--- title: "BCD code and 3-bit binary-to-Gray converter" subject: "Digital Logic" chapter: 1 key: "semester_3/digital_logic:bcd-code-gray-converter" source: https://www.bctengineer.com/q/semester_3/digital_logic/bcd-code-gray-converter site: BCT Engineering --- # BCD code and 3-bit binary-to-Gray converter Digital Logic · Chapter 1: Introduction - Asked in: 2081 Bhadra, 2+4 marks - Video: [Binary Code to Gray Code Converter: Circuit, Truth Table, and Designing](https://www.youtube.com/watch?v=Pk9YouWUwHI) — Engineering Funda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/digital_logic/bcd-code-gray-converter - Whole chapter: https://www.bctengineer.com/md/semester_3/digital_logic/chapter-1 ## Question What is a BCD code? Design a 3-bit binary to Gray code converter. ## Short answer BCD encodes each decimal digit in 4-bit binary (8421), valid 0000–1001. A 3-bit binary to Gray converter: G2 = B2, G1 = B2⊕B1, G0 = B1⊕B0, implemented with two XOR gates. Example: binary 101 yields Gray 111. ## Answer #### BCD code **BCD (Binary Coded Decimal)** represents each decimal digit separately by its 4-bit binary value (8421 weights). Only 0000–1001 are valid; 1010–1111 are invalid. Example: $(93)_{10}$ = 1001 0011 (BCD), while in binary $93 = 1011101_2$. BCD is used in calculators, clocks and 7-segment displays. #### 3-bit binary to Gray code converter Inputs: binary $B_2B_1B_0$ ($B_2$ = MSB). Outputs: Gray $G_2G_1G_0$. In Gray code, successive values differ in only one bit. #### Truth table (3-bit binary to Gray) | $B_2$ | $B_1$ | $B_0$ | $G_2$ | $G_1$ | $G_0$ | |---|---|---|---|---|---| | 0 | 0 | 0 | 0 | 0 | 0 | | 0 | 0 | 1 | 0 | 0 | 1 | | 0 | 1 | 0 | 0 | 1 | 1 | | 0 | 1 | 1 | 0 | 1 | 0 | | 1 | 0 | 0 | 1 | 1 | 0 | | 1 | 0 | 1 | 1 | 1 | 1 | | 1 | 1 | 0 | 1 | 0 | 1 | | 1 | 1 | 1 | 1 | 0 | 0 | #### Simplification (K-maps) - $G_2 = \sum m(4,5,6,7) = B_2$ - $G_1 = \sum m(2,3,4,5) = \bar B_2 B_1 + B_2\bar B_1 = B_2 \oplus B_1$ - $G_0 = \sum m(1,2,5,6) = \bar B_1 B_0 + B_1\bar B_0 = B_1 \oplus B_0$ K-map for $G_1$ (rows $B_2$, columns $B_1B_0$): ```text B1B0 B2 00 01 11 10 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | ``` #### Logic circuit ```text B2 ---+------------------------ G2 | +-------+ +--->| | | XOR |----------- G1 B1 ---+--->| | | +-------+ | +-------+ +--->| | | XOR |----------- G0 B0 ------->| | +-------+ ``` The converter needs only **two XOR gates**: the MSB passes straight through and each other Gray bit is the XOR of two neighbouring binary bits. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is BCD? A. The whole decimal number is converted to binary B. Each digit is coded as digit + 3 C. Each decimal digit is coded separately by its 4-bit binary value D. A code where successive values differ by one bit Answer: C. BCD uses 8421 weights, valid codes 0000-1001. (Key point: BCD definition) 2. Which equations give the 3-bit binary to Gray converter? A. G2 = B2 XOR B1, G1 = B1 XOR B0, G0 = B0 B. G2 = B2, G1 = B1, G0 = B0 C. G2 = B2 AND B1, G1 = B1 AND B0, G0 = B0 D. G2 = B2, G1 = B2 XOR B1, G0 = B1 XOR B0 Answer: D. The MSB passes straight through and each other Gray bit is the XOR of neighbouring binary bits. (Key point: 3-bit converter equations) 3. For binary 100 (B2B1B0), what is the Gray output? A. 110 B. 100 C. 010 D. 101 Answer: A. G2=1, G1=1^0=1, G0=0^0=0. (Key point: Truth table) 4. How many gates does the 3-bit binary to Gray converter need? A. Three XOR gates B. Two XOR gates C. Two AND gates and one OR gate D. Three NOT gates Answer: B. G2 is a direct wire; G1 and G0 each need one XOR gate. 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