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Official PYQ · 1991

GATE CS 1991 Q5-c — Sequential Circuits

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Find the maximum clock frequency at which the counter in the figure below can be operated. Assume that the propagation delay through each flip flop and each AND gate is 10  ns10\;\text{ns}. Also, assume that the setup time for the JKJK inputs of the flip flops is negligible.

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Answer: 33.33

In a standard synchronous binary counter with cascaded carry logic, the inputs to the successive flip-flops are generated using a chain of AND gates. For a 4-bit synchronous counter, the inputs to the most significant flip-flop (FF3FF_3) are given by J3=K3=Q0Q1Q2J_3 = K_3 = Q_0 Q_1 Q_2, which is implemented using two cascaded AND gates:

J3=K3=(Q0⋅Q1)⋅Q2J_3 = K_3 = (Q_0 \cdot Q_1) \cdot Q_2

The critical path (longest delay path) starts from the clock edge triggering the first flip-flop (FF0FF_0) and propagates through the two cascaded AND gates to reach the inputs of the last flip-flop (FF3FF_3).

The minimum clock period TminT_{\text{min}} required for stable operation is:

Tmin=tpd,FF+2×tpd,AND+tsetupT_{\text{min}} = t_{pd, FF} + 2 \times t_{pd, AND} + t_{\text{setup}}

Given:

  • Propagation delay of each flip-flop, tpd,FF=10 nst_{pd, FF} = 10\text{ ns}
  • Propagation delay of each AND gate, tpd,AND=10 nst_{pd, AND} = 10\text{ ns}
  • Setup time, tsetup=0 nst_{\text{setup}} = 0\text{ ns}

Substituting these values:

Tmin=10 ns+2×10 ns+0=30 nsT_{\text{min}} = 10\text{ ns} + 2 \times 10\text{ ns} + 0 = 30\text{ ns}

Thus, the maximum clock frequency fmaxf_{\text{max}} is:

fmax=1Tmin=130 ns≈33.33 MHzf_{\text{max}} = \frac{1}{T_{\text{min}}} = \frac{1}{30\text{ ns}} \approx 33.33\text{ MHz}

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