55. Code Conversion Logic

Question.4

The circuit is intended to convert 4-bit Binary X3X2X1X0 into Gray code Y3Y2Y1Y0, but one input of the XOR gate producing Y1 is disconnected. Which signal should be connected to the open input?

Four-bit Binary-to-Gray converter with X3 connected directly to Y3 and one missing connection at the XOR gate producing Y1.

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Code conversion circuits change digital data from one code representation to another while preserving the represented value.

Input Code → Combinational Converter → Output Code

Code Reference Tables

For decimal digits 0–9, Binary and BCD use the same 4-bit patterns.

Decimal	Binary	BCD	Gray	Excess-3	Hex
0	0000	0000	0000	0011	0
1	0001	0001	0001	0100	1
2	0010	0010	0011	0101	2
3	0011	0011	0010	0110	3
4	0100	0100	0110	0111	4
5	0101	0101	0111	1000	5
6	0110	0110	0101	1001	6
7	0111	0111	0100	1010	7
8	1000	1000	1100	1011	8
9	1001	1001	1101	1100	9

Decimal 10–15

Binary, Gray, and Hex use a single 4-bit/digit representation; BCD and Excess-3 encode each decimal digit separately.

Decimal	Binary	BCD	Gray	Excess-3	Hex
10	1010	0001 0000	1111	0100 0011	A
11	1011	0001 0001	1110	0100 0100	B
12	1100	0001 0010	1010	0100 0101	C
13	1101	0001 0011	1011	0100 0110	D
14	1110	0001 0100	1001	0100 0111	E
15	1111	0001 0101	1000	0100 1000	F

Binary and BCD Conversion

For a single decimal digit 0–9, 4-bit Binary and BCD use the same bit pattern, so no conversion logic is required. For values greater than 9, BCD encodes each decimal digit separately.

Decimal 5: Binary = 0101, BCD = 0101
Decimal 12: Binary = 1100, BCD = 0001 0010

Binary and Gray Conversion

Binary-to-Gray

Copy the MSB directly. XOR each pair of adjacent input bits.

G3 = B3
G2 = B3 ⊕ B2
G1 = B2 ⊕ B1
G0 = B1 ⊕ B0

Binary 0010 → Gray 0011

Binary to Gray converter using a direct MSB path and three XOR gates

Gray-to-Binary

Copy the Gray MSB directly. Generate each remaining output bit by XORing the previous output bit with the next Gray bit.

B3 = G3
B2 = B3 ⊕ G2
B1 = B2 ⊕ G1
B0 = B1 ⊕ G0

Gray 0011 → Binary 0010

Gray to Binary converter using cascaded XOR gates from MSB to LSB

BCD and Excess-3 Conversion

Excess-3 represents each decimal digit by adding binary 0011 (decimal 3) to its corresponding 4-bit BCD value.

BCD + 0011 → Excess-3
Excess-3 − 0011 → BCD

BCD-to-Excess-3

Excess-3 = BCD + 0011
0011 + 0011 = 0110

For BCD inputs B3_B2_B1_B0, define:

X = B1 + B0

E3 = B3 + B2 · X
E2 = B2 ⊕ X
E1 = (B1 ⊕ B0)'
E0 = B0'

BCD to Excess-3 converter using minimized combinational logic

Excess-3-to-BCD

BCD = Excess-3 − 0011

0110 − 0011 = 0011

For inputs E3_E2_E1_E0, define:

Y = E1 · E0

B3 = E3 · (E2 + Y)
B2 = (E2 ⊕ Y)'
B1 = E1 ⊕ E0
B0 = E0'

Excess-3 to BCD converter using minimized combinational logic

BCD-to-7-Segment Decoder

A BCD-to-7-segment decoder converts a BCD digit 0–9 into seven control signals a–g used to display the corresponding decimal digit. The following uses an active-HIGH common-cathode display.

Seven-segment display diagram showing segment labels a, b, c, d, e, f, and g

Common-cathode display: 1 = Segment ON, 0 = Segment OFF

7-segment code order: a b c d e f g

Decimal	BCD	abcdefg
0	0000	1111110
1	0001	0110000
2	0010	1101101
3	0011	1111001
4	0100	0110011
5	0101	1011011
6	0110	1011111
7	0111	1110000
8	1000	1111111
9	1001	1111011

Minimized Active-HIGH Segment Equations

a = B1 + B3 + B2 · B0 + B2' · B0'
b = B2' + B1 · B0 + B1' · B0'
c = B0 + B2 + B1'
d = B3 + B1 · B0' + B1 · B2' + B2' · B0' + B2 · B0 · B1'
e = B1 · B0' + B2' · B0'
f = B3 + B2 · B0' + B2 · B1' + B1' · B0'
g = B3 + B1 · B0' + B1 · B2' + B2 · B1'

Note: BCD inputs 1010–1111 are unused for decimal digits and may be treated as don't-care conditions when simplifying the logic.

BCD-to-7-Segment Circuit

BCD 0011 (3) → 7-Segment 1111001

Active-HIGH BCD-to-7-segment decoder driving a common-cathode display.

Note: For a common-anode display, the segment-control logic is inverted: 0 = Segment ON and 1 = Segment OFF.

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