when a pn junction diode is forward biased, the electrons in n-region get attracted towards the p-region. after it crosses the the juction it recombines with the holes and thereafter under the influence of the applied bias voltage, it jumps from vacancies to vacancies thereby creating movement of holes. but how does depletion region gets narrowed. there is negative ions present in p-region which should get attracted to the positive terminal of the battery. so the depletion region width should increase.?????

 

when a pn junction diode is forward biased, the electrons in n-region get attracted towards the p-region. after it crosses the the juction it recombines with the holes and thereafter under the influence of the applied bias voltage, it jumps from vacancies to vacancies thereby creating movement of holes. but how does depletion region gets narrowed. there is negative ions present in p-region which should get attracted to the positive terminal of the battery. so the depletion region width should increase.?????

This assumption is where you are struggling. The P-region has net positive charge. When the diode is forward biased the negative terminal is on the N-side of the junction and the net-positive charge (holes) in the P-side are attracted across the junction, in part contributing to the narrowing of the depletion region.

I have given a previous member a detail explanation of PN-junctions before (http://forum.allaboutcircuits.com/showthread.php?t=8521) have a read an let us know if the concepts are clearer. There is also a link in that thread to our section on this in the e-book.

Dave

 

actually you misunderstood the negative ions i am talking about. i am talking about the negative, immobile ions present in the depletion region....
please let me know what you think...

 

actually you misunderstood the negative ions i am talking about. i am talking about the negative, immobile ions present in the depletion region....
please let me know what you think...

Ok. The charged ions in the depletion region are because of the net charge of the semiconductor after the process of recombination - importantly there is no mobile charge in the depletion region. Therefore it is not responsive to an applied voltage in the same way mobile charge which gives the net charge distributions in the N- and P-sides. Under forward-bias conditions the behaviour of the depletion is a result of mobile charge action.

Dave