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.
---
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] |
| +<-------+
+<---[AMMETER]--->[POT] [VOLTMETER]
| +<-------+
[R2] |
| ___ |
+-----------O O-----+
| S2
GND
What you do is set the ammeter on its highest range, adjust the pot
(any convenient value which won't load down E1 much will do) to
about midrange, close S1 and S2, and read the voltage across the pot
with the voltmeter.
Rotate the pot until the ammeter reads zero, then switch in the next
lowest current range and rotate the pot until the ammeter reads zero
again, then repeat until you run out of current ranges.
Write down the voltage indicated by the voltmeter and open S1 and
S2.
Disconnect the voltmeter and measure the resistance from the arm of
the pot to the S1 end. Call that resistance R1. Measure the
resistance from the arm to the S2 end of the pot. Call that
resistance R2.
You now have:
E1
|
[R1]
|
+---E2
|
[R2]
|
GND
Where E1 is the voltage you recorded earlier and E2 is what you'll
get when you solve:
E1*R2
E2 = -------
R1+R2
Notice that since no current was drawn from the circuit under test
at null, there'll be no loading error due to the resistance of the
voltmeter used originally. There are other errors, though, and if
you like I'd be happy to discuss them.
