Voltage divider

Two reasons:

1) You resistances are not identical. There will be error.

2) Your probe imposes an effective resistance in order to measure the voltage. The higher your resistance on your voltage divider, the more this effective resistance impacts your measurement.

BAJ

While you are using two identical resistors, you forgot to count the resistance of the meter. If you use two resistors that are 10 meg each, there will be 6 volts from the junction to either side of the power supply. If your meter has a resistance of 10 meg also, then you have put 10 meg and

10 meg in parallel for a total of 5 meg. So you really have a voltage devide of 10 meg and 5 meg. If you resistors of 100 ohms each, when you parallel one with the meter you get 100 ohms and 10 meg in parallel which is still almost 100 ohms.

You need to consider parallel resistors. The effect that you are seeing is the parallel (equivalent) resistance of the resistor in your divider and that of your meter ... your meter acts like a resistor.

So, if your meter resistance is 10 megohms, and is connected across a 10 megohm resistor, the parallel equivalent is 5 megohms and the voltage divider is now based on a 5 megohm resistance across those two nodes.

Ok , I think that might be it, DMM problem.Because I checked the resistances and i choose two that were identical. I guess with an analog multimeter, the problem will be less apparent, i will try it now.

thanks

k

Charles Schuler a =E9crit :

is

because your DMM is 10 Meg input most likely and is loading it down. this would be it unless you are not doing what i think you are.

--
Real Programmers Do things like this.
http://webpages.charter.net/jamie_5

The problem is likely to be far more apparent with an analog meter, which typically has a far lower internal resistance than a DMM.

Ed

--- Please bottom post.

Whether you use an analog or digital voltmeter has nothing to do with it. The voltage at the measured point will change depending on how much the meter loads the circuit at that point.

Your voltage divider looks like this:

E1 | [R1] | +---E2 | [R2] | GND

and the voltage at E2 will be:

E1*R2 E2 = ------- R1+R2

So, if E1 is 10 volts and R1 and R2 are both 10 megohms:

10V * 10E6R E2 = --------------- = 5 volts 10E6R + 10E6R

But, if you connect your [analog or digital] voltmeter into the circuit you'll have this, R3 being the resistance of the voltmeter:

E1 | [R1] | +------+----E2 | | [R2] [R3] | | GND GND

Note that the total resistance of R2 and R3 will be:

R2*R3 Rt = ------- R2+R3

So, if your meter resistance (R3) is 10 megohms, then we have:

10E6R * 10E6R Rt = --------------- = 5E6R = 5 megohms, 10E6R + 10E6R

the circuit will look like this:

E1 | [R1] | +---E2 | [Rt] | GND

and the voltage at E2 will be:

10V * 5E6R E2 = --------------- ~ 3.33 volts, 10E6R + 5E6R

which is an error of about 1.67 volts too low.

One way to eliminate the error would be to do this:

E1 | ___ +-----------O O-----+ | S1 | [R1] | | +