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Old March 30th 04, 10:32 PM
John Jardine
 
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Nick Kennedy wrote in message
...
OK, but remember that the voltage drop across the inductor plus that

across
the source resistance don't add directly to the source voltage. With your
example of 10 mV across L and 100 mV open circuit voltage, the voltage
across the 50 ohm source resistance is 99.5 mV. That's because they're 90
degrees out of phase and add in a right triangle fashion.

Maybe you considered that, and that's why you chose a small value of drop
across the inductor. When that's the case, the hypotenuse and x-side of

the
triangle are pretty close to equal. But users of this method need to
understand this simplification and the implied restrictions.

I've used something similar to measure the inductance of chokes too large
for my meter, such as transformer windings and power supply chokes. You

can
put a known resistor in series with the choke, put 60 Hz AC across it, and
measure the two drops. Then calculate the inductance using a method

similar
to what you've described. But keep the sum of squares relationship in

mind.
(Supply voltage squared equals drop across resistance squared plus drop
across inductance squared.) If the inductor has significant resistance,

it
should be added to the value of the series resistor.

Regards,

Nick, WA5BDU
in Arkansas

"John Jardine" wrote in message
...

[clip]

Yes. This is true. Keep the O/C S/C ratio as wide as you're happy with.
Fortunately this is usually easy to do and the method allows a quick and
(slightly) dirty measurement without the extraneous baggage that
accompanies other methods. I built an (accurate) test unit using this method
that goes down to a FSD of 300nH and though I have a digital LCR bridge the
homebrew item is *always* used to give final conformation and confidence in
an L's value. Measuring an inductor somewhere near it's operating frequency
is essential.
You clearly understood where I was coming in from on this topic. The usual
reaction is some kind of patronising look and words like "you can't do that,
don't you know anything of electronics?" :-)
regards
john



 
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