Question

An ideal p-n junction diode can withstand currents up to $ 10\,mA $ under forward bias. The diode has a potential difference of $ 0.5V $ across it which is assumed to be independent of the current. What is the maximum voltage of the battery used to forward bias the diode when the resistance of $ 200\Omega $ is connected in series with it?
A) 5V
B) 4V
C) $ 2.5\,V $
D) $ 1.25\,V $

Answer 1

Hint : The maximum voltage of the battery can be such that the current flowing in the circuit will be less than the limit of the current that can be withstood by the ideal p-n junction. For a forward bias P-N junction diode the positive terminal is connected to the P-type semiconductor and the negative terminal is connected to the N-type semiconductor.

Formula used: In this solution, we will use the following formula
- Ohm’s law: $ V = IR $ where $ V $ is the net potential of the circuit, $ I $ is the current flowing the circuit, and $ R $ is the net resistance.

Complete step by step answer:
We’ve been told that an ideal P-N junction diode is connected in forward bias such that the potential drop across it is $ 0.5V $ and we want to find out the maximum voltage of the battery used to forward bias the diode when the resistance of $ 200\Omega $ is connected with the diode.
The net resistance of the circuit will only be due to $ 200\Omega $ . Now, the maximum voltage of the battery will be that corresponding to which $ 10\,mA $ of current will flow in the circuit. If the current exceeds this value, the diode will break down so the current in the circuit can be $ 10\,mA $ at max.
The net potential of the circuit will be: $ V = {V_{{\text{battery}}}} - 0.5 $ as there will be a potential drop across the diode when the battery is connected to it. Using ohm’s law,
 $ V = IR $
We can calculate the maximum voltage of the cell as corresponding to $ 10\,mA = 10 \times {10^{ - 3}}\,A $ and $ 200\Omega $ resistance as
 $ {V_{{\text{battery}}}} - 0.5 = 10 \times {10^{ - 3}} \times 200 $
Solving for $ {V_{{\text{battery}}}} $ , we get
 $ {V_{{\text{battery}}}} = 2.5\,V $
This corresponds to option (C).

Note:
As mentioned in the question, the potential difference across the diode is assumed to be constant irrespective of the current in the circuit. However, the potential difference across the diode changes with the current flowing through it in reality.