LCR meter

Q is a measure of the sharpness of tuning and the "temporal response" of a system as I recall it. Example: A tuned RF circuit: How "sharp" is the frequency response curve? A higher Q will be "sharper" with a narrower "band".

HTH .. and that it applies to your question.

"Perfect" how? Hysteresis losses?

The Q of an inductor is the measure of its quality. The more 'perfect' the inductor, the higher its Q value.

Best

Daveb

Bart, You're in an area I don't know a lot about, even though I've done lots of motor work. My suggestion would be to test known good motors of various types and sizes and make yourself a chart. Then if you can find a known bad one you could test it to see where it falls in your chart. I did this with ohm meter readings and AC motors many years ago. My basic chart gave me a good idea of where the phase resistance falls for say a 5 hp motor. One that has open or shorted coils was then quite easy to pick out. Use a megohm meter also to test insulation resistance and you could almost guarantee a motor was good or bad.

Gary H. Lucas

Happy Thanksgiving Cliff,

In reality there is no perfect inductor (he is talking motors) so hystersis is always considered.

All inductors we deal with on a day to day basis have a resistive component, in addition to many other variables

In some permanet magnet motors, hystersis losses are almost non existent ( in advertising only) but you know what I mean.

This is a topic that there is no end .

Regards

. Daveb

Might not the lead/lag of voltage & current have an impact? And that might depend on what else is on the powerline ..... ?

--- I don't believe that's true with air-wound coils, at any frequency, since it's impossible to saturate air.

There is skin effect, though, and the propensity for charge to flow closer and closer to the outside of a conductor as frequency increases is what causes losses to increase as frequency increases.

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--- I don't believe that's true either, because since eddy currents will be time-variant, if anything, they'll have an effect on the reactance of the coil, not on its resistance.

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--- ???

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-- John Fields Professional Circuit Designer

In any inductor being a single coil or dual cores (transformer) there are always hystersis losses, one of the causes of heat in a coil.

Hystersis losses are frequency related, as the higher the frequency the larger the losses.

Also eddy-current losses change the resistive element of an inductor.

This could go on and on ,but the basis fact is when you have a magnet these conditions are part of the deal.

But as stated before by Don Young:

Basically, "Q" is the ratio of inductive reactance to resistance, so an inductor generating 1000 ohms of inductive reactance and having a resistance of 100 ohms has a Q of ten.

But, lol I still dont have a clue how the OP is to use his meter.

Regards

Daveb

Your correct, I thought we were talking iron-core

Again, iron core type inducters start to heat up as frequency increases, (ferrite cores are probably the worse)

Eddy current can develop in the cross section of the wire itself at high frequency. Thats why they make Litz wire to help this condition.

I think the term "Effective Resistence" is used when dealing with the combination of all the fore mentioned losses

bad term-should have said mf.

Daveb

brushes,

way

motor

bad

Motors usually fail with a short or open, or by de-magnitizing. Typical inductances will be from tens of microHenrys to ten or so milliHenrys. DC motors will be lower than induction motors. Resistances are usually low enough to be hard to read with a DVM. Use your LCR meter at 120 Hz since this approximates line frequency. The best bet is to test a few typical motors you use and get an idea what to expect.

Nope, All measurements were made using the DC current put out by the meter. I was only measuring static conditions.

Gary H. Lucas

Hystersis losses cause the coils to heat up? Resistive losses do but .... any hystersis losses in the coils are rather minor indeed I suspect.

Eddy-current losses are an issue in the coils?

Ferrites would be better as they have almost no conductive paths to produce eddy-current losses.