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JEE Main 2016 · Physics · Semiconductors and Communication · medium

Four NAND gates are connected as follows: inputs A and B feed gate N1. The output of N1 and the input A feed gate N2. The output of N1 and the input…

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Four NAND gates are connected as follows: inputs AA and BB feed gate N1N_1. The output of N1N_1 and the input AA feed gate N2N_2. The output of N1N_1 and the input BB feed gate N3N_3. The outputs of N2N_2 and N3N_3 feed gate N4N_4, whose output is YY. The circuit is equivalent to a
  1. A. OR gate
  2. B. XNOR gate
  3. C. AND gate
  4. D. XOR gate
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Answer: (D) XOR gate

Write the outputs using Boolean algebra: N1=AB‾N_1=\overline{AB}; N2=A⋅AB‾‾=A‾+AB=A‾+BN_2=\overline{A\cdot\overline{AB}}=\overline{A}+AB=\overline A+B; N3=B⋅AB‾‾=B‾+AN_3=\overline{B\cdot\overline{AB}}=\overline B+A. Then Y=N2N3‾=(A‾+B)(B‾+A)‾=AB+A‾ B‾‾=AB‾+A‾BY=\overline{N_2N_3}=\overline{(\overline A+B)(\overline B+A)}=\overline{AB+\overline A\,\overline B}=A\overline B+\overline AB. Truth table check: A=B=0⇒Y=0A=B=0\Rightarrow Y=0; A=1,B=0⇒N1=1,N2=0,N3=1,Y=1A=1,B=0\Rightarrow N_1=1,N_2=0,N_3=1,Y=1; A=B=1⇒Y=0A=B=1\Rightarrow Y=0. Thus YY is high only when the inputs differ, which is the exclusive-OR function.
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Original question written in the JEE Main 2016 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake