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A 4 bit modulo–6 ripple counter uses JK flip-flop. If the propagation delay of each FF is 50 ns, the maximum clock frequency that can be used is equal to
- a)5 MHz
- b)10 MHz
- c)4 MHz
- d)20 Mhz
Correct answer is option 'A'. Can you explain this answer?
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A 4 bit modulo–6 ripple counter uses JK flip-flop. If the propag...
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A 4 bit modulo–6 ripple counter uses JK flip-flop. If the propag...
Answer:
To determine the maximum clock frequency that can be used for a 4-bit modulo-6 ripple counter using JK flip-flops, we need to consider the propagation delay of each flip-flop.
Propagation Delay:
Propagation delay is the time it takes for a signal to propagate through a circuit element or system. In this case, the propagation delay refers to the time it takes for the output of a flip-flop to stabilize after a clock edge.
Given that the propagation delay of each flip-flop is 50 ns, we need to ensure that the counter has enough time for all the flip-flops to stabilize before the next clock edge is applied.
Ripple Counter:
A ripple counter is a type of counter circuit where the output of one flip-flop serves as the clock input for the next flip-flop in the sequence. In a 4-bit ripple counter, there are four stages of flip-flops connected in series.
Calculating Maximum Clock Frequency:
To calculate the maximum clock frequency, we need to determine the worst-case scenario where the output of the first flip-flop takes the longest time to propagate through all the flip-flops.
In this case, the worst-case scenario occurs when the initial state of the counter is 1111 (decimal 15). In this state, the next state will be 0000 (decimal 0) after the next clock edge.
The time taken for the counter to transition from 1111 to 0000 is equal to the propagation delay of each flip-flop multiplied by the number of flip-flops in the counter. In this case, it is 50 ns * 4 = 200 ns.
The maximum clock frequency is the reciprocal of the time taken for this transition, i.e., 1 / 200 ns = 5 MHz.
Therefore, the maximum clock frequency that can be used for the 4-bit modulo-6 ripple counter is 5 MHz.
To determine the maximum clock frequency that can be used for a 4-bit modulo-6 ripple counter using JK flip-flops, we need to consider the propagation delay of each flip-flop.
Propagation Delay:
Propagation delay is the time it takes for a signal to propagate through a circuit element or system. In this case, the propagation delay refers to the time it takes for the output of a flip-flop to stabilize after a clock edge.
Given that the propagation delay of each flip-flop is 50 ns, we need to ensure that the counter has enough time for all the flip-flops to stabilize before the next clock edge is applied.
Ripple Counter:
A ripple counter is a type of counter circuit where the output of one flip-flop serves as the clock input for the next flip-flop in the sequence. In a 4-bit ripple counter, there are four stages of flip-flops connected in series.
Calculating Maximum Clock Frequency:
To calculate the maximum clock frequency, we need to determine the worst-case scenario where the output of the first flip-flop takes the longest time to propagate through all the flip-flops.
In this case, the worst-case scenario occurs when the initial state of the counter is 1111 (decimal 15). In this state, the next state will be 0000 (decimal 0) after the next clock edge.
The time taken for the counter to transition from 1111 to 0000 is equal to the propagation delay of each flip-flop multiplied by the number of flip-flops in the counter. In this case, it is 50 ns * 4 = 200 ns.
The maximum clock frequency is the reciprocal of the time taken for this transition, i.e., 1 / 200 ns = 5 MHz.
Therefore, the maximum clock frequency that can be used for the 4-bit modulo-6 ripple counter is 5 MHz.
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A 4 bit modulo–6 ripple counter uses JK flip-flop. If the propagation delay of each FF is 50 ns, the maximum clock frequency that can be used is equal toa)5 MHzb)10 MHzc)4 MHzd)20 MhzCorrect answer is option 'A'. Can you explain this answer?
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