Question

The maximum clock frequency in MHz of a 4-stage ripple counter, utilizing flip-flops, with each flip-flop having a propagation delay of 20 ns, is

• A. 12
• B. 12.5
• C. 13
• D. 13.5

Hint:

For a ripple counter, \(f_{max} = 1 / (N \times t_{pd})\). For a synchronous counter, the delays are not cumulative in the same way, and the maximum frequency is typically higher, limited by the delay of a single flip-flop plus some gate delay, i.e., \(f_{max} = 1 / (t_{pd} + t_{gate})\).

Answer 1

The Correct Option is B

Step 1: Understand the operation of a ripple (asynchronous) counter. In a ripple counter, the output of one flip-flop serves as the clock input for the next. This means the total time required for a change to propagate through the entire counter is the sum of the individual delays.
Step 2: Calculate the total propagation delay (\(T_{total}\)). For an N-stage ripple counter, the total delay is N times the propagation delay of a single flip-flop (\(t_{pd}\)). \[ T_{total} = N \times t_{pd} \] Given N = 4 stages and \(t_{pd} = 20 \text{ ns}\): \[ T_{total} = 4 \times 20 \text{ ns} = 80 \text{ ns} \]
Step 3: Determine the maximum clock frequency (\(f_{max}\)). The minimum clock period (\(T_{min}\)) must be at least as long as the total propagation delay to ensure the counter settles to its correct state before the next clock edge. \[ T_{min} = T_{total} = 80 \text{ ns} \] The maximum frequency is the reciprocal of the minimum period: \[ f_{max} = \frac{1}{T_{min}} = \frac{1}{80 \text{ ns}} = \frac{1}{80 \times 10^{-9} \text{ s}} \] \[ f_{max} = 12,500,000 \text{ Hz} = 12.5 \text{ MHz} \]