electronic elementary questions

[vaidhyanathan]

what happens exactly in a p-n diode during reverse recovery time? can anyone explain me in terms of minority and majority charge carriers? thanks in advance

 

[goldsmith]

Hello Dear vaidhyanathan
Simple thing will happen ! We will have two effect : Delay , and capacitive effect . Delay : if the current wen from p to n , how much time need to come back those carriers to the first situation ?! and the other effect : we have PN junction , and we have a place without conductivity between N and P . So isn't it like a capacitor ( two conductor and a nonconductor )?
Ok , this capacitor will destroy all thing ( at high frequencies it will become short circuit . for instance , try a 1n4007 diode and series resistance , and see the voltage across the resistor , and give an square wave with 5MHZ frequency as input ! what will happen ?! the out put will be alike with input ! with both cycle . Thus we need to use fast diodes and schottky diodes.
Best Wishes
Goldsmith

 

[LvW]

Hi Vaidhyanathan,

during transition from the "on" condition to "off" condition the carrier distribution along the p and the n-region is important, respectively.

On: In the p region there are peaks of n-carriers (p-carriers) very near to the junction. This is due to the limited "lifetime" of the carriers until the can recombine and disappear. This carrier distribution leads to the definition of the diffusion capacity Cd.

OFF: After changing the voltage polarity it takes some time until these carriers (left and right to the junction) have disappeared (recombination) . This effect leads to the definition of the switching time, because there is a current in the reverse direction during this short time period (until the diode is really in the off-condition)

I hope this answers your question.

 

[vaidhyanathan]

dear LVW,
can i interpret your answer like this?
In a forward biased pn diode, the depletion width is low and current(due to majority carriers) flows from p to n. when it is suddenly reverse biased, current(due to majority carriers) of same magnitude flows in the reverse direction for a short time. During this short time all the majority charge carriers get recombined and depletion layer becomes large. Thereafter the current is only due to minority carriers which is low.

[MODERATOR - threads merged - reply to the thread, don't create a new one]

 

[LvW]

I am afraid, I was not clear enough in my former explanation.
(a) Forward bias: majority carriers form the forward current. However, the positive carriers do not recombine in the n-region immediately. Instead, very close to the junction their concentration is still relatively high because of their "lifetime" (time necessary for recombination). The same is true for the negative carriers within the p-region.
This effect leads to a minority "peak" left and right to the junction. Very far from the junction the minorities decrease to their "normal" level. This minority carrier concentraton is equivalent to a charge storage and acts as a capacitive effect (forward diffusion capacity).
(b) Reverse bias: It takes some time until these carrier peaks disappear again. Because of polarity reversal they go back to the region they originate from. For a short time (switching time) this causes a current in the reverse direction. OK?

 

[vaidhyanathan]

i got it now.Thanks a lot Mr.LVW for your explanation.