5,014 views The delays of NOR gates, Multiplexer and Inverters are 2ns, 1.5ns and 1ns respectively.

If all the inputs P, Q, R, S and T are applied at the same time instant, Then the Maximum propagation delay (in ns) of the circuit is _______________

$6ns$.

@Magma lets think at time =0ns     T=0 so first  NOR gate will be selected now at time=2ns T changed and become =1 so in the second mux again NOR (in the time data cross mux T will cross its own inveter ) will be selected so maximum delay become 7ns

@anyother dont negative vote first try to understand

this case can only be possible we we are allowed to change T in between (and we can do this because we have to maximize delay ) .

6 seems to be the right answer . Not sure how . I got 7 as the  answer tho !

As you know that the function of multiplexer is to select any one input based on the given input via its select lines. So, the following cases will arise. It is but obvious when the select lines labeled S0(MUX1) will have value 0 input 0 will be selected. Similarly,when S0(MUX1) will have value 1,input 1 will be selected. The same holds for S1(MUX2) as well.

Now consider these situations:-

Case 1:- Suppose T=0, S0(MUX1)=0 and S1(MUX2)=1. The Propagation delay is as under:-

Delay due to input 0(Nor GATE selected since input=0,S0(MUX1) has input 0)+Delay due to MUX1+Delay due to input(Since S1(MUX2=1)+Delay due to MUX2

So,propagation delay=2ns+1.5ns+0ns+1.5ns=5ns.

Case 2 :-

Suppose T=1, S0(MUX1)=1 and S1(MUX2)=0. The Propagation delay is as under:-

Delay due to input 1(Not GATE selected since input=0,S0(MUX1) has input 1)+Delay due to MUX1+Delay due to input(Since S1(MUX2=0)+Delay due to MUX2

So,propagation delay=1ns+1.5ns+2ns+1.5ns=6ns.
Here you select the maximum value of the two inputs which is 6ns. I hope this helps you. :)

why at any given time only one nor gate is active ?
@rajesh

when T=1, is it need to evaluate QR associated NOR gate ?

when T=0, is it need to evaluate P associated NOR gate ?
Got it.
thanks sir.
In mux output will depend upon select line, when S = 0, input line 0 will be selected, when select line, S =1, input line 1 will be selected.

Considering S0 = select line of first mux, S1 = select line of second mux.

Case 1 : when T = 0, S0=0 and S1 = 1
Propagation delay= Delay due to input 0 (Nor gate, Since S0=0) + Delay due to MUX1 + Delay due to input 1(Since S1 =1) + Delay due to MUX2
So propagation delay = 2ns + 1.5ns + 0ns+ 1.5ns = 5ns

Case 2 : when T =1, S0 =1 and S1 = 0
Propagation delay = Delay due to input 1 (Not gate, Since S0=1) + Delay due to MUX1 + Delay due to input 0(Nor gate,Since S1 =0) + Delay due to MUX2
So propagation delay = 1ns + 1.5ns + 2ns+ 1.5ns = 6ns

Maximum propagation delay = max(case 1, case 2) = max ( 5ns, 6ns) = 6ns.
by

nice explanation @stblue Loyal.
why not to consider the inverter gate present for calculation of selection line of second multiplexer?

There will be 2 cases for this question because the values received at the select lines by the MUX changes the inputs it waits for.

So, on the left is when Value for left MUX is 1 and right MUX as 0
Right has the opposite. SO the maximum is max(5,6) = 6.

edited
Thanku

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