[moved] Bridge measurement method question

[SK245230]

I am trying to measure capacitors and inductors using the bridge method. (There might be other ways to do this but I want to understand this method)

There is one thing that I don't understand. We measure an inductor L1 and it's serial résistor R4 by tuning a capacitor C1 and a resistor R2, but the capacitor has also it's own ESR. So how is it possible to measure R4 without knowing R2 because of the ESR?:bang:
Moreover R4 should be somewhere near 0.1ohm or 1ohm, so how do I tune such a small résistor?

 

[BradtheRad]

There are ways to measure ESR, and its different values at various frequencies. If ESR is high, then it is easier to measure.

The idea is to select R1 so it is much greater than R2 (esr). You want R2 to contribute very little effect. To get usable readings, I suppose it is enough if R1 is at least ten times R2.

R4 is part of the inductor, and it throws off readings. However R4 is easy to measure. Perhaps it is easier to incorporate in the L/R time constant, added to R3.

There is a way to minimize the effect of R2 & R4. Create an LC loop, so the AC is filtered through it. If you look at the formula for LC center frequency, resistance is not a factor in it, in theory anyway.

Your wheatstone bridge does not have LC acting in a loop. However it appears to be intended for taking measurements of L or C, or experimenting with phase relationships, etc.

 

[KlausST]

Hi,

Use capacitors with low ESR, so it should have (about) no influence on the result.

If the R2 is in the range of 100 or 1000 of ohms, then a ESR in the range of millions won't hurt.

Klaus