Thread: PSU load tester
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Old June 29th 06, 10:38 PM posted to alt.comp.hardware
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Default PSU load tester

On Thu, 29 Jun 2006 15:40:33 -0400, "Skeleton Man"
wrote:

Hi all,

I'm soon to be reviewing a new 750W psu but I don't have anything for load..
I have seen heavy mention of wire wound power resistors of 'various values'
but no mention of what those values are. I can get various load resistors
of around 100ohm/10W from ebay.. but oviously 100ohm isn't gonna be
sufficient.. I'm not sure if 10W is heavy enough either..



There are quite a few ways to create the load but if you
want to use load resistors, so be it. How did you plan on
mounting and cooling them? Merely hooking up this load for
a few minutes isn't going to tell you much rather than
running long term and this means you have significant heat
(750W) to be rid of. There are resistors inside extruded
aluminum casings to help 'sink them but that's a massive
heatsink if you were to try to passively remove 750W, I'd
suggest something more like a large aluminum tube with a fan
blowing through the middle, or oil cooling or (your
inventiveness, budget and skill in fabricating something
will determine what your (subjective) best approach is.

Ohm's law will get you started,

V = I * R (or)

I = V/R

So suppose your PSU is rated for 24A on the 12V rail. The
series resistance to load it to 24A is;

24 = 12 / R

R = 0.5 Ohm

So you'll have to find either a .5 Ohm resistor rated for 12
x 24A = 288W (which is uncommon), or use series and/or
parallel resistors to reach 0.5 Ohm, and based on how many
are in series you calculate out how many watts rating each
would need, except that it is generally good to over-spec
power resistors, you shouldn't really run a resistor rated
for 20W at 20W continuously if you want it to last and be
easier to cool.

Same applies to the other power rails, you just calculate
out what resistance you need to achieve the rated current on
each, and the wattage rating (plus a margin) of each member
in that reistor array.

You may find that eBay doesn't allow the selection of
resistor wattage or resistance you need (want) for best
results. You might also find that it's easier to device the
housing, board, cooling method, etc, and go from there
towards determining what resistors you'd use.



Any suggestions on the values I should use to create a good load... I'm
talking like 600W+ at least.. and if I can setup multiple banks of
resistors that would be good too so I can have 100W, 200W, 350W, 500W, etc..


I'd think about running the resistors at roughly half their
wattage rating... so for 12V rail, "if" you used 10W
resistors you might pick 22Ohm so you had 12(V)/22(Ohm)=
0.55A, 0.55A * 12V = 6.6W each. The number of resistors you
put in parallel then determines how many amps of load on the
12V rail. Suppose you wanted a closer fractional current
addition per resistor, let's say each reistor is additional
0.5A load instead of 0.55A.

12 / 0.5 = 24 (Ohm).

or perhaps you want 1A per resistor,

12 / 1 = 12 Ohm, but to keep some margin you'd want 20W
instead of 10W resistors.



Just to be clear I want a _load tester_ not a $10 PSU tester that tells me
if it's dead or alive..


Understood, but remember that a resistive load tester will
only qualify the PSU for running a constant, resistive load.
A computer is a highly variable capacitive load too. A
constant resistive load is *easier* for a PSU if it has the
capacity to do it at all.




I have a good DMM and will probably buy a second so I can measure V and A
simultaneously.. I also plan on getting a Kill-A-Watt P3 meter.. so the load
tester is the only real concern I have..



We can expect a typical 750W PSU to have well over 10A on at
least 2, probably 3 or more rails. Make sure your (DMM
with current feature) is rated for this current. Most
aren't, on consumer grade DMM 10A is a common limit.

The other option would be to meaure the voltage drop across
one of the resistors and calculate it out (Ohms law again).

Suppose you had a 0.5 ohm resistor and 0.21V difference in
voltage at one end of the resistor to the other. It matters
not if there are other resistors in series with it (on the
same rail).

V = I * R

0.21 = I * 0.5

I = 0.42A

The best combination of resistors to use can be calulated as
described above but will also depend on what you are willing
to sacrifice in convenience of implementation to get cheaper
parts (going on eBay to find them, for example).

If you want better resistor selections then consider some of
the major online electronics houses like

http://www.alliedelec.com/
http://www.digikey.com/
http://www.jameco.com/
http://www.mouser.com/
http://www.newark.com/

You can sometimes get lower prices at some of the
electronics surplus 'sites but it is hit-or-miss what they
have available. Something like the following might be
convenient, especially if you have some large heatsinks
lying around (or reasonably thick aluminum sheeting, say
0.1" or thicker) and can put it all onto your own PCB
(though if screwed or bolted down I suppose you could
air-wire it instead but it might look a bit messy).

http://www.bgmicro.com/prodinfo.asp?prodid=RES1406

As for switching between different current levels, again
just do the math for how many resistors you'd group together
for the target current, and get some heavy duty switches
rated for this high a DC current. Regular light switches
available at a local hardware store might be as cost
effective as anything but are a bit larger than some
alternative due to the bracket on them, but the bracket does
allow easier mounting if you were to put this all on a board
or something like that instead of a proper chassis.