a)5 ns

1 vote

For what minimum value of propagation delay in each filp-flop will a **10** bit ripple counter skip a count, when it is clocked at **10 MHz**?

a) 5 ns

b) 10 ns

c) 20 ns

d) 40 ns

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.