VCore up, temp down???

[svol]

Normal if you raise the VCore your CPU will run hotter, but why is this.
According to a formula the heat produced by a resistor or wire, etc. is calculated like this:
Q=(I^2)*R*t
Q=warmth in watts or joules
I=ampere
R=resistance (ohm)
t=time (seconds)

In theory if you higher your VCore there will be less ampere running through your wire, so according to this formula the produced heat will be LOWER, strange isn't it?
So why does the temp is higher when raising the VCore?

The only explenation I have is that this formula doesn't work by the components a CPU is made of, so which components are this and why does their produced heat raise when lowering the ampere?

BTW - sorry for my bad English .

My case has so many fans that it hovers above the ground 😱 .

 

[Guest]

Your a little off with your theory. Just because you raise the voltage of something does not mean that you are running less amperage. However if you want to stay at the same wattage then you would have either lower your amperage while increasing the voltage or increase your amperage while lowering your amperage.

In the case of the cpu's you are not staying at the same wattage. As you increase the voltage and your Power supply should keep the same amperage then you are increasing the wattage that your cpu is consuming, thus producing more heat.

Hope that helps you.

IT Tip of the Month:"C:\>ipconfig /all >>C:\Ip.txt" Look in root for txt file.

 

[svol]

So the PSU will continue to give the same amperage although you've increased your voltage, which will increase the wattage and the temp. Stupid PSU, otherwise you'll can run cooler by increasing the voltage and lowering the amperage.

My case has so many fans that it hovers above the ground 😱 .

 

[koke]

Svol:

I see your stated formula as something to employ to determine Heat generation but the issue is confused by associating this formula with the relationship between Voltage, Current and Power dissipation (Watts).

Electricity 101 teaches that a constant and proportional relationship exists between Voltage(V), Current(I) and Power(P).

It matters not that we are dealing with a CPU or a wire or a resistor. These relationships are CONSTANT.

Therefore, IF (V) is increased.....a directly proportional increase in (I) will result.

The increase in (I) will "physically" generate more heat. This is due to the greater electromotive force (i.e.(V)) which draws a greater flow of electrons (i.e.(I)) through the circuit. This in turn physically creates friction and HEAT.

Components, including CPU's, are designed with specific materials and to operate within, and dissipate, certain levels of (P). In most cases, this dissipation takes place in the form of Heat. (Unless of course, things get really hot at which point some (P) may be dissipated in the form of light which is not a healthy situation for a CPU 🙂).

The trick, therefore, is in the ability to dissipate the heat. What we really need is big heat sinks and fans on each of those millions of transistors in that CPU 🙂 If dissipation is inefficient, the heat will change the physical properties of which the components are constructed thereby rendering it incapable of performing the function it was originally designed for. Some people refer to this as "damage". 🙂

Happy O/C'ing...
/koke
Energy is neither created nor destroyed. It is only transformed.

 

[svol]

I've learned that on school, but it can be that I've not learned everything yet, Im still a student you see.

Say you've a device which uses 100 Watt on a 10 V line so the amparege is 10 A (10*10=100W). If you increase the voltage to 20 and the device still uses 100 Watt your amparege will be 5, because 5*20=100 Watt, right?

So if you increase the voltage and the wattage stays the same youre amparege will drop because of P(Watt)=U(V)*I(A). And so the heat output will be less because of the formula, right?

It could be that I'm wrong or not learned enough yet .

My case has so many fans that it hovers above the ground 😱 .

 

[koke]

OK - I see where you are coming from....I have a question for you:

-In your hypothetical circuit: What do you think is limiting the flow of current to only 10 Amps?

The hypothetical device has been designed to safely operate within certain parameters.

These parameters are fixed and state that the device will operate efficiently with a 10 Volt supply and it can/will dissipate/consume 100 Watts of power.

By calculating, we know that the device will be running on 10 Amps of current.

There is a direct relationship between Voltage and Current. In other words, if you increase the Voltage, you WILL increase the current. This should be logical. (Imagine a garden hose connected to a slightly opened tap and water running through it. If you open the tap full, more water will try to flow through the hose - right?)

The resistive properties of the hypothetical device does in fact limit the current to 10 Amps and at the same time consumes 100 Watts of electrical power but it is not designed to dissipate more than this and it cannot physically limit the flow of current to more than 10 Amps without consuming/dissipating more than 100 Watts of power.

If the Voltage is increased, the current will exceed 10 Amperes and the device will be faced with dissipating more than 100 Watts of power. If the device cannot change it's physical qualities, it will therefore suffer damage and possibly fail due to excessive heat.

Back to hose....if behind the tap is a pressurized tank that can achieve say 2,000 psi (pounds per square inch) but the garden hose is only rated to 200 psi, What will happen to the garden hose if the tap is opened full??

>>So if you increase the voltage and the wattage stays the same youre amparege will drop because of P(Watt)=U(V)*I(A). And so the heat output will be less because of the formula, right?<< Wrong...

/koke

 

[svol]

Yes, I see it now, you're right.

My case has so many fans that it hovers above the ground 😱 .

 

[labdog]

P = R i²

but u = ri

so

P = ui

if you raise u or i P raise and give more heat

so if you vCore is 10V, you have to buy another cpu



EasyInfo 😎
I would like to Invest for my PC !!
ok, buy nothing.

 

[labdog]

i think you take the pb at reverse side

P isInFonctionOf(U I) & not reverse

if u used more amperes, P raise
if u used less amperes, P decrease

you have a radiator for home of 1000W

this is the power in watts that it can be deliver
obviously more watts it has, more it can heat

but to have those 1000W, on standard constant 220V (at home in france, i don't know what is your ref in USA)
it consume for exemple about 10A

& a 2000W radiator will consume more amp about 15A
(not simply the double, because I² & not I on the formula)



EasyInfo 😎
I would like to Invest for my PC !!
ok, buy nothing.

 

[svol]

Yes I know I forgot some other formula, and that my statement isn't right. If I had take a longer thought about it I think I also found out that it doesn't work in reality. I was just wondered that amparage (accordning to my formula, which came out of a physics book) has also something to do with heat production, beside the P (Watt) production of heat.
Afterall (as you and the others said)the following formula also have to do with heat production:
P=U*I

For a heat producing electrical component also counts:
U=I*R and E=U*I*t
That can also be written as:
E=(I*R)*I*t and in short that's the same as E or Q=I^2*R*t.
So it also had something to do with heat, but isn't the most important formula which is P=U*I.

Hope that my English makes sence, because I also life in Europe and English is not my original language...

My case has so many fans that it hovers above the ground 😱 .

 

[labdog]

i think, the more complicated is to make a good relation, a rapprochement between Theory & Reality, between Formulas & Practice & not thinking each apart


note: i'm French & as you, i have to translate every thing in English & reverse

but "Babylon" Soft is a good one to translate in real time words you don't know in French as in Spanish, Corean, or even Chinese or Russian

maybe, you could use it to simplify your translation

EasyInfo 😎
I would like to Invest for my PC !!
ok, buy nothing.<P ID="edit"><FONT SIZE=-1><EM>Edited by labdog on 12/19/01 02:41 AM.</EM></FONT></P>