- Asked 2/22
- 2080 Baisakh · 2+6 marks
Setup/hold time; SR to JK conversion
Short answer
Setup time is the minimum time data must be stable before the active clock edge; hold time is the minimum time after the edge. To convert an SR flip-flop to a JK flip-flop, use excitation tables and K-maps to derive S = J Q_n' and R = K Q_n, then connect these to the SR inputs.
What is the Setup time and hold time of a flip-flop? With the help of excitation table and K-map, convert SR flip-flop into JK flip-flop.
Setup time and hold time
- Setup time (): the minimum time the data input (D, J, K, S, R) must be stable before the active clock edge so that the flip-flop reads it correctly.
- Hold time (): the minimum time the data input must stay stable after the active clock edge.
Data ===X======STABLE=======X====
|<-tsu->|<-th->|
CLK ______________|‾‾‾‾‾‾‾‾‾‾‾‾
^ active edge
If either is violated the flip-flop may store a wrong value or go metastable. Both are given in the data sheet; for LS-TTL flip-flops they are of the order of tens of ns () and a few ns ().
SR flip-flop into JK flip-flop
We add a combinational circuit with inputs J, K and that drives S and R.
SR excitation table
| S | R | ||
|---|---|---|---|
| 0 | 0 | 0 | X |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 1 |
| 1 | 1 | X | 0 |
Conversion table
| J | K | S | R | ||
|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | X |
| 0 | 0 | 1 | 1 | X | 0 |
| 0 | 1 | 0 | 0 | 0 | X |
| 0 | 1 | 1 | 0 | 0 | 1 |
| 1 | 0 | 0 | 1 | 1 | 0 |
| 1 | 0 | 1 | 1 | X | 0 |
| 1 | 1 | 0 | 1 | 1 | 0 |
| 1 | 1 | 1 | 0 | 0 | 1 |
K-maps
S: KQn=00 01 11 10 R: KQn=00 01 11 10
J=0 0 X 0 0 J=0 X 0 1 X
J=1 1 X 0 1 J=1 0 0 1 0
Circuit
J --[AND]-- S +--------+--+-- Q
Q'--[ ] | SR FF | |
| CLK | |
K --[AND]-- R | |--|-+-- Q'
Q --[ ] +--------+ | |
(feedback of Q, Q' to the AND gates)
Check: , : , , so Q resets (toggle). The two AND gates never make , so the invalid state is removed.
✓ Answer checked by Claude Opus 5.5 (high effort)
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