Question
Download Solution PDFConsider a 4-bit Johnson counter with an initial value of 0000. The counting sequence of this counter is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFCircuit for Johnson counter:
From LSB bit complement (Q̅3 ) is fed into MSB(Q0)
Q0 is MSB and Q3 is LSB
|
CLK |
Q0 |
Q1 |
Q2 |
Q3 |
Decimal |
|
0(initial) |
0 |
0 |
0 |
0 |
0 |
|
1 |
1 |
0 |
0 |
0 |
8 |
|
2 |
1 |
1 |
0 |
0 |
12 |
|
3 |
1 |
1 |
1 |
0 |
14 |
|
4 |
1 |
1 |
1 |
1 |
15 |
|
5 |
0 |
1 |
1 |
1 |
7 |
|
6 |
0 |
0 |
1 |
1 |
3 |
|
7 |
0 |
0 |
0 |
1 |
1 |
|
8 |
0 |
0 |
0 |
0 |
0 |
Count sequence of 0 → 8 → 12 → 14 → 15 → 7 → 3 → 1 → 0 → (repeat)
Confusion Point:
If Q3 is MSB and Q0 is LSB
|
CLK |
Q3 |
Q2 |
Q1 |
Q0 |
Decimal |
|
0(initial) |
0 |
0 |
0 |
0 |
0 |
|
1 |
0 |
0 |
0 |
1 |
1 |
|
2 |
0 |
0 |
1 |
1 |
3 |
|
3 |
0 |
1 |
1 |
1 |
7 |
|
4 |
1 |
1 |
1 |
1 |
15 |
|
5 |
1 |
1 |
1 |
0 |
14 |
|
6 |
1 |
1 |
0 |
0 |
12 |
|
7 |
1 |
0 |
0 |
0 |
8 |
|
8 |
0 |
0 |
0 |
0 |
0 |
Count sequence of 0 → 1 → 3 → 7 → 15 →14 →12 → 8 → 0 → (repeat)
Important Points:
In Anser key. option 4 is given as correct
Last updated on Jan 8, 2025
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