Electric OU: Oscillations on the Current Viewing Resistor
Here, instead of looking at the voltage on the mosfet common drains, I am looking at the voltage drop across the current viewing resistor, aka shunt. The resistor’s value is 0.25 Ohms, so by Ohm’s Law I = V / R, so if the voltage drop is +0.6 volts and the resistance is 0.25 Ohms, the current through the CVR is 2.4 amps. This is a peak current value in one direction, followed by a peak of the same value in the opposite direction. But what about the average values? One way is to look at the integral of the waveform, that is the area under or over the curve with respect to the baseline. This area is the integral over time of the oscillating signal. Then the total integral value is divided by the time to give the average value of the current. I don’t quite go that far in this video; my intention is not to teach calculus but rather to explain how even an old analog scope can be used to make quantitative and reliable measurements.