Can undervolting actually raise temps?

[skunkx1]

I have an Opteron 168 at stock volts (which is pretty low, everest reports 1.328) that I clocked up to 2.7ghz last night and ran prime95 on both cores for three hours at which one core called it quits. Temps were hovering at about 51 degrees. It might be a stupid question but, can undervolting actually raise the stress on a cpu and create more heat. In my head, physics would tell me this is impossible... but i've been wrong before...

[deathadder]

It would make it unstable but I can't see it making it run hotter. I have a mobile here that runs at lower voltages. If I lower them without increasing the operating frequency it runs cooler. Also If I just raise the voltage and keep frequency constant the temps will rise.

[dr_bowtie]

I would say i wouldn't think so.... but when the cpu needs amp to do work and it's being shorted of those amp...well things can heat up...

 

Kinda like pulling amps thru a wire thats rated for less amps than it's pulling... when what ever requires the amps... is not getting the correct supply...the that not only gets hot but so does the wire...

 

But see for yourself...crank up the vcore to 1.45 or a little higher and retest and post back...I would say it's more your cooling...but I could be wrong also

[MalibuSS]

I would say i wouldn't think so.... but when the cpu needs amp to do work and it's being shorted of those amp...well things can heat up...

 

Kinda like pulling amps thru a wire thats rated for less amps than it's pulling... when what ever requires the amps... is not getting the correct supply...the that not only gets hot but so does the wire...

 

But see for yourself...crank up the vcore to 1.45 or a little higher and retest and post back...I would say it's more your cooling...but I could be wrong also

 

Not quite, but I understand why you would come to that conclusion. Here's how it really works:

 

Power (in this case, heat) = Current x Voltage

 

Less Voltage = less power = Less heat.

 

The reason that the wire heats up in the example above, is because:

 

Voltage = Current x Resistance.

Smaller wire = more resistance.

More resistance = More Voltage.

More Voltage = More power (again in this case, heat)

 

Hey, maybe that degree in Electronics Technology was useful after all.

[dr_bowtie]

Ok I feel ya on that one but....

 

How does more resistance = more voltage...?

 

Lets say the voltage is constant...?

 

lets just use for an example an appliance needs a constant 12volts at 60 amps...(using a 10gauge wire)

 

now lets for a minute say the person that installed the appliance used less than sufficient wire...lets say 16gauge wire....

 

now under load the appliance is still getting 12volts but cannot get the full 60amps supply needed...lets say it's only getting 20amps (if thats possible using 16gauge wire)

 

Now under load the appliance as well as the wire is gettting hot due to the resistance of not getting the right amount of amps.. but it's still getting the right amount of volts...

 

Or in easy terms it's trying to pull more amps than the wire can carry... thus making the appliance work harder and effectively over heating itself as well as maybe burning the wire supplying it...?

 

in this case how does the resistance of the wire equal more volts when there is only 12volts...

 

even in this case if you had the right size wire (10gauge) and supplied the right amount of amps but lowered the voltage to say 10volts the appliance would still build a fairly good amount of heat from being under volted under load....maybe not as much as being under amped at the correct voltage but it still will....Not arguing with your point...just using real word experience...

[ReelFiles]

Doc, that principle doesn't apply here;

 

lower voltage = less heat

higher resistance = more heat

 

You are just lowering the voltage, the actual resistance of the CPU remains unchanged.

 

So it cannot get hotter from less volts.

[MalibuSS]

Ok I feel ya on that one but....

 

How does more resistance = more voltage...?

 

Lets say the voltage is constant...?

 

lets just use for an example an appliance needs a constant 12volts at 60 amps...(using a 10gauge wire)

 

now lets for a minute say the person that installed the appliance used less than sufficient wire...lets say 16gauge wire....

 

now under load the appliance is still getting 12volts but cannot get the full 60amps supply needed...lets say it's only getting 20amps (if thats possible using 16gauge wire)

 

Now under load the appliance as well as the wire is gettting hot due to the resistance of not getting the right amount of amps.. but it's still getting the right amount of volts...

 

The appliance isn't getting the full 12v. The wire is (because it's to small and has become a resistor). There is only voltage where there is resistence.

 

And to further clarify, Most people say that voltage causes heat (including myself). This is really over simplifying it. Voltage doesn't create heat. Resistance to voltage creates heat.

 

Or in easy terms it's trying to pull more amps than the wire can carry... thus making the appliance work harder and effectively over heating itself as well as maybe burning the wire supplying it...?

 

in this case how does the resistance of the wire equal more volts when there is only 12volts...

 

It doesn't equal more volts than is supplied (in your example 12v). The wire is just "stealing" the required voltage from the appliance.

 

even in this case if you had the right size wire (10gauge) and supplied the right amount of amps but lowered the voltage to say 10volts the appliance would still build a fairly good amount of heat from being under volted under load....maybe not as much as being under amped at the correct voltage but it still will....Not arguing with your point...just using real word experience...

 

Current = voltage / resistance. If you are using the correct guage wire you "in theory" have no resistance, in which case the formula becomes:

 

Current = voltage. The above example is impossible because you can't keep current constant while lowering voltage in this case.

 

 

Lets compare electricity to water:

 

You've got a pipe, some water flowing through it, and a water spigget to keep water flowing through the pipe (to keep water pressure in the pipe).

 

The pipe = resistance, The water = current, the spigget supplying pressure = voltage.

 

If you make the pipe smaller (more resistance), there will be less water flowing (less current), unless more pressure (voltage) is applied to the pipe.

 

In other words, it takes more "work" to keep the same current. This is where power comes in. Power is the rate at which work is done. And in simple terms Power = heat

 

Now let's apply this to your last example:

 

You've got the correct sized pipe. (it can flow as much water as the spigget can provide). If you turn the water spigget to "half open" let's say, then it's impossible to have the same flow of water(current) in the pipe.

 

And don't worry, I'm not taking your post as arguing with me, just as this post shouldn't be considered arguing with yours.

 

I'm just clarifying this to everyone as it seems to be a common mis-conception.

 

[MalibuSS]

Doc, that principle doesn't apply here;

 

lower voltage = less heat

higher resistance = more heat

 

You are just lowering the voltage, the actual resistance of the CPU remains unchanged.

 

So it cannot get hotter from less volts.

 

Exactly.

[dr_bowtie]

Yep....Sounds good...

 

Now he has the full detailed answer... see how I made you do all the work...

 

I know lower volts wont cause more heat on a cpu...but now the original poster sees both side of the coin to eliminate any posible confusion....

 

That was worded rather well technically....I couldnt have said it like that....for being a dummy with NO degree...

 

Thanks for the reply....good job...

[MalibuSS]

Yep....Sounds good...

 

Now he has the full detailed answer... see how I made you do all the work...

 

I know lower volts wont cause more heat on a cpu...but now the original poster sees both side of the coin to eliminate any posible confusion....

 

That was worded rather well technically....I couldnt have said it like that....for being a dummy with NO degree...

 

Thanks for the reply....good job...

 

Pretty sneaky dr_bowtie...You'll have to teach me that trick sometime.

[dr_bowtie]

LOL...

 

I feel ya man...sometimes I just feel...well....like being goofy...I know I'm not being a big help...but sometimes showing someone both sides of the coin helps them understand better...

 

plus sometime I just like to test people...I drew your number...

 

just kidding on the later part....have a nice weekend...

[RGone]

Now that we go all that sorted go see what they say about "under-power'' logic switches...and what they can create...that one is most fun...

 

RGone...:confused: