8:1 MUX from 4:1 and 2:1 MUXes

An 8:1 MUX has 8 data inputs $I_0$–$I_7$ and 3 select lines $S_2S_1S_0$. It can be built as a multiplexer tree from two 4:1 MUXes and one 2:1 MUX.

Design

• Level 1: two 4:1 MUXes. MUX-1 takes $I_0$–$I_3$, MUX-2 takes $I_4$–$I_7$. Both use the lower select bits $S_1 S_0$.
• Level 2: one 2:1 MUX picks MUX-1 output (when $S_2 = 0$) or MUX-2 output (when $S_2 = 1$).

           S1 S0
         +-------+
 I0 ---->|       |
 I1 ---->| 4:1   |--P--+
 I2 ---->| MUX-1 |     |    S2
 I3 ---->|       |     |  +-----+
         +-------+     +->|0    |
         +-------+        | 2:1 |---> Y
 I4 ---->|       |     +->|1    |
 I5 ---->| 4:1   |--Q--+  +-----+
 I6 ---->| MUX-2 |
 I7 ---->|       |
         +-------+
           S1 S0

Function table

Working

$$P = S_1'S_0'I_0 + S_1'S_0 I_1 + S_1S_0'I_2 + S_1S_0I_3$$ $$Q = S_1'S_0'I_4 + S_1'S_0 I_5 + S_1S_0'I_6 + S_1S_0I_7$$ $$Y = S_2'P + S_2 Q$$

Substituting gives $Y = \sum_{k=0}^{7} m_k I_k$, the 8:1 MUX equation, where $m_k$ is the minterm of $S_2S_1S_0$.

Example: $S_2S_1S_0 = 110$. Both 4:1 MUXes select their input 2, so $P = I_2$ and $Q = I_6$. The 2:1 MUX with $S_2 = 1$ passes $Q$, so $Y = I_6$. Correct.