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While preparing a measuring setup for a project, I've realized that the multimeter our lab can provide only supports resistance measurements up to 10^8 Ohm. Since the materials we want to measure go way beyond that, I intended to connect the material in parallel to another resistance and simply add one calculation step to get the desired value, but what if the samples cannot be prepared that way, are there other alternatives to appropriating the measurement for the multimeters range?

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  • \$\begingroup\$ If you can connect a multimeter then you can connect a parallel resistor. Think about it. \$\endgroup\$
    – Andy aka
    Commented Oct 7, 2021 at 11:25
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    \$\begingroup\$ If you have a resistor which you think is near the value of the resistance you are trying to measure, you could use a Wheatstone bridge. \$\endgroup\$ Commented Oct 7, 2021 at 11:26
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    \$\begingroup\$ Megaommeter is used for those tasks. \$\endgroup\$
    – user263983
    Commented Oct 7, 2021 at 11:30
  • \$\begingroup\$ Hi everyone, thank you for the comments and help. I should have specified that we are working inside an isolated fume hood, thus having the setup as small as possible is essential. That's what motivated the question to look for alternatives to additional elements. \$\endgroup\$
    – Lara
    Commented Oct 7, 2021 at 11:36
  • \$\begingroup\$ In a pinch, an Insulation Tester might also work. Note that these typically output a large voltage (up to thousands of volts, careful!) in order to measure high resistances. \$\endgroup\$
    – rdtsc
    Commented Oct 7, 2021 at 11:57

4 Answers 4

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If the things you want to measure have a resistance that goes way beyond what you can measure with a DMM, then paralleling a shunt resistor to bring it into range will rapidly turn small measurement errors into big device under test errors.

You would do better to go back to first principles and devise your own measurement system. For modest number of GΩ you can buy GΩ level resistors and use a very low bias current CMOS opamp as a buffer. Measuring the voltage Wheatstone style will also work.

For even higher resistances, measuring the discharge rate of a good PTFE capacitor (ie sample the voltage, disconnect, then sample again a minute later) shunted by the sample will allow you to observe very small currents.

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  • \$\begingroup\$ If you measure the discharge rate of a good PTFE capacitor connected to a 1 GΩ resistor, a voltmeter with 10 MΩ input resistance will discharge the capacitor during a second more than the 1 GΩ resistor during a minute. Discharge is a 100 times faster. \$\endgroup\$
    – Uwe
    Commented Oct 7, 2021 at 12:45
  • \$\begingroup\$ @Uwe Did you read my last paragraph, which says (ie sample the voltage, disconnect, then sample again a minute later) ? The 'disconnect' bit is important, but perhaps I should have emphasised 'disconnect the meter' \$\endgroup\$
    – Neil_UK
    Commented Oct 7, 2021 at 13:12
  • \$\begingroup\$ Of course I read your last paragraph, which says (ie sample the voltage, disconnect, then sample again a minute later). If you need a second to sample, the interval between two samples should be several minutes long to get a precise result. The discharge rate of PTFE capacitor alone should be much smaller than the rate with the resistor in parallel. \$\endgroup\$
    – Uwe
    Commented Oct 7, 2021 at 13:33
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Lets do the math. A 1 GΩ resistor in parallel to a 100 MΩ resistor, the result is 90.90 MΩ. A 2 GΩ resistor instead is 95.238 MΩ.

A large change of the GΩ resistor will result in a small change of the parallel connection. A precise measurement of the GΩ resistors is impossible this way.

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Measure the current and voltage instead. And calculate the resistance with R=U/I.

(Atleast if the current and voltage are in range of the DMM)

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Here's a way, using a DC voltage source (< 50 V), a DMM having an input impedance around 10 MΩ (to measure current using its voltage function), and a calculator to calculate the resistance of the specimen.

enter image description here

Measuring gigaohms with a simple multimeter

Courtesy: https://www.giangrandi.org/electronics/gigaohm/gigaohm.shtml#voltage_source

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