K
kony
<snip>
I meant " .. would be (slightly LOWER) due to the element
not getting as hot.
<snip>
I meant " .. would be (slightly LOWER) due to the element
not getting as hot.
K
kony
Ok, but then you need a way to mount them and/or either to
solder or sockets.
I'm not arguing that bulbs couldn't work, but rather the
resulting project could be a little nicer looking, sturdier,
more compact, by avoiding bulbs. It may also be easer to
select different current going another route (depending on
the control method).
Yes if he has some, but if not, the project budget is
getting pretty tight as it is.
You can still find 6V lamps if you like to "see the light" on the 5V
rail.
Google for "automobile lamps" to find amperage specs. The "inrush
current" of ~1A taillight is ideal for simulating the 12V start-up of a
hard drive motor. (Don't look for the 12V rail droop with a DVM, in
spite of what w_tom says!) ;-)
I'd think the inrush current would be better than nothing,
but not long enough duration (at an elevated level) to be a
good simulation.
R
Rod Speed
I meant your specific reference to the 12V rail.
There arent too many 100W class resistors readily buyable.
Dunno, most obviously with surplus bulbs from car wreckers.
Lower, actually.
Dont have to cost much tho if you get them from a car wrecker.
Only if you use sealed beam headlights.
Cant see that its going to be any bigger than those
TO220 resistors if you dont use sealed beam headlights.
You must have noticed that headlights have sockets for the bulbs.
Thats what a car wrecker is.
You dont pay anything like that at a car wrecker.
Pity about the car wreckers.
Nope, just get a box full from the wreckers.
You can obviously get them from wreckers.
Wreckers dont charge much.
There arent too many 100W class resistors readily buyable.
Dunno, most obviously with surplus bulbs from car wreckers.
Lower, actually.
Dont have to cost much tho if you get them from a car wrecker.
Only if you use sealed beam headlights.
Cant see that its going to be any bigger than those
TO220 resistors if you dont use sealed beam headlights.
You must have noticed that headlights have sockets for the bulbs.
Thats what a car wrecker is.
You dont pay anything like that at a car wrecker.
Pity about the car wreckers.
Nope, just get a box full from the wreckers.
You can obviously get them from wreckers.
Wreckers dont charge much.
S
Skeleton Man
Okay.. time to throw in some more details I think..
The 750W PSU I will likely be testing is this one:
http://www.silverstone-usa.com/products-75zf.htm
If the company decides to send me a 650W model instead, it will likely be
this one:
http://www.silverstone-usa.com/products-65zf.htm
I just need a way to load either of those supply's close to their maximum
ouput rating.. 600W is probably sufficient though.. (80% of 750W)
Budget is the number one consideration.. I want to building something to
generate a suitable load, but I don't want to spend a fortune.. If I have to
spend like $50 to build something suitable then so be it.. as long as I can
re-use it..
I don't mind at all how I accomplish the end result.. resistors just seemed
'the norm' for this kinda thing.. if I can use headlamps from a wreckers or
resistance wire then so be it.. whatever is cheapest and least complex..
Bottom line: I will be recieving a brand spanking new power supply for
absolutely nothing, and I don't want to screw up my relationship with the
company by doing a lame review with no real stats.. (ie. idle voltage
readings and reading at 100% cpu aren't gonna cut it all.. I need much more
info and much more load)
Chris
The 750W PSU I will likely be testing is this one:
http://www.silverstone-usa.com/products-75zf.htm
If the company decides to send me a 650W model instead, it will likely be
this one:
http://www.silverstone-usa.com/products-65zf.htm
I just need a way to load either of those supply's close to their maximum
ouput rating.. 600W is probably sufficient though.. (80% of 750W)
Budget is the number one consideration.. I want to building something to
generate a suitable load, but I don't want to spend a fortune.. If I have to
spend like $50 to build something suitable then so be it.. as long as I can
re-use it..
I don't mind at all how I accomplish the end result.. resistors just seemed
'the norm' for this kinda thing.. if I can use headlamps from a wreckers or
resistance wire then so be it.. whatever is cheapest and least complex..
