# How do I calculate this?

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If an AMD processor with no special power-saving features is drawing 200 Watts
while running at 2 GHz with a 200Mhz Front side buss, How many watts will it
draw while running at 1 GHz with a 100Mhz FFB? 100W? Is that the formula,
half the frequency means half the heat generated? Or is it more complicated
than that?

Re: How do I calculate this?

That is very close for CMOS logic, such as
comprises most of modern microprocessors.

Not enough so to worry about.

--
--Larry Brasfield
email: donotspam_larry snipped-for-privacy@hotmail.com
We've slightly trimmed the long signature. Click to see the full one.
Re: How do I calculate this?
On Thu, 21 Apr 2005 10:02:30 -0700, "Larry Brasfield"

Except that at 1 GHz, the processor would be able to run at a lower
core voltage. The power dissipation is proportional to the core
voltage squared.

Paul

Re: How do I calculate this?

Watts

The OP strongly implied that unmentioned variables
were unchanged.  His question makes little sense
otherwise.  If we must mention all those variables,
of the code being run, its time dependencies, etc.

--
--Larry Brasfield
email: donotspam_larry snipped-for-privacy@hotmail.com
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Re: How do I calculate this?

200 Watts
will it
formula,

Do they do that?

I know that some microprocessor power supply contollers are supplied with
D/A converters to adjust the output voltage.

Sort of makes sense that they might be able to do it dynamically according
to the sums being done.

Clever stuff.

DNA

Re: How do I calculate this?
[...]

Sure.  For the range of processors being talked about here, whenever
power consumption of CPUs is seriously considered (e.g. because it's a
mobile computer running off a battery whose weight you have to
justify), core voltage modulation is a routine operation in modern
designs --- the capability of doing this dynamically is the only real
difference between e.g.  Mobile and Desktop versions of the same x86
processor family.  For the AMD processors mentioned by the OP, google
up "PowerNow!".

--
Hans-Bernhard Broeker ( snipped-for-privacy@physik.rwth-aachen.de)
Even if all the snow were burnt, ashes would remain.

Re: How do I calculate this?
donotspam_larry_ snipped-for-privacy@hotmail.com says...

Watts

The processes that modern microprocessors use leak like hell (both gate
oxide tunneling and sub-threshold) and it's getting worse.  Much of the
power dissipation is DC and is unaffected by the clock frequency.  Only
the dynamic power component is proportional to frequency.

Complete bullshit.

--
Keith

Re: How do I calculate this?

Watts

Given that as a fact, one has to wonder why so much
trouble is taken with clock speed reduction in mobile
processors.

True by definition and not at issue.

Since you are so all-knowing, how about listing the DC
current versus dynamic current for a few AMD processors
to give the OP an idea of what kind of error he can expect
to incur by using the Idc == 0 approximation?  (In other

--
--Larry Brasfield
email: donotspam_larry snipped-for-privacy@hotmail.com
We've slightly trimmed the long signature. Click to see the full one.
Re: How do I calculate this?
donotspam_larry_ snipped-for-privacy@hotmail.com says...

Watts

it

Reduce the clock speed and you can also reduce the voltage.  Reduce the
voltage and the leakage decreases too (by at least the third power).
It's not hard to see why one reduces the clock speed in mobile
applications.

No, not by definition.  The dynamic power is also a second order
function of the voltage, is that the definition of "dynamic power"?
It *is* the issue, since the question was asked about (total) power,
not a component of the power.

At 90nm and fast-sort processors, it's upwards of a 50% error (assuming
no Vcore modulation).  Is that enough of an error to admit you haven't
been paying attention for a few years?

--
Keith

Re: How do I calculate this?

Watts

will it

Interesting.

Dynamic power is the power dissipated due to
charging capacitances.  There was no voltage
change in the OP's question.  Maybe that should
be assumed on his behalf.

The question was how power varies when the clock
speed is changed.

Yes.  The last time I looked into this, the short
channel and gate tunnelling effects were still on
the horizon as a predicted limit to Moore's law.

It looks like the OP needs to look at a spec for
the actual AMD part he has in mind.

--
--Larry Brasfield
email: donotspam_larry snipped-for-privacy@hotmail.com
We've slightly trimmed the long signature. Click to see the full one.
Re: How do I calculate this?

200 Watts

will it

formula,

Will you never have a high enough IQ to be able to rent a clue?  What,
pray tell, do you think the buzz has been about for the last couple of
years?  Moore's law has come to a *big* speed-bump, and that's *LEAKAGE*.
Sheesh!

