Question

1. A processor running at 2.5 GHz consumes 60 W of dynamic power and 15 W of leakage power. It briefly enters Turbo-boost mode and operates at a frequency of 3.0 GHz. How much dynamic power and leakage power does the processor consume in Turbo-boost mode?

Answer 1

1. 
   `P'_(dynamic) = 72.0\ W`
2. 
   `P'_(leakage) = 15\ W`

Solution:

As per the question:

Initial frequency of the processor, f = 2.5 GHz =
`2.5* 10^(9)\ Hz`

Final frequency attained by the processor, f' =
`3.0* 10^(9)\ Hz`

Initial dynamic power,
`P_(dynamic) = 60\ W`

Leakage Power,
`P_(leakage) = 15\ W`

Now,

To calculate the Dynamic power,
`P'_{dynamic`and leakage power consumed by the processor:

We know that the dynamic power is in direct proportion to the frequency of the processor:

`P_(dynamic)` ∝ frequency, f

Thus we can write:

`(P'_(dynamic))/(P_(dynamic)) = (f')/(f)`

`P'_(dynamic) = (f')/(f)* P_{dynamic`

Substituting the appropriate values in the above expression:

`P'_(dynamic) = (3.0* 10^(9))/(2.5* 10^(9))* 60 = 72.0\ W`

Now,

We know that the leakage power does not depend on the frequency of the processor and hence remains same.

Thus

`P'_(leakage) = P_(leakage) =15\ W`