Question

The Pentium 4 Prescott processor, released in 2004, had a clock rate of 3.6 GHz and voltage of 1.25 V. 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.

1.8.1 For each processor find the average capacitive loads.

1.8.2 Find the percentage of the total dissipated power comprised by static power and the ratio of static power to dynamic power for each technology.

1.8.3 If the total dissipated power is to be reduced by 10%, how much should the voltage be reduced to maintain the same leakage current? Note: power is defined as the product of voltage and current

Short answer

1.8.1)

average capacitive loads = 2.904 * 10^-8

1.8.2)

Ratio of the static power to the dynamic power for each of the technology is 0.75

1.8.3)

Reduction in voltage is a 9.8 % reduction from 0.9 V

Step-by-step solution

Determine the formulas

Write the formula for capacitive load

$Capacitiveload=\frac{Dynamicpower}{0.5*V^2*Clockrate}$ ..... (1)

Write the formula for Pentium r prescoot processor

$Pentiumrprescootprocessor=\frac{Static}{Dynamic}$

….. (2)

Write the formula for percentage reduction is

$\frac{\left(Newvoltage-Oldvoltage\right)}{Newvoltage}\times 100$

Write the formula for Total power.

$Totalpower=Staticpower+Dynamicpower$ ..… (4)

Determine the average capacitive load

The average capacitive load for both of the processors is given by the formula:

$Capacitiveload=\frac{DynamicPower}{0.5*v^2*clockrate}$

For Pentium 4 Prescoott Processor, Capacitive Load

role="math" localid="1650926173043" $$\begin{gathered} =\frac{90}{0.5·{\left(1.25\right)}^2·3.6\times {10}^9} \\ =3.2\times {10}^{-8} \end{gathered}$$ (Since 1 GHz - 10⁹ Hz)

For core i5 capacitive load

$$\begin{gathered} =\frac{40}{0.5·{\left(0.9\right)}^2·3.4\times {10}^9} \\ =2.904\times {10}^{-8} \end{gathered}$$

So, we can see that the capacitive load on the core i5 processor is less than that of the Pentium 4 Prescoot processor.

Find the percentage of the total dissipated power

1.8.2)

Ratio of the static power to the dynamic power for each of the technology is given as:

For pentium 4 prescoot processor

$$\begin{gathered} =\frac{Staticpower}{DynamicPower} \\ =\frac{10}{90} \\ =0.11 \end{gathered}$$

For core i5

$$\begin{gathered} =\frac{Staticpower}{DynamicPower} \\ =\frac{30}{40} \\ =0.75 \end{gathered}$$

Find after dissipated power is to be reduced by 10%

1.8.3)

We know , Total Power=Static Power+Dynamic Power

And that leakage current is due to static power.

Since P=IV ,

I (leakage current) = P/V

Because the total power dissipated is diminished by 10%,

$P_2=\left(1-0.1\right)P_1=0.9P_1$

where P₁is the total power dissipated before the 10% reduction and P₂is the new power dissipated after the 10% reduction in total dissipated power.

Let new total dissipated power,

For the Pentium 4 Prescott processor,

localid="1650927872055" $$\begin{gathered} P_2=Newstaticpower\left(I_2{V}_2\right)+newdynamicpower\left(\frac{1}{2}\times {V_2}^2\times f_2\right) \\ {P}_2=\left(I_2{V}_2\right)+\left(\frac{1}{2}\times {V_2}^2\times f_2\right)=0.9P_1 \end{gathered}$$

Since I₂ (leakage current)=$\frac{Staticpower}{Voltage}$

localid="1650928232901" $$\begin{gathered} =\frac{10Watt}{1.25V} \\ =8A \end{gathered}$$

(because the leakage current is fixed), we have

$8A\times V_2+\frac{1}{2}\times 32\times 10F\times {V_2}^2\times 3.6\times 10Hz=0.9\times 100$

$$\begin{gathered} 8V_2+57.6V^2=90 \\ V=1.18Vor-1.32V \end{gathered}$$. As a result, the quadratic equation is formed.

By the quadratic formula,

The new voltage is 1.18 V if you select the positive option.

The percentage reduction is=

$$\begin{gathered} \frac{newvoltage-oldvoltage}{newvoltage}\times 100\% \\ =\frac{1.18-1.25}{1.18}\times 100\% \end{gathered}$$

.Which is a 5.9 % reduction from 1.25 V

Since

I(leakage current)=$\frac{staticpower}{voltage}=\frac{30W}{0.9v}=33.33A$

(because the leakage current is fixed)

$$\begin{gathered} 33.33A\times V_2+\frac{1}{2}\times 29.05\times 10F\times {V_2}^2\times 3.4\times 10Hz=0.9\times 70 \\ 33.33V+49.385V^2=63 \end{gathered}$$

As a result, the quadratic equation is formed.

$33.33V+49.385V^2-63=0$

By the quadratic formula,

V=0.82V or -2.30V

The new voltage is 0.82 V if you select the positive option

. Percentage reduction=$\frac{\left(0.82-0.9\right)}{0.82}\times 100\%=-9.8\%$

Which is a 9.8 % reduction from 0.9 V