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I'm trying to measure the reverse leakage current of some diodes with the method proposed HERE - the diode is connected in reverse bias to 5V through 1 megaohm and voltage is measured across the resistor, the current is calculated from the voltage. However I need to do this as accurately as I can, and as I understand the input impedance of the measurement device will distort my results greatly because it is close to the resistor's value. Instead of a voltmeter I'm using an oscilloscope (1M input impedance, 10x attenuation probe). My questions:

  1. How would one compensate the change in measured voltage due to the voltmeter input impedance?
  2. Is there any difference to doing this while using an oscilloscope?

If I'm doing something wrong here it would be nice to know as well. Any input is greatly appreciated!

EDIT: As I suspected, the way to go here is to consider the voltmeter/oscilloscope to be a resistor in parallel of the load to be measured - as Andy suggested. You should also check EM Fields' answer for measuring methods. HERE'S a well written paper about the effect of measurement device input impedance on your circuit.

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3 Answers 3

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If the input impedance of the measurement device is (say) 10 Mohms then you can assume that this 10 Mohms is in parallel with the 1 Mohm circuit resistance. This effectively lowers the 1 Mohm to 0.9090909 Mohms.

Try researching resistors in parallel.

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  • \$\begingroup\$ Thank you for your answer. I do understand how resistors in parallel work - using the parallel resistor - voltmeter impedance value instead of only the resistor does yield a result. What threw me off is that the voltmeter and oscilloscope show different values for the same measurements, so I'm not sure what I should be doing different here. \$\endgroup\$
    – I have no idea what I'm doing
    Oct 14, 2014 at 8:30
  • \$\begingroup\$ How different are the measured values? maybe the voltmeter has different impedance to the o-scope? Maybe they are both working within their respective specs but one is inherently more accurate than the other. \$\endgroup\$
    – Andy aka
    Oct 14, 2014 at 8:37
  • \$\begingroup\$ The voltmeter has an input impedance of 10M, the oscilloscope - 1M plus the 10x probe, which as I understand has 9M, adding to 10M as well. The difference between readings is around 10%. For example if the voltmeter gives me 9mV, the oscilloscope shows 10.2mV. Could this simply be an accuracy issue? \$\endgroup\$
    – I have no idea what I'm doing
    Oct 14, 2014 at 8:43
  • \$\begingroup\$ You need to be able to find the specified accuracy for each piece of equipment. Without that it's impossible to say. \$\endgroup\$
    – Andy aka
    Oct 14, 2014 at 8:59
  • \$\begingroup\$ Well, according to the manuals the accuracies can't be worse than +- a couple percent, 10% seems way too high. \$\endgroup\$
    – I have no idea what I'm doing
    Oct 14, 2014 at 10:42
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Use your oscilloscope to measure the voltage in a circuit that has the diode replaced by a known large resitor (e.g. another 1MOhm). Ideally (input impedance of scope is infinite) you would get 5V / 2 = 2.5 V. In reality you will get a lower result. Use this result to calculate the real input impedance of your oscilloscope.

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  • \$\begingroup\$ Using this method I get 2.38V for the voltmeter (which should mean 9.58M input impedance, if I'm correct) and 2.46V for the oscilloscope. 2.46V would mean that the input impedance is 35.81M, which is a really odd value if you ask me. Also, a variation of 0.01V means a difference of 10M in impedance, so it doesn't seem like an accurate way to measure. \$\endgroup\$
    – I have no idea what I'm doing
    Oct 15, 2014 at 8:29
  • \$\begingroup\$ You are right, it is not easy to measure very high input impedance accurately. This is because it does not much affect the circuit. But this means also that it doesn't much affect you original measurement. \$\endgroup\$
    – Curd
    Oct 15, 2014 at 10:58
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Use a bridge, as shown below, to get what amounts to a voltmeter with an infinite impedance across R1.

In use, when you adjust R2 until the microammeter reads zero, the voltage across R1 will be indicated by the voltmeter.

1000 ohms or 10k would be OK for R2, but the higher the resistance the greater the microammeter's null will be degraded when the bridge is being balanced.

enter image description here

Or. you could use a DMM with a 10 megohm input resistance as the resistance in series with the diode, as shownn below, and get 100pA sensitivity on the 200mV range.

enter image description here

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  • \$\begingroup\$ The second schematic is pretty much what I'm trying right now, I will try the bridge, thank you for your answer! \$\endgroup\$
    – I have no idea what I'm doing
    Oct 15, 2014 at 8:33
  • \$\begingroup\$ You're welcome :-) A caveat: to make sure the reading you get from the second schematic is valid, check the voltmeter specs for your DMM to make sure you know what its resistance is on the range you'll be using. \$\endgroup\$
    – EM Fields
    Oct 15, 2014 at 8:48

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