JK flip-flop operating modes On a standard JK flip-flop, under which input condition is the flip-flop in the hold (no change) state?
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AJ = 0, K = 0
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BJ = 1, K = 0
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CJ = 0, K = 1
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DJ = 1, K = 1
Answer
Correct Answer: J = 0, K = 0
Explanation
Introduction / Context:The JK flip-flop unifies SR behavior while avoiding the invalid state. Its truth table includes four operating modes: hold, reset, set, and toggle. Recognizing the hold condition is essential when analyzing synchronous state machines and designing gating networks that intentionally suppress state changes.
Given Data / Assumptions:
- Device: edge-triggered JK flip-flop.
- Inputs: J and K are evaluated at the active clock edge (subject to setup/hold times).
- We seek the input combination that preserves the current state.
Concept / Approach:The JK truth table indicates: J = 0, K = 0 → no change (hold); J = 1, K = 0 → set; J = 0, K = 1 → reset; J = 1, K = 1 → toggle. Thus the explicit hold condition is when both J and K are 0 during the sampling edge.
Step-by-Step Solution:
Ensure the clock edge occurs with J = 0 and K = 0 valid at the inputs.At the sampling edge, the device logic selects “no change,” leaving Q and Q̄ unchanged.Subsequent edges will maintain the stored state as long as J = K = 0.Any deviation (for example, J = 1 or K = 1) will invoke set/reset/toggle modes.Verification / Alternative check:Vendor datasheets and standard textbooks list the JK table, confirming the hold behavior for J = K = 0.
Why Other Options Are Wrong:
- J = 1, K = 0: Set mode.
- J = 0, K = 1: Reset mode.
- J = 1, K = 1: Toggle mode (divide-by-two action under continuous clock).
Common Pitfalls:Confusing JK with SR (which has a forbidden input) or forgetting that the JK = 11 case toggles, not holds.
Final Answer:J = 0, K = 0