Can anyone explain me in simple words that why does the depletion zone decrease when forward biased and increase when reverse biased.

The picture I have in mind is that when forward biased, the majority charged carriers of both p and n move towards the junction and recombine. This should actually increase the depletion region.

Should it not?? and if not then why?

 

It is not a permanent thing. You are basically right, they do move closer together. They also move further apart when the reverse charge is applied, and become an insulated layer.

 

You mean that after we remove the biasing it become what it was originally?? (as in the depletion layer size becomes the same?)

But even then, does the depletion layer thickness actually decrease when forward biased?

 

A diode only has a measurably different depletion region DURING BIASING. Once the biasing is removed (0v) the diode materials reforms to an equilibrium.

 

howsoever small the difference may be, but why does the thickness of depletion zone decrease in forward biasing?

why does the recombination of major carriers at the junction not lead to increase in depletion layer thickness??

 

Un biased condition:

At the junction the electrons in the n-substrate diffuse to P-substrate creating positive charge zone in the n-substrate and the holes in the P-substrate diffuses to n-substrate causing a negative charge zone in the P-substrate till equilibrium. Now the P-substrate has negative barrier and the N-substrate has positive barrier.

Hope you concur till here!

Forward Biased condition:

Where N-substrate is connected to negative terminal of the supply and p-substrate connected to positive terminal of the supply.

Negative terminal of the supply will infuse electrons into n-substrate these electrons will combine with the holes in the positive barrier of the n-substrate, hence neutralising the charges in that zone, the increase in neutralisation will reduce the size of the barrier. At one point all the positive charges will be neutralised so that there will be no barrier at all.

Same thing happens in the P-substrate where all the negative charges will be neutralised.

The voltage thus required for neutralising all the charges in the barrier is called as knee voltage, or forward conductiong voltage or froward breakover voltage.

 

Un biased condition:

At the junction the electrons in the n-substrate diffuse to P-substrate creating positive charge zone in the n-substrate and the holes in the P-substrate diffuses to n-substrate causing a negative charge zone in the P-substrate till equilibrium. Now the P-substrate has negative barrier and the N-substrate has positive barrier.

Hope you concur till here!

Forward Biased condition:

Where N-substrate is connected to negative terminal of the supply and p-substrate connected to positive terminal of the supply.

Negative terminal of the supply will infuse electrons into n-substrate these electrons will combine with the holes in the positive barrier of the n-substrate, hence neutralising the charges in that zone, the increase in neutralisation will reduce the size of the barrier. At one point all the positive charges will be neutralised so that there will be no barrier at all.

Same thing happens in the P-substrate where all the negative charges will be neutralised.

The voltage thus required for neutralising all the charges in the barrier is called as knee voltage, or forward conductiong voltage or froward breakover voltage.

And this state is unstable. It can only hold its state while there is power.

Once the power is withdrawn, it reverts to its stable un-biased state.

Kind of like an electro-magnet. It is only magnetic while power is applied.

Once the power is removed, it reverts back to its stable state.

 

i put this on u-tube a year or so ago.
it is fun to watch from time to time

it is forward biased but too much so.

http://www.youtube.com/watch?v=zaIC6q1pxCo&feature=related

 

So if the depletion region decreases in a forward bias state, why does the voltage drop increase with current? More bias should mean smaller depletion region = less effort getting across region.

 

The voltage drop is more of a quantum effect as well as Ohm's Law leaking through. Even though diodes don't follow Ohm's Law they have internal resistance, which does.