Digital Electronics - Shift Registers - Discussion
Discussion Forum : Shift Registers - General Questions (Q.No. 17)
17.
An 8-bit serial in/serial out shift register is used with a clock frequency of 2 MHz to achieve a time delay (td) of ________.
16 
s
s8 s
s4 s
s2 s
sDiscussion:
6 comments Page 1 of 1.
Ramesh Bhagwat said:
1 decade ago
One clock period is .5 microseconds.
So the total delay of .5*8, ie 4 micro seconds to transmit information of 8 bits.
So the total delay of .5*8, ie 4 micro seconds to transmit information of 8 bits.
Kris said:
10 years ago
T = (n/f).
T = (8/200 MHz).
T = 4 micro-sec.
T = (8/200 MHz).
T = 4 micro-sec.
Chaitanya said:
8 years ago
@Kris.
In Que, Frequency is 2MHz.
8/2MHz. = 4 micro -sec.
In Que, Frequency is 2MHz.
8/2MHz. = 4 micro -sec.
Mayur said:
7 years ago
Can anyone answer this-?
Find the time required to load 8-bit data serially in a register if the duration of the clock pulse is 1 usec?
Find the time required to load 8-bit data serially in a register if the duration of the clock pulse is 1 usec?
Ritvik Khandelwal said:
6 years ago
Since,
Q=N*T.
Where,
Q = delay of flip flop.
T = time or 1/f.
N = number or register so,
So,
8/2=4us.
Q=N*T.
Where,
Q = delay of flip flop.
T = time or 1/f.
N = number or register so,
So,
8/2=4us.
Madu said:
2 years ago
One clock period = 0.5 microseconds.
Tot delay 0.5*8 = 4us.
Tot delay 0.5*8 = 4us.
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