Propagation delay

Question

In the circuit shown below, the propagation delay of each NOT gate is 2 nsec (2 
nano sec), then the time period of generated square wave is –

Answer is 2*5*2*10^-9 which is 20ns.

Can anyone please tell mewhere is this formula coming from (2*n*propagation delay)?  Can someone give some reasoning behind this formula?

Answer 1

Let us understand it with one inverter (NOT gate) in feedback circuit..

So in 1 propogation delay we get 1 --> 0 assuming 1 as the initial value of the waveform..

Now as this NOT gate is in feedback circuit hence now 0 is served as input to NOT gate..

Hence we will get  0 --> 1 in 2nd propogation delay time interval..

Hence the value "1" is repeated after 2 propogation delay time units..Hence time period with one inverter in feedback circuit is 2 propogation delay units.

Extending the same reasoning to "n" inverters in feedback circuit :

Time period of wave    =   2*n*t_pd

Comments

@Shubhanshu , Questions asks for the time period i.e, time in which it retains its initial state (1 to 1 or 0 to 0)...

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only for feedback circuit this formula is valid,right ??

Time period = 2* propagation delay of ckt

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Why the propagation delay due to 6th not gate not counted?

For the output to be obtained ,the delay of 6 th inverter must be counted!!