--- title: "Full adder using a 2×4 decoder" subject: "Digital Logic" chapter: 4 key: "semester_3/digital_logic:full-adder-2x4-decoder" source: https://www.bctengineer.com/q/semester_3/digital_logic/full-adder-2x4-decoder site: BCT Engineering --- # Full adder using a 2×4 decoder Digital Logic · Chapter 4: Data Processing Circuits - Asked in: 2080 Baisakh, 8 marks; 2078 Bhadra, 4 marks (asked 2 times) - Video: [Implementation of full adder using two 2 to 4 decoders and nand gate](https://www.youtube.com/watch?v=9QWQtvtUeWY) — Digital2Real Tutorials - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/digital_logic/full-adder-2x4-decoder - Whole chapter: https://www.bctengineer.com/md/semester_3/digital_logic/chapter-4 ## Question Realize Full Adder Circuit using a 2×4 decoder and using logic gates. ## Short answer Use a 2×4 decoder with A,B as inputs to produce minterms Y0–Y3. Sum S = C_in XOR (Y1 + Y2) and carry C_o = Y3 + C_in·(Y1 + Y2). This requires one XOR, two OR, and one AND gate. ## Answer A **full adder** adds three bits A, B and carry-in $C_{in}$ and gives sum S and carry-out $C_o$. #### Truth table | A | B | $C_{in}$ | S | $C_o$ | |---|---|---|---|---| | 0 | 0 | 0 | 0 | 0 | | 0 | 0 | 1 | 1 | 0 | | 0 | 1 | 0 | 1 | 0 | | 0 | 1 | 1 | 0 | 1 | | 1 | 0 | 0 | 1 | 0 | | 1 | 0 | 1 | 0 | 1 | | 1 | 1 | 0 | 0 | 1 | | 1 | 1 | 1 | 1 | 1 | $$ S = \Sigma m(1,2,4,7), \qquad C_o = \Sigma m(3,5,6,7) $$ #### Realization with logic gates From K-maps / algebra: $$ \begin{aligned} S &= A \oplus B \oplus C_{in} \\ C_o &= AB + BC_{in} + AC_{in} = AB + C_{in}(A \oplus B) \end{aligned} $$ ```text A, B ────────[XOR]── P = A⊕B P, Cin ──────[XOR]──────────────── S P, Cin ──────[AND]── Cin·P ──┐ ├─[OR]── Co A, B ────────[AND]── AB ─────┘ ``` (Two XOR, two AND and one OR gate.) #### Realization with a 2×4 decoder A 2×4 decoder has only two inputs, so it cannot generate all 8 minterms. Feed A and B to the decoder; its outputs are the minterms of A, B: $$ Y_0 = \overline{A}\,\overline{B},\quad Y_1 = \overline{A}B,\quad Y_2 = A\overline{B},\quad Y_3 = AB $$ Write S and $C_o$ for each AB combination in terms of $C_{in}$: | AB | Decoder output | S | $C_o$ | |---|---|---|---| | 00 | $Y_0$ | $C_{in}$ | 0 | | 01 | $Y_1$ | $\overline{C_{in}}$ | $C_{in}$ | | 10 | $Y_2$ | $\overline{C_{in}}$ | $C_{in}$ | | 11 | $Y_3$ | $C_{in}$ | 1 | $$ \begin{aligned} S &= C_{in}(Y_0+Y_3) + \overline{C_{in}}(Y_1+Y_2) = C_{in} \oplus (Y_1+Y_2) \\ C_o &= Y_3 + C_{in}(Y_1+Y_2) \end{aligned} $$ ($Y_1 + Y_2 = A \oplus B$, so these are the same equations as the gate realization.) ```text +----------+ A ──────>| I1 Y0 |── (not used) B ──────>| I0 Y1 |── Y1 | 2x4 Y2 |── Y2 | DEC Y3 |── Y3 +----------+ Y1, Y2 ──────[OR]── P (= A⊕B) P, Cin ──────[XOR]─────────────── S P, Cin ──────[AND]── Cin·P ──┐ ├─[OR]── Co Y3 ──────────────────────────┘ ``` Gates needed besides the decoder: two OR gates, one XOR gate and one AND gate. **Check:** A=1, B=1, $C_{in}$=0: $Y_3 = 1$, $P = 0$, so $S = 0 \oplus 0 = 0$ and $C_o = Y_3 = 1$; correct ($1+1 = 10_2$). (With a 3×8 decoder, the usual method $S = Y_1+Y_2+Y_4+Y_7$, $C_o = Y_3+Y_5+Y_6+Y_7$ would need no XOR gate.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why can a 2×4 decoder not directly give all minterms of a full adder? A. It has only two inputs, so it cannot generate all 8 minterms of three variables B. It is slower than gates C. It has no enable D. It gives maxterms only Answer: A. Three inputs need a 3×8 decoder; here A, B feed the decoder and Cin is handled with gates. (Key point: Decoder limitation) 2. What is S in terms of decoder outputs and Cin? A. Cin ⊕ (Y1 + Y2) B. Cin + Y3 C. Cin · (Y1 + Y2) D. Y0 ⊕ Y3 Answer: A. S = Cin for AB = 00, 11 and Cin' for AB = 01, 10, so S = Cin ⊕ (Y1 + Y2). (Key point: Sum expression) 3. What is Co? A. Y1 + Cin·Y3 B. Y3 + Cin(Y1 + Y2) C. Y0 + Cin D. Y3 · Cin Answer: B. For AB = 11 carry is 1, for 01 and 10 it equals Cin, and for 00 it is 0. (Key point: Carry expression) 4. When AB = 00 (Y0 active), what are S and Co? A. S = Cin', Co = Cin B. S = Cin, Co = 0 C. S = 0, Co = Cin D. S = 1, Co = 0 Answer: B. 0 + 0 + Cin gives sum Cin and no carry. 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