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

Answer:

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 = $2.5\times 10^{9}\ Hz$

Final frequency attained by the processor, f' = $3.0\times 10^{9}\ Hz$

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:

$\frac{P'_{dynamic}}{P_{dynamic}} = \frac{f'}{f}$

$P'_{dynamic} = \frac{f'}{f}\times P_{dynamic$

Substituting the appropriate values in the above expression:

$P'_{dynamic} = \frac{3.0\times 10^{9}}{2.5\times 10^{9}}\times 60 = 72.0\ W$

Now,

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

Thus

$P'_{leakage} = P_{leakage} =15\ W$