Why does the reverse current shows a sudden increase at the critical voltage?

Why does the reverse current shows a sudden increase at the critical voltage?

When the reverse voltage across the p-n junction reaches a critical voltage, the reverse current suddenly rises to a large value. It is due to sudden increase in the number of minority charge carriers, resulting breakdown of the diode.

When a reverse bias is applied to a diode what happens to potential barrier?

In reverse bias applied to a p-n junction diode raises the potential barrier because p-type material conncted to the negative terminal and pullesthe holes away from the junction.

Does the reverse bias current depends on voltage?

In the reverse bias region the current has almost a constant value with a slight slope until the breakdown voltage. In the reverse bias region it is the minority carriers which contribute to the reverse saturation current. This flow is diffusive and depends on the temperature and not on the applied voltage.

Why current is reverse bias independent?

(i) The reverse current is almost independent of critical voltage because the current in a reverse-biased diode is due to the drifting of the minority charge carriers from one region to another through the junction. Hence a small amount of voltage is enough to carry on the sweeping of the minority charge carriers.

Why is the current under reverse bias?

At any temperature the number of minority carriers is constant so there is the small current at any applied potential. This is the reason for the current under reverse bias to be almost independent of applied potential. Thus there is a sudden increase in the current at the critical voltage.

When a reverse bias is applied to a diode?

In a standard diode, forward biasing occurs when the voltage across a diode permits the natural flow of current, whereas reverse biasing denotes a voltage across the diode in the opposite direction. However, the voltage present across a diode during reverse biasing does not produce any significant flow of current.

When we apply reverse bias to a junction diode is?

lowers the potential barrier. raises the potential barrier.

Why is reverse current independent of voltage?

Why does current increase at critical voltage under reverse bias?

At any temperature the number of minority carriers is constant so there is the small current at any applied potential. This is the reason for the current under reverse bias to be almost independent of applied potential. Thus there is a sudden increase in the current at the critical voltage.

What is the effect of reverse bias on potential barrier?

(i) Under the reverse bias condition the holes of p-side are attracted towards the negative terminal of the battery and the electrons of the n-side are attracted towards the positive terminal of the battery. This increase the depletion layer and the potential barrier.

Why does the critical voltage increase when the bond is broken?

Thus the bond is finally broken and the valence electrons move into the conduction band resulting in enormous flows of electrons and thus formation of hole-electron pairs. Thus there is a sudden increase in the current in the critical voltage. Was this answer helpful?

What is reverse saturation current in PN junction diode?

In a PN junction diode, the reverse saturation current is due to the diffusive flow of minority electrons from the p-side to the n-side and the minority holes from the n-side to the p-side. when voltage breakdown occurs,current remains constant and not increase eventhough there is increase of voltage.