# Mailinglist Archive: opensuse (4053 mails)

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##### Re: [SLE] AMD running too hot
• From: Dave Smith <Dave.Smith@xxxxxx>
• Date: Wed, 22 Aug 2001 09:34:28 +0100
• Message-id: <20010822093428.D653@xxxxxxxxxxxxxx>
On Wed, Aug 22, 2001 at 12:46:51AM +0100, laurence@xxxxxxxxxxxxxxx wrote:
> At 12:52 21/08/01 -0700 you scribbled:
> >Hi Laurence,
> >
> >You are missing one crucial point, and that is you are
> >ignoring the impedance(Resistance) of the actual
> >processor involved!
>
> Are you saying then that the impedance of the processor involved changes?

No, he's not. That's the whole point. If you increase the voltage, the
current does not remain constant, since you are unable to rely on the
voltage staying constant, it is better to rely on a different equation where
the other factor, resistance (impedance) *is* constant, i.e. P=V^2/R

> I am not saying that with higher voltage you get higher power,
> I am saying that with higher voltage you get lower current at the same power.

This is true, but the power is not constant. Someone (I can't remember who)
was claiming that it was. To maintain constant power, the resistance must
be proportional to the square of the voltage across it. This is not the case.

> >using p = i^2 * r, you should find that the second
> >current is higher than the first. Current cant and
> >wont stay constant. If you have a different proof, I
> >would like to see it.
>
> Are you saying that as the processing load increases,
> the power (wattage) increases
> or that the impedance (resistance) increases?

Neither, really. As the processing load increases, more gates are changing.
A gate which is not changing state takes virtually no current at all (pA for
an entire chip). It is the point of switching where a small burst of current
is required. The cumulative effect of all these gates switching makes up the
current through the chip. The more gates which have to switch, and the
more frequently they switch, the higher the current. At a macroscopic level,
the impedance of the chip will appear to decrease (P=V^2/R, remember), but
the individual transistors are not changing in impedance.

--
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