Question

Reverse bias applied to a p-n junction diode.

(A) lowers the potential barrier

(B) raises the potential barrier

(C) increases the majority carrier current

(D) increases the minority carrier current

Answer 1

Hint:- Reverse bias applied to a p-n junction diode raises the potential barrier because p-type material connected to the negative terminal and pulls the holes away from the junction. Similarly, n-type material connected to the positive terminal and pulls the electrons. Therefore the depletion region widens and this leads to the rise in the potential barrier

Complete step by step solution
When a positive terminal of a voltage source is connected to the n-type region and the negative terminal of the source is connected to the p-type region then the pn junction is said to be in reverse biased condition.

When there is no voltage applied across the p n junction, the potential developed across the junction is 0.3 volts at room temperature for germanium pn junction and 0.7 volts for silicon p n junction.

The polarity of this potential barrier is the same as the polarity of the voltage source applied during reverse biased condition.

Now if reverse biased voltage across the pn junction is increased the barrier potential developed across the pn junction is also increased. Hence, the pn junction is widened.

Hence , In reverse biasing, the conduction across the p-n junction takes place due to minority carriers, therefore the size of depletion region (potential barrier) rises.

So option B is correct

Note:- Forward Biased PN Junction
When we connect the p-type region of a junction with the positive terminal of a voltage source and n-type region with the negative terminal of the voltage source, then the junction is said to be forward biased. At this condition, due to the attraction of positive terminal of source, electrons which participated in covalent bond creations in p-type material, will be attracted towards the terminal.