Truth Table to Decode Binary into 7-Segment
Decimal Number |
Inputs | Outputs | |||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|---|
D | C | B | A | a | b | c | d | e | f | g | |||
0 1 2 3 45 6 7 8 9 |
0 0 0 0 00 0 0 1 1 |
0 0 0 0 11 1 1 0 0 |
0 0 1 1 00 1 1 0 0 |
0 1 0 1 01 0 1 0 1 |
1 0 1 1 01 1 1 1 1 |
1 1 1 1 10 0 1 1 1 |
1 1 0 1 11 1 1 1 1 |
1 0 1 1 01 1 0 1 1 |
1 0 1 0 00 1 0 1 0 |
1 0 0 0 11 1 0 1 1 |
0 0 1 1 11 1 0 1 1 |
Karnaugh Mapping for Segment a:
The Truth Table:
BA | 00 | 01 | 11 | 10 | |
---|---|---|---|---|---|
DC\ | |||||
00 | 1 | 0 | 1 | 1 | |
01 | 0 | 1 | 1 | 1 | |
11 | x | x | x | x | |
10 | 1 | 1 | x | x |
BA | 00 | 01 | 11 | 10 | |
---|---|---|---|---|---|
DC\ | |||||
00 | 1 | 0 | 1 | 1 | |
01 | 0 | 1 | 1 | 1 | |
11 | x | x | x | x | |
10 | 1 | 1 | x | x |
Red cells are represented by B; this is a 4x2 supercell.
BA | 00 | 01 | 11 | 10 | |
---|---|---|---|---|---|
DC\ | |||||
00 | 1 | 0 | 1 | 1 | |
01 | 0 | 1 | 1 | 1 | |
11 | x | x | x | x | |
10 | 1 | 1 | x | x |
Green cells are represented by D; this is a 2x4 supercell.
So far then we have:
BA | 00 | 01 | 11 | 10 | |
---|---|---|---|---|---|
DC\ | |||||
00 | 1 | 0 | 1 | 1 | |
01 | 0 | 1 | 1 | 1 | |
11 | x | x | x | x | |
10 | 1 | 1 | x | x |
where grey boxes indicate cells included in more that one supercell. The shaded boxes are represented by B+D.
Focusing on the remaining cells with ones:
BA | 00 | 01 | 11 | 10 | |
---|---|---|---|---|---|
DC\ | |||||
00 | 1 | 0 | 1 | 1 | |
01 | 0 | 1 | 1 | 1 | |
11 | x | x | x | x | |
10 | 1 | 1 | x | x |
Blue cells are represented by A.C; this is a 2x2 supercell.
BA | 00 | 01 | 11 | 10 | |
---|---|---|---|---|---|
DC\ | |||||
00 | 1 | 0 | 1 | 1 | |
01 | 0 | 1 | 1 | 1 | |
11 | x | x | x | x | |
10 | 1 | 1 | x | x |
Cyan cells are represented by A'.C'; this is a 2x2 wrap-around supercell.
BA | 00 | 01 | 11 | 10 | |
---|---|---|---|---|---|
DC\ | |||||
00 | 1 | 0 | 1 | 1 | |
01 | 0 | 1 | 1 | 1 | |
11 | x | x | x | x | |
10 | 1 | 1 | x | x |
All cells are represented by B + D + A.C + A'.C', as shown for segment a.
Karnaugh Mapping for Segment c:
The Truth Table:
BA | 00 | 01 | 11 | 10 | |
---|---|---|---|---|---|
DC\ | |||||
00 | 1 | 1 | 1 | 0 | |
01 | 1 | 1 | 1 | 1 | |
11 | x | x | x | x | |
10 | 1 | 1 | x | x |
BA | 00 | 01 | 11 | 10 | |
---|---|---|---|---|---|
DC\ | |||||
00 | 1 | 1 | 1 | 0 | |
01 | 1 | 1 | 1 | 1 | |
11 | x | x | x | x | |
10 | 1 | 1 | x | x |
The red cells are represented by C'.B.A'; of course the complement is required for the final realization. Thus (C'.B.A')' is the combination of gates that is realized from the Karnaugh mapping of zeroes. Application of De Morgan's theorem results in the combination of gates shown for segment c:
(C'.B.A')' = C + B' + A
Decoder for segment a
|
D | C | B | A | b | |||
---|---|---|---|---|---|---|---|
0 1 2 3 45 6 7 8 9 |
0 0 0 0 00 0 0 1 1 |
0 0 0 0 11 1 1 0 0 |
0 0 1 1 00 1 1 0 0 |
0 1 0 1 01 0 1 0 1 |
1 1 1 1 10 0 1 1 1 |
|
Karnaugh |
|
D | C | B | A | e | |||
---|---|---|---|---|---|---|---|
0 1 2 3 45 6 7 8 9 |
0 0 0 0 00 0 0 1 1 |
0 0 0 0 11 1 1 0 0 |
0 0 1 1 00 1 1 0 0 |
0 1 0 1 01 0 1 0 1 |
1 0 1 0 00 1 0 1 |
BA | 00 | 01 | 11 | 10 | ||||
---|---|---|---|---|---|---|---|---|
DC\ | ||||||||
00 | 1 | 0 | 0 | 1 | ||||
01 | 0 | 0 | 0 | 1 | ||||
11 | x | x | x | x | ||||
10 | 1 | 0 | x | x |
Decoder for segment f
D | C | B | A | f | |||||||||||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
0 1 2 3 45 6 7 8 9 |
0 0 0 0 00 0 0 1 1 |
0 0 0 0 11 1 1 0 0 |
0 0 1 1 00 1 1 0 0 |
0 1 0 1 01 0 1 0 1 |
1 0 0 0 11 1 0 1 1 Decoder for segment g
|
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