--- title: "Encoder and octal-to-binary (8-to-3) encoder" subject: "Digital Logic" chapter: 4 key: "semester_3/digital_logic:octal-to-binary-encoder" source: https://www.bctengineer.com/q/semester_3/digital_logic/octal-to-binary-encoder site: BCT Engineering --- # Encoder and octal-to-binary (8-to-3) encoder Digital Logic · Chapter 4: Data Processing Circuits - Asked in: 2079 Bhadra, 1+5 marks; 2074 Asoj, 1+4 marks (asked 2 times) - Video: [Encoder in Digital Electronics | Working, Application and Logic Circuit of Encoder](https://www.youtube.com/watch?v=NWiPVMDh7GE) — ALL ABOUT ELECTRONICS - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/digital_logic/octal-to-binary-encoder - Whole chapter: https://www.bctengineer.com/md/semester_3/digital_logic/chapter-4 ## Question What is an encoder? Explain 8 to 3 line encoder with circuit diagram and truth table. ## Short answer An encoder converts one active input into its binary code. The 8-to-3 encoder assumes exactly one input high; outputs are A2 = D4+D5+D6+D7, A1 = D2+D3+D6+D7, A0 = D1+D3+D5+D7. D0 is inferred when all outputs are 0; a valid bit resolves the ambiguity. ## Answer An **encoder** is a combinational circuit that converts an active signal on one of its $2^n$ (or fewer) input lines into an n-bit binary code. It does the reverse of a decoder. #### 8-to-3 line (octal-to-binary) encoder It has 8 inputs $D_0$–$D_7$ (one per octal digit) and 3 outputs $A_2A_1A_0$. It is assumed that only one input is 1 at any time. | $D_0$ | $D_1$ | $D_2$ | $D_3$ | $D_4$ | $D_5$ | $D_6$ | $D_7$ | $A_2$ | $A_1$ | $A_0$ | |---|---|---|---|---|---|---|---|---|---|---| | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | #### Output expressions Each output is the OR of the inputs whose code has a 1 in that bit: $$ \begin{aligned} A_2 &= D_4 + D_5 + D_6 + D_7 \\ A_1 &= D_2 + D_3 + D_6 + D_7 \\ A_0 &= D_1 + D_3 + D_5 + D_7 \end{aligned} $$ #### Circuit diagram ```text D0 D1 D2 D3 D4 D5 D6 D7 │ │ │ │ │ │ │ │ x │ │ │ ├──┼──┼──┼──[OR]── A2 (D4,D5,D6,D7) │ ├──┼──┼──┼──┼──┼──[OR]── A1 (D2,D3,D6,D7) ├──┼──┼──┼──┼──┼──┼──[OR]── A0 (D1,D3,D5,D7) ``` Three 4-input OR gates are enough; $D_0$ is not connected (no output bit is 1 for it). **Limitations** - If no input is active, the output is 000, the same as for $D_0$; a valid-output line is needed to tell them apart. - If two inputs are active together, the output is wrong (e.g. $D_3$ and $D_5$ give $011 + 101 = 111$, which reads as 7). A **priority encoder** solves this by encoding only the highest active input. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which equation gives A0 for the 8-to-3 encoder? A. D1 + D3 + D5 + D7 B. D4 + D5 + D6 + D7 C. D2 + D3 + D6 + D7 D. D0 + D2 + D4 + D6 Answer: A. A0 is 1 for the codes with LSB = 1, namely 1, 3, 5 and 7. (Key point: Output equation) 2. What problem does the output 000 create? A. Both D0 active and no input active give 000 B. It damages the encoder C. It means D7 is active D. It indicates two inputs active Answer: A. D0 has no OR gate input, so a valid-output line is needed to tell the two cases apart. (Key point: Limitation with no input) 3. If D3 and D5 are active together, what is the output? A. 011 B. 101 C. 111 D. 110 Answer: C. The ORs combine 011 and 101 into 111, giving a wrong result; a priority encoder avoids this. 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