Consider the circuit in below figure which has a four bit binary number b_3 b_2 b_1 b_0 as input and a five bit binary number, d_4 d_3 d_2 d_1 d_0 as output. The circuit uses a 4-bit binary adder where the second input is fixed at 0110 (i.e., 6) and carry-in is 0, producing a 5-bit output (carry-out is d_4).
A. Binary to Hex conversion
B. Binary to BCD conversion
C. Binary to Gray code conversion
D. Binary to radix-12 conversion
GATE 1996 · Digital Logic · Circuit Output · medium
Answer: The circuit implements Binary to BCD conversion. Answer: B.
Identify the adder configuration: The circuit is a 4-bit adder with one input = b_3b_2b_1b_0 and the other hardwired to 0110 (decimal 6), carry-in = 0. The 5-bit output is d_4d_3d_2d_1d_0 where d_4 is the carry-out.
Relate to BCD conversion: In BCD arithmetic, after adding two BCD digits, if the result > 9 (or carry = 1), we add 6 to correct it. This circuit always adds 6 — it converts binary to BCD by applying the excess-6 correction. For inputs 0-9, the output still needs the upper nibble check; for 10-15, adding 6 correctly creates the two-digit BCD representation via the carry-out d_4.
Examples:
- Input = 0 (0000): 0+6=6 (00110) — but for input < 4, output is in single-digit range after adjustment
- Input = 9 (1001): 9+6=15 (01111), d_4=0
- Input = 10 (1010): 10+6=16 (10000), d_4=1, lower nibble = 0000 → BCD = 0001 0000 = 16
- Input = 15 (1111): 15+6=21 (10101), d_4=1, lower = 0101 → BCD = 0010 0001 = 21
This matches BCD representation of the original binary values, confirming Binary to BCD conversion.
Eliminate other options: Binary to Hex: trivial, no conversion needed since binary IS hex already.
Binary to Gray code: Gray code uses XOR of adjacent bits, not addition of 6.
Binary to radix-12: adding 6 does not produce base-12 encoding.
Only Binary to BCD conversion uses the +6 correction step.