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Measuring tiny unknown resistance?

Discussion in 'Electronic Basics' started by kevwalsh, Jan 28, 2005.

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  1. kevwalsh

    kevwalsh Guest

    Hi,

    Anyone know a simple way I can measure a small, unknown resistance? I
    am guessing it is about between 1ohm and 0.5ohms, roughly. It is part
    of a car, so I can't bring it inside to test it.

    I have several little multimeters, but none go down anywhere near as
    low as half an ohm -- the cheap analog one goes as low as about 100ohms
    at the very tinies graduation, and I don't think my digital ones go
    much lower than that either.

    I can also put 12v to the thing and try to measure current draw, but
    obviously 10A or more is too much for just sticking my little
    multimeter inline with the resistance.

    I have access to plenty of simple parts, resistors and such, to build
    some kind of simple measuring circuit. Any tips?

    Thanks,
    Kevin

    BTW -- it is a rear window defroster, if you didn't guess that already.
    My problem is that the new rear window draws too much current (how much
    I am trying to discover) and melts things, like the on-off switch.
     

  2. You would have some very unusual meters if they can't measure resistances of
    less than 100 ohms. As digital multimeters go they almost universally (from
    the very cheapest to the most expensive) have a minimum resistance scale of
    200 ohms. On the 200 ohm scale the maximum resistance that can be measured
    is 199.9 ohms. The minimum resistance that can be measured is limited by
    the accuracy of that scale. On the 200 ohm scale, this means the minimum
    resistance that can be measured is 0.1 ohm. The last digit is not very
    accurate, so it could be off by a little. Measuring a resistance between
    0.5 ohm and 1 ohm will not be too especially accurate, but it will likely be
    accurate to about as good as within 0.1 ohm. Sometimes the meter will not
    be calibrated to zero and will thus produce a fixed offset to your
    measurement. You can manually compensate for this by shorting the meter
    leads together on the 200 ohm resistance scale and observing the result. It
    may for instance measure 0.2 ohms, but for a hard short circuit it should be
    0.0 ohms. So, take you measurement of your 0.5ohm to 1 ohm resistance and
    then subtract 0.2 ohms from the displayed value.

    To get better accuracy and to further extend the measurement range down to
    lower values you need another technique. One way is to build a 100mA
    constant current source using an LM317 linear regulator. Then apply the
    100mA constant current to the unknown low value resistance and measure the
    voltage across it. Ohms law predicts V=IR, so for I=100mA, the voltage that
    appears will be one tenth of the resistance. IE: if the meter reads 50.0mV,
    then the resistance is 0.500 ohms.
     
  3. Spajky

    Spajky Guest

    try to measure current thru it & voltage across it & calculate the
    resistance than ...
     
  4. Gareth

    Gareth Guest

    If you have a low value, high power resistor you could put it in series
    with your unknown resistance. Apply a voltage then measure the voltage
    across the know resistance and across the unknown resistance. Since the
    current through both is the same, the ratio of the two resistances will
    be equal to the ratio of the two voltages. Therefore:

    R1 = R2*V1/V2

    Where V1 = voltage across R1, V2 = voltage across R2

    To get an accurate measurement the known resistor value should be a
    similar order of magnitude to the unknown resistance, then you will get
    sensible voltages across both resistances.

    Obviously, your known resistor must be capable of handling the current.

    Gareth.

    --
     
  5. Bob Masta

    Bob Masta Guest

    To add to what others have posted, note that the trick to
    accurate low-resistance measurements is to measure the
    voltage drop across the unknown resistance with
    *different leads* than those driving the current
    through it. (Unlike the the conventional ohmmeter
    arrangement.) So, if you use the suggestion of
    a series power resistor, you can measure the
    voltage across that to determine the current,
    then measure the voltage across the unknown and
    compute its resistance.

    Best regards,


    Bob Masta
    dqatechATdaqartaDOTcom

    D A Q A R T A
    Data AcQuisition And Real-Time Analysis
    www.daqarta.com
     
  6. john jardine

    john jardine Guest

    Stick a 100ohm resistor in series with the defroster lead. Switch on and
    measure the DC voltage across the heater. A 1ohm defroster would give 100mV.
    A 0.5ohm defroster would give 50mV etc.
    regards
    john
     
  7. "john jardine" <> skrev i melding

    Or, like this:

    o-------| R1 |--------| R2 |-----------|!' Gnd
    Ucc

    Ur2 (voltage across R2) = (Ucc * R2) / (R1 + R2) // The basic formula.

