The minimum number of NAND gates used to construct an OR gate is\

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Professor PC Thomas Classes · Answer

The output $Y$ of the circuit is\ $\mathrm{Y}=\overline{\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}}=\overline{\overline{\mathrm{A}}}+\overline{\overline{\mathrm{B}}}=\mathrm{A}+\mathrm{B}$\ It is a Boolean expression of an OR gate.\ Hence, three NAND gates are required to make an OR gate.\