Bottom line: I will be recieving a brand spanking new power supply for
absolutely nothing, and I don't want to screw up my relationship with the
company by doing a lame review with no real stats.. (ie. idle voltage
readings and reading at 100% cpu aren't gonna cut it all.. I need much more
info and much more load)
Chris
K
kony
That you can plug the 110W lamp into 12V and expect close to
110W, but have to recalc for lower wattage on 3.3 or 5V
rails.
How many types does he need? I linked a few, but 100W isn't
needed per se, but high wattage per $ is, if it is to be
within the budget. That's part of why I linked that surplus
'site resistor because it is not only inexpensive (if he has
a way to 'sink it) but also a smaller contraption to built
which unless this is only a one-time test, means less costly
wood/metal/whatever enclosure to hold all these parts. If
it's going to be built, it might as well remain a whole
modular device rather than just a string of bulbs spread out
across a workbench. That makes storage, transportation, and
setting it up again much easier too.
Yes, that'd be a good way to cut costs if he is inclined to
go that route.
Yes I noted my error in a followup post.
Maybe, he'll have to try it and see. I don't know that
they'll be gathering up old light bulbs for resale though,
it might take a lot of work to collect bulbs out of cars.
It's not going to be all that small no matter what
headlights you used. "maybe" but it all depends on the
source, and the cost, other parts, etc... the whole design.
If you say so. You must only be referring to the 10 ohm
resistors though, right? There were 100W in TO247 package
too. Further he may be able to create a load but it'll be
fiddly to hit any exact load with a random assortment of
bulbs... he might end up using resistors anyway to get that
last few % accuracy.
yep, all this is still a lot more work. Maybe easier, maybe
not.
Maybe, he'll have to call a few up and see. Some people
might be fortunate to have an accomdating junkyard nearby
and others might not.
Probably not, but there's still a lot of time involved, and
not knowing what that wrecker has, we can't draw a
conclusion yet on what support parts he'd need.
You figure those guys are going around collecting bulbs for
resale? I kinda doubt it. They'll be in cars still and if
he wants to spend an evening collecting them, so be it... or
if he had to pay the cost of having someone else go around
collecting them (for insurance reasons of no other), the
cost goes up some. Probably still cheaper than new bulbs
but it ends up being a lot of time and BS to go through just
to save $10 for some resistors or whatever.
I guess we have completely different ideas about the best
way to tackle a project.
S
Skeleton Man
www.xbitlabs.com has a detailed description of its PSU testing
Where does it have all this info ? I can't find two words about their test
rig !
Chris
Where does it have all this info ? I can't find two words about their test
rig !
Chris
K
kony
Well you probably already have a hint about the direction
I'd go... I'd use TO220 or TO247 resistors arrays. I'd use
a circuit board and mount PSU connectors on it (though come
to think of it I don't have any 4-pin (12V) connectors like
motherboard use, or at least I don't know where they'd be,
but probably Digikey would have them or you might be able to
get samples from Molex/etc).
You wouldn't necessarily need a circuit board for anything,
but, I'd think you would at least want the whole *whatever*
mounted somehow whether it be on a circuit board, on a piece
of plywood or other wood planking or whatever.
I've built quite a few one-off projects and can tell you
that the incidental expenses add up quick, depending on what
parts you have around already. I tend to buy not only what
I need for a project but whatever else I might need someday
if/when the prices are good, as it can cut down costs in the
long run not to have to mail-order a 15 cent resistor or
whatever to finish something. If you already had the wire,
screws/bolts/nuts/etc, it helps towards your budget but if
not there are lots of incidentals that can end up costing
near what a bunch of resistors would. Incidentally here's
another example of some potentially useful resistors from a
surplus 'site,
http://www.goldmine-elec-products.com/prodinfo.asp?number=G14510
With those, unlike the TO220 parts I'd linked previously, if
you give them a little margin you don't have to try to
heatsink them. For example on the 12V rail with paralleled
series of those you'd have each series as 4 of them, so 3A
through that series is 36W / 4 resistors = 9W each.
Since you have 18A rails, that's 6 paralleled groups of 4 in
series per each 12V rail. The entire PSU has 60A 12V limit
though instead of 72A, so the total cost to load all 12V
rails is 80 resistors, $8. Personally I always buy some
spares of parts *just_in_case* something went wrong but
that's up to you.
