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problem with digital multimeter

Discussion in 'General Electronics Discussion' started by scientifico, Mar 13, 2012.

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  1. scientifico

    scientifico

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    Mar 13, 2012
    Hello, i've the multimeter DT9205A and I measured the voltage of a new AA battery it says 1.57 V, then I connected the battery in series with a 990 ohm resistor so the current should be 1.57/990 = 1.58 mA but measuring the current with the multimeter it says 1.40 mA that is a bad measure, but why it happen?

    I read that in my calculations i should add to the 100 ohm internal resistance of the multimeter so 1.57/1090 = 1.44 mA or to subtract about 0.2 V because the multimeter internal voltage drop when measure current so 1.37/990 = 1.38 mA but what is the real problem?
    Is the voltage measure 1.57 V correct or this multimeter isn't even reliable for tension measures?

    Thank you!
     
  2. duke37

    duke37

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    Jan 9, 2011
    The current is less than expected since the meter has some resistance as you say but also because the battery has an internal resistance so its terminal voltage will drop when a current is drawn and will sink over time.

    You could try measuring the battery terminal voltage without and with the resistive load and see how fast the voltage recovers when the resistor is disconnected. Your mutimeter is likely to be perfectly good.
     
  3. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Your meter on a current range has a voltage burden -- that is, the voltage drop across the meter. Depending on the meter, this can be modelled as a simple resistance.

    See here, here, and here. Oh, and of course here.

    Most meters have a burden voltage defines in terms of millivolts per microamp or something similar. Note that V/I is simply a restatement of resistance. If you're given a burden voltage of a particular meter it *may* be at full scale.

    The current is also affected by the internal resistance of the battery. Whilst this is low, it may be noticeable if the battery is a little flat.

    It may be far better to measure the resistance of your resistor, then the voltage across it and from that calculate current.

    Remember that measuring something changes it, so even measuring the voltage will change the voltage due to the meter's impedance across the resistor under test.

    Reading small currents at low voltages is the worst case for multimeters though.
     
  4. scientifico

    scientifico

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    Mar 13, 2012
    so if I put my multimeter in an unknown circuit will I get a lower current than the real especially in low power circuits?

    How affidable is the voltage measurement with that multimeter? Will I real higher voltage?

    Thank you!
     
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    "affidable"? Do you mean accurate?

    Voltage measurements are always (well almost always) less than the actual voltage because of the current drawn by the meter. Normally this is insignificant due to the relative impedance of the meter.

    In this case, measuring the voltage across the resistor with a typical DMM is going to result in a reading some small number of parts per million too low (well under the resolution of the meter).
     
  6. scientifico

    scientifico

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    Mar 13, 2012
    Yes I mean the nearest possible to the real voltage.
    If a new AA battery should be 1.5 V why my multimeter says 1.57 V ?
     
  7. duke37

    duke37

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    Jan 9, 2011
    Because a new AA battery will have a terminal voltage of 1.57V. Try an old one and see what you get then.
     
  8. scientifico

    scientifico

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    Mar 13, 2012
    An old one have lower voltage than 1.5V but why they write in a new battery 1.5 instead of 1.6 V that is more real?
     
  9. duke37

    duke37

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    Jan 9, 2011
    No, what is more real is the voltage when the battery is supplying power.

    Did you try measuring the voltage when the battery is loaded with your 990 ohm resistor as I suggested earlier?
     
  10. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Last edited: Mar 18, 2012
  11. scientifico

    scientifico

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    Mar 13, 2012
    I tryed to do what you say and I observed that the smaller the resistance is the lower is the voltage that appears on the multimeter.

    Thank you!
     
  12. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Excellent!

    What you're seeing is that the voltage on the battery falls under load.

    Here are some other things you might notice:

    It will fall more or less depending on the type of battery. (Try a D cell compared to an AAA cell, or an expensive AA cell compared to a real cheapie)

    If you keep a heavy load on the battery the voltage will continue to fall slowly as the battery goes flat.

    The voltage recovers almost immediately you remove the load (but it may not recover 100% and it may recover a little more if you leave it unloaded for a while)
     
  13. duke37

    duke37

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    Jan 9, 2011
    Good work.

    Now that you know some characterisics of your battery, what nominal voltage would you use to describe it?
     
  14. scientifico

    scientifico

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    Mar 13, 2012
    1.4-1.5 V, I think it's a good average
     
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