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Voltage Divider Help

 
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Oct14-07, 04:13 PM   #1
 
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Voltage Divider Help


I blew another 12V (13.5V actually) power supply running my telescope the other night. Primary power comes from a 17 a-h lead-acid battery in a power tank. I run the telescope and a dew heater from it and over the course of a night (10-12 hours), will drain the power tank completely. I keep the charger plugged-in to try to make up some of the difference, but a couple of nights ago, I guess I drained it so far that the charger overloaded and burned-out. It is a 750ma charger. I actually have a couple of issues:

First one way or another, I need a new charger for my power tank (assuming I didn't destroy the battery). From what I understand, "12v" adapters are rarely ever 12V and probably can't be if they are meant for equipment compatible with car batteries, which are actually more like 13.5V. The assumption I'm making here is that virtually any "12v" adapter will be able to charge my power tank. Is this true?

Second, since I don't feel like spending $75 for an adapter that will power everything, I'm thinking of either using a spare computer power supply I have laying around. I think they would work for giving true 12V power to my stuff, but probably wouldn't work as a charger. Correct?

I have a couple of laptop a/c power supplies lying around that I could use. They are 19 or 20V. I understand the concept of a voltage divider, but don't know how to design one or even if it is really a feasible option: does it sound like a good idea and how do I pick the resistors? The voltage divider concept tells you the relative size of the resistors, but not their actual size. Do they get sized based on expected amp draw? Do I need to calculate the resistance of my loads (and what happens when that changes?)? Should I be doing an external voltage divider or is it possible to swap-out a resistor inside the power supply to change the voltage?

Thanks, guys.
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Oct14-07, 06:03 PM   #2
 
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There are two types of walwarts nowadays.
The old nominal 12v type is a transformer with a diode bridge.
Thus peak voltage is around 16 volts.
Enough to charge a lead acid battery.
You can probably get away without paying attention to current limiting as long as you just have the battery on the unit.

The other type is a mini switch power supply. They produce the voltage stated.

Using the computer power supply should work for the scope.
The battery will put out about 12v with a load and no charger on it.
So the voltage is the same. You may need to bridge the 5v output of the computer power supply with a resistor so that it will start. These units require some load to run and I'm thinking that the scope draws current intermittently.

For charging the battery you don't want a divider. You need a current limiter (ok) or voltage regulator (better). A resistor will work as a current limiter, but a real charging circuit is better and is only a few components. If you take the +12 and -5 line from your computer supply you can make a charger out of it.
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