Mailinglist Archive: opensuse (4053 mails)

< Previous Next >
Re: [SLE] AMD running too hot
Hi

You are right about power in one way.

If the total wattage stays the same, increasing voltage decreases
current.

But with CPU, the resistance where You dump the voltage is not rising
with it, so it will take more power, because You are feeding the same
resistance with higher voltage.

Rise/fall times are the signal rise/fall times.. ie. how fast the
signal-lines transits from 0 to 1.. actually from about zero voltage
close to Vcc (the CPU core voltage).

The faster the transition is, the more power it will consume. Also if
the transition lasts the same amount of time, but needs to transit
larger voltage range, it means faster volt/second value, and thus
consumes more power.

Pentium-class CPU has over 300 lines that transits between zero and one
hundreds of millions of times per second, so a tiny increase in power
loss in one of such pin means a lot of increased total power loss, that
will dissipate in heat and RF-radiation.

Then there is also internal bus structures that acts the same way, but
don't play such a big role as external pins.

So summa-summarum, with CPU, increasing voltage increases
power-consumptions, increases heat dissipation, and RF-radiation.

I hope this clears it up.

Sorry I'm not english, but I hope You understand.

Jaska.



Viestissä Maanantai 20. Elokuuta 2001 13:34, Laurence Orchard kirjoitti:
> 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!!!

< Previous Next >