    Turning the formula to get the value of R2 is quite easy:
    UR2( R1 + R2 ) = Ucc * R2
    ( (UR2 * R1) / R2 ) + UR2 = Ucc


    // This is the important stuff
    So finding R2 is to calculate this formula:
    R2 = (UR2 * R1) / (Ucc - UR2)



    Hope this helps : )
     
  8. Bill Bowden

    Bill Bowden Guest

    Anyone know a simple way I can measure a small,
    You could use a 12 ohm resistor is series with the 12 volt battery
    so you get about 1 amp of current. It will heat up with about 12 watts
    of heat, so you need a large resistor. Or, use a tail lamp bulb
    which is a about 2 amps. Then use your multimeter to measure the
    exact current. Also measure the voltage across the unknown resistance
    and divide this by the known current.

    So, if you read 0.6 volts across the resistance, and the current
    is 1 amp, the unknown resistance will be 0.6/1 = 0.6 ohms.

    -Bill
     
  9. "Bill Bowden" <> skrev i melding

    You could use a 12 ohm resistor is series with the 12 volt battery
    so you get about 1 amp of current. It will heat up with about 12 watts
    of heat, so you need a large resistor. Or, use a tail lamp bulb
    which is a about 2 amps. Then use your multimeter to measure the
    exact current. Also measure the voltage across the unknown resistance
    and divide this by the known current.

    So, if you read 0.6 volts across the resistance, and the current
    is 1 amp, the unknown resistance will be 0.6/1 = 0.6 ohms.

    Disagree. When he measures 0.6 volts across the resistance, there is no
    longer 1 amp through the circuit any more because the total resistance will
    be more then 12 ohms. The higher the "unknown" resistance is, the higher
    error on the measurment.

    If using my formula (erlier post, same thread) the resul will be:
    "unknown resistance" = (0,6V * 12ohm) / (12V - 0.6V) = 0.63 ohm
     
  10. Bill Bowden

    Bill Bowden Guest

    Disagree. When he measures 0.6 volts across the
    I don't quite follow the logic. It doesn't really matter what
    the total resistance is, or the battery voltage. We only need
    to know the current through the circuit, and the voltage across the
    unknown resistance. If we measure 1 amp through the circuit, and 0.6
    volts across the unknown resistance, the value of the resistance will
    be *exactly* 0.6 ohms, unless you are considering the error introduced
    by the meter resistance, in which case you need to know the meter
    resistance, which wasn't given.

    -Bill
     
  11. That's also possible to find.

    If you couple two resistors in series and measures the voltage over one of
    them, say f.ex that you have R1 and R2 in series and measure the voltage
    over R2. If you know the exact resistance for both R1 and R2, then the
    formula for resistance in the meter is as follow:

    Rmeter = (UR1 * R1 * R2) / ( Ucc*R2 - UR1*R1 - UR1*R2 )
     
  12. Bill Bowden

    Bill Bowden Guest

    Rmeter = (UR1 * R1 * R2) / ( Ucc*R2 - UR1*R1 - UR1*R2 )

    Yes, very good formula. But the meter resistance measuring
    voltage is quite high, probably several megohms, so it
    has little effect in parallel with 1 ohm or less.
    Probably less than .001%

    I would imagine the series meter resistance measuring current
    would be of more concern, since the resistance in series
    is very low and the meter resistance could be a significant
    part of the total.

    I think only one known resistor is needed to figure this out,
    but I haven't worked it out.

    -Bill
     
  13. Don Kelly

    Don Kelly Guest

    Ohmmeters are notoriously lousy for measuring low resistances. While the
    nominal range may allow readings to 0.1 ohms, it is likely that at this
    level the error may be, in fact, much higher - esesentually making the
    reading no more than a continuity test. Such things as a thin oxide coating
    on the terminals . can blow the reading to hell as the ohmmeter will not
    have the "oomph" to break down this layer as happens in normal use. You
    need to be able to supply enough voltage (something near nominal.) to drive
    a reasonable current - and measure V and I. Someone mentioned use of a known
    resistance in series (check its power rating ) and measuring the voltage
    drop across this (getting, by the way, the current fromV/R) and that across
    the unknown. This works (and in fact is related to the way most multimeters
    handle current measurements)
    Use of a such a shunt (low resistance) in series with the load, and
    measuring the voltage across the shunt is a lot cheaper than making a
    constant current source in the 5+ A range. 100ma won't do it for proper
    measurement of the resistance of a 12V device drawing 12 to 24A.
     
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