Above is a circuit diagram with 2 sources V1 and V2, a switch SW, a diode D1 and a resistor R1. The points X,Y,A,B,C,D,E,F are terminals of the components indicated with red dots.
If the switch is OPEN as in the figure, the diode D1 is reverse biased and no current will flow.
Lets say the switch is closed(SW CLOSED) and opened in a very short amount of time which is enough to penetrate the p-n junction so that V1-V2 = 10V is applied in that duration to the ends of diode and the diode is forward biased for that short time.
And then lets assume suddenly the switch is opened(SW OPEN) again and kept always opened from then on. Here is my question:
Will the p-n junction remain penetrated and will a current flow through the points C,B,A,D,E,F in order? Or will it block the current since it looks like it is conencted as reverse biased?
Best Answer
A diode has a limited bandwidth, reversing voltage will take time to react. However it is not that big current is flowing in the opposite direction, it is more like the capacitance of the diode is charging in this time, limiting the bandwidth.
The story about the DC motor switching is just that the charge within the inductor of the motor needs to go somewhere, the current can not be suddenly stopped. If you do not use the diode the voltage will peak because of the inductance of the motor possibly blowing the transistor. Because of the diode the charge of the motor inductance will be "shortcircuited" over the motor-coil itself and the voltage will just ripple a little.