For the 5V rail, it's only 30A and that takes up most of the
current available for the combined 3V +5V rating too (180W
total combined rating), so for a thorough test you'd have to
unload the 5V rail some so you can then have 24A on 3.3V
(almost 80W) and only 20A on the 5V to pull that addt'l
100W.
So with 1 ohm resistors on 5V rail, paralleled series of 2
in each series, 2.5A per series and 6.25W per resistor.
You'd need 12 paralleled series of 2, 24 resistors is $2.40
or $3 in groups of 10/$1.
Same math applies for 3.3V rail, 1 resistor (no series
needed since 3.3V/1= 3.3, * 3.3 = 11W, still enough margin
on one 20W resistor. So for 24A, 7 resistors in parallel.
So you'd need 111 resistors or $12 worth. The extra 9 might
come in handy to load the other lower current rails or even
buy another 10 or 20 more while you're at it (I'm too lazy
to do the math as to how many you'd need for the other low
current rails but with so many USB devices these days, I'd
at least be sure to test the full rated 5VSB current.
That's a lot of resistors of course, but it is only $12
total cost (check my math though, I might have easily made
an error) and while the number seems awefully high, it also
allows a finer granularity for testing not only this PSU but
any others you want to put a different load on (which would
be practically any PSU, since most are not 750W).
Light bulbs would work though, I just won't go that way
myself. I have oddles of leftover heatsinks though,
aluminum sheeting, a metal brake, etc, so I can easier
forsee what I'd do... if you feel more comfortable mounting
na bunch of light bulbs on a piece of plywood, then explore
that angle more. If you were going to be posting a picture
of your PSU loading setup, I'd tend to think going with
resistors or another electronics-parts method would look a
little more professional than a bunch of lights, but
inevitably no matter what you do, *somebody* will be finding
fault with an article while the majority will instead
welcome the information you provide whether it is, or isn't,
as thorough as it could be- some information always beats
_no_ information, so long as the wrong conclusions are drawn
prematurely. For example, running a PSU at 600W load for a
few hours is a better method of indicating a problem rather
than declaring the PSU is suited for long term use at that
wattage. Even a poor generic may have sufficient output
into a resistive load for a day but in an actual system
might fail in a few weeks time. I don't suppose you really
want to have a 750W project running for months though, so
back to that part about any info being better than none.
I tried to download that PSU spec sheet but it seems
unavailable or perhaps the link is mangled, they didn't seem
to spell "download" right on the link but the correct
spelling didn't work either. Anyway, with all the
independant 12V rails it would be good to make sure they're
up to spec. Likely they're all essentially duplicates of
each other internally so it may not matter as much that all
are simultaneously tested for 18A (indeed, they also mention
the 60A max for 12V so it couldn't apply), but putting at
least 50-60 amps of load on the rail would be a useful test,
as I doubt anyone would want to pay a premium for the PSU if
it doesn't actually perform better than a lower cost,
550-or-so watt alternative.
L
larry moe 'n curly
Skeleton said:
I searched their website for "power supply" and found this article from
Jan 16, 2005:
www.xbitlabs.com/articles/other/display/psu-methodology.html
Here's the page where the test hardware is shown:
www.xbitlabs.com/articles/other/display/psu-methodology_11.html
P
Paul
"Skeleton Man" said:
750W +3.3V +5V +12V1 +12V2 +12V3 +12V4 +5VSB -12V -5V
Max.(Amps) 24A 30A 18A 18A 18A 18A 3.0A 0.5A -
Min.(Amps) 1.5A 1.0A 0.8A 0.8A 0.5A 0.5A 0.1 0 -
combined +3.3, +5V 180W
combined +12V 60A/720W
There are (3) 3.3V and (4) 5V wires on a 20 pin ATX power connector.
There are (4) 3.3V and (5) 5V wires on a 24 pin ATX power connector.
You are allowed 6 amps per pin. I would aim for 5 amps per pin
for a small safety factor. If I load up the 3.3V safely, then use
the remainder of the combined limit for the 5V load, that gives
3.3V@20A (on four wires) and 5V@22A (five wires, 4.5A ea) for
a total power of 66+110=176W (slightly less than 180W).
