32:1 MUX from 16:1 and 2:1 MUXes

A 32:1 MUX has inputs $I_0$–$I_{31}$ and five select lines $S_4S_3S_2S_1S_0$. It is built from two 16:1 MUXes and one 2:1 MUX (a multiplexer tree).

Design

• Level 1: MUX-1 (16:1) takes $I_0$–$I_{15}$; MUX-2 (16:1) takes $I_{16}$–$I_{31}$. Both share the lower selects $S_3S_2S_1S_0$.
• Level 2: a 2:1 MUX with select $S_4$ passes MUX-1 output when $S_4 = 0$ and MUX-2 output when $S_4 = 1$.

             S3 S2 S1 S0
           +-----------+
 I0-I15 -->| 16:1 MUX-1|--P--+      S4
           +-----------+     |   +------+
                             +-->|0     |
           +-----------+         | 2:1  |---> Y
 I16-I31 ->| 16:1 MUX-2|--Q----->|1     |
           +-----------+         +------+
             S3 S2 S1 S0

Output equation

$$Y = S_4'P + S_4Q$$

where $P = \sum_{k=0}^{15} m_k I_k$ and $Q = \sum_{k=0}^{15} m_k I_{k+16}$, with $m_k$ the minterms of $S_3S_2S_1S_0$.

Example

$S_4S_3S_2S_1S_0 = 11001$ (= 25). MUX-1 gives $I_9$ and MUX-2 gives $I_{25}$ (both select input 9). The 2:1 MUX with $S_4 = 1$ passes $Q$, so $Y = I_{25}$. Correct.

Alternative: if the 16:1 MUXes (e.g. 74150) have enable inputs, connect $S_4'$ to the enable of MUX-1 and $S_4$ to MUX-2, and OR the two outputs instead of using the 2:1 MUX.