--- title: "DEMUX vs decoder; 4×16 decoder from 2×4" subject: "Digital Logic" chapter: 4 key: "semester_3/digital_logic:decoder-4x16-from-2x4" source: https://www.bctengineer.com/q/semester_3/digital_logic/decoder-4x16-from-2x4 site: BCT Engineering --- # DEMUX vs decoder; 4×16 decoder from 2×4 Digital Logic · Chapter 4: Data Processing Circuits - Asked in: 2081 Baisakh, 2+5 marks - Video: [4*16 decoder design using 2*4 decoder](https://www.youtube.com/watch?v=_xCZ21Z3sQo) — Dr. Dhiman (Learn the art of problem solving) - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/digital_logic/decoder-4x16-from-2x4 - Whole chapter: https://www.bctengineer.com/md/semester_3/digital_logic/chapter-4 ## Question Differentiate between de-multiplexer and decoder. Construct 4×16 line decoder using 2×4 line decoders with enable. ## Short answer A demux has a data input plus select lines, output equals data; a decoder only selects one output based on code (fixed active level). A 4×16 decoder is built with five 2×4 decoders: the first decoder (A,B) enables one of four output decoders (C,D), each producing four outputs. ## Answer #### DEMUX vs decoder | Point | Demultiplexer | Decoder | |---|---|---| | Inputs | 1 data + $n$ select lines | $n$ code inputs (+ enable) | | Outputs | $2^n$ | $2^n$ | | Job | Routes the data bit to one output | Activates the output for the input code | | Output value | Equals the data input | Fixed active level (1 or 0) | | Typical use | Data distribution | Address decoding, code conversion | A decoder with enable becomes a DEMUX when the enable pin is used as the data input; e.g. 74138 serves both roles. #### 4×16 decoder from 2×4 decoders with enable A 4×16 decoder has inputs $A B C D$ ($A$ = MSB) and outputs $Y_0$–$Y_{15}$. Use **five** 2×4 decoders: - **Decoder 0 (control):** inputs $A, B$; enable tied to 1 (or to an overall enable $E$). Its outputs $O_0$–$O_3$ enable one of the four output decoders. - **Decoders 1–4:** inputs $C, D$; each enable comes from one output of decoder 0. Each produces four of the final outputs. ```text A B C D +--------+ +--------+ E ->|E O0 |------------>|E DEC1 |--> Y0 .. Y3 | O1 |---------+ +--------+ | DEC0 | | +--------+ | O2 |------+ +-->|E DEC2 |--> Y4 .. Y7 | O3 |---+ | +--------+ +--------+ | | +--------+ | +----->|E DEC3 |--> Y8 .. Y11 | +--------+ | +--------+ +-------->|E DEC4 |--> Y12 .. Y15 +--------+ (C, D go to all of DEC1 - DEC4) ``` Connections: $O_0 \to$ DEC1, $O_1 \to$ DEC2, $O_2 \to$ DEC3, $O_3 \to$ DEC4 (all enables). | AB | Enabled decoder | Outputs for CD = 00…11 | |---|---|---| | 00 | DEC1 | $Y_0$–$Y_3$ | | 01 | DEC2 | $Y_4$–$Y_7$ | | 10 | DEC3 | $Y_8$–$Y_{11}$ | | 11 | DEC4 | $Y_{12}$–$Y_{15}$ | #### Working Input $ABCD = 1101$: decoder 0 sees $AB = 11$ and enables DEC4 only; DEC4 decodes $CD = 01$ and makes its second output high, i.e. $Y_{13}$. The disabled decoders keep all their outputs at 0, so exactly one of the 16 lines is active, giving $Y_k = m_k(A,B,C,D)$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is a key difference between a demultiplexer and a decoder? A. A decoder has a data input; a demux does not B. A demux has a data input plus selects and passes the data to one output; a decoder only decodes an input code C. They are identical in all respects D. A demux has fewer outputs Answer: B. A demux routes data to one output; a decoder activates the line for the input code. (Key point: DEMUX vs decoder) 2. How many 2x4 decoders are needed for a 4x16 decoder? A. 4 B. 3 C. 5 D. 8 Answer: C. One control decoder plus four output decoders. (Key point: Number of decoders) 3. What does decoder 0 do? A. Takes C,D and drives Y0-Y3 B. Generates all 16 outputs C. Inverts the inputs D. Takes A,B and its outputs enable one of four output decoders Answer: D. O0-O3 are tied to the enables of DEC1-DEC4 so only one group works. (Key point: Control decoder) 4. For ABCD = 1101 which output is active? A. Y13 B. Y11 C. Y7 D. Y14 Answer: A. AB = 11 enables DEC4 (Y12-Y15) and CD = 01 picks its second output, Y13. 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