Hi again,
Ok that looks better so now we have something to work with.
You'll note that the current waveform is somewhat like a ramp up and ramp down, with noise in between, and the voltage waveform is a square wave.
The only strange thing is that the voltage waveform seems to be inverted from what it would look like if we used two separate scopes. Perhaps something is still not right there yet, but we can see that the duty cycle is not 50 percent so using that we'll estimate the inductance using the positive voltage pulse and the low going current ramp until we clear up the possible sync problem.
Now keep in mind that normally you would use the positive voltage pulse and the positive going ramp for example.
I've attached the two scope pictures with the quantities labeled on the picture so you can see how there measurements are made. Once we take the measurements we then go ahead and estimate the inductance based on the well known inductance equation:
v=L*di/dt
So looking at the first attachment which is the voltage, we see the voltage goes above zero by 5.5 volts. I assume you have the zero adjustment of the scope done properly. Looking at the pulse width shown there also, we see the pulse width is 8us.
Now looking at the second attachment which is the current, we see the voltage fall from about 0.0063 to 0.0003 which means we have a differential of 0.006 volts, and 0.006 divided by 0.1 ohms (i assume you are using 0.1 ohms for the current measurement) we get 0.060 amps.
Solving the inductance equation for L we get:
v=L*di/dt
L=v*dt/di
Now we know v is 5.5, dt is 0.000008, and di is 0.060, so we plug these values in and we get:
L=v*dt/di
L=5.5*0.000008/0.060
so
L=733uH approximately.
So that's the way to get the inductance value. Note again that we used the positive voltage pulse and the low going current ramp, but normally we would use the positive voltage pulse and the high going current ramp. See if you can get both the voltage and the current on the scope face at the same time using the dual channel feature. It might be better NOT to invert any waveshapes for now.