# How will the diode act after its junction is penetrated?

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?

• If the switch is closed you will instantly blow the diode. The diode will let current flow trying to reduce the 15V to 0.7V (a bit higher). Unless you add a resistance in series with the PSU you will just see smoke. Oct 26, 2014 at 18:39
• thnx for the input i made an edit. how would be ur answer now? Oct 26, 2014 at 18:43
• Initially, current will flow. After a short time (actually quite long for a 1N4007) known as its reverse recovery time, it will block the current. (Assuming you include some current limiting in series with V1. Otherwise, ... blown diode) Oct 26, 2014 at 18:55
• i ask this question because it is similar to pwm controlled transistor switching a DC motor like here: i.stack.imgur.com/Mcmfg.png some guys wrote that during the switch is switching from ON to OFF a large inverse voltage will be applied due to inductor so it will bias the diode forward. i observe that back emf of the dc motor doesn't change its polarity but the they say the current will flow through diode in OFF mode. i was wondering if diode is forward biased with great voltage for a very short time will back emf cause a current even though it is still reverse biased. Oct 26, 2014 at 19:03

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.

• about i.stack.imgur.com/s00OA.png. thats the story about inductor behaviou coming from -Ldi/dt. but there is another issue here. DC motor is generating a back emf when rotating. and when pwm is OFF, it is still rotating same direction and the back emf is same polarity. but the diode is reverse biased. the question is will the current flow from Vgen's + to Vgen's -. Oct 26, 2014 at 20:33
• Hmm, as far as I know when the transistor opens the inductor current will forward bias the diode and the motor will start working as a generator so this nett current will also generate a forward current. The diode will just open and work like a brake for the motor (and protect you transistor). Oct 26, 2014 at 20:57
• i didnt get anything from what u write Oct 26, 2014 at 21:14
• in short, the diode will short-circuit the engine and get rid of inductor current and kinetic induced current. Oct 26, 2014 at 22:18
• why dont you just say the path of the current from Vgen? just say the loop it will follow. Oct 26, 2014 at 22:26