My circuit has 4Hz 50% duty cycle. How do I measure the average current using a DMM?

I have a Fluke 87 DMM. It has an average setting but for 100 ms which less than one complete duty cycle.

My circuit takes abouts 13 uA in sleep and 20uA in active.

  • \$\begingroup\$ did you mean 20 uA or 20 mA? \$\endgroup\$
    – Russell McMahon
    May 8, 2019 at 8:58
  • \$\begingroup\$ I meant micro Amp (uA) \$\endgroup\$
    – RckR
    May 8, 2019 at 14:11
  • \$\begingroup\$ If duty cycle is a fixed 50% and currents are 13 and 20 uA then Imean = 16.5 uA. Or you can use my 2 caps and a resistor method. C2 = 1000 uF ideally but 100 uF may be good enough. For 0.1V across Rm at 20 uA, Rm =~ 5K. \$\endgroup\$
    – Russell McMahon
    May 8, 2019 at 14:30

3 Answers 3


If you know the duty cycle, just measure the current in both states and do the average by hand.

If you don't know the duty cylce, get a tool with which you can figure out: an oscilloscope in high input impedance mode, attached across a shunt resistor would be a good tool.


50% is not a "very low duty cycle", it's a very low frequency.

A low duty cycle would be something like 1mA for 10ms and 100uA for 240ms.

If your meter is not giving you a consistent reading because of sampling, you can make your own current shunt with a resistor and an RC filter. Say you use a 5K shunt resistor with an RC filter of 10uF ceramic and 200K, which should give you 40uA full scale on your meter 200mV DC voltage range.

  • \$\begingroup\$ Sorry. I meant low freq. \$\endgroup\$
    – RckR
    May 8, 2019 at 14:12

My circuit has 4Hz 50% duty cycle. How do I measure the average current using a DMM?

Easier to do than to describe :-).


Provide 2 large filter capacitors. "Large" = will change in voltage by a minimal amount during a cycle.
Connect a resistor Rm between them.
Connect Vin to C1, Vout to C2.
Measure V across Rm to determine Im



Ground capacitor negatives, connect C1 +ve to power supply, connect C2 +ve to load. A capacitor Cm will change in voltage by ABOUT Vm in period Tm for a current of Im where
Cm = Tm x Im / Vm
- Im as max current liable to be drawn when circuit is active.
- Vm as allowable change in capacitor voltage
- Tm = period of high drain - or some suitable period if unknown.

eg Vm = 0.01V, Tm = 1 second, Im = 20 mA
ie Vcap must not sag more than 10 mV in 1 second at 20 mA drain
So C = Tm x Im / Vm = = 1 x 0.020 / 0.010 V = 2 F.
If this is larger than acceptable, revisit allowed values.

C2 is more important than C1. C2 provides a reservoir into which Im flows. C1 provides a stable source .

Connect a resistor Rm between the two capacitors.
Size Rm such that mean current drain will cause a voltage drop that is as large as possible BUT not so large as to significantly alter load operation.
eg if 0.1V is an acceptable drop and mean current is expected to be say 25 uA then.
Rm = Vdrop/Imean = 0.1 V / 25 uA = 4K -> say 3K9 or similar.
If a larger dV drop is acceptable make Rm larger. eg here maybe 10K.


Connect load to circuit.
Temporarily short out Rm to allow C2 to charge to Vc1.
Remove short.
Measure voltage across Rm. This allow I average to be determined.
Iavg = V/R = V_rm/Rm


simulate this circuit – Schematic created using CircuitLab


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