Q.
The given combination represents the following gate
ManipalManipal 2008Semiconductor Electronics: Materials Devices and Simple Circuits
Solution:
The given combination can be represented as
All the gates in the combination are NAND gates. Output of gate-1, $ {{Y}_{1}}=\overline{A\cdot A} $ Output of gate-2, $ {{Y}_{2}}=\overline{B\cdot B} $ Output of gate-3, $ Y=\overline{{{Y}_{1}}\cdot {{Y}_{2}}} $
$ \therefore $ $ Y=\overline{\overline{A\cdot B}\cdot \overline{B\cdot B}} $ .. (i)
By Demorgans theorem, we have
$ \overline{A\cdot A}=\overline{A} $ and $ \overline{B\cdot B}=\overline{B} $
Therefore, Eq. (i) becomes,
$ Y=\overline{\overline{A}\cdot \overline{B}} $
Again from Demograns theorem
$ \overline{\overline{A}\cdot \overline{B}}=\overline{\overline{A}}+\overline{\overline{B}} $
$ \therefore $ $ Y=\overline{\overline{A}}+\overline{\overline{B}}=A+B $ $ (as\,\,\overline{\overline{A}}=A) $
Hence, the combination behaves as $ OR $ gate.
