How to know how much power (watts) a computer needs?

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larkmore

I've been wondering for a while now how one goes about sizing a power
supply for the components in a computer? Mostly I just play it safe
and get a beefy 420W or 500W these days, but I probably don't really
need that I'm guessing. A quick search or two online didn't help me
find any power specifications for things like my DVD burner or hard
drives. I don't recall information like that being on the side of the
box (though I buy a lot of stuff OEM so often I don't even get a box,
or if I do it's just blank cardboard). Is there some repository I
could go to for the information? If I ask the manufacturers nicely
would they be likely to release that information? Should I just stick
with the guessing for now? How do you guys deal with this
uncertainty?
-Will
 
I've been wondering for a while now how one goes about sizing a power
supply for the components in a computer? Mostly I just play it safe
and get a beefy 420W or 500W these days, but I probably don't really
need that I'm guessing. A quick search or two online didn't help me
find any power specifications for things like my DVD burner or hard
drives. I don't recall information like that being on the side of the
box (though I buy a lot of stuff OEM so often I don't even get a box,
or if I do it's just blank cardboard). Is there some repository I
could go to for the information? If I ask the manufacturers nicely
would they be likely to release that information? Should I just stick
with the guessing for now? How do you guys deal with this
uncertainty?
-Will

I cheated a little bit, and typed in the answer to a search engine, to
find an instance :-) This page gives an example of a CDRW power requirement.
The drive doesn't really require this amount of power, but it would seem
that a lot of manufacturers use this "boiler plate" value. 12V @ 1.5A
and 5V @ 1.5A. The measured value, for example, on my current CDROM
drive, is 12V @ 1A while the media is spinning. Note that some other,
more complicated optical devices, have a higher boiler plate value
that is used.

http://b2b.sony.com/Solutions/produ...ageName=B2bUnified:Home:Storage:Optical+CD-RW

A little toy I own, is a clamp-on DC ammeter. This is a device that uses a
Hall probe, and measures the magnetic field around a wire. On the CDROM
drive, for example, I clamp the jaws around the 12V wire, and the meter reads
out the "1 amp" figure. On the main motherboard power cable, I can measure
all rails in a minute or two. The meter even adds the current in the wires
together. So, for example, you can clamp the meter around three orange
wires, and get the sum total of current through them. (This is both an advantage
and disadvantage - put the meter around an ordinary line cord, and the two
conductors cancel out.) The meter supports both AC and DC current (but you
buy them based on the ability to measure DC, as supporting just AC is a bit
easier for a meter maker).

(Device on lower right. 40A range is good for PCs. Meter will be quite
wrong for non-sinusoidal AC currents - gives ridiculous values for standby
AC power for ATX supplies. I've never tested it on a PFC supply, but it should
be better in that case.)

http://www.extechproducts.com/products/extech/380941_942_947.pdf

Hard drives are available on the manufacturer's site. Power varies with
drive type. Laptop drives are 2.5W to 5W or so (using 5V only). Desktops might
be 12V @ 0.6A plus 5V @ 1A, with some drives now requiring a bit less than
that. On the other hand, startup current has changed a bit in the other
direction. For the longest while, desktop drives had 12V @ 2.0 amp srartup
requirements, and then a couple manufacturers broke the rules and moved to
2.5A. That is important during the first 10 seconds of system operation,
and is especially important for homebrew systems having no "staggered spin"
capability.

For processors, you look up the power on places like processorfinder.intel.com.
Assume TDP/12V * (1/0.90) to get a Vcore input current requirement, assuming
the Vcore circuit is about 90% efficient. An 89W processor, would need 8.24A
from the 12V rail. Naturally, not all Vcore designs have the same efficiency,
so the 90% figure is a place holder, to show that efficiency counts.

For video cards, I rely on the data collected by Xbitlabs.com . They have a
search engine, and with enough fiddling around, I can find one of their video
card review/benchmark articles, where they measure power. They have a couple
specially equipped motherboards, with current shunts installed, so they can
measure slot power and aux connector power.

For motherboards, I assume 50W total power for the 3.3V and 5V rails total.
As time passes, it seems the chipset power goes up, and the RAM power goes
down.

