Connect with us

Potential Divider: Unexpected Voltage Readings

Discussion in 'Electronics Homework Help' started by Jairs, Oct 28, 2015.

Scroll to continue with content
  1. Jairs

    Jairs

    2
    0
    Oct 28, 2015
    I carried out an experiment involving four different rated resistors, lined up in pairs in a potential dvider setup (see diagram for clarification). The resistors were: 10KΩ, 100KΩ, 1MΩ and 10MΩ. There were two of each resistor.

    For the measurements we used two different multimeters: one stand-alone and one handheld

    The first thing we did was to meaure the 'true' resistances of the resistors and as expected they weren't the exact nominal reading; but still within the tolerance. The handeld multimeter was unable to measure accross the 10MΩ resistors but I'm assuming this is because it was out of its range. However the stand-alone multimeter did give an accurate result.

    The issues happened when taking voltage accross the potential divider. I know that in a potential divider setup having two of the same valued resistors will half the input voltage, and with the lower rated resistors this was true. However once using the 10MΩ resistors the voltage readings were very different from the expected readings.

    The input voltage was 10.0V. Therefore using two 10MΩ resistors (measured to be 10.17MΩ and 10.04MΩ) the output voltage should be around 5.0V however the meaurement was 3.459V which clearly is significantly different. I have tried to account for tolerances on both the resistors and the power source however it makes very little difference to the expected result of 5V.

    I know that fundementaly I must be missing something to account for why the output voltage was so low, (also worth noting that the voltage was also taken from the 'other side' - BC, but that was equally low).

    I would appreciate any advice on what to look at to help explain these readings.

    A theory I have is that the 10MΩ pushes the multimeter to the top of its range or at least beyond its accurate range. But that is just speculation and would appreciate someones input.

    Disclaimer: I am a mechanical engineer by trade and electrical is completely new to me so appologies for any misnomers or wrong assumptions I have made! question.png
     
  2. Martaine2005

    Martaine2005

    3,554
    966
    May 12, 2015
    Just a guess, but perhaps has something to do with the meter having about 10MΩ.
    So in effect 2 x 10M in parallel?
    I assumed that both your meters were DVM. So I could be way off if not.

    Martin
     
  3. hevans1944

    hevans1944 Hop - AC8NS

    4,615
    2,154
    Jun 21, 2012
    Yup, I agree with @Martaine2005. The input resistance of the meter is loading the 10 Megohm divider, effectively decreasing the ohmic value of whichever resistor it is measuring across.

    One simple solution to this "loading problem" is to make a balanced potentiometric bridge measurement of the divider voltage, using a sensitive null meter to determine when the potentiometer voltage equals the divider voltage. You then measure the voltage from the potentiometer using a conventional meter. Since the potentiometer will have a very low resistance compared to your divider resistance, it will produce less of a load on the conventional meter and result in a more accurate reading. The null meter must have adequate sensitivity, typically a few microamperes full scale, to accurately determine when the voltage divider voltage has been nulled by the potentiometer voltage.

    There are digital voltmeters that use this potentiometric technique to measure voltages in very high impedance circuits without loading the circuits. See this Google page for more information.
     
    Last edited: Oct 29, 2015
    duke37 and Martaine2005 like this.
  4. Martaine2005

    Martaine2005

    3,554
    966
    May 12, 2015
    That's a very informative post as usual Hop.
    Thank you, I learnt (for the Yanks learned) something else!
    I now need a brain expansion slot fitted and a file system for quick access.:D

    Martin
     
  5. duke37

    duke37

    5,364
    772
    Jan 9, 2011
    As Hop says it is all to do with the meter loading of the circuit. You have not given details of your meters. It does not matter if they are handheld or fixed to a skyhook.
    I have a cheap digital meter with a 1M input resistance, trying to measure a voltage with a source resistance of 5M would obviously give very low readings.

    A good analog meter may have a sensitivity of 20 000Ω/V so on the 1k range it would have a resistance of 2M and on the 10V range it would have a resistance of 200k. It would be reasonable measuring the 10k resistors but a very long way out on 100k resistors.

    Valve radio circuit diagrams often give voltages as measured with a 20 000Ω/V meter and when they are measured with a 10M meter the voltages may appear to be excessive.

    To get very high input resistance is not easy and involves things like guard rings to reduce leakage across insulators.

    Trevor
     
  6. Jairs

    Jairs

    2
    0
    Oct 28, 2015
    Thanks a lot with your help everyone. The multimeter I am focused on is a model ut804 by uni-t. I have found the user manual so I'm assuming this should tell me the input resistance?
     
  7. duke37

    duke37

    5,364
    772
    Jan 9, 2011
    You can find the input resistance easily. Measure the voltage of a 9V battery. Place a resistance in series with a meter lead and measure again. If you measure 4.5V, the the input resistance is the same as the resistor.
    Try 10M and 1M
     
    Merlin3189 and hevans1944 like this.
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-