Semiconductor Electronics: Materials Devices & Simple: JEE Main Previous Year Questions (2021-2026)

 Page 1


JEE Main Previous Year Questions (2021-2026): 
Semiconductor  
 
(January 2026) 
 
Q1: Two p-n junction diodes D
1
 and D
2
 are connected as shown in figure. A and B are 
input signals and C is the output. The given circuit will function as a ________. 
 
(a) AND Gate 
(b) OR Gate 
(c) NOR Gate 
(d) NAND Gate 
Ans: (a) 
Sol:  
For the input voltages at A and B we assume a positive logic system where 0 V represents logic 
0 (LOW) and 5 V represents logic 1 (HIGH).  
The supply voltage V
dc
 = 5 V connected through a resistor R to the output C.  
Diodes D
1
 and D
2
 are connected with their n -sides (cathodes) toward the inputs A and B. They 
act as switches.  
A = 0V, B = 0V (Both LOW)  
Both diodes are forward-biased because the p -side (connected to V
dc
) is at a higher potential 
than the n -side (connected to 0 V). Current flows through the diodes to the ground. This pulls 
the potential at C down to approximately 0V, i.e., output C = 0 (LOW). 
Page 2


JEE Main Previous Year Questions (2021-2026): 
Semiconductor  
 
(January 2026) 
 
Q1: Two p-n junction diodes D
1
 and D
2
 are connected as shown in figure. A and B are 
input signals and C is the output. The given circuit will function as a ________. 
 
(a) AND Gate 
(b) OR Gate 
(c) NOR Gate 
(d) NAND Gate 
Ans: (a) 
Sol:  
For the input voltages at A and B we assume a positive logic system where 0 V represents logic 
0 (LOW) and 5 V represents logic 1 (HIGH).  
The supply voltage V
dc
 = 5 V connected through a resistor R to the output C.  
Diodes D
1
 and D
2
 are connected with their n -sides (cathodes) toward the inputs A and B. They 
act as switches.  
A = 0V, B = 0V (Both LOW)  
Both diodes are forward-biased because the p -side (connected to V
dc
) is at a higher potential 
than the n -side (connected to 0 V). Current flows through the diodes to the ground. This pulls 
the potential at C down to approximately 0V, i.e., output C = 0 (LOW). 
 
 
Diode D
1
 is forward-biased and conducts current to the 0 V input. This shorts point C to ground 
through the conducting diode. Diode D
2
 is reverse-biased and does not conduct. Because D
1
 is 
conducting, point C remains at approximately 0V , i.e., output C = 0 (LOW). 
 
 
Diode D
2
 is forward-biased and conducts current to the 0V input at B. This pulls point C down to 
ground level, i.e., output C = 0 (LOW). 
 
 
Both n -sides and p -sides of the diodes are at 5 V (or the n -side is higher than the p -side). The 
diodes do not conduct, i.e., they are OFF. Since no current flows through the resistor R, there is 
no voltage drop across it. Therefore, point C stays at the supply voltage of 5 V , i.e., output C = 
1 (HIGH). 
 
The resulting truth table for the given circuit is, 
Page 3


JEE Main Previous Year Questions (2021-2026): 
Semiconductor  
 
(January 2026) 
 
Q1: Two p-n junction diodes D
1
 and D
2
 are connected as shown in figure. A and B are 
input signals and C is the output. The given circuit will function as a ________. 
 
(a) AND Gate 
(b) OR Gate 
(c) NOR Gate 
(d) NAND Gate 
Ans: (a) 
Sol:  
For the input voltages at A and B we assume a positive logic system where 0 V represents logic 
0 (LOW) and 5 V represents logic 1 (HIGH).  
The supply voltage V
dc
 = 5 V connected through a resistor R to the output C.  
Diodes D
1
 and D
2
 are connected with their n -sides (cathodes) toward the inputs A and B. They 
act as switches.  
A = 0V, B = 0V (Both LOW)  
Both diodes are forward-biased because the p -side (connected to V
dc
) is at a higher potential 
than the n -side (connected to 0 V). Current flows through the diodes to the ground. This pulls 
the potential at C down to approximately 0V, i.e., output C = 0 (LOW). 
 
