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4-bit Asynchronous Counter Problem

 

4-bit Asynchronous Counter Problem

4-bit Asynchronous Counter (Propagation Delay Problem)

Given Data

  • Number of flip-flops (n) = 4
  • Propagation delay of each D flip-flop = 10 ns
  • Counter type = Asynchronous (Ripple) counter

Step 1: Understand the Concept

In an asynchronous counter, each flip-flop is triggered one after another. So, propagation delays are added cumulatively.

Step 2: Total Propagation Delay

Total delay = number of flip-flops × propagation delay

t_total = n × t_pd

t_total = 4 × 10 ns = 40 ns

Step 3: Minimum Clock Period

To avoid errors, clock period must be at least equal to total propagation delay:

T_min = 40 ns

Step 4: Maximum Frequency Calculation

Frequency formula:

f_max = 1 / T_min

f_max = 1 / (40 × 10⁻⁹)

f_max = 25 × 10⁶ Hz

Final Answer

Maximum allowed frequency = 25 MHz

Important Note

  • Ripple counters suffer from cumulative delay
  • More flip-flops → more delay → lower speed
  • Synchronous counters are faster because all flip-flops are clocked together

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