the voltage it gives you is higher!

[brett]

Although it is a nuisance, if you know the impedance involved and the impedance of you meter,you can back-calculate to the tru voltage.  The formula is Vtrue = Vmeasured x (circuit impedance + meter impedance) / meter impedance

Having a meter with an impedance close to 1M makes it easy to calculate the right multiplier.  330k = x 1.33  470k = x 1.47  1 M = x 2  3.3 M = 4.3   4.7M = x 5.7.  See the pattern?
Above 4.7M you'll need to calculate the impedance of the meter exactly and use the formula, because small deviations away from 1M makes a large difference to the multiplier.
cheers 

[Auke Haarsma]

thanks for the tips, I'll check my DMM soon!

[slacker]

Another solution that I've seen mentioned here but never tried is to make a simple noninverting buffer out of a high impedance opamp like a tl072. Then connect a wire to the + input of the opamp and use that to probe the circuit and connect the probe of your DMM to the output of the opamp.
The opamp presents a much higher impedance to the circuit than your DMM so it should improve the accuracy of the readings.