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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".
Ok so i propose to up it to charge a backup battery, sort of a rudimentary UPS . So all i would need to do is design a charge controller and it would work, in theory? Here are the specs for the supply i would use.
I don't know that much about computer power supplies. One concern is that a mostly discharged lead-acid battery requires a current-limited power supply'; both to protect the battery, as well as preventing blowing up the supply. I suspect that a computer supply would blow-up if overloaded.
If you can tweak up the 12V output to ~14.2V, then it could make a charger, provided you can limit the current to no more than about 20A.
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.
Any computer power supply I've ever worked with had a common ground for the 12v and 5v outputs. This means you can't tie them together for a total of 17v.
I will look into a current limiter design. I'm pretty sure i can adjust the voltage with a solar cell charge controller circuit. Thanks for the advise.
A typical computer power supply only puts out 12 volts. To charge a 12 volt battery you will need around 13.8 - 14. 5 volts.
Also on a computer power supply the regulation system reads the 5 volt and 3.3 volt outputs and the rest of the outputs float along with rather poor regulation if a high load is placed on them without one being placed on the rest.
Go spend $25 and buy an actual automotive battery charger and be done with it. You and your battery will be far better off.
Here is what I'm trying to do, to clarify my intentions. Everything runs on 12v in my incubator except the heater. I was wanting to replace my 120v 60 ohm ni-chome wire with a 12v 0.6 ohm Ni-chrome wire to eliminate the 120v inside of the cabinet. The idea to use a battery is for UPS reasons. If there is a power failure by my calculations the incubator would continue running for 8 hours on a deep cycle battery. Before you retort to this message the 240watts that the ni-chrome uses isn't a constant load, it is switched on/off with the thermostat. The battery charging mode would just need to be a float charger really. However if the battery does drain do to a power failure, when the power is returned, i need it to charge back to it's full capacity.
Maybe if i were to just wire in the power supply, alter it to give out 13.5v (my electronics have LM7812's to bring it back down to 12v) and have a current limiter ,let's say 3A, inline from the source to the battery load. It would be sufficient enough to charge the depleted cell over a extended period of time.
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".
Err, modern computer power supplies have moved over primarily to putting power on 12V rails as per the newest ATX power specifications. My power supply, for instance, has 20A @ 12V available on 4 rails. The 5V line has become rather wimpy now. Just wanted to chime in.
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.
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.
Thanks for the input. I am now thinking that I'll just try to crank the 12v up instead of using the 5+ or 5- whichever the case would be. Maybe there's a pot for trimming on it or maybe a 7812 i can swap out. I will find out more when i get the supply that i ordered. If i can't make it work then i'll use it for a bench power supply. Then i'll just build on from scratch assuming that i can find a big enough transformer. I found a schematic on a ham radio that produces 20A @ 13.8v. I will have to modernize the design cause most of the components aren't sold anymore, No big deal. The problem will be the 30A @ 20v transformer that they were using.
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.
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.
Thanks for taking the time to type this out. There is a lot of info here. It sounds the same as the ham radio circuit. If your interested here is the link to the schematic. **broken link removed**
i studied this and traced all the components down to understand the circuit. It looks sound. This is the basis of my design if i can't get the CPS to work. The problem sound like it's going to be the amps I currently use 240watts to power my heater. This will have to 12v for the UPS to work which puts me at 20A that is the limit of the supply. I am hoping that is RMS well see. I need 0.6 A for the 12v fan and 0.6A for the led lighting and the circuitry. This will put me at 12v @ 21.2A minimum. I might be able to steal the 1.2A from the heater and get it down to 20A but that is assuming that i can get 20A out of the CPS. I should hope with the new ATX supply's that they already have a current limiter. I'll chase the traces and build a schematic when i get the supply. I'll post the schematic and post my proposed fix for the supply when i get it. Thanks again for the input.
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