Technical Tidbit:

You have a long wire of let's say length L. And it has the net delay of 100ps. Now you split the wire into two equal halves of length 0.5L each and insert a buffer at the center of the wire. The delay of buffer is 25ps. What is the total delay of the system: (half wire+ buffer + half wire)?

Option a: 50+25+50= 125ps

Option b: 25+25+25=75ps

Option c: 25+25+50=100ps

Option d: 50+25+25=100ps

Option b: 25+25+25=75ps

Option c: 25+25+50=100ps

Option d: 50+25+25=100ps

Choose the right option, and please give a short explanation.

Thanks!

Option b. Currently studying this so I wish to try it out. I remember wire delay's proportional to the square of the length (0.38RCL^2) so if we split the wire into half the delay goes down from 100ps for L to 25ps for 0.5L. Then the total delay is just the sum of the buffer plus two 0.5L wires.

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DeleteYes, I do agree with Chirag’s comment

DeleteIf R0 is the unit resistance and C0 is the unit capacitance

Total delay of wire of unit length = R0 L x C0 L

= R0C0 L^2

As this buffer dividing the net into equal half’s i.e. L/2

Now delay offered by wire = R0C0 (L/2)^2 + R0C0 (L/2)^2

= R0C0 (L^2/4) + R0C0 (L^2/4)

= 25 + 25

Total Delay of system = 25 + 25 + 25

This comment has been removed by the author.

ReplyDeleteOver all system delay, depends on kind of buffer you add too, in this case a 100ps buffer will increase the overall delay.

DeleteAlso, you cannot divide R and C, its the length that should get divided.

DeleteLets take this,

ReplyDeleteT=R.C

R= kL/A

so T directly proportional to R & R is directly proportional to L

so T is directly proportional to L

so 50+25+50=125ps

Give me explanation if I am wrong..!

Your approach is right! But you are missing C component change in your equation. If you consider that too, answer would come out as option b.

Deleteyes R prop to L..den c prop to L..so T prop to L^2..

ReplyDeleteso option b. .

ReplyDelete