PN junction depletion region

[nikogian]

Hi all,
I don't know if this is the correct forum for this question but i will give it a try:
Consider a PN junction in forward bias, the width of the depletion region contains a term Vb - V where V is the applied voltage and Vb is the build in potential. My question is what happens when the applied voltage equals the build in potential. Is the depletion region elliminated? If it is so, how can injection electroluminescence be explained, since the area which there is an abudance of electrons and holes does not exist any more. The build in potential takes values around 0.7V so it is very easy to be reached especially with the applied voltages when we want electroluminescence.

Thanks in advance

 

[electronics_kumar]

in the apllied voltage should be greater that builtin potential of depletion region..after that only we can flow of electrons..if it is less than no flow of electrons ...builtin potential will act as barrier hill for flow of electrons...for climbing the hill ,the potential applied should be greater than barrier voltage

we can use the word 'overcome' not the word 'eliminated'

 

[hnejati]

To find a good solution to this problem let's model this PN junction to a potential well in this case as you know electrons can pass through it with tunneling mechanism or easily by applying bigger voltage relating to built in potential in the first mechanism as the applied voltage is less than the built in potential some carriers pass through this barrier but this amount of carriers are negligible (if you see the I-V characteristic of PN junction you can see this) so as the applied voltage reaches the built in potential the depletion region decreases.

 

[v_c]

I think this page gives pretty good information about the depletion region for a pn-junction **broken link removed**

I also recommend the entire web book as well -- pretty good online reference.
**broken link removed**