The Pentium 4 Prescott processor released in 2004, had a clock rate of 3.6 GHz a

Question

The Pentium 4 Prescott processor released in 2004, had a clock rate of 3.6 GHz and a voltage of 1.25V. Assume that, on average, it consumed 10 W of static power and 90 W of dynamic power. The Core i5 Ivy Bridge, released in 2012, had a clock rate of 3.4 GHz and voltage of 0.9 V. Assume that, on average, it consumed 30 W of static power and 40 W of dynamic power for each processor find the average capacitive loads. Find the percentage of the total dissipated power comprised by static power and the ratio of static power and dynamic power for each technology. If the total dissipated power Is to be reduced by 10%, how mu Ch should the voltage be reduced to maintain the same leakage current? power Is defined as the product of voltage and current.

Explanation / Answer

1. Ans:-
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Formula is :-
capacitive load=Power/(voltage^2xfrequency switched)
and here, frequency switched = clock rate

So, average capacitive load = Dynamic power/(voltage^2xclock rate)

a) ACL for Pentium 4 is:-
       CL(P4) = 90/(0.5 X V^2 X ClockRate)
           = 90/(0.5 x 1.25^2 x 3.6x10^9)
           = 3.2x10^-8

b) ACL for Core i5 Ivy:-
       CL = 40/(0.5 x 0.9^2 x 3.4x10^9)
       = 2.90x10^-8

2 Ans:-
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a) For for Pentium 4 processor:-
Total Dissipated Power = SP + DP
       = 10 + 90 = 100 W
percentage of total dissipated power comprised by static power = 10/100 x 100
                           = 10 %
Ratio of Static and dynamic = 10/90 = 0.1111

b) For Core i5 Ivy processor:-
Total Dissipated Power = SP + DP
       = 30 + 40 = 70 W

percentage of total dissipated power comprised by static power = 30/70 x 100
                           = 42.85 %

Ratio of Static and dynamic = 30/40 = 0.75

3 Ans:-
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P=VI or I=P/V
thus , Leakage current = Power/voltage

a) For for Pentium 4 processor:-

Static power = 10 and voltage = 1.25
Leakage current = 10/1.25 = 8A

b) For Core i5 Ivy processor:-

Static power = 30 and voltage = 0.9
Leakage current = 30/0.9 = 33.33A

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