Making your own ohm meter – help needed!

Discussion in 'The Projects Forum' started by ja7me, Sep 17, 2012.

  1. MrCarlos

    Senior Member

    Jan 2, 2010
    Hello ja7me

    Let us see things in parts:

    Is the LM317 configured as a current regulator?
    NO as a voltage regulator but current regulator.
    If so, note that the output current will pass through the potentiometer setting.
    The pot supports that current?

    Note that in the loop formed by: the current regulator and the resistor under test (RUT) must have some very good connections.
    Otherwise vary the millivoltmeter reading. in other words: the reading fluctuates.
    Because the readings are in the range of low voltage, fluctuations in the reading may also occur by ambient electromagnetic noise.
    The poor quality of the RUT can produce also instability in the readings.

    You already looked at the data sheets of the LM317 ?
    Probably it is exceeding a parameter and is therefore not regulate the current as expected.

    As per you writings I can see that your current regulator is NOT working fine.
    A current regulator must maintain it’s output current no matter how you change the RUT.

    Of course there is a range for the maximum value for this RUT.
    This depends on how much voltage, the current regulator has, to apply the current against the RUT.
    And also the electrical characteristics of the LM317.

    your writings:
    my writings:
    Regulator = LM317
    Ra = Adjustment resistor
    Rload = Resistor to be measured
    Vs = 6V

    Ra = 120Ohm
    Rload = 2.2Ohm
    Vload = 0.024
    Panel meter reads 1.87
    V/R= I, 0.024 / 2.2 = 10.91 mAmp.
    Vin – Vout, 6 – 0.024 = 5.976 Volts.

    Ra = 120Ohm
    Rload = 4.4Ohm
    Vload = 0.047
    Panel meter reads 4.10
    V/R=I, 0.047 / 4.4 = 10.68 mAmp.
    Vin – Vout, 6 – 0.047 = 5.53 Volts.

    Also I tried a variable resistor for Ra, so that I could adjust the the panel meter display to read the correct resistance:
    Ra = Variable resistor
    Rload = 2.2Ohm
    Vload = 0.028V
    Panel meter reads 2.20
    V/R=I, 0.028 / 2.2 = 12.72 mAmp.
    Vin – Vout, 6 – 0.028 = 5.972 Volts.

    Ra = Variable resistor
    Rload = 4.4Ohm
    Vload = 0.054V
    Panel meter reads 4.80
    V/R=I, 0.054 / 4.4 = 12.27 mAmp.
    Vin – Vout, 6 – 0.054 = 5.946 Volts.

    Ra = Variable resistor
    Rload = 6.8Ohm
    Vload = 0.082V
    Panel meter reads 7.53
    V/R=I, 0.082 / 6.8 = 12.058 mAmp.
    Vin – Vout, 6 – 0.082 = 5.918 Volts.

    at your service
    ja7me likes this.
  2. Audioguru


    Dec 20, 2007
    The original project uses an LM317L but the OP is using an ordinary LM317 that has a different minimum load current.

    The resistor in the original project is 12.5 ohms but the OP is using 120 ohms.

    So the circuits are completely different.
  3. ja7me

    Thread Starter New Member

    May 24, 2012
    I decided to change the resistor to 120 ohms as i wanted a source current of 10ma, where as before there was a current of 100ma. The main reason for this change is because i wanted to measure in a higher resistance range, 10ma allows me to measure up to 20 ohm.

    20ohm X 0.01amps = 0.2volts.

    0.2 volts is the limit of The panel display meter.

    The circuit is in current source mode, and has been laid out on a good quality bread board. Even the resistor to be measured is pushed into the breadboard. The lm317 i am upping supports higher currents, even has a heat sink on it. I have also tired a different regulator that is designed to be stable at very low currents, between 1 and 10ma. For some reason i was gettin a minus reading on the panel display with this when using 2.2ohm resistor.
  4. ja7me

    Thread Starter New Member

    May 24, 2012

    Your right, your calculations do seem to indicate that it is not regulating the way it should, i will experiment today, and check all my connections and look through the data sheet again
  5. MrChips


    Oct 2, 2009
    Here is my solution that I would like to offer:


    This circuit is a 1mA current source into the unknown resistance R.
    The output in mV would be equal to R in ohms.
    (I have not tested this.)
    ja7me likes this.
  6. Audioguru


    Dec 20, 2007
    I think the intermittent contacts of the breadboard are messing up the circuit.
    Breadboards ALWAYS mess up circuits.
    ja7me likes this.
  7. bertus


    Apr 5, 2008

    The circuit in post # 25 will output a negative voltage.
    The opamp must have a dual power supply to work.

    ja7me likes this.
  8. ja7me

    Thread Starter New Member

    May 24, 2012
    Hi everyone, I am really thankful for everyones input. It turns out it was my breadboard connections. I just made fewer and more secure connections, this seemed to increase the accuracy of the measurements alot.