Test Set-Up for Inductance

in article snipped-for-privacy@posting.google.com, jemilio at snipped-for-privacy@hotmail.com wrote on 5/4/04 2:46 PM:

Again, this seems to be one of those poorly worded posts that I see all too often.

My guess is that you are supposed to connect a low impedance 10 V (20V P-P) square wave generator across the inductor and look at the current drawn with a suitable probe. With a square wave input, the current waveform should be triangular.

From L*(di/dt) = 10 V, the rate of current change should be 10/L or 0.917 A/µs. At 200kHz, a half period is 5 µs. During such a half period, the current should change from -2.3 to +2.3 A. This sounds like a fairly hefty square wave generator to me.

My having said that, what can you add?