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
...


0.27uH is 5ohms at 5MHz.

Set to 5MHz and wind up your 50ohm signal generator to max output (say a
100mV or so). Measure it's voltage using the scope.
Short the inductor across the sig genny terminals and measure the voltage
across the inductor.
Ignore all irrelavencies such as phase angles, stray capacitances, Q
factor,
series resistance,skin effects, self resonances etc. Just treat the
inductor
as if it's a straight forward potential divider resistance in series with
the 50 ohms of the genny.

Eg. Sig genny open circuit output at 5megs is 100 mV peak-peak
Inductor connected and voltage measured is now 10mV peak-peak.
Therefore 1/10th of the available voltage is across the inductor and 9/10
lost across the genny internal 50ohms.
So inductor looks like 5 ohms therefore,
5ohms = 2 x Pi x 5megs x L?
=270nH. Simple !
Using this method it's easy to get down to the odd few nH.
Exact same method as used by a nice 'inductance meter' design that turned
up
in Radcom about 12 years ago

regards
john