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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 _______________

@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 ) .

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11 votes

Best answer

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. :)

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. :)

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6 votes

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.

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.