2073 Shrawan · 6 marks

1:32 DEMUX tree from 1:8 and 1:2 DEMUXes

Short answer

Build in three levels: one 1:2 DEMUX with S4 feeds two 1:2 DEMUXs with S3, each feeding two 1:8 DEMUXs with S2-S0, producing 32 outputs. No additional gates are required.

Design 1:32 demultiplexer tree using 1:8 DEMUXs and 1:2 DEMUXs only.

A 1:32 DEMUX has one data input $D$ , five select lines $S_4S_3S_2S_1S_0$ and 32 outputs $Y_0$ – $Y_{31}$ . Since $32 = 2 \times 2 \times 8$ , it can be built with three 1:2 DEMUXes and four 1:8 DEMUXes in three levels.

Design

Level	DEMUXes	Select	Outputs
1	one 1:2 (DM-A)	$S_4$	2 lines
2	two 1:2 (DM-B, DM-C)	$S_3$	4 lines
3	four 1:8 (DM1–DM4)	$S_2S_1S_0$	32 outputs

                        S3                   S2 S1 S0
                    +-------+
             +----->| DM-B  |------> [1:8 DM1] --> Y0..Y7
             |      |  1:2  |------> [1:8 DM2] --> Y8..Y15
     S4      |      +-------+
  +-------+  |
D>| DM-A  |--+
  |  1:2  |-----+
  +-------+     |       S3
                |   +-------+
                +-->| DM-C  |------> [1:8 DM3] --> Y16..Y23
                    |  1:2  |------> [1:8 DM4] --> Y24..Y31
                    +-------+

DM-A output 0 (when $S_4 = 0$ ) feeds DM-B; output 1 feeds DM-C.
DM-B sends data to DM1 ( $S_3 = 0$ ) or DM2 ( $S_3 = 1$ ); DM-C to DM3 or DM4.
Each 1:8 DEMUX decodes $S_2S_1S_0$ to one of its eight outputs.

$S_4S_3$	Active 1:8 DEMUX	Output range
00	DM1	$Y_0$ – $Y_7$
01	DM2	$Y_8$ – $Y_{15}$
10	DM3	$Y_{16}$ – $Y_{23}$
11	DM4	$Y_{24}$ – $Y_{31}$

Example

$S_4S_3S_2S_1S_0 = 11010$ ( $= 26$ ): DM-A ( $S_4 = 1$ ) → DM-C; DM-C ( $S_3 = 1$ ) → DM4; DM4 ( $S_2S_1S_0 = 010$ ) → its output 2, which is $Y_{24+2} = Y_{26}$ . So $D$ appears at $Y_{26}$ and all other outputs stay 0.

Total: 3 DEMUXes of 1:2 + 4 DEMUXes of 1:8 = 7 ICs, no extra gates.

✓ Answer checked by Claude Opus 5.5 (high effort)

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1:32 DEMUX tree from 1:8 and 1:2 DEMUXes

Design

Example

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