Master–Slave J–K Flip-Flop — Toggle Condition and Timing In a master–slave J–K flip-flop, the toggle condition (J = 1 and K = 1) implies that Q and Q' will switch to their ________ state(s) at the ________. Choose the option that correctly completes the statement for conventional edge-triggered operation.
-
Aopposite, active clock edge
-
Binverted, positive clock edge
-
Cquiescent, negative clock edge
-
Dreset, synchronous clock edge
Answer
Correct Answer: opposite, active clock edge
Explanation
Introduction / Context:The J–K flip-flop is a versatile bistable device widely used in counters and state machines. A particularly important operating case is the toggle condition, which occurs when both inputs J and K are set to logic 1. Understanding exactly when Q and its complement Q' change state, and how that relates to the clock edge in a master–slave arrangement, is essential for robust synchronous design.
Given Data / Assumptions:
- Device: master–slave J–K flip-flop, functionally equivalent to an edge-triggered J–K FF.
- Toggle condition: J = 1 and K = 1.
- Edge sensitivity: the device responds at the active clock edge (rising or falling, per symbol).
Concept / Approach:In the toggle condition, the next state is the logical complement of the present state. In a master–slave architecture, the master latch samples inputs on one clock level; the slave transfers the master’s state to the output on the opposite level, yielding an effective edge-triggered response. Therefore, the externally observed change of Q and Q' happens at the active clock edge and results in both outputs switching to their opposite states.
Step-by-Step Solution:Assume present state Q(n). With J = K = 1, the next state is Q(n+1) = Q(n)'.In a master–slave implementation, the master captures the toggle intent during its transparent phase.At the active clock edge, the slave updates the external output to the captured next state.Thus, Q and Q' switch to their opposite states at the active clock edge.
Verification / Alternative check:Examine the truth/characteristic table for the J–K FF: with J = K = 1, Q(next) = Q(prev)'. Timing diagrams confirm that the visible change aligns with the specified triggering edge, not with the entire clock level.
Why Other Options Are Wrong:
- inverted, positive clock edge: “Inverted” is ambiguous and redundant; the precise phrase is “opposite” state, and the edge may be rising or falling depending on the device.
- quiescent, negative clock edge: “Quiescent” suggests no change, contradicting toggling.
- reset, synchronous clock edge: Toggle does not force reset; it complements the state.
Common Pitfalls:Confusing internal latch levels with the observed edge-triggered behavior, or assuming toggling happens throughout a clock level rather than at the edge.
Final Answer:opposite, active clock edge