2
\$\begingroup\$

I'm building a resistive probe using four 10 megaohms resistors to measure the output of a high impedance (> 2Meg)/high voltage bias supply without loading it too much. The idea is that the jury-rigged probe will form a 5:1 divider with the lower leg being the input impedance of the (consumer-grade) DVM. So the voltage reading is directly dependent on the value of the DVM input impedance itself; hence the question.

I have an oscilloscope I can use If the DVM input impedance isn't precise enough, but the same question applies to its 1M input impedance and/or the series resistance of a 10X probe.

As an alternative, I can measure the actual series resistance of the probe fairly precisely using the admittance function on my Fluke meter (good up to 100 M according to manual). If I then measure it using the regular ohmeter function, I could calculate the actual (paralleled) input impedance from the lower resistance reading right?

Thanks in advance

\$\endgroup\$
3
  • \$\begingroup\$ Depending upon the (large and unknow) internal resistance (Ri) of the HV supply you would have to accept a large measurement error even if you know the exact input resistance of the multimeter. How about measuring current instead? Once with a known series resistance of 30 MOhm and a second time with e.g. 50 MOhm. From there you can calculate both the voltage and Ri of the HV source. \$\endgroup\$ Commented yesterday
  • \$\begingroup\$ Interesting alternative. Then Ri = (dV/dI) - dR with dR = (50M - 30M) in your example right? \$\endgroup\$ Commented 21 hours ago
  • \$\begingroup\$ Ri = (R2 * I2 - R1 * I1) / (I1 - I2) with I1 @ R1 = 30M and I2 @ R2 = 50M. Open circuit voltage Vo of the HV source can then be calculated with either Vo = I1 * (Ri + R1) or Vo = I2 x (Ri + R2). \$\endgroup\$ Commented 19 hours ago

2 Answers 2

Reset to default
4
\$\begingroup\$

You should be able to simply connect the Fluke in Ohms mode across the voltmeter input (both meters on) and it will display the input resistance (and the voltmeter will display the test voltage from the Fluke). If it’s a typical 10M input that should be more accurate than the uS ranges.

This does not account for the (typically negligible) bias current, but you can verify that is indeed negligible by simply leaving the voltage meter input open and observing that it stil reads zero.

\$\endgroup\$
2
  • 1
    \$\begingroup\$ Thanks speff, that works. I don't know why I assumed that the test voltage would foul things somehow. Turns out the input impedance of all my meters is over 10% higher than specified at 11.12 Meg. One day I'll get proper equipment.... \$\endgroup\$ Commented yesterday
  • \$\begingroup\$ @Joe Your instincts are good. It's not silly to think about it. If the test voltage from the Fluke was high enough to drive the other over range to the point where some kind of (nonlinear) internal clamping took place you could get an inaccurate (low) reading. But that's not at all likely the way DMMs are actually designed and if the voltage meter is not reading over range during the test then it should not be clamping. \$\endgroup\$ Commented yesterday
3
\$\begingroup\$

To calibrate using even fewer assumptions, measure a stable (doesn't even need to be known) voltage source with and without an input resistor in series with the DVM.

Beware that DVM input resistance may be different on different ranges. Be sure to calibrate it on the range you intend to use for your final measurements. This is a typical range-switching input for a DVM with a nominal 10 MΩ input resistance. You can see how if you calibrate it on the x10 range, but use it on the x1000 range, the input resistances will be different.

schematic

simulate this circuit – Schematic created using CircuitLab

If you have a manually-ranged DVM, or auto-ranging with a range-hold override, then you can simply set the calibration range to the same as the measurement range. If you have an auto-ranging meter that cannot be overridden, then you must use a calibration voltage that sends it to the right range.

\$\endgroup\$
2
  • \$\begingroup\$ Of course! Then use voltage divider equation to find DMM impedance. Thanks \$\endgroup\$ Commented 21 hours ago
  • \$\begingroup\$ @Joe I've added a second paragraph. \$\endgroup\$ Commented 19 hours ago

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.