Question

Processor A has a clock rate of 3.6 GHz and voltage 1.25V. Assume that, on average, it consumes 90W of dynamic power. Processor B has a clock rate of 3.4 GHz and voltage of 0.9V. Assume that, on average, it consumes 40W of dynamic power. For each processor find the average capacitive loads.

Answer 1

Processor A :
`C=3.2* 10^(-8)\ F`

Processor B :
`C=2.98* 10^(-8)\ F`

Step-by-step explanation:

We know that

`Dynamic\ power=(1)/(2)* C* V^2* f`

C=Average capacitive loads.

Processor A :

f= 3.6 GHz

V= 1.25 V

P=90 W

`Dynamic\ power=(1)/(2)* C* V^2* f`

`90=(1)/(2)* C* 1.25^2* 3.6* 10^9`

`C=(90* 2)/(1.25^2* 3.6* 10^9)\ F`

`C=3.2* 10^(-8)\ F`

Processor B :

f= 3.4 GHz

V= 0.9 V

P=40 W

`Dynamic\ power=(1)/(2)* C* V^2* f`

`40=(1)/(2)* C* 0.9^2* 3.4* 10^9`

`C=(40* 2)/(0.9^2* 3.4* 10^9)\ F`

`C=2.98* 10^(-8)\ F`