The maximum clock frequency in MHz of a 4-stage ripple counter, utilizing flip-flops, with each flip-flop having a propagation delay of 20 ns, is _______. (round off to one decimal place).

The correct answer is 12.5.

Step-by-Step Explanation:

In an asynchronous or ripple counter, the flip-flops are connected in a cascaded series. The clock signal is applied only to the first stage flip-flop, and the output of each flip-flop serves as the clock input for the next stage. Because of this, the propagation delays of all the stages accumulate.

For an n-stage ripple counter to operate reliably, the total propagation delay of the counter must be less than or equal to the time period of the clock signal ($T_{CLK}$).

Given details:
- Number of stages (bits), n = 4
- Propagation delay of each flip-flop, t_pd = 20 ns

The total propagation delay ($t_d$) of the 4-stage ripple counter is the sum of the propagation delays of all four flip-flops:
$t_d=n\times t_{\text{pd}}$
$t_d=4\times 20\text{ns}=80\text{ns}$

For proper counter operation, the minimum clock period ($T_{CLK\text{(min)}}$) is equal to the total propagation delay:
$T_{\mathrm{CLK}\left(\text{min}\right)}=t_d=80\text{ns}$

The maximum clock frequency ($f_{CLK\text{(max)}}$) is the reciprocal of the minimum clock period:
$f_{\mathrm{CLK}\left(\text{max}\right)}=\frac{1}{T_{\mathrm{CLK}\left(\text{min}\right)}}$
$f_{\mathrm{CLK}\left(\text{max}\right)}=\frac{1}{80\times {10}^{-9}\text{s}}$
$f_{\mathrm{CLK}\left(\text{max}\right)}=\frac{{10}^9}{80}\text{Hz}$
$f_{\mathrm{CLK}\left(\text{max}\right)}=12.5\times {10}^6\text{Hz}=12.5\text{MHz}$

Thus, the maximum clock frequency of the ripple counter is 12.5 MHz.