Use one ground wire per power wire, as the ground pins will also
have the same 6 amp limit. I.e. Connect all the resistors being
used in parallel, to four red wires and four black wires, so
that no Molex pin is overstressed. Using one black wire will burn
the pin on that wire for sure.
If the total load is to be 600W, then we put 420W on the 12V
outputs. The spec above allows 15A per output, but we only need
8.75A each, to reach 420W (4*12V*8.75A).
The unit has minimum load requirements, so all outputs should
have enough load to meet that minimum. When the minimums are
met, then the voltages should read correctly.
650W +3.3V +5V +12V1 +12V2 +12V3 +12V4 +5VSB -12V -5V
Max.(Amps) 33A 24A 13A 18A 16A 8A 2.0A 0.5A -
Min.(Amps) 1.5A 5.0A 1.5A 1.5A 1.5A 1.5A 0.1 0 -
combined +3.3, +5V 170W
combined +12V 42A/504W
(7) When the combined current for the +12V outputs is 30A to 38A,
the +5V minimum load is 10A.
(8) When the combined current for the +12V outputs is 38A to 42A,
the +5V minimum load is 15A.
(10) The total peak load is 710W for 10 second.
Notes 7,8, and 10 come from the downloadable manual (228KB).
I happened to download this previously. Link appears bad now.
Contact the lamers at Silverstone and get the manual. It will
help you. Note that the manual for the 650W, details which
pins have the 12V1, 12V2, 12V3, and 12V4. You cannot guess
at that, and must use the docs.
Loading this thing will be a bit more of a pain. The 12V outputs
should be loaded proportionally, since the outputs do not
have equal capabilities.
For a 170W load on the lower voltage rails, we could use
3.3V@20A (on four wires) and 5V@20A (five wires, 4A each).
The 20A load on the 5V rail, means we meet the minimum in notes
(7) and (8). On a real computer, it may be difficult to meet
those minimum numbers, a bit silly really. For example, an FX60
and a couple 1800XTX might use 30A from 12V while gaming, but
what is going to draw 10A from the 5V rail ? Hard drives typically
only use 1A a piece for the controller board, and there may not be
a lot of other 5V loads on the motherboard.
If we aim for a 500W total load, and draw 170W via the lower
rails, that leaves 330W from 12V. That is 27.5A. The four outputs
total 13+18+16+8=55A, and proportions are 13/55, 18/55, 16/55, 8/55
or 23.6%, 32.7%, 29%, 14.5%. Taking those percentages of 27.5A gives
6.49A, 8.99A, 7.98A, 3.99A, for a total of 27.45A.
To be able to test either power supply, we settle on the same load
for both supplies on the 3.3V and 5V rails. That means the
3.3V@20A (on four wires) and 5V@20A (five wires, 4A each) applies.
http://www.mouser.com/catalog/626/487.pdf (lower right, axial mount)
(6) 1 ohm 25W in parallel, across 3.3V (~11W power each) draws 19.8A
(8) 2 ohm 25W in parallel, across 5V (~12.5W power each) draws 20A
For the 12V, the nicest quanta is 1 ampere loads. A 12 ohm resistor
on 12V will draw 1 ampere, and have 12W power dissipation. We are
using half of the allowed power rating of 25W, to try to keep the
surface temperature of the resistor down (needs the forced air).
(7) 12 ohm 25W in parallel, across 12V1 (~12W power each)
(9) 12 ohm 25W in parallel, across 12V2 (~12W power each)
(8) 12 ohm 25W in parallel, across 12V3 (~12W power each)
(4) 12 ohm 25W in parallel, across 12V4 (~12W power each)
Throw (2) 12 ohm resistors in parallel across +5VSB and GND
(0.83A load).
Put a 100 ohm resistor across -12V and GND, for 0.12A load,
which is well less than the 0.5A limit. Since the power is
V**2/R = 144/100 = 1.4W, even a 5W should do for that one.
Or don't bother loading the output at all.
Total, roughly speaking, is (6) 1 ohm for 6*$1.09 = 6.54
(8) 2 ohm for 8*$1.09 = 8.72
(30) 12 ohm for 30*$0.99 = 29.70
total = 44.96
The nice thing about using a large number of resistors like
this, is you can use more or fewer in parallel, to get other
load values. For example, a total of (60) 12 ohm resistors
would draw the 60A max listed for the 750W supply, so buying
a few more 12 ohm resistors prepares you for a range of
measurements. 60 resistors costs you roughly 60 bucks, plus
the solid copper house wire and solder to assemble the loads.
