Topic: Digital Circuits
The inputs to a digital circuit are signals A, B and C. $\overline{A}$, $\overline{B}$ and $\overline{C}$ are not available. The +5V supply and the ground can be used as logic '1' and '0' inputs respectively. To implement a function $\overline{B}\overline{C}+A\overline{C}$ using only 2:1 multiplexer, minimum number of MUX required is
- A. 2
- B. 3
- C. 4
- D. 5
Answer: A. 2
Explanation: The boolean function is $F = \overline{B}\overline{C} + A\overline{C} = \overline{C}(\overline{B} + A)$.
Using a 2:1 MUX, output $Y = S'I_0 + S I_1$.
First MUX: Set Select $S=B$, $I_0=1$, $I_1=A$. The output is $Y_1 = B'(1) + B(A) = \overline{B} + A$.
Second MUX: Set Select $S=C$, $I_0=Y_1$, $I_1=0$. The output is $Y_2 = C'(Y_1) + C(0) = \overline{C}(\overline{B} + A) = \overline{B}\overline{C} + A\overline{C}$.
Thus, only 2 MUXes are required.