Digital Electronics - Shift Registers - Discussion
Discussion Forum : Shift Registers - General Questions (Q.No. 16)
16.
With a 200 kHz clock frequency, eight bits can be serially entered into a shift register in ________.
Discussion:
8 comments Page 1 of 1.
NEELESH BIRADAR said:
9 years ago
Time required to store in shift reg = Total number of flipflops/clock frequency.
Aniket joshi said:
9 years ago
T=(n/f).
T=(8/0.2) convert khz to mhz.
T=40 micro sec.
T=(8/0.2) convert khz to mhz.
T=40 micro sec.
Kris said:
10 years ago
T = (n/f).
T = (8/200x10^3).
T = 40 micro-sec.
T = (8/200x10^3).
T = 40 micro-sec.
Krishna said:
1 decade ago
f = 200 KHZ;
T = (1/200) m sec;
T = (1/0.2) micro-sec;
T = 5 micro-sec;
After 8 clock cycles only 8 bit will be loaded=8*5=40 micro-sec;.
T = (1/200) m sec;
T = (1/0.2) micro-sec;
T = 5 micro-sec;
After 8 clock cycles only 8 bit will be loaded=8*5=40 micro-sec;.
(1)
Krishna said:
1 decade ago
Freq = 200 KHZ.
T = 1/200 KHZ = 0.5 micro-sec.
For 8 bits serial in 8 clock cycles.
So, Total time = 8*T = 8*0.5 micro-sec = 40 micro-sec.
T = 1/200 KHZ = 0.5 micro-sec.
For 8 bits serial in 8 clock cycles.
So, Total time = 8*T = 8*0.5 micro-sec = 40 micro-sec.
(1)
Belew said:
1 decade ago
1 bit in 1/f sec=1/(200KHz)==0.005ms.
8 bits in ==?
So, 8bit * 0.005ms/bit==0.04ms==40 micro-sec.
8 bits in ==?
So, 8bit * 0.005ms/bit==0.04ms==40 micro-sec.
Jasbir Singh said:
1 decade ago
Time for one clock cycle=(1/200000) that equals 0.5 * 10E-5
So 0.5*8=4 i.e. our answer comes 40 * 10E-5 seconds
=> 40 micro seconds.
So 0.5*8=4 i.e. our answer comes 40 * 10E-5 seconds
=> 40 micro seconds.
Kasiviswanathan said:
1 decade ago
(1/200 k)*8 which is equal to 0.04 ms.
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