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modifying a multimeter

Discussion in 'General Electronics Discussion' started by DrGuy, Jul 21, 2012.

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  1. DrGuy


    Jul 21, 2012
    Hello guys,

    Newbie here with some questions :)

    I need to make some resistance measurements, but my DMM has to high of an amplification for the sensitive parts I am trying to measure. Even on a good fluke DMM It still has a relatively high current. From my research a micro-ohm low resistance meter would be ideal, but it's a bit costly and overkill for what I need. I was thinking of making a circuit to reduce the measuring current that my DMM uses for Ohm measurements.

    Any tips from the pros?
  2. john monks

    john monks

    Mar 9, 2012
    What is "amplification for the sensitive parts" mean?
    What are you trying to measure?
    Is it particularly voltage sensitive or current sensitive?
    If your trying to measure way up into many megohms you can place a charged capacitor on the device and see how fast the charge dissipates.
  3. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    Nov 28, 2011
    Your description isn't very clear. Are you saying the current your DMM generates for testing resistance is too high? It might damage the components you're testing?

    I think it would be crazy to try to muck around inside the multimeter. I would use an external current source, with a suitably low current, to provide the test current, and use the multimeter on voltage range. If your components have a low resistance, the low current will mean that the voltage drop is low; if the multimeter can't measure voltages that low, you would also need an instrumentation-type amplifier.

    In any case you should use the four-terminal method if possible - connect the current source near the ends of the component's leads, and the voltage monitor near the middle of the leads, so that any resistance in the connection between the current source and the component doesn't cause the measured voltage to increase.

    Google will find circuits for current sources and DC amplifiers. If you can clarify your needs, please do so here.
  4. DrGuy


    Jul 21, 2012
    I apologize for not explaining it well.

    Let me try to elaborate.

    I am trying to measure the resistance of some preamplifier's that are very sensitive.
    The preamplifier's are voltage sensitive (not amp's sorry) and a Fluke DMM will send approx, 2V during resistance measurement. Which can and has killed some preamplifier's for me in the past. Measurement range should be between 20 - 100 Ohm's, so there is no need to measure megaohm's.
  5. DrGuy


    Jul 21, 2012
    Ideally I would like to get my DMM down to about 20mV for this. The current should be approx. 100μA. I agree that modding the multimeter itself is probably not the greatest idea.
    Last edited: Jul 21, 2012
  6. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    Nov 28, 2011
    "resistance of preamplifiers"? Do you mean input resistance? Are these preamps for moving coil cartridges or something?

    In your case it's probably easiest to apply a known voltage and measure the current, or supply a voltage through a resistor and measure both current and voltage. Here is an easy example.

    Assume your preamp has a DC input resistance of no more than 100 ohms. It must be DC coupled, i.e. no DC blocking capacitor in the input path. If it's for a microphone or a cartridge it probably won't have a capacitor so you'll be OK.

    Let's say we want to put 100 mV maximum into the input, and it has a resistance of 100 ohms maximum. Ohm's law says I = V / R, where I is current in amps, V is voltage in volts, and R is resistance in ohms. This gives a current of 1 mA maximum to prevent damage to the preamp.

    Let's say you want to use a 1.5V cell as your power source. An AAA, AA, C or D cell is fine. You need to put a resistor in series with the cell, to limit the current that's fed into the preamp. You need to calculate the value of that resistor using Ohm's law, rearranged into R = V / I where V will be the voltage across the resistor, which is the cell voltage minus the voltage across the preamp input (because the resistor will be in series with the preamp input, so the voltages will add together).

    ------ 1.5V cell --------------/\/\/\/\/\/\--------------------
    | . . . . . . . . . . . . . . . . . resistor . . . . . . . . . . to preamp

    With V=1.4 and I=0.001 (remember, I is in amps, not milliamps), R = 1400 ohms. The closest higher preferred value is 1500 ohms or 1.5 kilohms.

    If you make up that circuit using a 1.5K resistor, you will get a voltage across the preamp input, which you can measure with your DMM, and a voltage across the resistor, which you can also measure. From the second measurement, using Ohm's law, you can calculate the current flowing in the resistor, which must be equal to the current flowing in the preamp input, because they're in series. Then you can use Ohm's law again to calculate the resistance of the preamp input.

    Ohm was a pretty dull guy, but his law sure does come in handy :)
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