Voltage divider

Discussion in 'Electronic Basics' started by lerameur, Aug 24, 2006.

1. lerameurGuest

Hello,

In theory, if I take 12v supply, put it through two identical
resistance in series,which are grounded. I should get 6v between the
two resistors. I do get that suing low resistance, but the higher I go
100kohm, or 1Mohm I get 4.5v, higher at10M I get 3.9v.
why is this ? it is really messing my digital probe circuit

thanks
ken

2. Byron A JeffGuest

Two reasons:

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

2) Your probe imposes an effective resistance in order to measure
more this effective resistance impacts your measurement.

BAJ

3. Ralph MoweryGuest

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.

4. Charles SchulerGuest

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

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.

5. lerameurGuest

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 écrit :

6. JamieGuest

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

7. ehsjrGuest

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

8. John FieldsGuest

---

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.

9. jasenGuest

your voltmeter has a finite resistance it will change the voltage of the
circuit you apply it to.

Bye.
Jasen