Semiconductors

JEE Main level, single correct, medium

136 of 209

Question 136 of 209

Find the correct combination of A,B,C\mathrm{A}, \mathrm{B}, \mathrm{C} and D inputs which can cause the LED to glow.

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Answer: (A) 1101

The LED will glow if it is forward biased, which means the potential at the p-side must be higher than the potential at the n-side.
Let Y1Y_1 be the output of the upper NAND gate (p-side) and Y2Y_2 be the output of the lower NOR gate (n-side).
For the LED to glow, we need Y1=1Y_1 = 1 and Y2=0Y_2 = 0.
The upper circuit consists of two NOT gates (NOR gates with shorted inputs) followed by a NAND gate.
Output of first NOT gate is Aˉ\bar{A}, and second is Bˉ\bar{B}.
Y1=Aˉ⋅Bˉ‾=Aˉ‾+Bˉ‾=A+BY_1 = \overline{\bar{A} \cdot \bar{B}} = \overline{\bar{A}} + \overline{\bar{B}} = A + B (using De Morgan's Law).
The lower circuit is a NOR gate with inputs CC and DD.
Y2=C+D‾Y_2 = \overline{C + D}.
Condition 1: Y1=1  ⟹  A+B=1Y_1 = 1 \implies A + B = 1. This means at least one of AA or BB must be 1.
Condition 2: Y2=0  ⟹  C+D‾=0  ⟹  C+D=1Y_2 = 0 \implies \overline{C + D} = 0 \implies C + D = 1. This means at least one of CC or DD must be 1.
Checking the options:
(1) A=1,B=1,C=0,D=1A=1, B=1, C=0, D=1: Y1=1+1=1Y_1 = 1+1 = 1 and Y2=0+1‾=0Y_2 = \overline{0+1} = 0. (Satisfied)
(2) A=0,B=0,C=1,D=1A=0, B=0, C=1, D=1: Y1=0+0=0Y_1 = 0+0 = 0. (Not satisfied)
(3) A=1,B=0,C=0,D=0A=1, B=0, C=0, D=0: Y2=0+0‾=1Y_2 = \overline{0+0} = 1. (Not satisfied)
(4) A=0,B=1,C=0,D=0A=0, B=1, C=0, D=0: Y2=0+0‾=1Y_2 = \overline{0+0} = 1. (Not satisfied)
Thus, the correct combination is 1101.

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