Question

A 4-bit ripple counter uses flip-flops with a propagation delay of 50 ns each. The maximum clock frequency that can be used is:

• A. \(5\text{ MHz}\)
• B. \(10\text{ MHz}\)
• C. \(20\text{ MHz}\)
• D. \(50\text{ MHz}\)

Hint:

For ripple counters, remember that delays add up linearly: \( f_{\text{max}} = \frac{1}{n \cdot t_{pd}} \). For synchronous counters, the maximum clock frequency is independent of the number of stages and depends only on a single flip-flop delay and combinational logic delay: \( f_{\text{max}} = \frac{1}{t_{pd} + t_{\text{logic}}} \).

Answer 1

The Correct Option is A

Concept: In an asynchronous counter (also known as a ripple counter), the clock signal is supplied only to the first flip-flop. The subsequent flip-flops are clocked by the outputs of the preceding stages. Because of this chain structure, the total propagation delay accumulates through each stage. For the counter to operate safely and correctly, the total time required for a change to ripple through all the stages must be less than or equal to the time period of the clock signal. Key formulas used here:
• Total propagation delay: \( t_{\text{total}} = n \times t_{pd} \)
• Minimum clock time period: \( T_{\text{clk}} \geq t_{\text{total}} \)
• Maximum clock frequency: \( f_{\text{max}} = \frac{1}{T_{\text{clk(min)}}} = \frac{1}{n \times t_{pd}} \)

Step 1: Identify the given values from the problem statement.
We are given the following data points:
• Number of bits in the ripple counter (\( n \)) = \( 4 \)
• Propagation delay of a single flip-flop (\( t_{pd} \)) = \( 50\text{ ns} = 50 \times 10^{-9}\text{ s} \)

Step 2: Calculate the total accumulated propagation delay.
Since it is a 4-bit ripple counter, there are 4 flip-flops connected in series. The output of the first flip-flop drives the clock input of the second, the second drives the third, and so on. In the worst-case scenario, a change at the input must propagate through all 4 stages to produce a stable configuration at the output. \[ t_{\text{total}} = n \times t_{pd} \] \[ t_{\text{total}} = 4 \times 50\text{ ns} = 200\text{ ns} \]

Step 3: Determine the maximum operating frequency.
To ensure proper counting behavior without catching erroneous intermediate transient states, the minimum clock period \( T_{\text{clk}} \) must be at least equal to this total cumulative propagation delay. \[ T_{\text{clk}} \geq 200\text{ ns} \] The maximum frequency \( f_{\text{max}} \) corresponds to the reciprocal of the minimum time period: \[ f_{\text{max}} = \frac{1}{t_{\text{total}}} = \frac{1}{200 \times 10^{-9}\text{ s}} \] \[ f_{\text{max}} = \frac{10^9}{200}\text{ Hz} = \frac{1,000,000,000}{200}\text{ Hz} = 5,000,000\text{ Hz} \] Converting Hertz to Megahertz (\( 1\text{ MHz} = 10^6\text{ Hz} \)): \[ f_{\text{max}} = 5\text{ MHz} \] Thus, the maximum clock frequency that can be safely applied to this ripple counter is 5 MHz.