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 ripple counter operation.In a ripple counter, a flip-flop's output clocks the subsequent flip-flop. The total propagation delay is the sum of delays through each stage.
Step 2: Calculate total propagation delay (\(T_{total}\)).For an N-stage ripple counter, \(T_{total}\) equals N multiplied by a single flip-flop's propagation delay (\(t_{pd}\)).\[ T_{total} = N \times t_{pd} \]Given N = 4 and \(t_{pd} = 20 \text{ ns}\):\[ T_{total} = 4 \times 20 \text{ ns} = 80 \text{ ns} \]
Step 3: Determine maximum clock frequency (\(f_{max}\)).The minimum clock period (\(T_{min}\)) must be greater than or equal to the total propagation delay to ensure correct settling before the next clock edge.\[ T_{min} = T_{total} = 80 \text{ ns} \]The maximum frequency is the inverse of the minimum period:\[ f_{max} = \frac{1}{T_{min}} = \frac{1}{80 \text{ ns}} = \frac{1}{80 \times 10^{-9} \text{ s}} \]\[ f_{max} = 12,500,000 \text{ Hz} = 12.5 \text{ MHz} \]