A 4bit synchronous counter uses flip-flops with a propagation delay time of 25ns each. The maximum possible time required for change of state will be
25 ns
This question asks about the time it takes for a 4-bit synchronous counter to change its state, considering the propagation delay of the flip-flops used.
In a synchronous counter, all flip-flops change their state simultaneously during the active edge of the clock pulse. This means the clock signal is fed to all flip-flops at the same time.
The propagation delay (often denoted as $t_{pd}$) of a flip-flop is the time interval between the application of a clock edge and the point at which the output of the flip-flop becomes stable and valid. Each flip-flop in this 4-bit counter has a propagation delay of 25ns.
When a clock pulse arrives in a synchronous counter:
Given:
Since all flip-flops are triggered simultaneously by the clock and have the same propagation delay:
Maximum time for state change = Maximum propagation delay among all flip-flops
Maximum time = Maximum ($t_{pd\_FF1}$, $t_{pd\_FF2}$, $t_{pd\_FF3}$, $t_{pd\_FF4}$)
Maximum time = Maximum (25 ns, 25 ns, 25 ns, 25 ns)
Maximum time = 25 ns
The maximum time required for the state change in this 4-bit synchronous counter is dictated solely by the propagation delay of a single flip-flop, as they all operate in parallel triggered by the same clock edge. Thus, the time is 25 ns.
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