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I've got an issue that appears to be caused by damaged resistors that are either open circuited or of too low a value due to contamination. The problem is that they're gigaohm resistors, so to a multimeter, they're always open-circuited. How can I measure the resistance, or, at least, test the continuity?

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  • \$\begingroup\$ Beware that you must test the insulation at a voltage close to working. What appears to be insulated at 500V might show kohms resistance at 1000v. \$\endgroup\$ – Kristoffon Aug 6 '12 at 3:55
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    \$\begingroup\$ @Kristoffon: Working voltage is less than 1 V in this case. :) Just the leakage current of an FET gate times the resistor's value, max. \$\endgroup\$ – endolith Aug 19 '13 at 17:09
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Many Fluke meters (e.g. 87,287) have a nanoSiemens conductivity range which will measure up to 100 GigaOhms - it needs to be manually ranged up from the ohms range. \$\mathrm{1 G \Omega = 1 nS}\$, \$\mathrm{10 G \Omega = 0.1 nS}\$.

Alternatively, most DMMs have with a 10M input impedance (easily checked with a second meter), so a resistor with value R in series with the millivolt range will form an R+10M/10M voltage divider. So applying 10 volts through a 1 gigohm resistor will read around 99 millivolts. A close-enough approximation for high value resistors from a 10V supply would be resistance in gigohms = 100/millivolts.

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  • \$\begingroup\$ The divider method is fast and easy with a 9 V battery. R = Rmultimeter*(Vbattery - Vdivided)/Vdivided. Just don't touch more than one of the metal parts with your fingers. \$\endgroup\$ – endolith Dec 11 '14 at 17:45
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You need insulation testers. The ones I've seen had 2 GOhm range. Not necessary Flukes, there are cheaper ones.

And for the future, I would try to add some protective insulation on top of such nasty things :-)

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  • \$\begingroup\$ What kind of protective insulation? \$\endgroup\$ – endolith Feb 18 '11 at 2:09
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I'll assume that you're able to isolate the resistor from the rest of the circuit.

You probably need to construct a high-impedance analog buffer. It doesn't need to be super-fast, but it does need to be high impedance. A very high impedance amplifier is National's LMP7721, requiring only 3 femtoamps of bias current.

Once you have your buffer, get another resistor with a resistance comparable to the one you want to test (a known value). Connect one side of this resistor to ground, and the other to a probe and to your buffer. Then, apply a voltage to one side of your resistor, and connect your buffered probe to the other side. Measure the voltage at the output of your buffer and solve the voltage divider to determine the unknown resistance

You may not need a buffer if your meter has extremely low impedance when measuring voltage.

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  • \$\begingroup\$ 1V across 1Gohm is 1nA of current rather than 1pA. I think you'd have to be very careful with your buffer design and make sure it has strong high frequency rejection. Its not hard to generate currents on the level of 1nA from stray EMI especially with probe leads in the mix. \$\endgroup\$ – Mark Feb 16 '11 at 23:46
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    \$\begingroup\$ Higher voltage will definitely help in that case. You need to go lower than 1 pA though. Check out national.com/pf/LM/LMP7721.html, especially some of the application circuits. You will have to be VERY careful about contamination on your board, any kind of flux will create a leakage path. Also, you'd be much better off with an alternate circuit than a voltage divider. Noise will dominate your measurement. Check out a transimpedance amplifier. \$\endgroup\$ – Chris Gammell Feb 17 '11 at 1:01
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    \$\begingroup\$ @Chris - Thanks for the advice! My answer was just a first shot at solving the problem, and unfortunately I didn't know anything about transimpedance amplifiers before tonight. Want to throw an answer up? \$\endgroup\$ – Kevin Vermeer Feb 17 '11 at 2:47
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"If you use a battery operated DMM, and keep it isolated, you can use 1000s of volts for the test."

DONT TRY THIS !!!

Most of GigaOhm resistors, including 200 GigaOhm resistors in glass tubes have a rating of maximum 500 volts, and the maximum voltage for a digital voltmeter is 1000 volts. Thousands of volts across such a resistor will only be sparking around the resistor and instantly fry your digital voltmeter!

