--- title: "Full adder using decoder; (57−43) by 2's complement" subject: "Digital Logic" chapter: 4 key: "semester_3/digital_logic:full-adder-decoder-subtract" source: https://www.bctengineer.com/q/semester_3/digital_logic/full-adder-decoder-subtract site: BCT Engineering --- # Full adder using decoder; (57−43) by 2's complement Digital Logic · Chapter 4: Data Processing Circuits - Asked in: 2078 Kartik, 3+3 marks - Video: [2's complement subtraction| very easy](https://www.youtube.com/watch?v=JKXs4qjVkWw) — RAUL S - 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-decoder-subtract - Whole chapter: https://www.bctengineer.com/md/semester_3/digital_logic/chapter-4 ## Question Realize full adder circuit using decoder and gates. Subtract (43)₁₀ from (57)₁₀ using 2's complement method. ## Short answer Full adder: Use a 3-to-8 decoder with inputs A,B,C. OR minterms: S = Y1+Y2+Y4+Y7, Cout = Y3+Y5+Y6+Y7. Subtraction: 57-43 in 7-bit: 57 = 0111001, 2's complement of 43 = 1010101; sum = 10001110, discard carry → 0001110₂ = 14₁₀. ## Answer #### Full adder using a decoder and gates A full adder adds $A$, $B$ and carry-in $C$: | A | B | C | 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 | $$ S = \Sigma m(1,2,4,7), \qquad C_o = \Sigma m(3,5,6,7) $$ A 3-to-8 decoder (inputs $A, B, C$) produces every minterm; OR the required lines: $$ S = Y_1 + Y_2 + Y_4 + Y_7, \qquad C_o = Y_3 + Y_5 + Y_6 + Y_7 $$ ```text +-------+ A ----->| 2 Y0 | B ----->| 1 Y1 |---> to S C ----->| 0 Y2 |---> to S | Y3 |---> to Co |3:8 Y4 |---> to S | Y5 |---> to Co | Y6 |---> to Co | Y7 |---> to S and Co +-------+ Y1,Y2,Y4,Y7 --[OR 4]--> S Y3,Y5,Y6,Y7 --[OR 4]--> Co ``` #### (57)₁₀ − (43)₁₀ by 2's complement Use 7 bits (57 needs 6 bits; one extra bit for sign safety). $$ 57 = 0111001_2, \qquad 43 = 0101011_2 $$ 1's complement of 43: $1010100$. Add 1: 2's complement $= 1010101$. ```text 0111001 (+57) + 1010101 (2's complement of 43) --------- 1 0001110 ^ end carry discarded ``` An end carry appears, so the result is positive and equals the remaining bits: $0001110_2 = 8 + 4 + 2 = 14$. **Answer:** $(57)_{10} - (43)_{10} = (0001110)_2 = (14)_{10}$ ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which minterm lists give the Sum and Carry of a full adder? A. S = Σm(1,2,4,7), Co = Σm(3,5,6,7) B. S = Σm(3,5,6,7), Co = Σm(1,2,4,7) C. S = Σm(0,3,5,6), Co = Σm(1,2,4,7) D. S = Σm(1,2,4,7), Co = Σm(1,2,3,4) Answer: A. Sum is 1 when an odd number of inputs are 1 (1,2,4,7); carry is 1 when at least two are 1 (3,5,6,7). (Key point: Full adder minterms) 2. How is a full adder built from a 3-to-8 decoder? A. Connect the decoder outputs directly to S and Co B. OR the required minterm outputs for S and for Co using two 4-input OR gates C. Use AND gates on the decoder inputs D. Use the decoder outputs as XOR inputs only Answer: B. The decoder produces each minterm; OR gates collect Y1,Y2,Y4,Y7 for S and Y3,Y5,Y6,Y7 for Co. (Key point: Decoder plus OR gates) 3. What is the 7-bit 2's complement of 43 (0101011)? A. 1010100 B. 1010101 C. 1010111 D. 0101100 Answer: B. The 1's complement is 1010100; adding 1 gives 1010101. (Key point: 2's complement of 43) 4. After adding 57 and the 2's complement of 43, an end carry appears. What does this mean? A. The result is negative and must be complemented B. The result is positive; discard the carry and keep the remaining bits C. The addition overflowed and is invalid D. The carry must be added back to the result Answer: B. In 2's complement subtraction an end carry means the result is positive; it is discarded, leaving 0001110 = 14. 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