Question

What is the voltage across an ideal p-n junction diode for the shown circuit?

A) 0.7 V
B) 1 V
C) 2 V
D) 0 V

Answer 1

Hint : The semiconductors are of two types. The first type is p-type semiconductor, and the second is the n-type semiconductor. The p-type semiconductor has a deficiency of electrons or a greater number of holes, and the n-type semiconductor contains the electrons in excess. The combination of both types of semiconductor forms a p-n junction diode. The p side is the positive terminal, and the n side is the negative terminal.

Complete Step By Step Answer:
The electrons in the circuit move from the negative terminal to the positive terminal. The negative terminal of the battery in the circuit is connected to the p side of the p-n junction, so the diode is reverse biased. The p-n junction diode cannot pass the current from the junction in reverse bias due to high resistance. The p-n junction in the circuit behaves like an open circuit, so the voltage drop across the ideal p-n junction diode remains equal to the voltage of the battery.
Therefore, the voltage across an ideal p-n junction diode for the shown circuit is $2\;{\text{V}}$ and the option (C) is correct.

Note :
The flow of the current happens due to the electrons and holes. The electrons are called the majority charge carrier of n-type semiconductors, and holes become the majority charge carrier in a p-type semiconductor. The p-n junction diode is called in forward bias if the p side of the p-n junction diode is connected with the positive terminal of the battery. The resistance of the p-n junction diode for forward bias becomes low and current passes from the diode. Be careful in observing the connections between the terminals of the battery, and p and n sides of the p-n junction diode because current can only pass in the forward bias.