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ESR vs Capacitive reactance

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**bearman**:

I am confusing myself. I want to build an ESR meter. It soulds like a very handy tool for troubleshooting 25 plus year old circuits that may have dried up caps.

I started digging around and reading about ESR then looked into capacitive reactance and now I can't see what the difference is.

Capacitive reactance looks like you use a sin wave for measurment. Reactance is frequency dependant.

All of the circuits I have found talk about a square wave (100 or 150 khz) signal injected into an RC circuit to measure ESR. Isn't ESR frequency dependant?

Why did they choose 100 to 150 khz signal?

What am I not seeing?

HELP.

John

**jimmc**:

Good explanation here... http://midwestdevices.com/pdfs/Tnote3.pdf

There are links to practical circuits at the end of the Wikipedia article...http://en.wikipedia.org/wiki/ESR_meter.

For a digital device look here...http://members.ozemail.com.au/~bobpar/k7214.pdf

Jim

**hans**:

ESR is a series resistance in series with the capacitor. Put in mind that you would place a 1 ohm resistor in series with a capacitor to decouple your power supply. Would you think that would improve the effect of that capacitor? There is a reason to avoid capacitors with such high ESR values.

A capacitor is designed to block DC (high impedance) and short high frequencies (low impedance). Impedance is a complex number that consists of resistance and reactance. It is written as Z=R + jX. Here is R the 'real part' and X the 'imaginary part' (note the j symbol).

A resistor has an impedance of Z=R. A capacitor has an impedance of Z=1/jwC. As you can see, a capacitor is has an imaginary component. This is for a perfect capacitor. Here is C the capacitance value, and w the angular velocity (=frequency times 2*PI).

But we are measuring imperfect capacitors (that's what an ESR meter is for), so the impedance will be Z = R_ESR + 1/jwC. R_ESR is the resistor in series with the capacitor. If you want to know the 'resistance' (better to say: impedance) of the component at a certain frequency (capacitors are frequency dependant), you need to calculate the magnitude:

|Z| = sqrt(R^2 + X^2)

So, for a capacitor that is:

|Z| = sqrt(R_ESR^2 + (1/wC)^2)

If you apply a sine wave of 1Vpp on a capacitor and measure the current at different frequencies, you should get different values of current. That is because the impedance of the capacitor is moving from very high (blocking DC currents) to very low (shorting high frequency AC).

->> Now after this boring theory, it comes down to this: If you apply an AC voltage and you make w (the frequency) very high, it will mean that the piece 1/wC of the impedance becomes very small. If you divide 1 by a big number, you get a small number. If you raise this frequency so much that the part 1/wC is ignorable, you can measure the ESR value (because the impedance of what you're measuring only shows the ESR). After all, if you have something like:

|Z| = sqrt(0.1^2 + 1/(2*PI*150k*4700u)^2)

|Z| = sqrt(0.01 + 5.096383*10^-8)

with R ESR = 0.1 ohms, C = 4700uF, test frequency of 150kHz. The impedance is:

= 0.1000000254 (ohms, for 150kHz signals).

If you want to see the effect of the capacitor at 150kHz (and C=4700uF), you need atleast a 8-digit multimeter. Goodluck ;)

If you would do this at like say, 500Hz:

|Z| = sqrt(0.1^2 + 1/(2*PI*500*4700u)^2)

|Z| = sqrt(0.01 + 0.00458674)

=0.120776 (ohms, for 500Hz).

Which is a value we could expect, because a capacitor tries to block low frequent signals.

**bearman**:

Hans

Great explanation.

I couldn't see the forest because the trees were in the way.

That was the KEY I was missing. I couldn't quite see the fact that the point was the reactance becomes negligable and ESR is the dominant value being measured.

I read the theory in my old circuit analysis text books but it clouded my view of the application of it.

I'm off to build one of the ESR circuits from the links provided.

Thank you and all others that replied.

J.P.

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