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    \$\begingroup\$ even 1/4 watt carbon resistors had a 500V rating. Normally they are longer and rated > 1~10 kV Since we are talking long after the question. I think the accepted answer missed the hidden more important point of doing root cause failure analysis and simply answered how to measure a normal resistor. Failures occur from non-linear V vs I characteristics that lead to failure as you indicated @Marc. Zapp! by contamination is a major flaw. the material must be well sealed and moisture proof. THat takes variable Hipot testing with a current limiting R to protect the device and uA meter to measure it \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 16 '12 at 21:41
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There is special equipment for this. Couple of weeks ago someone showed me one that can do > 500G and in this particular case was used to test 10kV breakers. It was called a Megger. Basically what it does is measure resistance, but where your multimeter does this with 3V, these thing slowly increase the voltage to test in the range of kV's. https://en.wikipedia.org/wiki/Megger I expect there are other vendors for similar equipment.

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What you would want is a megaohmmeter. These are just another permutation of the V=IR Meter which exploit high voltage to produce a measurable current across a high resistance. If you have access to a high voltage source and a DMM with a current mode, you could measure the resistance but placing the resistor, DMM, and high voltage in series, then math-ing it out.

If you use a battery operated DMM, and keep it isolated, you can use 1000s of volts for the test. I used to calibrate the leakage current readings of 1-200KV Hi-Pots using just a normal fluke DMM with this method.

You can find megaohmmeters on ebay as "Hi-Pots", "insulation tester", "oil tester", "dielectric tester".

Also, the opposite of a megaohmmeter is a digital low resistance ohm-meter(DLRO), these use a high current (1-100+ amps) to measure very low resistances.

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I have just tried to measure 10 Gigaohm resistors with my DMM and a 10 volt power supply with success.

My DMM is a 4 1/2 Digit with a stated 10 Megaohm imput impedance. The DMM has an accurancy of 0.05 % for voltages measures. I first adjusted my power supply so that the voltage shown an my DMM was exactly 10.000 volts, then put the 10 Gigaohm resistor in serial with the DMM on it's 200 mV range. The reading was 11.35 mV.

In fact, the only thing that is not stated as precise with my DMM is it's imput impedance! I tried to measure it with another multimeter (not digital) and found that the real imput impedance of my DMM is in fact above 11 megaohm, so there are about 10 % error.

The 10 Gigaohm resistors I measured (I have 4 of them) have only a 5 % tolerance, but they all gave me about the same reading on my DMM. If I had one of 0.1 % tolerance, I could adjust my power supply so that the DMM would read exactly 10 mV to compensate for it's 11.35 Megaohm impedance, in this case the voltage from the power supply would be adjusted to 8.81 V and I would have a precise gigaohm meter.

Another thing to note is that the probes of the DMM have lot of leakages. I had to put the DMM on a separate table with the probes and the resistor to be measured hanging on the air. I then tried to put the 10 volts from the power supply across the PVC part of each probes and had a voltage reading of 0.05 mV on the DMM, corresponding to a resistor of about 2 Teraohm...

Time to buy teflon insulated wires...

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I great trick I learned from reading the HP 3478A DMM Service manual (section 3-119 extended ohms operation) is to first measure a 10M resistor, and then put the 10M in parallel with the unknown high resistance and measure the parallel value. The formula unknown=(reference value * measured parallel value) / (reference value - measured parallel value) does the trick. As an example, say you used a 10 ohm reference, and say you are measuring a 10 ohm unknown. The two 10 ohm resistances in parallel would measure 5 ohm, so running the formula gives 10 * 5 = 50 and 10 - 5 = 5, and 50 / 5 = 10 ohms. This works for any reference value, and the measured value will always be less than the reference value. Some of the other answers point out some of the limitations of any high resistance measurement. You also run out of digits of measurement precision at some point.

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    \$\begingroup\$ Recalculate your measured 1 Gohm resistance with min and max tolerance of your measurement and see how wide the uncertainty range is for said 1 Gohm resistor, then report back. \$\endgroup\$ – winny Mar 31 '17 at 8:53

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