Question

In the circuit shown below, the AND gate has a propagation delay of 1 ns. The edge-triggered flip-flops have a set-up time of 2 ns, a hold-time of 0 ns, and a clock-to-Q delay of 2 ns. The maximum clock frequency (in MHz, rounded off to the nearest integer) such that there are no setup violations is (answer in MHz).

 

Hint:

When calculating the maximum clock frequency, always consider the total delay, which includes propagation delays, setup times, and clock-to-Q delays.

Answer 1

To calculate the maximum clock frequency without any setup violations, we must ensure that the total delay time for each flip-flop does not exceed the clock period. 
The total delay for the signal is the sum of: 
1. The propagation delay of the AND gate (\( t_{prop} \)) = 1 ns
2. The setup time of the flip-flop (\( t_{setup} \)) = 2 ns
3. The clock-to-Q delay of the flip-flop (\( t_{CQ} \)) = 2 ns
Thus, the total delay (\( t_{total} \)) is: \[ t_{total} = t_{prop} + t_{setup} + t_{CQ} = 1 \, {ns} + 2 \, {ns} + 2 \, {ns} = 5 \, {ns} \] The clock period \( T \) must be greater than or equal to the total delay: \[ T \geq t_{total} = 5 \, {ns} \] The maximum clock frequency (\( f_{max} \)) is the reciprocal of the clock period: \[ f_{max} = \frac{1}{T} = \frac{1}{5 \times 10^{-9}} = 200 \, {MHz} \] Thus, the maximum clock frequency is 200 MHz.