Difficulty: Easy
Correct Answer: Connect Q̅ (the complemented output) back to D
Explanation:
Introduction / Context:A D (data) flip-flop copies its D input to Q on the active clock transition, subject to setup and hold times. A common technique is to turn a D flip-flop into a T (toggle) flip-flop. This is useful for building divide-by-2 counters and frequency dividers with minimal hardware.
Given Data / Assumptions:
Concept / Approach:If the D input always receives the complement of the current output, then at the next clock edge Q must change to the opposite value. The simplest way to enforce this is to connect Q̅ (the inverted output) directly to D. If Q̅ is 1, D is 1 and Q becomes 1 at the next edge; if Q̅ is 0, D is 0 and Q becomes 0 at the next edge. Because Q toggles at each qualifying edge, the flip-flop behaves like a T-FF with T implicitly at logic 1.
Step-by-Step Solution:
Start with Q at an arbitrary state.Connect Q̅ to D so D = not(Q).On the next active clock edge, Q takes D = not(Q), thus flipping state.Repeat each clock: the device toggles continuously, dividing the clock by 2.Verification / Alternative check:Simulate the flip-flop with D = Q̅; you will observe Q toggling on every active edge, producing a 50% duty cycle square wave at half the input clock frequency (assuming symmetric timing).
Why Other Options Are Wrong:
Connect Q to D: The device becomes a 1-bit memory with no toggling (Q holds state).Tie D LOW: Q asynchronously stays LOW after the first clock that captures 0; no toggling.Tie D HIGH: Q becomes HIGH after the first clock that captures 1; no toggling.Common Pitfalls:Neglecting setup/hold timing or using the wrong clock edge; forgetting that asynchronous clears/presets override the toggling behavior.
Final Answer:Connect Q̅ (the complemented output) back to D
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