(1 ohm is out of stock. Use twice as many 2 ohm resistors instead.
Brings order up to about $50.) These ceramic style, cement filled
resistors are the same style used in my load box.
http://www.mouser.com/index.cfm?han...uctid=802530&e_categoryid=358&e_pcodeid=02825
(2 ohm)
http://www.mouser.com/index.cfm?han...uctid=763167&e_categoryid=358&e_pcodeid=02825
(10 ohm)
http://www.mouser.com/index.cfm?han...uctid=264866&e_categoryid=358&e_pcodeid=02825
Other parts needed, may be in your spare parts box:
(1) 80mm fan, preferably above 35CFM (connect across a 12V output)
arrange resistors in a "wind tunnel", if possible. This
project is mostly mechanical, in the sense that you have
to come up with a cheap way of mounting and cooling this
stuff.
14 gauge solid copper wire (house wiring) at hardware store
this is selected for its rigidity, as a mechanical framework,
plus you can reflow solder to it, with a high wattage soldering
iron or solder gun (the old Weller guns will do, just remember
they have a <20% duty cycle and burn out easily on a job like
this). I get my wire at Canadian Tire, from the rack.
soldering iron (I have an 80W iron I use for this - the best I can
find at Radio Shack, is the old style soldering gun...
http://www.radioshack.com/product/index.jsp?productId=2062752
(Device must be allowed to cool off between making connections!
I know, because I burned mine out. Solder for one minute,
rest for five minutes, or as directed in the manual. A
soldering iron has no limit like that, and can be left
powered and hot while you work. But big irons, like my 80W
are harder to find.)
solder (from electronics store, about $5 worth)
Assorted extension cables for ATX PSUs (this is how you
access the output). Snip off one end, connect resistors to
the bundle of wires as appropriate. Finding the extension cables
is actually the hardest part of your project, so I'll let
you track those down
That should be enough to get you started. While the resistors have
an obvious cost of $50, the sundries will easily eat up another
$50. You cannot do electronics without tools. And no matter
which soldering iron or soldering gun you get, it is basically
a throwaway, as other electronics is normally soldered with
more delicate 25-35W irons and tips.
Light bulbs as loads are not recommended, as they draw twice
the current when they are cold, as they do when they are hot.
If you aimed for a 10A load when the bulbs are hot, then they
would draw 20A at the instant you first turn on the supply,
and a supply equipped with overcurrent protection would likely
shut down if your cold current load is too much.
I already mentioned a contact-free method of measuring current
in my other posting. The one in the lower right corner
does both AC and DC currents, without any connection to the
circuit. When your review site is the size of Anandtech, you
can buy one of these.
http://www.extechproducts.com/products/extech/380941_942_947.pdf
HTH,
Paul
W
w_tom
1) Article cited by larry moe 'n curly provides useful background
information such as how voltage may vary both with load (wattage) on
this voltage as load (wattage) changes on other voltages. That is a
static load test where voltage is measured and compared to Intel ATX
specs. Also pictured is a dynamic load tester. Power supplies must
not maintain voltage just for a static load. Supply must also respond
fast enough for dynamic loading. A meter excellent for static testing
will not be sufficient for dynamic testing.
However your description only asks for static testing under full
load.
2) Motor load inside a computer is all but nil. Motors have a large
startup current. Power supplies (circuit breakers, fuses, etc) are
specially designed for this startup current. But computer supplies are
not because that large startup current does not exist. Some mistakenly
think a disc drive is a large load. Those motors draw so little as to
make startup current trivial.
However, if loading a power supply 12 volts with maybe 200 watt
lights (16 amp load) may first demand more than 100 amps (for a short
period). This might put a power supply into current foldback limiting
- a protective lockout function. IOW a power supply would fail test
even though that supply is perfectly OK. A partial load created by
incandescent lights would be better if part or majority of that load is
resistors or transistors; to create a smaller startup current.
3) BTW, if paralleling transistors (as shown in the picture), then
use FETs (not bipolar) transistors for reasons only provided if
relevant.
