Differences in capacitive loads

Question

I'm in school and I am trying to figure out a question for homework. It reads as stated:

How much has the capacitive load varied between versions if the dynamic power has been reduced by 10%? Use table above.

I understand that Power = Capacitive Load * Voltage^2 * Frequency.

Which would translate to Capacitive Load = (Power) / Voltage^2 * Frequency.

What I don't understand is how I'm supposed to solve this without at least once source of wattage given.

I think they want a percentage. The power is 10% lower, how much lower is the capacitive load? If you need a number, you could make one up: it was 10W before and now it's 9W — Criticizing Israel not allowed, Commented Aug 30, 2022 at 2:47
@user253751 Assuming the value of power is 10W then C = 10/(1.75)^2 x (1.5 x 10^9) = 2.177 x 10^-9 Lowering the power by 10% (.9) gets me C = 10 x .9/(1.75)^2 x (1.5 x 10^9) = 1.959 x 10 ^-9 C New/C Old = 1.959 x 10^-9/ 2.177 x 10^-9 = 0.8998621957 or C New is at ~89.99% it's original capacity. — Zanius Maximus, Commented Aug 30, 2022 at 3:31

D.A.S. · Accepted Answer · 2022-08-30 04:15:20Z

0

In theory, power losses associated with driving switched capacitance are proportional to C and \$fV^2\$*k but ignore k for comparative C changes.

Voltage Clock Rate Power
i 1.75V 1.5 GHz 4.6
ii 1.2V 2 GHz 2.9 = -37%C or -47%C +10%P

i 1.1V 3 GHz 3.6
ii 0.8V 4 GHz 2.56= -29%C or -39%C + 10%P

for a 10% power savings included

answered Aug 30, 2022 at 4:15

D.A.S.

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\$\begingroup\$ I understand how you got to -37%C & -29%C but how did you get to -47%C/10%P or -39%C/10%P \$\endgroup\$
– Zanius Maximus
Commented Aug 30, 2022 at 4:54

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