AGAIN FOR RAM :D

A depletion layer looks like what is given below, right?

So won't the positive ions in n junction attract the free electrons in n junction and negative ions in p junction attract holes from the p junction? So the holes and electrons come near the depletion layer and again get neutralised don't they, in turn creating more immobile ions? So finally what should happen is the p-jn becomes a collection of immobile negaitve ions and n-jn becomes a collection of positive ions.

FOR RAM:
Idhula edhanaavadhu velangalanaa asingama thittadha[4]. Inniku raathiri pesikkalam. Aana pls try to understand the question.

7 Answers

11
Anirudh Narayanan ·

???

1
Pavithra Ramamoorthy ·

hey.. in n region maj carrier iis electron.. les no of holes....

in p region maj carrier is holes.. less no of electron..

11
Anirudh Narayanan ·

As I predicted, no one understood my question. Won't the electrons from n and holes from p neutralise each other till there r no free charge carriers left?

1
Pavithra Ramamoorthy ·

dats wat i said on dat day....

only d e-1 n holes near d jun ll recombine......

right than a barrier is formed...

11
Anirudh Narayanan ·

Barrier consists of pos ions in n side and neg ion s in p side. So these ions will attract the charge carriers in their respective regions. This attraction pulls them so close together that they neutralise. This happens on till all the free charge carriers are exhausted and both sides contain only immobile ions.

PLS TRY 2 UNDERSTAND MY DOUBT DA MACHAN

1
Pavithra Ramamoorthy ·

already d ions which u r sayin is immobile......

11
Anirudh Narayanan ·

bUT THERE ARE STILL FREE CHARGE CARRIERS. THEY ARE ATTRACTED BY THE IMMOBILE IONS

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