In article <12d4v2t2bqhls54@corp.supernews.com>, Yugo <nobody@nowhere.com>
wrote:
> Bazzer Smith wrote:
> > "Yugo" <nobody@nowhere.com> wrote in message
> > news:12d4mod9ti425a3@corp.supernews.com...
>
> > Hi, I am not sure precisely what you are asking (I take it English is not
> > your
> > first language), but anyway....
>
> I didn't think my english was that bad :) I will reformulate my question.
>
> "Is the Ahtlon 64 expected to work at 2200 MHz with 35 or 62W?"
>
> And the answer is in the link you provided:
>
> > http://arstechnica.com/news.ars/post/20060516-6840.html
>
> The answer is no. 35 and 62W are "the power you have to design the cooling
> solution to dissipate". As a processor with Cool and Quiet shouldn't work at
> full speed all the time, the cooling solution doesn't have to be adapted to
> maximum wattage.
>
> OTOH, if Cool and Quiet is not enabled, the temperature is hot and you do
> heavy processing...
>
> > What puzzeled my when looking at that processor was that you see two
> > processors which are identical in everyting apart from power consumption.
> > Well there is a slight voltage difference but not enough to explain the
> > power consumption figures.
> > It just seems they are tested for power consumption
>
> Exactly. That's what Paul explained in my preceeding thread:
>
> "So, to make processors with different wattages, you can sort through
> all the chips, and take the ones with low power consumption, and
> make those your "low power" devices."
>
> <nospam-0208060423300001@192.168.1.178>
>
> Thanks for your link. It answered my last question.
The 35W and 62W for the two processors, are when the processors
are running at 100% load. They will draw less power when idle and
Cool N' Quiet is being used. I had hoped to find a document with
details as to the power at different frequencies, but AMD hasn't
released that document for AM2 yet.
When it comes to the use of electricity, it all eventually
gets converted to heat. I'm going to draw a magic black box (say
there is an electric kettle inside). Now, if the box stores
neither potential nor kinetic energy, then when the electric
power goes into the box, it all has to come out as heat.
I=10 amps +-----------+
V=120 --------------->| magic | ~~~
volts | black | ~~~ 1200W of heat
--------------->| box | ~~~
+-----------+
The 35W and 62W processors will dissipate 35W and 62W of heat
respectively. They will need a heatsink and fan good enough
to remove those amounts of heat energy. The fan on the 62W
processor, will be spinning a bit faster and moving more air,
to maintain the processor at the same temperature.
(In fact, when digital signals flow from one logic device
to another, they carry a small amount of power as well. The
power is dissipated in the device receiving the signal and
other places as well. These effects are small and I choose
to ignore them, for the purposes of this explanation.)
For a basic CMOS logic gate, the formula for power dissipation is:
P = F*C*V1*V2
F is the frequency that the output moves at (engineers call it
the "toggle rate"). C is the capacitance (charge holding ability)
of the structure on the output of the gate. V1 is the supply voltage
for the gate (1.2V for the processor below). If the output signal
of the logic gate is the same voltage as the supply, then V2 will
be 1.2V as well. If the height of the signal (amplitude) is smaller,
then the V2 value would be smaller. Since for a lot of circuits,
V1 and V2 have the same numeric value, the equation can be written
in slightly simplified form:
P = F*C*V_squared
Now, we plug in some numbers. The CPU is millions of gates, so
this is sort of a composite example, adding the contributions of
the millions of gates together.
Frequency 2200Mhz
Voltage 1.20V/1.25V
Thermal Power 35W
35W = 2200*10**9 x C x 1.2V x 1.2V
Well, in this case, we don't know the value of C, it is a constant
of proportionality for all we know, in a black box sense.
Now, say I am an overclocker. I turn up the frequency above 2200.
To overclock, usually I need a little more voltage as well. The
result is the power increases faster than linearly, due to the
voltage increase. At least with that formula, you can see some
of the elements that go into determining how much current and
voltage are used, and the resulting power dissipated as heat.
Now, we'll try another experiment. I'm going to draw
a basic storage element. All this thing is doing right now,
is updating the output on the rising edge of the clock,
so there is logic 1 on the second cycle of the output and
a logic 0 on the third cycle of the output.
___1___ ___1___
_| |___0___| +--------------+ ________| |___0___|
----------| DATA Output |-------------
| |
| |
----------| CLK |
___ ___ +--------------+
_| |___| |___|
This storage element could be considered to be 100% busy, because
it cannot move its output faster than the diagram above. The
state can only change on each rising edge of CLK.
Now, if this thing was hooked to Cool N' Quiet, we could change
the frequency of the clock signal from 2200MHz to 1000MHz when
the computer is idle. Based on the formula above, the computed
power drops by at least 1000/2200 = 45%.
Now, I'm going to reduce the work the storage element is doing.
I'm going to connect a "flat" signal to the storage element.
__________________ +--------------+ ____________________________
----------| DATA Output |-------------
| |
| |
----------| CLK |
___ ___ +--------------+
_| |___| |___|
The above F*C*V_squared equation actually takes as the frequency
value, the frequency of the output. In my first example, if the
clock was 2200MHz, the output would be moving (toggling) at 1100MHz.
Now, in the second example, the output isn't moving at all. So
when the CPU isn't doing any work, it doesn't dissipate any power!
Sounds wonderful. But if we measure the actual power in the lab,
it is about one tenth of the "full power value". The storage
element has multiple logic gates inside, and some of the gates
are connected to the CLK line, and they do work as long as the
CLK is running. So, the electricity flow has not stopped.
So, what can we conclude from this poorly constructed analogy ?
1) When the computer does no useful work (no information is
flowing through the functional units, so the functional
units are idle), the power should drop to zero. But since
parts of the circuits are still wiggling their signals,
power is still needlessly wasted. That is the nature of
this particular kind of storage element. There are other
kinds, but they are harder to work with.
2) So, now we know, that if the computer is idle, it will still
be burning up some power. (There are many kinds of defects in
processors that waste power, but I'm not going to explain
them, as it distracts from the simplified principles.)
3) But if we drop the CLK frequency at the same time, the amount
of power used, drops to 45% of its former value.
In terms of the above logic gate power formula I have been
abusing, what can we say about the 35W and 62W processors.
Yes, the voltage is slightly different. I'll ignore that
for a second. We never really discussed "C" the proportionality
constant. We could say that the 35W processor has a lower
value of C than the 62W processor. That would account for
some of the difference between the processors. The voltage
drop of a fraction of a volt, also accounts for a small
percentage as well.
Now, I don't expect this explanation will serve any useful
purpose, but it is my attempt to explain when and why a
processor draws power, and all the power it draws is
converted into heat.
HTH,
Paul
Still one more question about AMD processors 
Linear Mode