 
Diode D
1
 is forward-biased and conducts current to the 0 V input. This shorts point C to ground 
through the conducting diode. Diode D
2
 is reverse-biased and does not conduct. Because D
1
 is 
conducting, point C remains at approximately 0V , i.e., output C = 0 (LOW). 
 
 
Diode D
2
 is forward-biased and conducts current to the 0V input at B. This pulls point C down to 
ground level, i.e., output C = 0 (LOW). 
 
 
Both n -sides and p -sides of the diodes are at 5 V (or the n -side is higher than the p -side). The 
diodes do not conduct, i.e., they are OFF. Since no current flows through the resistor R, there is 
no voltage drop across it. Therefore, point C stays at the supply voltage of 5 V , i.e., output C = 
1 (HIGH). 
 
The resulting truth table for the given circuit is, 
 
Hence, the logic gate is an AND Gate, where the output is HIGH only when all inputs are HIGH. 
Hence, the correct option is (a). 
 
Q2: Assuming in forward bias condition there is a voltage drop of 0.7 V across a silicon 
diode, the current through diode D
1
 in the circuit is ______ mA.  
(Assume all diodes in the given circuit are identical) 
 
(a) 17.6 
(b) 18.8 
(c) 20.15 
(d) 11.7 
Ans: (b) 
Sol: 
For diodes D
1
 and D
3
 , the p -side (anode) of these diodes is connected toward the positive 
terminal of the 12 V source. Therefore, D
1
 and D
3
 are forward biased.  
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JEE Main Previous Year Questions (2021-2026): 
Semiconductor  
 
(January 2026) 
 
Q1: Two p-n junction diodes D
1
 and D
2
 are connected as shown in figure. A and B are 
input signals and C is the output. The given circuit will function as a ________. 
 
(a) AND Gate 
(b) OR Gate 
(c) NOR Gate 
(d) NAND Gate 
Ans: (a) 
Sol:  
For the input voltages at A and B we assume a positive logic system where 0 V represents logic 
0 (LOW) and 5 V represents logic 1 (HIGH).  
The supply voltage V
dc
 = 5 V connected through a resistor R to the output C.  
Diodes D
1
 and D
2
 are connected with their n -sides (cathodes) toward the inputs A and B. They 
act as switches.  
A = 0V, B = 0V (Both LOW)  
Both diodes are forward-biased because the p -side (connected to V
dc
) is at a higher potential 
than the n -side (connected to 0 V). Current flows through the diodes to the ground. This pulls 
the potential at C down to approximately 0V, i.e., output C = 0 (LOW). 
 
 
Diode D
1
 is forward-biased and conducts current to the 0 V input. This shorts point C to ground 
through the conducting diode. Diode D
2
 is reverse-biased and does not conduct. Because D
1
 is 
conducting, point C remains at approximately 0V , i.e., output C = 0 (LOW). 
 
 
Diode D
2
 is forward-biased and conducts current to the 0V input at B. This pulls point C down to 
ground level, i.e., output C = 0 (LOW). 
 
 
Both n -sides and p -sides of the diodes are at 5 V (or the n -side is higher than the p -side). The 
diodes do not conduct, i.e., they are OFF. Since no current flows through the resistor R, there is 
no voltage drop across it. Therefore, point C stays at the supply voltage of 5 V , i.e., output C = 
1 (HIGH). 
 
The resulting truth table for the given circuit is, 
 
Hence, the logic gate is an AND Gate, where the output is HIGH only when all inputs are HIGH. 
Hence, the correct option is (a). 
 
Q2: Assuming in forward bias condition there is a voltage drop of 0.7 V across a silicon 
diode, the current through diode D
1
 in the circuit is ______ mA.  
(Assume all diodes in the given circuit are identical) 
 
(a) 17.6 
(b) 18.8 
(c) 20.15 
(d) 11.7 
Ans: (b) 
Sol: 
For diodes D
1
 and D
3
 , the p -side (anode) of these diodes is connected toward the positive 
terminal of the 12 V source. Therefore, D
1
 and D
3
 are forward biased.  
In diode D
2
 the n -side (cathode) of this diode is connected toward the positive terminal. 
Therefore, D
2
 is reverse biased and acts as an open circuit (no current flows through it).  
Since D
1
 and D
3
 are in forward bias and are silicon diodes, the voltage drop across them is fixed 
at V
d
 = 0.7V. Because they are in parallel, the voltage at the junction after resistor R
1
 is 0.7V 
relative to the ground. 
 
