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What is the time delay (td) of an 8-bit serial in/serial out shift register with a clock frequency of 4 MHz?
- a)0.2 μs
- b)8 μs
- c)4 μs
- d)2 μs
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
What is the time delay (td) of an 8-bit serial in/serial out shift reg...
Ms
To calculate the time delay (td) of an 8-bit serial in/serial out shift register, we need to consider the number of bits being shifted and the clock frequency.
The time delay can be calculated using the formula:
td = (Number of bits x Clock cycle time)
Given that we have an 8-bit register and a clock frequency of 4 MHz (4 x 10^6 Hz), we can calculate the time delay as follows:
td = (8 bits x 1/(4 x 10^6 Hz)) = 8 x 0.25 microseconds = 2 microseconds
Therefore, the time delay (td) of an 8-bit serial in/serial out shift register with a clock frequency of 4 MHz is 2 microseconds.
To calculate the time delay (td) of an 8-bit serial in/serial out shift register, we need to consider the number of bits being shifted and the clock frequency.
The time delay can be calculated using the formula:
td = (Number of bits x Clock cycle time)
Given that we have an 8-bit register and a clock frequency of 4 MHz (4 x 10^6 Hz), we can calculate the time delay as follows:
td = (8 bits x 1/(4 x 10^6 Hz)) = 8 x 0.25 microseconds = 2 microseconds
Therefore, the time delay (td) of an 8-bit serial in/serial out shift register with a clock frequency of 4 MHz is 2 microseconds.
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Community Answer
What is the time delay (td) of an 8-bit serial in/serial out shift reg...
Concept:
An n-bit serial in serial out register using D flip flops is as shown:
For an n-bit shift register, we require n D flip flops.
Let the propagation delay of a single D flip flop = t ns.
Propagation delay for the n bit SISO (serial in serial out) shift register will be:
tpd= (n × t) ns
Calculation:
Given:
n = 8
f = 4 MHz
t = 1 f = 1/4
t = 0.25 μs (for one flip flop)
For 8 flip flops
tpd = 8 × 0.25 μs
tpd = 2 μs
Hence option (4) is the correct answer.
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