# How to correctly use a LC meter/bridge

I have a LC meter at hand that I'd like to use it to measure the inductance (dominated by L1) of a circuit branch like this : simulate this circuit – Schematic created using CircuitLab

The bridge I'm using has the following settings:

• Frequency: 100Hz or 1kHz
• Assumption: parallel or series
• Bias ON/OFF: I don't know what that one does, when activated the unit freezes and the screen shows "Bias"...
• Q ON/OFF: Certainly to display the quality factor instead of the value of inductance/capacitance

First, in my case R1 is much lower so I intend to use the series assumption, but in the general case where R2 is the same order of magnitude of R1, what should be used: parallel, or series?

Second, the inductance is supposed to be independent on the frequency, so what's the effect that forces us to test at particular frequencies?

Edit: On request, here are the results of a quick test I've just run:

Series, 100Hz: 18.8mH

Series, 1kHz: 18.3mH

Parallel, 100Hz: 1.4H

Parallel, 1kHz: 33mH

Please note that I have a LED in series with R2 and a diode "head to toe" across that LED (to protect it against negative voltages), which will certainly change the "seen" resistance of R2. In this particular test case, R1=96Ohms and R2 = 1.5kOhms.

• Why not do the measurements on all the ranges and if you can't make sense of something report the numbers back here? – Andy aka Mar 16 '15 at 14:45
• Parallel capacitance (eg. as you approach the SRF) will affect the reading at high frequencies. At low frequencies you may not have enough signal to get a good reading if the inductance is very low. If XL is 1m$\Omega$ at the test frequency, you're obviously not going to have much voltage, no? You should be comparing R1 to XL and R2 to XL at the test frequency. – Spehro Pefhany Mar 16 '15 at 15:31
• @AndyAka: Updated my question with test results. – Mister Mystère Mar 16 '15 at 15:37
• @SpehroPefhany: very true, thanks. Care to expand in a bit more details how that fits in the measurement method? – Mister Mystère Mar 16 '15 at 15:38
• What was R1 and R2 do you think? – Andy aka Mar 16 '15 at 15:43

An LC meter basically measures impedance at a given frequency as shown in the diagram. simulate this circuit – Schematic created using CircuitLab

I've calculated that your inductor is around 21.3 mH, its reactance is therefore $X=2\pi·100·21.3m \Omega$ . If you look at the impedance of your whole circuit at 100 Hz, it becomes: $$\frac{(R1+jX)·R2} {R1+R2+jX}=90.32+j·11.82$$.

If LC meter is operated at series mode at 100 Hz, it sees $L1=\frac {11.82} {2\pi·100}=18.8 mH$.

Let's now calculate the admittance of this impedance: $$Y=\frac {1} {90.32+j·11.82}=0.0109-j·0.0014$$

The admittance of the parallel circuit (at the right of the image) is given by $$1/R2-j1/X$$ where X is the reactance of L2. Therefore, when LC meter is operated in parallel mode, it will find $X=1/0.0014=714.3 \Omega$. Therefore $L2=714.3/(2\pi·100)=1.14 H$.

The important point is that your original circuit is equivalent to the circuits in the images only at the given frequency. And that the LC meter is either finding L1 or L2, depending on the selected mode.

If you repeat the calculations for f=1000 Hz, you'll find: L1=18.7 mH (series) and L2=32.5 mH (parallel). All those values are rather close to your LC meter readings. The difference might be attribuable to the presence of the LED and other factors not being taken into account.

First, in my case R1 is much lower so I intend to use the series assumption, but in the general case where R2 is the same order of magnitude of R1, what should be used: parallel, or series?

If the impedance you're testing doesn't fit a series neither a parallel model. Then the inductance value given by the LCR meter, in both modes, will be far away from the "embedded" inductance value and highly dependent on the test frequency.

You can therefore use any of the modes and, by circuit analysis, calculate what is the value of the "embedded" inductor. This is what I did to get the 21.3 mH value above. I looked at the imaginary part of $\frac{(R1+jX)·R2} {R1+R2+jX}$ and solved the unknown X that would make it equal to the reactance of 18.8 mH (as per your LCR readings).

Second, the inductance is supposed to be independent on the frequency, so what's the effect that forces us to test at particular frequencies?

As seen before, unless you're testing an inductor that perfectly fits one of the models (series or parallel), the reading will depend on the frequency. Even if you're testing an inductor that perfectly fits the model, your LCR meter might have different accuracy depending on the range and the test frequency. The datasheet will give the accuracy numbers for different ranges and test frequencies.

It is also worth noting that some parameters, like the core losses of an inductor, depends on the frequency. The losses are related to the inductor quality factor Q. Finally, at higher frequency, you might approach the self resonance of the inductor, in which case the L reading will be substantially different to the L seen at low frequency.

• Excellent answer, kudos. – Mister Mystère Mar 17 '15 at 11:05