I bought a very cheap RLC meter, mainly to measure inductance. It has a single LCD display, so it shows only one value at a time, unlike more expensive RLC meters that measure both inductance and resistance at the same time. There is only one test frequency for each measurement range, so you can't choose the test frequency.

The exact meter model is PeakTech 3730.

The meter is about as cheap as you can find. For example it lacks auto-range which is very common in decent multimeters. Instead, there is a large dial switch with separate positions for each measurement range. Apparently, the selecting the correct range from the dial isn't enough because there is also an "L/C" button which must be in the correct position for the measurement, if measuring inductance or capacitance ("L" for inductance, "C" for capacitance). At all but the 20 Henry range, it is using a fixed test frequency of 1 kHz for measurement that can't be chosen from a number of options (20 H range uses 100 Hz).

I tested how well I could separate the effects of resistance from the effects of inductance. I took an axial inductor with 56 µH nominal value and 5% accuracy. Assuming that the 56 µH nominal value is correct, it would have an inductive reactance of 0.35 ohms at that frequency.

The meter read 61 µH (actually it read 63 µH, but a measurement without any inductor with test leads shorted showed 2 µH which I subtracted).

Then I measured the resistance. It was 3.4 Ohms (actually it read 3.6 Ohms, but a measurement with test leads shorted showed 0.2 Ohms which I subtracted).

So apparently, for this inductor, total impedance is 3.4 Ohms + j*0.35 Ohms if we can believe the 56 µH nominal value to be correct. This impedance has an absolute value of 3.42 Ohms, which corresponds to an inductance of 544 µH for a cheap 1 kHz meter that's measuring only the absolute value of impedance and not the phase angle at all.

So it appears the meter is either measuring inductance by some other method than measuring impedance, or that it is measuring impedance but can somehow remove effects of resistance ten times larger than inductive reactance. The latter sounds incredible for such a cheap meter, if it's so cheap that selecting inductance from the dial isn't enough but I must press a "L/C" toggle switch too.

How does such a cheap meter measure true inductance without mixing in effects of resistance?

  • 1
    \$\begingroup\$ maybe it also measures DC resistance \$\endgroup\$
    – jsotola
    Oct 21, 2022 at 18:40
  • \$\begingroup\$ 1kHz sine or square? The latter has many frequencies and thus allows one to derive L and R. \$\endgroup\$
    – tobalt
    Oct 21, 2022 at 18:48
  • \$\begingroup\$ A guess (hence not an answer) - some meters like this employ an oscillator, whose frequency is affected by the reactance of the externally-applied inductance. Frequency can be measured fairly precisely. Short-term frequency changes due to temperature should be small. The frequency change due to a 56uH inductor is likely very small, but easily measured. You'd likely get better results if oscillator frequency was much higher, for this small inductance. \$\endgroup\$
    – glen_geek
    Oct 21, 2022 at 18:59
  • \$\begingroup\$ Auto-range can be a blessing or a curse. It can be a curse because the meter is slow to change ranges and if it constantly changes ranges you might not even get a reading. I'd put that in the bucket of optional extras and not necessarily a sign of a low quality meter. (which is not to say that you don't have one) \$\endgroup\$
    – user253751
    Oct 21, 2022 at 20:05
  • \$\begingroup\$ I suspect that oscillator with frequency depending on external L/C is the most plausible explanation. The L/C switch could choose whether the oscillator is tuned for external L or external C. The specs say 150 µA current for inductors. With 1 kHz square wave, for example detecting rise time of current in 56 µH inductor from 0 to 150 µA with any accuracy would require ~100 MHz clock (assuming the voltage source would be 15 mV), which I suspect such a cheap meter won't have. I'll have to check with a scope if the oscillation is sine wave, and if it varies with inductance value. \$\endgroup\$
    – juhist
    Oct 22, 2022 at 7:57

1 Answer 1


Ok, I measured this meter with a scope.

I used three inductors:

  1. nominal 100 µH (measured 110 µH), 0.3 Ohm
  2. nominal unknown (measured 1424 µH), 0.2 Ohm.
  3. nominal 56 µH (measured 64 µH), 3.4 Ohm

The meter definitely uses a sine wave. However, the sine wave on the scope output had so much noise that triggering was unreliable. Therefore, I had to stop the image on this digital storage oscilloscope (it has a phosphor feature so it remembers a lot of the signal and shows the most common value with stronger color).

This noise and unreliable triggering could be consistent with measuring both inductance with a sine wave and resistance with DC alternately.

All inductors had about 914 - 916 Hz sine wave frequency, so the frequency definitely was the same. (The 2 Hz difference was due to oscilloscope cursor accuracy.)

The first inductor (impedance 0.3 + j*0.6324 Ohm, absolute value 0.7 Ohm) had a certain voltage on the output.

The second inductor (impedance 0.2 + j*8.19 Ohm, absolute value 8.19 Ohm) had a much stronger voltage. Not by a factor of 8.19 / 0.7, though.

The third inductor (impedance 3.4 + j*0.37 Ohm, absolute value 3.42 Ohm) too had a much stronger voltage. Not by a factor of 3.42 / 0.7, though.

Based on this, it seems a complete mystery how the inductance is measured (so this isn't really an answer but a long comment). The frequency is the same. The current is not the same or else the voltage factors would be very much different than what they are.

Also, to me it seems to be impossible to determine the imaginary part of impedance of the inductor having 3.4 + j*0.37 Ohm impedance, if measuring only the absolute value and real part. Its absolute value is 3.42 Ohm, so you have only 0.02 Ohm difference between the real part and absolute value. Given the nature of Pythagorean theorem, the inaccuracy would be so larger that this measurement simply isn't possible.

One possibility is that it's using a variable-frequency oscillator where the frequency is tuned to be constant in closed loop control, and the tuning factors are used to tell what the inductance is. For example an electronically tuned capacitor could do it.

One supporting evidence for the electronically tuned oscillator is that the accuracy in measuring capacitance is 1%, but the accuracy in measuring inductance is 2%. If the frequency is 1/(2*pi*sqrt(L*C)), due to the square root, 1% accuracy in frequency translates to 2% accuracy in inductance.

Another possibility is that if this meter is measuring the real and imaginary parts of impedance at the same time, by measuring both absolute value and phase angle, and not only the absolute value of the impedance. This could explain how it can so reliably detect the inductance of inductors having a huge resistance.

Thinking about this again, the constant frequency and the small voltage variations (not directly proportional to impedance) that I saw are consistent with the meter having a sinusoidal voltage source with a small resistor (let's say 2 Ohms) on the negative side. The meter could be measuring voltage across that resistor (giving current according to Ohm's law), and at the same time voltage across the output of the sinusoidal voltage source (directly at output, before the small resistor). Then from current and voltage, the meter can calculate total impedance. If the meter is also measuring phase angle between current and voltage, it can calculate the real and imaginary parts of the impedance. From the imaginary part, it can get the inductance.


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