CBSE Class 12 · 2007 · SemiconductorsShort Answer

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The output of an OR gate is connected to both the inputs of a NAND gate. Draw the logic circuit of this combination of gates and write its truth table.

Solution

The OR gate gives

Y^{'} = A + B

.
Both inputs of the NAND gate are tied to

Y^{'}

, so

Y = \overline{Y^{'} \cdot Y^{'}} = \overline{Y^{'}} = \overline{A + B} .

This is the NOR operation; the truth table is

00 \to 1, 01 \to 0, 10 \to 0, 11 \to 0

Y = \overline{A + B} (NOR)

CBSE Class 12 Board · 2007 · Subjective

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