Mailinglist Archive: opensuse (4053 mails)

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Re: [SLE] AMD running too hot
  • From: Dave Smith <Dave.Smith@xxxxxx>
  • Date: Tue, 21 Aug 2001 21:22:06 +0100
  • Message-id: <20010821212206.B653@xxxxxxxxxxxxxx>
On Mon, Aug 20, 2001 at 11:34:04AM +0100, laurence@xxxxxxxxxxxxxxx wrote:
> At 11:39 20/08/01 +0300 you scribbled:
> >Hi
> >
> >Correct... And even more so, because the fact that with higher core voltage
> >the rise/fall times decreases, and thus consumes more power, generates more
> >heat, and generate more RF- and EMI
> >
> >Jaska.
> Don't understand about the rise/fall times decreasing?

Rise/Fall times actually increase with increase in temperature.

Hotter transistors have higher on-resistance, as the atoms are bouncing around
much more, getting in the way of the electrons (that's an oversimplified
explanation, but it's good enough. Higher resistance driving the same
capacitance equals slower edge. In theory, the higher resistance transistors
will dissipate less power, but the slower edge rate means that the following
transistors will spend more time in the semi-on state, which is where all
the power is used.

> Using the standard formula watts = volts * amps
> if the required power remains the same then if you increase the voltage the
> current must fall!

True, but the required power is not a constant. What you are saying is
true in the case of electrical transmission - to deliver the same amount
of power to the destination, a higher voltage means lower current, and
therefore less heat is dissipated in the wires. This is why power is
distributed along lines at hundreds of kV.

In the case of chips, most of the heat is being dissipated the transistors,
and their resistance remains approximately constant. So, the equation
P=V^2/R applies here, i.e power is proportional to the *square* of the

Trust me I know. Look at my .sig :-)

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