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,
1 answer:
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 =
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 =
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
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