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I am very very new to the electronic engineering side of things but I am taking a class on Computer Architecture. I have ran into a problem.

Power = Capacitive Load * Voltage ^ 2 * Frequency Switched.

I have a hard time understanding what capacitive load is. Let's say I have a CPU with the following specs.

Operates on 3.3 Volts,
Consumes 29.1 Watts,
When running at a clock rate of 200 MHz

How would I use these three bits of info to find a capacitive load? Can I? I'm not sure if I am going about this correctly. Help is appreciated! :)

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    \$\begingroup\$ While Dean has already provided the "correct" answer, it's a somewhat strange and unrealistic question. Normally you'd start by modelling the capacitance and use that to work out the dynamic power; then you need to address the fact that not all the CPU switches at once; then you need to consider static leakage power. \$\endgroup\$
    – pjc50
    Commented Sep 20, 2013 at 10:10
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    \$\begingroup\$ Yeah it is strange. I am working out of a book called Computer Organization and Design. I thought it would be good learning material but honestly they don't explain hardly anything. Questions in the book are very vague at times. \$\endgroup\$
    – James Manes
    Commented Sep 20, 2013 at 10:17
  • \$\begingroup\$ It's sure an interesting way of looking at things, I remember a very similar question some months back but unlike your question it was very poorly defined so I think it got closed / deleted. \$\endgroup\$
    – PeterJ
    Commented Sep 20, 2013 at 10:44
  • \$\begingroup\$ In simpler times you could empirically determine a processors "capacitance", which was really just the constant that best fit the relationship between power and clock frequency. Modern processors have leaky transistors and the relationship is not so simple. \$\endgroup\$
    – Joe Hass
    Commented Sep 20, 2013 at 11:19

1 Answer 1

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This is a simple arranging to resolve the equation for Capacitive load. Like so

\$ Energy_{dynamic}= Capactive \ load \times {Voltage^2 } \$

This equation is the energy of pulse of the logic transition of 0\$\rightarrow\$1\$\rightarrow\$0 or 1\$\rightarrow\$0\$\rightarrow\$1. The energy of a single transition (0\$\rightarrow\$1 or 1\$\rightarrow\$0) is then:

\$ Energy_{dynamic}= 1/2 \times Capactive \ load \times {Voltage^2 } \$

The power required per transistor is just the product of the energy of a transition multiplied by the frequency of transitions: \$ Power_{dynamic}= 1/2 \times Capactive \ load \times {Voltage^2 \times Frequency \ switched} \$

Therefore:

\$ \frac{2 \times Power_{dynamic}}{Voltage^2 \times Frequency \ switched} = Capacitive \ load \$

With the numbers you gave we get:

\$ \frac{2 \times 29.1W}{(3.3v)^2 \times 200*10^6} = 2.67\times 10^{-8} \mathrm{F} \$

F is farads the unit for capacitance.

I have assumed that the given equation takes power to be in Watts.

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    \$\begingroup\$ should be 200 \cdot 10^6 Hz, not ^3 \$\endgroup\$
    – Gunnish
    Commented Sep 20, 2013 at 10:13
  • \$\begingroup\$ Yes it is in Watts. :) Thank you very much. The book I was given doesn't show me anything other than the relative power formula, so I was a bit lost. \$\endgroup\$
    – James Manes
    Commented Sep 20, 2013 at 10:16

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