a)5 ns

The Gateway to Computer Science Excellence

+2 votes

Propagation delay is not affected even if the counter skips a count (number of flipflops remain same).

$f<=N*T_p$, where $f$ is the frequency, $N$ is the number of flipflops and $T_p$ is propagation delay.

$T_pmin=\frac{f}{N}=\frac{1}{T*N}sec=\frac{1}{10*10^6*10}sec=10 ns$. (Time period, $T=\frac{1}{f}$)

$f<=N*T_p$, where $f$ is the frequency, $N$ is the number of flipflops and $T_p$ is propagation delay.

$T_pmin=\frac{f}{N}=\frac{1}{T*N}sec=\frac{1}{10*10^6*10}sec=10 ns$. (Time period, $T=\frac{1}{f}$)

+1 vote

here CLK frequency is given = 10MHz

No of flip flops is given = 10

let Tpd be the propagation delay at each flip flop.

Now the clock frequency must be greater than the max delay in the ckt.

hence T(clk) >= No. of flip flop * Tpd ----- 1

f (clk) max = 1/ T(clk) --------2

from equation 1 and 2 :

tpd >= 1/(No of flip flop * f(clk))

by putting the values we will get tpd = 10ns.

No of flip flops is given = 10

let Tpd be the propagation delay at each flip flop.

Now the clock frequency must be greater than the max delay in the ckt.

hence T(clk) >= No. of flip flop * Tpd ----- 1

f (clk) max = 1/ T(clk) --------2

from equation 1 and 2 :

tpd >= 1/(No of flip flop * f(clk))

by putting the values we will get tpd = 10ns.

- All categories
- General Aptitude 1.9k
- Engineering Mathematics 7.6k
- Digital Logic 2.9k
- Programming and DS 4.9k
- Algorithms 4.4k
- Theory of Computation 6.2k
- Compiler Design 2.1k
- Databases 4.1k
- CO and Architecture 3.4k
- Computer Networks 4.2k
- Non GATE 1.4k
- Others 1.5k
- Admissions 595
- Exam Queries 573
- Tier 1 Placement Questions 23
- Job Queries 72
- Projects 18

50,832 questions

57,685 answers

199,268 comments

107,178 users