Re: Findings from a couple hours at the lab bench with a CIM and a Jaguar
 

Would you mind posting a quick hand-sketch schematic of the above? I'm trying to figure out what you mean by "matches my source's 50 ohm output".

I'm familiar with placing a reference resistance of known value in series with an inductor under test and finding the sinusoidal frequency at which the true RMS voltage drop across the coil equals the drop across the reference resistor, but that doesn't seem to be what you are describing.

Re: Findings from a couple hours at the lab bench with a CIM and a Jaguar
 

That's precisely what I'm doing, but I'm using the resistance of my source instead of a discrete resistor. Here's a link detailing the procedure.

Incidentally, we already had all the information necessary to compute inductance. Look at the current waveforms in the original post. During the on-time we apply ~13.8 volts for ~12.2 us and got a delta-current of ~2A comes out to about 84 uH. Of course, I'm not using cursors but rather eyeballing the PNGs I posted earlier, so it's completely expected that the number not come out the same. But it's right ballpark.

-Sasha

Re: Findings from a couple hours at the lab bench with a CIM and a Jaguar
 

In order not to confuse any students who may be following this discussion, it's not precisely what you're doing. In the method you used, the voltage across the coil is not equal to the voltage across the 50 ohm resistor.

Re: Findings from a couple hours at the lab bench with a CIM and a Jaguar
 

Well, not at the same time, but it's the same method. Basically I find a high-enough frequency where the voltage across at the output of the meter (post-50R source resistor) doesn't change when I attach the motor.

Then I lower the frequency until the voltage is exactly half that. At that point the voltage across the resistor and the voltage across the motor are the same, but I can't measure the voltage across the resistor directly. I have to trust that my instrument is consistent between the two measurements. It's probably a good assumption.

Re: Findings from a couple hours at the lab bench with a CIM and a Jaguar
 

They're not the same.

Consider:

When the voltage across the motor is half the voltage across the resistor+motor series circuit,
the frequency will be f = R/(2*pi*L*sqrt(3)).

At that frequency, the magnitude of the motor impedance will be R/sqrt(3), so the voltage across the resistor will be sqrt(3) times the voltage across the motor.



For the voltages across the resistor and motor to be equal, the frequency would have to be f = R/(2*pi*L). At that frequency, the magnitude of the motor's impedance would be R, and the voltage across the motor (and the resistor) would each be 0.707 times the voltage across the motor+resistor series circuit.