Cooling advice URGENTLY sought!!

[Wes Newell]

Well, DUH! I think I could have figured that out myself! THAT'S what I'm
trying to do!

Are you blonde by any chance? If it's 40C in your house and 25C outside,
would you have to ask how to cool your house? How about creating some
ventilation? And in case you didn't know, heat rises, so put exhaust fans
high, intake fans low. Simple common sense should apply.

 

[Wes Newell]

Just so we're clear - this http://tinyurl.com/2l4vq is the case in
question...There is only ONE case fan at the back - that's it, there isn't
room for any others.

Well, according to the link you provided, the case has mounts for dual
80x80x25mm intake fans in the front, and one exhaust fan in the rear. Now
where I come from that's 3, not 1. Now add the PSU exhaust and you have an
even match of 2 intake in front and 2 exhuast in rear. The only other
thing to make sure of is that the intake fans don't have restricted air.
Most cases DO restrict airflow. That's easily overcome by doing away with
the restrictions.

 

[kony]

kony's last words before the Sword of Azrial plunged through his body
were:

Depends on the sink, there are a few setups meant for larger fans.

What sink is meant for a fan larger than 120mm?
Motherboards simply aren't designed to have that large an area around the
socket, even if a 'sink were made to accomodate a 120mm fan, that fan
would mostly hang outside the fin area, where the greater portion of the
airflow didn't pass though the hotter portion of the fins.

Even Thermalright's SLK series, some of the very best air-coolers, which
will take a 92mm, show very little benefit beyond that seen with a similar
80mm fan. By similar, I mean the noise level isn't even lower with the
92mm fan unless both were running extremely slow, under 1000 RPM, at which
point the 92 mm is slightly better, but her system is looking like it
needs the opposite, an even higher RPM fan, _IF_ no other remedy suffices.

 

[kony]

What sink is meant for a fan larger than 120mm?

I meant, "What sink is meant for a fan larger than 92mm?"
There are some Zalman (flower?) coolers with the separate fan bracket, but
they are not suited to hotter running CPUs, for the same $ a CPU would be
cooler with something else, like one of the Thermalrights... which are
deeply discounted from time to time, the last SLK-947U I bought only cost
$20.

 

[General Schvantzkoph]

I have an FX-53 based system on an SK8V board in a CoolerMaster Wave Master
case. The HSF is a Swiftech MCX64-V with a Delta fan and the PSU is a 480W
Tagan.

The rest of the spec is as follows: -

1GB RAM
3xSATA HDs (2xRaptor in a RAID 0 array plus a WDC)
9800XT 256MB
Audigy 2 Platinum Pro XS
Plextor 708A
Samsung combi

As a temporary measure, until you add more fans to your case, reduce the
clock rate on your CPU. Reducing the clock rate will do two things for
you, 1) it will give you more timing margin. 2) It will reduce the power
consumed by the CPU. If you step the clock rate down enough you will also
be able to lower the core voltage. Power is V^2/R so reducing the core
voltage reduces the power dramatically. BTW that's what AMD is doing for
their 30W and 50W Opterons. The "low power" parts are just fast parts that
are run at a lower clock frequency and a lower core voltage.

 

[Ben Pope]

Power is V^2/R so reducing the core
voltage reduces the power dramatically.

I think that heat production in a chip is more about capacitance than
resistance. The voltage and capacitance (and switching time) will affect
the amount of current that flows, and that will determine the heat
production.

Ben

 

[rstlne]

I think that heat production in a chip is more about capacitance than
resistance. The voltage and capacitance (and switching time) will affect
the amount of current that flows, and that will determine the heat
production.

Ben

Capacitive Reactance (Xc)
If you had a pure Capacitive circut then the P = V^2/(Xc) might be correct
Of course, it's a ton more complicated than this in a processor..
and I would have to wonder just how much total capacitance there is in a
processor (I wouldnt guess it's a lot given it's size)
;P

 

[General Schvantzkoph]

Capacitive Reactance (Xc)
If you had a pure Capacitive circut then the P = V^2/(Xc) might be correct
Of course, it's a ton more complicated than this in a processor..
and I would have to wonder just how much total capacitance there is in a
processor (I wouldnt guess it's a lot given it's size)
;P

The leakage current is the resistive part, the capactance is reactive. The
capactive part dominated in earlier generations but as line widths have
become smaller the leakage has become much more significant. In both cases
the power is related to the square of the core voltage. The reactive part
is linearly related to the frequency, so reducing the clock frequency
reduces the reactive power. If you lower the clock frequency and lower the
core voltage you get a very large reduction in power.

 

[Ben Pope]

Capacitive Reactance (Xc)
If you had a pure Capacitive circut then the P = V^2/(Xc) might be correct
Of course, it's a ton more complicated than this in a processor..
and I would have to wonder just how much total capacitance there is in a
processor (I wouldnt guess it's a lot given it's size)
;P

No, but there are 50Million transistors being charged and discharged around
2Billion times/s = 100 * 10^15 charges and discharges per second. So they
might be small, but there are a considerable number of them. Thats rough
for a P4 working at 100%, which would probably melt it pretty quick. Still,
an order of magnitude doesn't make a whole load of difference

Ben

 

[CBFalconer]

I think that heat production in a chip is more about capacitance
than resistance. The voltage and capacitance (and switching time)
will affect the amount of current that flows, and that will
determine the heat production.

In CMOS logic the action consists of charging and discharging many
small capacitances at the clock rate. The charge q transferred is
proportional to V and C, i.e. dQ = CV. The time interval dT is
inversely proportional to frequence, so the effective current I is
dQ/dT, and the effective power is V*I. Since q is proportional to
V, that gives the usual power vs V squared relationship, but power
is also directly proportional to frequency.

Just keep your eye on the fundamental ball, power is voltage times
current.

 

[Ben Pope]

In CMOS logic the action consists of charging and discharging many
small capacitances at the clock rate. The charge q transferred is
proportional to V and C, i.e. dQ = CV. The time interval dT is
inversely proportional to frequence, so the effective current I is
dQ/dT, and the effective power is V*I. Since q is proportional to
V, that gives the usual power vs V squared relationship, but power
is also directly proportional to frequency.

Just keep your eye on the fundamental ball, power is voltage times
current.

I good point well made. Much better than I tried to make it.

Ben

 

[Arnfinn Haraldsen]

It shold not overheat like this in that case, it has adequate (but barely)
airflow with its two intake fans in front

http://www.modthebox.com/review287_4.shtml

did you remove any plastic protection and put on thermal paste?

maybe you can set a fan at top to get air out, but dont use the fine black
grid - it will reduce the airflow by 75%.

 

[Miss Perspicacia Tick]

The leakage current is the resistive part, the capactance is
reactive. The capactive part dominated in earlier generations but as
line widths have become smaller the leakage has become much more
significant. In both cases the power is related to the square of the
core voltage. The reactive part is linearly related to the frequency,
so reducing the clock frequency reduces the reactive power. If you
lower the clock frequency and lower the core voltage you get a very
large reduction in power.

I wish I'd paid more attention in Physics!