The total current flows through the resistor R
1
 = 0.3 kO. Using Ohm's Law: 
 
The total current total I
total
 reaches the parallel junction and splits between the two 
forward-biased diodes, D
1
 and D
3
. Since the diodes are identical, the current splits equally: 
Page 5


JEE Main Previous Year Questions (2021-2026): 
Semiconductor  
 
(January 2026) 
 
Q1: Two p-n junction diodes D
1
 and D
2
 are connected as shown in figure. A and B are 
input signals and C is the output. The given circuit will function as a ________. 
 
(a) AND Gate 
(b) OR Gate 
(c) NOR Gate 
(d) NAND Gate 
Ans: (a) 
Sol:  
For the input voltages at A and B we assume a positive logic system where 0 V represents logic 
0 (LOW) and 5 V represents logic 1 (HIGH).  
The supply voltage V
dc
 = 5 V connected through a resistor R to the output C.  
Diodes D
1
 and D
2
 are connected with their n -sides (cathodes) toward the inputs A and B. They 
act as switches.  
A = 0V, B = 0V (Both LOW)  
Both diodes are forward-biased because the p -side (connected to V
dc
) is at a higher potential 
than the n -side (connected to 0 V). Current flows through the diodes to the ground. This pulls 
the potential at C down to approximately 0V, i.e., output C = 0 (LOW). 
 
 
Diode D
1
 is forward-biased and conducts current to the 0 V input. This shorts point C to ground 
through the conducting diode. Diode D
2
 is reverse-biased and does not conduct. Because D
1
 is 
conducting, point C remains at approximately 0V , i.e., output C = 0 (LOW). 
 
 
Diode D
2
 is forward-biased and conducts current to the 0V input at B. This pulls point C down to 
ground level, i.e., output C = 0 (LOW). 
 
 
Both n -sides and p -sides of the diodes are at 5 V (or the n -side is higher than the p -side). The 
diodes do not conduct, i.e., they are OFF. Since no current flows through the resistor R, there is 
no voltage drop across it. Therefore, point C stays at the supply voltage of 5 V , i.e., output C = 
1 (HIGH). 
 
The resulting truth table for the given circuit is, 
 
Hence, the logic gate is an AND Gate, where the output is HIGH only when all inputs are HIGH. 
Hence, the correct option is (a). 
 
Q2: Assuming in forward bias condition there is a voltage drop of 0.7 V across a silicon 
diode, the current through diode D
1
 in the circuit is ______ mA.  
(Assume all diodes in the given circuit are identical) 
 
(a) 17.6 
(b) 18.8 
(c) 20.15 
(d) 11.7 
Ans: (b) 
Sol: 
For diodes D
1
 and D
3
 , the p -side (anode) of these diodes is connected toward the positive 
terminal of the 12 V source. Therefore, D
1
 and D
3
 are forward biased.  
In diode D
2
 the n -side (cathode) of this diode is connected toward the positive terminal. 
Therefore, D
2
 is reverse biased and acts as an open circuit (no current flows through it).  
Since D
1
 and D
3
 are in forward bias and are silicon diodes, the voltage drop across them is fixed 
at V
d
 = 0.7V. Because they are in parallel, the voltage at the junction after resistor R
1
 is 0.7V 
relative to the ground. 
 
The total current flows through the resistor R
1
 = 0.3 kO. Using Ohm's Law: 
 
The total current total I
total
 reaches the parallel junction and splits between the two 
forward-biased diodes, D
1
 and D
3
. Since the diodes are identical, the current splits equally: 
 
Hence, the correct option is (b). 
 
Q3: Identify the correct truth table of the given logic circuit.   
 
(a)  
 
(b)