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Voltage divider

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

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  1. lerameur

    lerameur Guest

    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 Jeff

    Byron A Jeff Guest

    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
     
  3. Ralph Mowery

    Ralph Mowery Guest

    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. 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.
     
  5. lerameur

    lerameur Guest

    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. Jamie

    Jamie Guest

    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.
     
  7. ehsjr

    ehsjr Guest

    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 Fields

    John Fields Guest

    ---
    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.
     
  9. jasen

    jasen Guest

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



    Bye.
    Jasen
     
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