--- title: "1:32 DEMUX tree from 1:8 and 1:2 DEMUXes" subject: "Digital Logic" chapter: 4 key: "semester_3/digital_logic:demux-1to32-tree" source: https://www.bctengineer.com/q/semester_3/digital_logic/demux-1to32-tree site: BCT Engineering --- # 1:32 DEMUX tree from 1:8 and 1:2 DEMUXes Digital Logic · Chapter 4: Data Processing Circuits - Asked in: 2073 Shrawan, 6 marks - Video: [1 X 32 Demultiplexer using 1 X 8 Demultiplexer | 1:32 Demux using 1:8 Demux](https://www.youtube.com/watch?v=njmMv2x1Kw8) — Techno Tutorials ( e-Learning) - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/digital_logic/demux-1to32-tree - Whole chapter: https://www.bctengineer.com/md/semester_3/digital_logic/chapter-4 ## Question Design 1:32 demultiplexer tree using 1:8 DEMUXs and 1:2 DEMUXs only. ## 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. ## Answer 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 | ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which DEMUXes build a 1:32 DEMUX using 1:8 and 1:2 only? A. Three 1:2 and four 1:8 B. One 1:2 and four 1:8 C. Four 1:2 and three 1:8 D. Two 1:2 and two 1:8 Answer: A. 32 = 2 x 2 x 8, so three 1:2 (1 + 2) and four 1:8 are used. (Key point: Component count) 2. Which select drives the first-level 1:2 DEMUX? A. S0 B. S2 C. S3 D. S4 Answer: D. The MSB S4 is at level 1, S3 at level 2, and S2S1S0 drive the 1:8 DEMUXes. (Key point: Level order and selects) 3. What do the four 1:8 DEMUXes decode? A. S4S3 B. S2S1S0 C. S3S2S1 D. S4S3S2 Answer: B. Each 1:8 DEMUX uses the three lower selects to choose one of its eight outputs. (Key point: Select of 1:8 DEMUXes) 4. For S4..S0 = 11010 (26), where does D appear? A. Y24 B. Y26 C. Y18 D. Y10 Answer: B. S4S3 = 11 picks DM4 (Y24-Y31), and S2S1S0 = 010 picks its output 2, giving Y26. 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