4) Paul makes an important point about maximum current per Molex pin.
Too much current through one pin can create excessive heating and
connector damage. What he has posted is informative - especially the
numbers.
5) How to measure load of more than 10 amps with a meter only rated
at ten amps: put a 14 AWG solid copper wire in series between power
supply connectors (all pins brought to a common point) and the load.
This is a current shunt. If that wire is about 5 feet long (ballpark
numbers from memory), then the DC voltage between both ends of that
wire might be 0.2 volts with a 16 amp current draw. Use a long wire so
that wire temperature changes are minimal. Better accuracy might be
obtained with 8 feet of 12 AWG wire.
Calibrate the current 'shunt' by loading 8 amps. The wire 'shunt'
might measure 0.1 volts. Therefore 1) no need to break and make
connection for measuring current (no additional connectors required),
and 2) your meter is not at risk for too much current.
6) Voltage using a volt meter is not only a great tool for testing a
power supply under full load (static load). Same meter is effective at
identifying a defective power supply when power supply remains inside a
computer - fastest method to confirm power supply integrity and without
disconnecting anything. Excessive AC ripple would also cause bad DC
voltage on meter - but only when power supply is loaded either by
computer or test fixture. IOW detect computer failures before those
failures start intermittently crashing the computer later.
information such as how voltage may vary both with load (wattage) on
this voltage as load (wattage) changes on other voltages. That is a
static load test where voltage is measured and compared to Intel ATX
specs. Also pictured is a dynamic load tester. Power supplies must
not maintain voltage just for a static load. Supply must also respond
fast enough for dynamic loading. A meter excellent for static testing
will not be sufficient for dynamic testing.
However your description only asks for static testing under full
load.
2) Motor load inside a computer is all but nil. Motors have a large
startup current. Power supplies (circuit breakers, fuses, etc) are
specially designed for this startup current. But computer supplies are
not because that large startup current does not exist. Some mistakenly
think a disc drive is a large load. Those motors draw so little as to
make startup current trivial.
However, if loading a power supply 12 volts with maybe 200 watt
lights (16 amp load) may first demand more than 100 amps (for a short
period). This might put a power supply into current foldback limiting
- a protective lockout function. IOW a power supply would fail test
even though that supply is perfectly OK. A partial load created by
incandescent lights would be better if part or majority of that load is
resistors or transistors; to create a smaller startup current.
3) BTW, if paralleling transistors (as shown in the picture), then
use FETs (not bipolar) transistors for reasons only provided if
relevant.
4) Paul makes an important point about maximum current per Molex pin.
Too much current through one pin can create excessive heating and
connector damage. What he has posted is informative - especially the
numbers.
5) How to measure load of more than 10 amps with a meter only rated
at ten amps: put a 14 AWG solid copper wire in series between power
supply connectors (all pins brought to a common point) and the load.
This is a current shunt. If that wire is about 5 feet long (ballpark
numbers from memory), then the DC voltage between both ends of that
wire might be 0.2 volts with a 16 amp current draw. Use a long wire so
that wire temperature changes are minimal. Better accuracy might be
obtained with 8 feet of 12 AWG wire.
Calibrate the current 'shunt' by loading 8 amps. The wire 'shunt'
might measure 0.1 volts. Therefore 1) no need to break and make
connection for measuring current (no additional connectors required),
and 2) your meter is not at risk for too much current.
6) Voltage using a volt meter is not only a great tool for testing a
power supply under full load (static load). Same meter is effective at
identifying a defective power supply when power supply remains inside a
computer - fastest method to confirm power supply integrity and without
disconnecting anything. Excessive AC ripple would also cause bad DC
voltage on meter - but only when power supply is loaded either by
computer or test fixture. IOW detect computer failures before those
failures start intermittently crashing the computer later.
S
saddam
Paul said:
M
meow2222
Skeleton said:
Forget using filament lamps. They eat in the region of 10x rated
current on start up, and would make your test results junk. And maybe
open you up to legal charges, depending on where you live.
A bit of sheet wood & metal with 3 screws and 2 bolts would do it, plus
resistance wire.
In fact you could likely do it with nothing more than 2 metal
rods/plates and a bucket of water. Water conducts some at 12v after
all. Or one rod and a metal bucket. Fully variable.
NT