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I'm trying to make a circuit that has a 12 volt battery backup that picks up if the 15 volt power supply fails, while also charging the battery while the power is on. I know I need a diode across the battery terminals which is in series with a resistor from the positive terminal of the battery to the positive terminal of the power supply. What would the proper resistor values (resistance and wattage) be?
And got the following answer:
The link below shows a battery backup circuit that gives a good idea of the basics for a 12V lead acid battery. This circuit does not provide a well regulated output. When there is voltage from the supply it charges the battery through R1 which limits the charge current. The diodes steer the current in the correct paths when there is power or only battery. They are essential. The voltage drop across a diode is like 0.7V, but can be up to 1 volt or more in some situations. These diodes are all rated for the currents involved.This diode voltage is reasonably constant. With a 12V battery, the maximum charge voltage may be 14V at the battery terminals, and an additional 0.7V is dropped across any diode so the supply must be at least 15V to fully charge the battery. The resistor is determined as: R = V/I where I is the maximum charge current, usually 1/10 of the battery capacity in Ah. V is the voltage across the resistor, which is Vsupply - Vdiode - Vbatty. The Vbatty in this case is for a flat battery, so 10-11V. The power in watts is V * I, using the same values, but note that at the rated power, resistors get very hot. A 12V tail lamp rated at 1A may also suit a 100Ah battery. If the battery is charged at 1/10 of the capacity it will overcharge, so an automatic cut off is needed when it reaches the correct voltage of 14V maximum. The second link shows one way to do this (different battery). A compromise to keep it simple is a resistor that gives 1/100 charge current. This is called a trickle charge, so the battery takes a long time to charge from empty (140 hours) but will not overcharge. This works well if the battery is rarely discharged more than a few percent. It might seem a good idea to add a regulator to the 12V output circuit. This output voltage is the input for the regulator. It will be between 10V with a flat battery and perhaps 15V on the mains supply. A 3 terminal regulator normally needs the input to be at least 2.5V more than its regulated output, so the regulator output would be 8V or less depending on the regulator dropout voltage. The regulator also provides protection for the diodes by limiting current. The diodes could be easily damaged by a short circuit or even charging a large capacitor. In some cases a 12V filament lamp in series with the load will limit the current (loads much less than the lamp current). I consider the regulator important. There also needs to be fuses with the battery and elsewhere, as in the article. As you can see it takes quite a lot of extra voltage at the supply to get to the well regulated output that will work till the battery is fully discharged.