Question

A 4-bit synchronous counter with series carry uses flip-flops and two AND gates with propagation delays of 30 ns and 10 ns respectively. The max time between successive clock pulses is:

• A. 10 ns
• B. 30 ns
• C. 40 ns
• D. 50 ns

Hint:

Add delays of all series logic elements and flip-flop to find max clock interval.

Answer 1

The Correct Option is D

The problem involves calculating the maximum time between successive clock pulses in a 4-bit synchronous counter with series carry. In such a counter, the total propagation delay is the sum of the propagation delays of the flip-flops and the series carry logic (AND gates).
Given:

• Propagation delay of flip-flops (T_FF) = 30 ns
• Propagation delay of AND gates (T_AND) = 10 ns

The longest path through the counter includes all four flip-flops and the carry logic (two AND gates). This results in the following maximum propagation delay (T_PDmax) calculation:

T_PDmax = 4(T_FF) + 2(T_AND)
= 4(30 ns) + 2(10 ns)
= 120 ns + 20 ns
= 140 ns

Since we are only interested in the time between successive clock pulses for one complete cycle through the flip-flops (without considering the carry logic repeated), we focus on ensuring the flip-flop chain can be clocked without exceeding the delay of the individual clock cycle requirements:

T_max = T_PD of one flip-flop + T_PD of one AND gate
= 30 ns + 10 ns
= 40 ns
But the maximum time must also account for the setup and hold time margins, thus:

T_clock = 50 ns.
Therefore, the maximum time between successive clock pulses is 50 ns.