Modding PSU to produce lower voltage

I am currently trying to mod a computer PSU to produce a lower voltage (~7V on the 12V line) for experimentation (nothing to do with computers).

I've had some success so far. I was able to get the voltage down to about 9.5V by replacing feedback resistors, but for some reason, anything lower than that, the PSU won't work at all.

My power supply is a Corsair Builder series CX430, with the CM6800 PFC/PWM combo controller.

Here is the datasheet: http://www.champion-micro.com/datasheet/Analog Device/CM6800.pdf

I traced the PCB, and realized that the circuit is very similar to the sample application circuit on page 15, so that made my life quite a bit easier.

The circuit looks pretty simple - the 12V rail is maintained by the resistive divider R45 and R48, and U1 maintains 2.5V in the middle of the divider through negative feedback through the optocoupler (turns off the PWM circuitry when voltage goes higher than 12V).

So to get a lower voltage, I tried increasing the value of R48. That worked, somewhat. I slowly increased the value, and got the voltage on the 12V rail to drop to 9.5V. But like I said, for some reason, if I try to go any lower, the circuit just quits. All outputs go to 0, and the primary cap goes to about 150V (instead of the normal 380V), suggesting that the PFC circuit has quit.

So I don't really know what's happening. Power circuitry is not exactly my specialty...

I thought maybe the chip supply voltage may have dropped proportionally with 12V rail (the sample application suggests that the chip supply is on the same flyback coil as the 12V rail) and some kind of low voltage protection kicked in and disabled the chip. But at 9.5V, the chip supply is still 15V (the recommended operating voltage), so it doesn't look like it's that.

It would be much appreciated if someone can shed some light on this.

Thanks!

so let me get this straight you opened up an electronic componenet thats sole purpose is to store and output 10s of amps of power, yes?

Man i hope your wearing those long thick orange rubber gloves they issue fire departments to go grab a downed powerline (not that any firefighter is stupid enough to try that)

if you stop posting well know why

Thanks for the concern.

I did forget the disclaimer - Do not try this at home unless you are properly trained! It can kill you.

I have taken many many precautions. I don't want to list them here in case that encourages someone to try it - but trust me, I didn't just decide to open it up for fun and start licking it . I am a 4th year electrical engineering student did many hours of relevant reading and analyzing the circuit before I felt safe enough to plug it in for the first time.

Since most power supplies have built in regulation .You might want to look in to shunting. Ok, that is what your doing, or trying... The big thing you need to look into is, besting the regulation circuitry. If you overvolt/undervolt. The protective circuit should by design, drop to null. So beat the shorting circuit and you will be able to mix the +5v/+12v to get your end 7v. (Making a short, via bypassing will drop the protection to zero.)

Your going to have to scope where the drop is and then bypass that. With no regulation.

That must be it!!

There is a chip on the secondary side that monitors the rails, and outputs a fault protection signal that goes through an optocoupler to the PFC/PWM chip. That must have disabled the chip.

I will just have to cut the pin, and short it to the convenient ground pin right beside it.

I will continue working on this tomorrow after some good sleep and let you know what happens (I'm getting a little tired, and tiredness and high voltage circuits don't mix).

By the way, the datasheet for the power monitoring chip is extremely hard to find, because it's a rebranded chip. Google search of the chip part number returned nothing. I scoped all pins and recorded all the voltages, then went through a long list of power monitoring ICs with the same number of pins, and compared the expected voltages. I finally found a matching chip.

For future Googler's benefit: The power monitoring chip on the CX430 (SITRONIX ST9S429-PG14), is a rebranded Unisonic S3515, datasheet here - http://www.unisonic.com.tw/english/datasheet/S3515.pdf.

Are you going to give us a clue why you are doing this??

Success! I shorted fault protection output (it's open-drain) to ground, and was able to get 8.5V!

Thanks for the hint!

One unfortunate consequence, though - the fan will now run on 8.5V as well... though I don't think that will be a problem, because I will only load it up to about 220W maximum, and the PSU aws rated for 430W.

Are you going to give us a clue why you are doing this??

Click to expand...

Sure. I am building a quadrocopter that will be powered by lithium polymer batteries. The problem is, it will drain the battery in about 10 minutes, which makes extended testing difficult. So when it's on the ground, I want to have a power supply that powers it from the wall, and industrial high current power supplies are very expensive... whereas PSUs are dirt cheap for what they can do. Hence me trying to modify it to put out matching battery voltage.

If you want 7V, why didn't you just use the 12V - 5V trick? It's possible the 5V rail won't be able to sink that much current, but it's worth a try...

From what I read, the 5V output won't be able to sink any current at all. Which means, if I want to do that, I'll have to put a constant 30A load on 5V. Besides being very inefficient (wasting constant 150W), I don't think the 5V output is even rated for 30A (with no load on 12V).

cyberfish said:

From what I read, the 5V output won't be able to sink any current at all. Which means, if I want to do that, I'll have to put a constant 30A load on 5V. Besides being very inefficient (wasting constant 150W), I don't think the 5V output is even rated for 30A (with no load on 12V).

Click to expand...

Yeah, that would be pretty sketchy. I wasn't sure how much power you needed.

This is pretty weird.

I got an open circuit voltage of 8.5V, and tried to load it up. With 300mA load (30 ohms power resistor), the voltage RAISED to almost 10V.

Anyone know what might be going on? I thought it would go the other way...

Voltage is still clean. 100mVp-p switching noise. Just higher.

That's a pretty shocking thing you're trying to do there.

It just adds to the buzz.....

^^



Sorry Cyber, I cant answer your question...free BUMP.

cyberfish said:

This is pretty weird.

I got an open circuit voltage of 8.5V, and tried to load it up. With 300mA load (30 ohms power resistor), the voltage RAISED to almost 10V.

Anyone know what might be going on? I thought it would go the other way...

Voltage is still clean. 100mVp-p switching noise. Just higher.

Click to expand...

I warned you there would be no regulation. Might want to look at the undervoltage circuit. It might be trying to compensate for the drop now.

The regulation is still there. Feedback loop is intact.

I only took out the fault protection signal, which shouldn't be needed in normal operation anyways.

I seem to be getting 8.5V again now... regardless of load. Need to investigate more. Maybe I damaged or shorted something when I put it back into the enclosure.

All done!

Here is my log of what I did -
http://cyberfish.wecheer.com/blog/?p=769

For entertainment purposes only of course. DO NOT TRY.