I have attached a simple schematic of this experiment. As you can see, there is a constant current phase in the curve known as storage phase. My question is, why
current is constant during storage phase? I have read in references that during this phase, injected minority carriers (in forward bias) will recombine.
it's OK, but I could not find any convincing answer for my question in related references. I would appreciate if you could kindly help me.
During the recombination or storage phase, minority charge carriers are still present so the junction is still conductive and the voltage across it remains at 0.7 v. The current through it is governed by Ur/Rr which is of course constant. Only when the last few minority carriers are recombining will the diode stop conducting and establish a depletion layer, causing the current to fade and the voltage to rise.
Thank you @Guy Inchbald. Why presence of minority carriers means conductivity (Ur/R means that the diode is operating like a wire in this phase)? In this phase, current is constant, OK, but immediately after switching minority carriers begin to recombine, why we don't see effect of this immediate recombining and consequently reduction of minority carriers on the current. I hope to clearly have explained my problem. Thank you again.
I apologize for the answer because it is not my specialty.
During conduction, electron-hole pairs continuously appear in the zone around the junction. Those caught on the "wrong" side of the P-N doping are minority carriers. Thus, minority carriers are always present when a diode conducts. When the voltage is reversed, it takes time for the pairs to recombine - to a little electron or hole, the junction zone is a very big place! During all this time the resistance of the junction is low and the current is determined by the series resistor. Only when they have nearly all combined will the depletion zone grow big enough for the junction resistance to rise and become comparable to the series resistor. Now, the current will begin to fall.