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Is it possible to use a ( computer power supply mod to a lab power supply) to charge a 12v marine battery? I read that one could tie the 5v+ and the 12v+ to get 17v+ and with a 500watt supply could have 20 amps @ 17v.
AFIK, the 12V output from a computer supply is very whimpy (~1 or 2A). Most of the Watts come out the 5V spigot.
Even a 1/2A supply will charge an automotive or marine battery if you are willing to wait long enough. The bigger batteries are ~70Ah, so it would take 140h to charge it with 1/2A. You need a supply that is intrinsically current limited (i.e. it won't catch fire if it is overloaded), that is accurately voltage limited to 14.4V +-0.1V during "charging", and then can be reduced to 13.2V for "floating".
Anyways, to the original post, tying the +12 and +5 on a PC power supply won't do much, I think you have to connect the -5 and the +12 to get 17V (assuming you can find a supply with -5, I think that was one of the casualties to the newer ATX specifications), but then you're limited by the wimpy rating on the -5V line.
Hi there,
As others have mentioned, the +12v rail on older power supplies does not allow too much output current, but then again you can charge with less if you like anyway.
The +12v rails on the modern supplies (also mentioned in previous posts) put out much more, like 14 amps and up. The real problem though is (as mentioned) the voltage is only +12v and you need about +14v to charge a 12v lead acid battery.
Luckily, the regulator circuits inside a typical AT or ATX power supply is not too complicated. They work like most other regulators in that the control chip senses the output voltage, divides that down with two resistors, then compares that resulting voltage with a very stable voltage reference. It tells the pulse widths to vary based on the result of this analog comparison. The nice thing is, since the circuit divides the output voltage down to a much smaller voltage for the comparison all you have to do is change one of the divider resistors and you change the output voltage. For example, increasing the upper resistor to a higher value will mean the output will have to go higher in order to supply the same divider voltage for the comparison, so that's the thing to do. You can also decrease the lower resistor as that will have the same effect. Because you are thinking of changing the output voltage by only about 10 percent it should not harm anything else like the output capacitors. There is one slight problem however, and that is that many of the somewhat newer supplies and even some of the older ones have another circuit that detects an over voltage condition and shuts down the whole supply if the voltage goes too high. The level varies a bit so it's hard to say just what voltage it will take before it shuts down. This means maybe you should try to decrease the lower resistor value by paralleling it with another much higher value (after calculating the final value) until you increase the output to say 13 volts, then use a somewhat smaller resistor to try to get up to 14v. You may even want to use a pot in series with a resistor to set the value and hence the output voltage level.
If in fact the output increases and you end up tripping the over voltage detection circuit, you'll have to modify that too in the same way...by changing the lower resistor to
a slightly less value.
Let's think about some resistor values and see what we end up with...
If your supply happens to use a 9k upper resistor (R1) and a 1k lower resistor (R2) then the reference voltage would have to be 1.2 volts, which means you would have to change the lower value (R2) to approx 844 ohms. To get 844 ohms as the lower resistor without having to unsolder that resistor you could parallel it with another resistor who's value is 5410 ohms (or close to that, erring on the low side if necessary).
The hard part because you probably dont have a schematic is finding what two resistors are used for this function.
An alternate method that might work is if you can follow the +12v sense line back to the circuit, you could cut that trace and insert three diodes in series. That would drop approx 2.1v and so would fool the circuit into thinking that there was really 12v on the output when there was really 14.1v.
This sense line shouldnt be too hard to find because it will be the line that goes from the output to the heart of the circuit, while the other lines would go to the output caps or the output wires. You may actually want to start with one diode and see that the output increased by about 0.7v and then go to two diodes and then finally three, just to make sure the over voltage circuit doesnt trip. IF it does trip, you'll have to find that circuit and modify it in the same way or with a change of lower resistor as above.
Of the two methods i suggested here though i would prefer to change the lower resistors rather than add diodes. Adding diodes does have the ill effect of possibly causing some instability if there are capacitors after the diodes. That would cause the output to go unstable which of course we dont want. The only way to fix this would be to move the caps to the other side of the diodes, and probably add a much smaller cap where the original one was.
If you find one of the ATX power supply schematics on the web you can take a look and find the two resistors for both circuits and that might help to find the resistors in your particular circuit.