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. For each processor find the average capacitive loads. 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. 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.

Answer 1

Answer:

1. Capacitive load for pentium 4 prescott processor = 32 nF. Capacitive load for core i5 ivy processor = 29.05 nF

2. Percentage of total dissipated power comprised by static power for Pentium 4 Prescott processor = 10 %. The ratio of static power to dynamic power = 0.11

Percentage of total dissipated power comprised by static power for Core i5 Ivy Bridge processor = 42.86 %. The ratio of static power to dynamic power = 0.75

3. Reduction in voltage for Pentium 4 Prescott processor = 5.9 % reduction

Reduction in voltage for Core i5 Ivy Bridge processor = 9.8 % reduction

Explanation:

1. We know that dynamic power, P ≈ 1/2 CV²f where C is the capacitive load of the transistor, v its voltage and f the frequency.

So C ≈ 2P/V²f

For the Pentium 4 Prescott processor, V₁ = 1.25 V, f₁ = 3.6 GHz and P₁ = 90 W. Let C₁ be its capacitive load. So, C₁ ≈ 2P/V²f = 2 × 90 W/ (1.25 V)² × 3.6 × 10⁹ Hz = 3.2 × 10⁻⁸ F = 32 × 10⁻⁹ F = 32 nF

For the Core i% Ivy Bridge processor, V = 0.9 V, f = 3.4 GHz and P = 40 W. Let C₂ be its capacitive load. So, C₂ ≈ 2P/V²f = 2 × 40 W/ (0.9 V)² × 3.4 × 10⁹ Hz = 2.905 × 10⁻⁸ F = 29.05 × 10⁻⁹ F = 29.05 nF

2. Total power = static power + dynamic power.

For the Pentium 4 Prescott processor, static power = 10 W and dynamic power = 90 W. So, total power, P = 10 W + 90 W = 100 W.

The percentage of this total power that is static power = static power/total power × 100 = 10/100 × 100 = 10 %

The ratio of static power to dynamic power = static power/ dynamic power = 10/90 = 0.11

For the Core i5 Ivy Bridge processor, static power = 30 W and dynamic power = 40 W. So, total power, P = 30 W + 40 W = 70 W.

The percentage of this total power that is static power = static power/total power × 100 = 30/70 × 100 = 42.86 %

The ratio of static power to dynamic power = static power/ dynamic power = 30/40 = 0.75

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

Since the total dissipated power is reduced by 10%, P₂ = (1 - 0.1)P₁ = 0.9P₁ where P₁ is the total power dissipated before the 10% reduction and P₂ is the new power dissipated after the 10% reduction in total dissipated power.

Let new total dissipated power, P₂ = new static power I₂V₂ + new dynamic power 1/2C₂V₂²f₂  = 0.9P₁.

For the Pentium 4 Prescott processor, P₂ = I₂V₂ + 1/2C₂V₂²f₂ = 0.9P₁. Since I₂(leakage current) = static power/voltage = 10 W/1.25 V = 8 A (since the leakage current is constant), we have

8 A × V₂ + 1/2 × 32 × 10⁻⁹ F × V₂² × 3.6 × 10⁹ Hz = 0.9 × 100

8V₂ + 57.6V₂² = 90. This leads to the quadratic equation

57.6V₂² + 8V₂ - 90 = 0. By the quadratic formula,

V₂ = $\frac{-8 +/- \sqrt{8^{2} -4X57.6 X -90} }{2X57.6} = \frac{-8 +/- \sqrt{64 + 20736} }{115.2} = \frac{-8 +/- \sqrt{20800} }{115.2}\\=\frac{-8 +/- 144.222}{115.2}$

V₂ = 1.18 V or - 1.32 V .

Choosing the positive answer, the new voltage is 1.18 V. The percentage reduction is (new voltage - old voltage)/new voltage × 100 % = (1.18 - 1.25)/1.18 × 100 % = -0.07/1.18 × 100 % = -5.9 % . Which is a 5.9 % reduction from 1.25 V

For the Core i5 Ivy Bridge processor, P₂ = I₂V₂ + 1/2C₂V₂²f₂ = 0.9P₁. Since I₂(leakage current) = static power/voltage = 30 W/0.9 V = 33.33 A (since the leakage current is constant), we have

33.33 A × V₂ + 1/2 × 29.05 × 10⁻⁹ F × V₂² × 3.4 × 10⁹ Hz = 0.9 × 70

33.33V₂ + 49.385V₂² = 63. This leads to the quadratic equation

49.385V₂² + 33.33V₂ - 63 = 0. By the quadratic formula,

V₂ = $\frac{-49.385 +/- \sqrt{49.385^{2} -4X33.33 X -63} }{2X33.33} = \frac{-49.385 +/- \sqrt{2438.8782 + 8399.916} }{66.66} = \frac{-49.385 +/- \sqrt{10838.794} }{66.66}\\=\frac{-49.385 +/- 104.110}{66.66}$

V₂ = 0.82 V or - 2.30 V .

Choosing the positive answer, the new voltage is 0.82 V. The percentage reduction is (new voltage - old voltage)/new voltage × 100 % = (0.82 - 0.9)/0.82 × 100 % = -0.08/0.82 × 100 % = -9.8 % . Which is a 9.8 % reduction from 0.9 V