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08-29-2002, 05:41 AM | #1 |
Cooling Neophyte
Join Date: Aug 2002
Location: Netherlands
Posts: 17
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Modding a PSU?
Several times i read on an internet site or even in a topic over here about modding a PSU so it can deliver high amps at +12V.
However nobody speaks about how this is done. Enybody ideas??? even a god link is ok !!
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At the moment Air cooling 7 fans, working towards watercooling |
08-29-2002, 07:02 AM | #2 |
Cooling Neophyte
Join Date: Aug 2002
Location: Netherlands
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sorry, a typo there
certainly i don't need a link to god i ment; a good link is ok
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At the moment Air cooling 7 fans, working towards watercooling |
08-29-2002, 11:20 AM | #3 |
Cooling Savant
Join Date: Jun 2001
Location: Connecticut
Posts: 140
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There are mods to increase the voltage, but I have never heard of a mod to increase amps.
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08-29-2002, 11:32 AM | #4 |
Responsible for 2%
of all the posts here. Join Date: May 2002
Location: Texas, U.S.A.
Posts: 8,302
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You could solder in another regulator, parallel to the existing one, and add a few transistors to the circuit, to allow for more current to be drawn, but I think gmat would be your man here.
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08-30-2002, 02:54 AM | #5 |
Cooling Neophyte
Join Date: Aug 2002
Location: Netherlands
Posts: 17
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Yesterday, i found an supply at work, 8 A at 15 V.
That should do for now, bench tested it with my WC and Pelt. Did get a nice minus 27 degrees at an ambiant of 23 degrees. I remember reading somewhere, that at these temps when you stick it on a AMD XP1800+ at stress load you should come out somewhere around 0 degrees. Would this be right, or am i rambling.? Oh, and i'm still intrested in a mod for an powersupply, for i'm planning a bigger pelt in the near future. And i hate to waste money when i can mod it.
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08-30-2002, 04:38 PM | #6 | |
Pro/Staff
Join Date: Oct 2001
Location: Klamath Falls, OR
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I wanted to reply to this project earlier but forgot to. Oops! Retrofitting a PSU to deliver more current can be done in several ways.
First, you might be able to cool the output transistors more aggressively. Usually, the output MOSFET's are rated in deliverable current based upon a certain heatsink rating, because they start to self-destruct at a certain output level. However, sometimes the FET's have current limiting shutdown protection so that you CAN'T overload them. I think it's more likely, though, it's a overheat thermal protection rather than over-current protection. Secondly, you can simply replace the output MOSFET's with beefier versions. Will this work? Maybe. Driving a heavier load might work this way at the expense of more noise in the voltage domain. Switching power supplies are designed to supply a max amount of current at a given max ripple. The bulky capacitors you see are used to smooth out the voltage, but they can only smooth out so much in between each cycle of supply current being switched in (see theory of how switched PSU's operate). You'd have to upgrade the output capacitors too. But, this might fail to work too! What about the inductors? They are designed for a given workload as well, since they can only store so much energy. Increase the inductor size though, and you might have to fiddle with the timing parameters of the switching feedback/control circuit (to avoid tripping the overcurrent protection for the sudden inrush when you turn it on, and to avoid other more subtle design "bugs"). Get the picture? A switched PSU is a designed SYSTEM, not a rack mount architecture. Weakest link breaks the chain. Now, don't get all pessimistic. I raised a bunch of bugaboos that might or MIGHT NOT happen. The easiest thing to do is to increase the heatsinking capacity on the PSU (or just increase the fan size or speed) and live with more ripple. TEC's don't like ripple, but they can live with it. They just end up not working as efficiently (i.e. reduced heat moving capacity for the given input energy, which means that the target doesn't get as cool as it should). If you are really excited about doing something hard core, just build a linear power supply. You need a beefy set of components, but it's not complicated. Here is a good excerpt from this website: Quote:
AC -> Bridge -> Capacitor(s) -> TEC More if you're interested. |
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09-02-2002, 07:14 AM | #7 |
Cooling Neophyte
Join Date: Aug 2002
Location: Netherlands
Posts: 17
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Wow, lot's of info to chew.
Thanks, i shall look in to it.
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11-12-2002, 02:56 AM | #8 |
Thermophile
Join Date: Aug 2002
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Did you mean like when you are running the 12v line only?, for fans,TEC,ect?...
Cause the 12v line is tied to the 5v line, unless the 5v line is drawing current the 12v lines is 'capped' at the lowest ampage. I think it's a resistor you have to solder on the 5v line to get it to draw current... I read an article or thread about it at coolhardware.co.uk a while back and can't remeber the details ... |
11-12-2002, 12:28 PM | #9 |
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Join Date: Oct 2001
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It's not that the 12VDC line is tied to the 5VDC line, really. What happens is that the PSU has a protection circuit on the 5VDC line. If the 5VDC line has no load, the PSU assumes that it is not connected to the computer and so it shuts down.
It goes back to switching power supply designs. It may not occur in the newer design styles (e.g. current sensing regulation instead of voltage sensing designs), but a zero or near zero load for a switching power supply is dangerous. The switch attempts to switch on and off so quickly that really high voltages are developed from the inductor, and PSU IC's can be damaged by that. The voltage spikes would not escape the PSU filtering capacitors, but the voltage regulation circuitry in the PSU might become damaged. To overcome that, simply put a resistor from +5V to ground. Different PSU's require a different threshold load, so you may be able to get away with a 1W 47Ohm resistor or you may need a 5W 10Ohm resistor. To pick the right resistor, remember Ohm's law (in a couple of variations): Current = Voltage / Resistance. Resistance = Voltage / Current Voltage = Current * Resistance For calculating the resistor size, choose a wattage above your actual power draw, or you will burn out the resistor. The higher the wattage resistor, the bigger and more expensive it will be. However, it will also be longer lasting. Power (Watts) = Voltage * Current For example, a 47Ohm Resistor on the 5VDC line would need to dissipate: Current = 5VDC / 47Ohm = about 0.106 Amps Power = 5VDC * 0.106 Amps = about 0.532 Watts So, pick a resistor of 1W or better. A 1/2W resistor would probably fry itself after a bit. Back to PSU's. If your PSU may require 100 milliamps (0.1 amps) of current on the 5VDC rail, try this: Resistance = 5VDC / 0.100Amps = 50 Ohms Looking at the above example, you can see you would need something like a 1W resistor in this case. If you needed more of a load current to get the PSU running correctly, you would drop the resistance and increase the wattage of the resistor. Remember, none of this dummy load stuff is needed on a normal computer because the hard drives and fans will load up the 5V line much more than 1A. |
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