While some power estimation sites give rather large figures for desktop RAM,
there are a few suppliers (Kingston) that might give a power estimate for RAM.
For DDR2, about 2W per stick might be good. For DDR, about 5W might be good.
Capacity matters, but not as much as you'd think.

http://www.valueram.com/datasheets/KVR667D2N5_2G.pdf

The second stick per memory channel, draws less power, because it would be
in standby (auto refresh or whatever). Power figures come from "industry
standard cycle mixes", and there is no assumption that 100% of clock cycles
involve the most expensive power consuming activities. So some gross assumptions
are made about how busy a typical memory stick will be.

FMDIMMs, another memory type, are a completely different animal, and draw
significant power. They would be more important to include in a calculation.

While I cannot calculate all power rails exactly, that gives some idea how you can
at least estimate the 12V draw, using freely available information.

Paul
 
wasbit said:

http://www.adecy.com/psu/ (No modern processors)
http://www.distortionwave.com/Power.html (No modern processors)
http://www.extreme.outervision.com/psucalculator.jsp (Computed 260W for 130W processor)
http://www.journeysystems.com/power_supply_calculator.php (No modern processors)
http://www.schrockinnovations.com/powercalc.php (Requires Java)
http://www.vbutils.com/power.asp (??? Won't waste my time)
Ups - http://www.aeceuro.co.uk/news/powercalculator.asp (No details given)

This is what I'm waiting for. Someone who will do one, in spreadsheet form.
View this page in Firefox. The information here is dated, but at least illustrates
how it should be done. Doing it this way, it is possible to critique the
content for correctness. Unlike the sites above. Hiding the information
does not build confidence that a correct answer will be given.

http://web.archive.org/web/20040411032947/http://www.takaman.jp/psu_calc.html?english

Paul
 
Paul said:
I cheated a little bit, and typed in the answer to a search engine, to
find an instance :-) This page gives an example of a CDRW power
requirement.
The drive doesn't really require this amount of power, but it would seem
that a lot of manufacturers use this "boiler plate" value. 12V @ 1.5A
and 5V @ 1.5A. The measured value, for example, on my current CDROM
drive, is 12V @ 1A while the media is spinning. Note that some other,
more complicated optical devices, have a higher boiler plate value
that is used.

http://b2b.sony.com/Solutions/produ...ageName=B2bUnified:Home:Storage:Optical+CD-RW

A little toy I own, is a clamp-on DC ammeter. This is a device that uses a
Hall probe, and measures the magnetic field around a wire. On the CDROM
drive, for example, I clamp the jaws around the 12V wire, and the meter
reads
out the "1 amp" figure. On the main motherboard power cable, I can measure
all rails in a minute or two. The meter even adds the current in the wires
together. So, for example, you can clamp the meter around three orange
wires, and get the sum total of current through them. (This is both an
advantage
and disadvantage - put the meter around an ordinary line cord, and the two
conductors cancel out.) The meter supports both AC and DC current (but you
buy them based on the ability to measure DC, as supporting just AC is a
bit
easier for a meter maker).

Wouldnt the bootup process take more power than the steady on state? Id
think the ammeter incapable of measuring a spike in power usage.
 
adam said:
Wouldnt the bootup process take more power than the steady on state? Id
think the ammeter incapable of measuring a spike in power usage.

I've measured processor power (ATX12V 2x2 cable) while sitting in the BIOS,
and it measures about half of max. The hard drive draws a bit more current
during the first ten seconds (motor current), but I'm going by the spec
from the drive manufacturer for that.

If you have a server, with ten drives drawing 2.5 amps a piece at spinup,
that represents a significant current. For a desktop with only one drive,
the drop in processor current, more than compensates for the hard drive
spinup current.

Also, while in the BIOS, the video card is not drawing full power either.
So with the exception of having a lot of disks, I don't need to work out
the startup condition as a second test case.

My meter does have a "peak" detection capability, but I expect it determines
that per sample, rather than using an analog peak detector. It works pretty
good, because I use it to measure the peak starting current of the starter
motor on my car. No, it won't detect or measure spikes. As I understand it,
a good Hall probe has a bandwidth of about 100KHz or so, so spikes can
easily escape it.


Paul
 
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