Hi, all this is my first post. I searched and could not find the answer to my question. So, here goes.

I have made a my own powersupply from an ATX computer supply. It is a 600W Coolmax, now it has 3 12V power outputs two that are rated at 18amps and one rated at 15amps. So, I have 3 12V outputs that I can connect to. Now when I apply a load to any one of the three the voltage on the other 2 drop to the same level.

And with a 5amp load the voltage drops to 11.5v with a 15amp load (spread across to outputs) it drops to 10.6v. I would like to know how one would be able to stablize the voltage to stay around 12.5v or higher under a 15 to 20amp load.

Here is a link to the powersupply that I have converted.
http://www.coolmaxusa.com/productDet...ategory=single

Thanks for any help.

 

You don't, the regulation is based on either the 3.3V or 5V rails - although you could probably redesign it to use the 12V rails instead.

All three rails drop because they are probably all the same rail, just joined together.

__________________
PIC programmer software, and PIC Tutorials at:
http://www.winpicprog.co.uk

 

Ok, so if they are all joined together would that mean that I would be able to pull 51amps total????

How would I be able to make it regulate off of the 12V side insted of the 3.3V/5.5V rail???

Thanks,

 

No, they are usually already joined together inside the supply.

Disconnect the feedback from the existing rail used, change the resistor values accordingly, and reconnect it to the 12V rail.

__________________
PIC programmer software, and PIC Tutorials at:
http://www.winpicprog.co.uk

 

OK, I understand that and yes they are all soldered at the same point on the PCB. But where do they get the amps for each leg???



How do I determine which rail is being used 3.3v, 5.0v or 12v??? And then what is a good way to determine the correct resistor values???

Thanks for the input Nigel.

 

The thickness of each wire, that's all.

You really need a circuit diagram in order to determine it.

__________________
PIC programmer software, and PIC Tutorials at:
http://www.winpicprog.co.uk

 

Well, if I can get a diagram from Coolmax I should be in business. Ok, I have another question while I am waiting on a diagram.

Now, when I wired up the 12V leads I wired them to individual fuses. Is this good or bad??? I did not put the fuse on the ground wire due to the fact that the out put specs were as follows

AC INPUT
100~132VAC or 200~264VAC

MODEL DC OUPUT +3.3V(24a) +5V(24a) +12V1(18a) +12V2(18a) +12V3(15a) -12V(0.5a) +5VSB(2.5a)


(600W) Max. Combined 160W 450W

By doing it this way I figured it would seperate the load between the outputs, but with them all coming from the same location this might be bad. I have seen some people install a 1amp fuse on the ground probe but, they only had one ground that they were using insted of 3, granted all of the grounds come out of the same location also. So, should I just remove the 3 fuse or just leave them???

 

leave the fuse they protect the wiring from power supply

 

Would increasing the load on the 3.3V or 5V rails cause it to put out more voltage to the 12V rail???

Is so that would be easy to solve with a couple resistors.

 

The easy way to use the supply is to put resistors on the 3.3 and 5 volt outputs. Figure 80% of the rating. .1375 ohms about 80 watts for the 3.3. v and .2 ohms about 120 watts for the 5 volt. The supply will now hold up the 12 volt outputs when you load them You can experiment by reducing the 3.3 and 5 volt loads to reduce power dissipation finding out where the minimum amps regulation point is.

 

Ok, I already have a 10ohm 50watt resistor on the 5V rail so the unit will power up when I turn on the powersupply (switch I installed). Do I need to put one on the 3.3V supply??? Where can I get that large of a resistor that is small enough to fit inside the unit???

 

it would be best to have the resistors out side the power supply. You can run a fan on them to keep them cool. A small 12 VDC fan would work. The 10 ohm resistor on the 5 volt line only draws 500 ma 5V/10 ohms = .5 amp this is not enough load to hold up the 12 volt line. We are talking .13 and .2 ohms. You might not need to load the 5 and 3.3 at the same time. See which one holds up the 12 v line.

 

Sorry I am a little confused. .13 and .2 ohms???? Do I need to install that small of resistor with more wattage??? Or do I need to install a higher ohm resistor and see which will hold the 12v line???

Sorry total newb here


Oh, what about installing a reostat or a potentiometer and just turn it up or down to see which stabilizes the voltage.

 

Sorry, I read up on Ohm's law and understand what you are talking about, but I am having trouble on finding resistor that low and with that high of wattage.

Any suggestions???

 

Generally you would have to make them yourself.

However, bear in mind doing this means your 12V outputs still aren't regulated, and will be too high with a small load.

__________________
PIC programmer software, and PIC Tutorials at:
http://www.winpicprog.co.uk