Re: [SLE] AMD running too hot

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Guy Van Sanden

14 Aug 2001 14 Aug '01

16:33

Hello Sinan Im' getting nowhere.. checked hardware, recompiled kernel, nothing... windows still runs 10° cooler :-( about that thermaltake, it produces a reasonable amount of noise (38 dba I think). For me the noise level was very important (that's why I didn't by the FOP), because my system is in the living room... But even than, it's acceptable. Good luck with your system... If I ever find the answer, I'll post it to the list. Guy

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Linux Fan

16 Aug 16 Aug

16:25

New subject: [SLE] AMD running too hot

Hello Guy, I'm not sure weather you should try this or not, but sometimes the processor might need more power than the default supplied. You either need to get a better power supply or increase the CPU core voltage. I never tried increasing the voltage to my CPU, but have tried decreasing it and worked fine under heavy use. Whats the specifications for your power supply? Is it a 250, 300, 400 power supply? Who is the manufacturer? Is it approved by AMD? What's the current ratings for 3.3, +12, +5 volt outputs? 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. I read in a thread that some guy was running an AMD 1.2GHz on a 250watt power supply, and his computer would crash under heavy use. That was until he got a 350watt PSU and his problems were solved. Good Luck Sinan Sinanian __________________________________________________ Do You Yahoo!? Make international calls for as low as $.04/minute with Yahoo! Messenger http://phonecard.yahoo.com/

Laurence Orchard

18 Aug 18 Aug

11:34

New subject: [SLE] AMD running too hot

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 If I over clock my AMD 800 i find that increasing the core voltage increases the stability of the cpu

...

I read in a thread that some guy was running an AMD 1.2GHz on a 250watt power supply, and his computer would crash under heavy use. That was until he got a 350watt PSU and his problems were solved.

I think the guy was lucky to even get it to boot with a 250W supply!! a 300W should be sufficient

...

Good Luck

Sinan Sinanian

__________________________________________________ Do You Yahoo!? Make international calls for as low as $.04/minute with Yahoo! Messenger http://phonecard.yahoo.com/

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Laurence ** if you want to know what the program really does, look at the code ** laurence@orchards.org.uk This Mail should NOT have an attachment, if it does it may have been created by a VIRUS DO NOT OPEN IT!!!

Mark Hounschell

20 Aug 20 Aug

08:39

New subject: [SLE] AMD running too hot

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 -- Mark Hounschell dmarkh@cfl.rr.com

Jaakko Tamminen

08:39

New subject: [SLE] AMD running too hot

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. Viestissä 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 Orchard

10:34

New subject: [SLE] AMD running too hot

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

Linux Fan

21 Aug 21 Aug

19:52

New subject: [SLE] AMD running too hot

Hi Laurence, You are missing one crucial point, and that is you are ignoring the impedance(Resistance) of the actual processor involved! 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: 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. Sinan --- Laurence Orchard 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

...
...
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

Viestiss Maanantai 20. Elokuuta 2001 11:39, Mark Hounschell kirjoitti: 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@orchards.org.uk

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

-- To unsubscribe send e-mail to suse-linux-e-unsubscribe@suse.com For additional commands send e-mail to suse-linux-e-help@suse.com Also check the FAQ at http://www.suse.com/support/faq and the archives at http://lists.suse.com

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Laurence Orchard

23:46

New subject: [SLE] AMD running too hot

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 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

...
...
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

Viestiss Maanantai 20. Elokuuta 2001 11:39, Mark Hounschell kirjoitti: 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@orchards.org.uk

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

-- To unsubscribe send e-mail to suse-linux-e-unsubscribe@suse.com For additional commands send e-mail to suse-linux-e-help@suse.com Also check the FAQ at http://www.suse.com/support/faq and the archives at http://lists.suse.com

__________________________________________________ Do You Yahoo!? Make international calls for as low as $.04/minute with Yahoo! Messenger http://phonecard.yahoo.com/

-- To unsubscribe send e-mail to suse-linux-e-unsubscribe@suse.com For additional commands send e-mail to suse-linux-e-help@suse.com Also check the FAQ at http://www.suse.com/support/faq and the archives at http://lists.suse.com

Dave Smith

22 Aug 22 Aug

08:34

New subject: [SLE] AMD running too hot

On Wed, Aug 22, 2001 at 12:46:51AM +0100, laurence@orchards.org.uk 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. -- David Smith Tel: +44 (0)1454 462380 (direct) STMicroelectronics Fax: +44 (0)1454 617910 1000 Aztec West TINA (ST only): (065) 2380 Almondsbury Home: 01454 616963 BRISTOL Mobile: 07932 642724 BS32 4SQ Work Email: Dave.Smith@st.com Home Email: David.Smith@ds-electronics.co.uk

Jaakko Tamminen

21 Aug 21 Aug

20:14

New subject: [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@orchards.org.uk

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

Dave Smith

20:22

New subject: [SLE] AMD running too hot

On Mon, Aug 20, 2001 at 11:34:04AM +0100, laurence@orchards.org.uk 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 voltage. Trust me I know. Look at my .sig :-) -- David Smith Tel: +44 (0)1454 462380 (direct) Silicon Design Engineer Fax: +44 (0)1454 617910 STMicroelectronics TINA (ST only): (065) 2380 1000 Aztec West Home: 01454 616963 Almondsbury Mobile: 07932 642724 Bristol Work Email: Dave.Smith@st.com BS32 4SQ Home Email: David.Smith@ds-electronics.co.uk

Linux Fan

01:48

New subject: OT [SLE] AMD running too hot

Hi Laurence, Well, I guess you're not in the electrical engineering field! Here is a free tutorial. Current = Voltage / Resistance Thus Higher voltage equals higher current! and Power equals voltage * current! Sinan Sinanian --- 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

If I over clock my AMD 800 i find that increasing the core voltage increases the stability of the cpu

...
I read in a thread that some guy was running an AMD 1.2GHz on a 250watt power supply, and his computer would crash under heavy use. That was until he got a 350watt PSU and his problems were solved.

I think the guy was lucky to even get it to boot with a 250W supply!! a 300W should be sufficient

...
Good Luck

Sinan Sinanian

__________________________________________________ Do You Yahoo!? Make international calls for as low as $.04/minute with Yahoo! Messenger http://phonecard.yahoo.com/

-- To unsubscribe send e-mail to suse-linux-e-unsubscribe@suse.com For additional commands send e-mail to suse-linux-e-help@suse.com Also check the FAQ at http://www.suse.com/support/faq and the archives at http://lists.suse.com

Laurence

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

laurence@orchards.org.uk

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

-- To unsubscribe send e-mail to suse-linux-e-unsubscribe@suse.com For additional commands send e-mail to suse-linux-e-help@suse.com Also check the FAQ at http://www.suse.com/support/faq and the archives at http://lists.suse.com

__________________________________________________ Do You Yahoo!? Make international calls for as low as $.04/minute with Yahoo! Messenger http://phonecard.yahoo.com/

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participants (6)

Dave Smith
Guy Van Sanden
Jaakko Tamminen
Laurence Orchard
Linux Fan
Mark Hounschell