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Edge triggered D flip-flop with preset, clear

Short answer

The edge-triggered D flip-flop copies D input to Q on the clock edge, with asynchrnonous preset and clear. Q(n+1)=D. It has no invalid state unlike the RS flip-flop, because internal S and R are complementary, eliminating the forbidden S=R=1 condition.

Draw a block diagram of edge triggered, with preset and clear facilities, D-flip flop and its truth table. State one important advantage over RS flip flop.

Block diagram of edge-triggered D flip-flop with preset and clear

          PR (preset, active low)
            |
        +---+---+
 D ---->|       |----> Q
 CLK -->|>      |
        |       |----> Q'
        +---+---+
            |
          CLR (clear, active low)
  • Preset (PR): when PR = 0, Q is forced to 1 asynchronously (without the clock).
  • Clear (CLR): when CLR = 0, Q is forced to 0 asynchronously.
  • For normal clocked operation, PR = CLR = 1.

Truth table (positive-edge triggered)

PRCLRCLKDQ(n+1)
00XXNot allowed
01XX1
10XX0
11↑00
11↑11
110 or 1XNo change

The output copies D at the active clock edge: Q(n+1) = D.

Advantage over the RS flip-flop

The D flip-flop has no invalid (forbidden) state. It has only one data input, which is applied internally as S = D and R = D', so S and R can never both be 1. The RS flip-flop has an undefined state at S = R = 1. The D flip-flop is also simpler to use (one input), and edge triggering makes the output change only at the clock edge.

✓ Answer checked by Claude Opus 5.5 (high effort)

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