hello there i want to find static power consumption of IC in following question

i know about static power and dynamic power

Static power is power consumed while there is no circuit activity


and dynamic power

PDP = ƒCVcc²

now here frequency and power is given . so we can find C.Vcc^2. then can anyone tell me how to find static power consumption??

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    \$\begingroup\$ Homework questions will be closed. \$\endgroup\$ – Marko Buršič Mar 14 '17 at 8:06
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    \$\begingroup\$ Surely you've done at least some regression analysis in your schoolwork up to this point, yes? \$\endgroup\$ – Ignacio Vazquez-Abrams Mar 14 '17 at 8:07
  • \$\begingroup\$ Here is a clue.... C.Vcc^2 is a constant.... and total power = PSP + PDP. \$\endgroup\$ – Trevor_G Mar 14 '17 at 8:10
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    \$\begingroup\$ You are missing the point. PDP curve is a straight line... extrapolate the sample points it back to ZERO Hz \$\endgroup\$ – Trevor_G Mar 14 '17 at 8:29
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    \$\begingroup\$ but i got it answer 20 mW \$\endgroup\$ – Beginner Mar 14 '17 at 8:33

This is clearly a homework question so this answer is not an answer, just guidance for you to take the next step to solving it yourself.

You are told that an increase of 5 MHz causes a power increase of 40 mW. You can use that to work out what the power is at 2 x 5 MHz less than your 10 MHz figure, giving you your static power.

You then have the zero-crossing point of a frequency vs power graph plot, and the question gave you the gradient of the graph plot, so you're there. You could either draw an actual graph or produce an equation in the form

Ptotal  =  Pstatic + (fMHz x PperMHz)

and get all your answers from that.

Good luck with solving it.

  • \$\begingroup\$ sorry that it is homework question. but i have not idea, so i asked here. and i always want hints here \$\endgroup\$ – Beginner Mar 14 '17 at 8:34
  • \$\begingroup\$ @Beginner, don't apologise, you've done the right thing asking for guidance. Learning is to be encouraged, not criticised and this is a help/learning site. (Some of us here get that, some others delight in being obstructive :-) ) Does what I described lead you through it \$\endgroup\$ – TonyM Mar 14 '17 at 8:40

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