Mailinglist Archive: opensuse (4053 mails)

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Re: [SLE] AMD running too hot
  • From: Laurence Orchard <laurence@xxxxxxxxxxxxxxx>
  • Date: Wed, 22 Aug 2001 00:46:51 +0100
  • Message-id: <4.3.2.7.2.20010822004153.00aacd10@xxxxxxxxxxxxxxxx>
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?


v = electrical potential(voltage)
i = current
r = resistance
p = power

v = i * r
p = v * i

or with substitution,
p = i^2 * r

Now, if you believe that with higher voltage, you will
get higher power, then you should also believe the
next few lines:

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.



Case 1
Voltage = V1
Power used = 30w
resistance(constant) = R2

Case 2
Voltage = V2 higher than V1
Power used = 35w
resistance(constant) = R2

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?

If you are saying that

Sinan

--- Laurence Orchard <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?
>
> Using the standard formula watts = volts * amps
> if the required power remains the same then if you
> increase the voltage the
> current must fall!
>
> if the power requirement does increase then with a
> higher voltage the
> current will not rise as high.
>
> 2 watts = 1volt * 2amps or
> 2 watts = 2volts * 1amp
>
> therefore
>
> 4watts = 1volt * 4amps or
> 4watts = 2volts * 2amps
>
>
> >Viestiss Maanantai 20. Elokuuta 2001 11:39, Mark
> Hounschell kirjoitti:
> > > Laurence Orchard wrote:
> > > > At 09:25 16/08/01 -0700 you scribbled:
> > > > >Hello Guy,
> > > > >
> > > > ><snip>
> > > > >
> > > > >Remember that increasing the core voltage to
> you CPU
> > > > >would increase the produced heat. Though, I
> don't
> > > > >think its the temperature that is causing
> your
> > > > >problem. Try e-mailing AMD and asking them if
> > > > >increasing the voltage to you cpu wouldn't in
> any way
> > > > >void your warranty since you would be running
> you CPU
> > > > >out of spec and that is not recommended.
> > > >
> > > > Increasing the core voltage does not increase
> the amount of heat produced
> > > > it should reduce it,
> > > > more volts = less amps = less current = less
> heat
> > >
> > > more volts with same impeddance = more amps
> > >
> > > 1volt into 1 ohm = 1amp
> > > 2volt into 1 ohm = 2amp
> > > 2volt into 2 ohm = 1amp
>
>
> Laurence
>
> ** if you want to know what the program really does,
> look at the code **
>
>
> laurence@xxxxxxxxxxxxxxx
>
> This Mail should NOT have an attachment, if it does
> it may have been
> created by a VIRUS
> DO NOT OPEN IT!!!
>
>
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Laurence

** if you want to know what the program really does, look at the code **


laurence@xxxxxxxxxxxxxxx

This Mail should NOT have an attachment, if it does it may have been created by a VIRUS
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