IFF they want you to know the answer.  If you're good (you're not) you
might be able to ferret out some information.  However, they're (AMD and
Intel) cozy with the facts.

--
Keith

Re: How do I calculate this?
On Thu, 21 Apr 2005 18:02:36 -0700, "Larry Brasfield"

This is not a Trivial Pursuit game with a single correct answer or a
homework help desk for lazy students, this is a news group that
traditionally has been used for open ended discussions, in which
participants can learn from by actively taking part in the discussion
or by just following the discussion.

As in engineering in general, there are usually not a single "correct"
answer, but it usually depends on many factors, which were not obvious
or stated in the original posting that stated a thread.

If this was a homework, I hope that the original poster has been
followed the whole thread and learned much more than from just a

Paul

Re: How do I calculate this?

The following '|>>|' quoted text is restored context
necessary to understand Paul's bare quotation.

|>>|>> > Only the dynamic power component is proportional to frequency.
|>>|>>
|>>|>> True by definition and not at issue.
|>>|>
|>>|> No, not by definition.  The dynamic power is also a second order
|>>|> function of the voltage, is that the definition of "dynamic power"?
|>>|
|>>|Dynamic power is the power dissipated due to
|>>|charging capacitances.  There was no voltage
|>>|change in the OP's question.  Maybe that should
|>>|be assumed on his behalf.
|>>|
|>>|> It *is* the issue, since the question was asked about (total) power,
|>>|> not a component of the power.
|>>|

I agree with what you are saying but object to your
use of a quote lifted out of its context to imply that I
need such a tutorial.  Without the context, you make
it appear that I attempted to constrain the discussion.
The (now restored) context shows otherwise.

That lesson, too, is worthy but already well taken.
...
--
--Larry Brasfield
email: donotspam_larry snipped-for-privacy@hotmail.com
We've slightly trimmed the long signature. Click to see the full one.
Re: How do I calculate this?

Once again, you don't know what the hell you're talking about...try
going to hell and staying there.

Re: How do I calculate this?
...

Oh, crap. I've just discovered that my plonkfile expires.

Bloggs, have you _ever_ posted _anything_ having anything to do with
electronics, or do you fill up your days merely spewing venom?

Thanks,
Rich

Re: How do I calculate this?
Dynamic average power = Ctotal*Vdd^2*fclk

so assuming all 200W is dynamic power(which it is not)
Ctotal is constant
Vdd^2 is constant

Pavg is proportunal to fclk.

Re: How do I calculate this?
On 21 Apr 2005 12:46:40 -0500, I.Hope.The. snipped-for-privacy@Purges.The.Pedophiles

One other thing that is relevant mainly in real-time control systems.

In such systems, it is quite common to perform a more os less constant
amount of work (say 500000 instructions or cycles) at very regular
intervals (say every millisecond) and the rest of the interval, the
processor is idle. In this example, those 500000 cycles would take
0.25 ms in every interval to execute at 2 GHz.

Dropping the clock frequency to 1 GHz, it would now take 0.5 ms to
execute the program in each 1 ms interval. Although the instantaneous
power dissipation drops to 1/2, the execution time doubles, so the
total energy dissipated remains constant during each interval and
hence the average power dissipation remains constant.

The only gains that you can get some power saving in such applications
is from such circuits that are running constantly regardless if the
CPU is doing any real work or not or by lowering the core voltage.

Other systems that needs to do a constant amount of work
(instructions) in a specified time (such as various media players) may
behave in a similar manner.

Paul

Re: How do I calculate this?
snipped-for-privacy@sci.fi says...

Yes, that's a very relevant observation.  It's actually worse since
doubling the clock doesn't get twice as much work done (still the same
wait for memory, for instance).  Also the Vcore can be throttled down
at a lower clock frequency, saving both dynamic and static power.  If
the actual needs are known ahead of time, a processor intended for a
slower clock can save even more because it can be
designed/processed/sorted (in any combination) for low power rather
than high clock rates.

Sure, but if you lower both the voltage and the clock rate you may be
able to save even more.

Sure.  In this case a processor chosen specifically for the application
is the best bet.

--
Keith

Re: How do I calculate this?
BTW, nice picture! ;-)

Re: How do I calculate this?

An interesting read on power consumption and leakage is this:
http://www.anandtech.com/printarticle.aspx?i23%43
Perhaps you can try and find related articles that show your specific
processor@2GHz containing the leakage level. That might give you an
indication of what power saving you can expect at a lower frequency.

Joop