Test Index

Semiconductor Electronics Part 3

Section: Physics

Question : 21 of 46

Marks: +1, -0

Solution:

Concept:

A NOR gate with its inputs tied together acts as a NOT gate.

Explanation:

First two NOR gates invert

A

and

B

to give

\overline{A}

and

\overline{B}

Third NOR gate has inputs

\overline{A}

and

\overline{B}

, so

Y_3 = \overline{\overline{A} + \overline{B}} = A \cdot B

(by De Morgan's law).

Fourth NOR gate inverts

Y_3

Y = \overline{Y_3} = \overline{A \cdot B}

The truth table below confirms this behaviour.

$\mathrm{A}$	$\mathrm{~B}$	$\mathrm{Y}_1=\overline{\mathrm{A}}$	$\mathrm{Y}_2=\overline{\mathrm{B}}$	$\mathrm{Y}_3=\overline{\mathrm{Y}_1+\mathrm{Y}_2}$	$\mathrm{Y}_4=\overline{\mathrm{Y}_3}$
1	1	0	0	1	0
1	0	0	1	0	1
0	1	1	0	0	1
0	0	1	1	0	1

The final output is

0

only when both

A

and

B

are

1

, which matches the NAND gate truth table.

\text { Input 1 }	\text { Input 2 }	\text { OR Output }	\text { NOR Output }
1	1	1	0
0	0	0	1

Answer:

Option D – NAND gate.

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