On Mon, Mar 22, 2004 at 02:40:47PM +0200, Lucian wrote: > I thought of using 2 LM1084 regulators (up to 5A for a maximum of 3A > load). I would prefer a linear power supply, because I think it is > cheaper, Cheaper may not be the issue, efficiency may be the controlling factor. Let's talk linear. Probably OK for the 12V supply, but problematic for the 5V one because presuming that you have a 12V battery, the 5V regulator will have to burn (waste) upwards of 21W of power at the 3A limit you're requesting. There are two possible paths that make sense: 1) Use a 12V lead acid battery which you keep topped off (more on that later). Use a LM1084 because it's low dropout for the 12V supply and use a switcher like the LM2576 or LM2596 for the 5V supply. By switching the supply with the greater voltage differential, you can conserve a lot of power. 2) Use a 6V lead acid battery which you also keep topped off. Use the LM1084 for the 5V supply (which now only will waste 3W or so of power) and use a switching step up (LM2577 or LM2597) to get the 12V supply. BTW you never stated the application. If it's a PC based system, then you may want to consider ditching whatever needs 12V, like using a laptop HD instead of a regular one. If you can get it to a single supply, then you can pick a linear solution that efficient. > but don't know how to connect the back-up battery, in order to > be able to charge it when powered from the wall outlet. Separate problem from the power supply. There's a bunch of info on battery charging around, and a bunch in the archives. But the basic rundown for sealed lead acid (SLA) is a follows: 1) Bulk: charge at constant current of C/4 where C is the Amp-Hour capacity of the battery until the battery reaches a terminal voltage of 2.4V/cell (7.2V for 6V, 14.4V for 12V SLA) 2) Absorption: charge at constant voltage of 2.4V/cell until the battery draws current of less than C/50. 3) Float: charge indefinitly at 2.3V/cell. Parts like the UC3906 (was Unitrode, now with TI) can handle this. Digikey carries the part for $5.50 US. It'll solve all of your charging issues. As a challenge, I'm taking on building my own chargers using a 16F88 and a LM324 quad op-amp to drive a LM317/pass transistor to do all three charge phases. Here's the nickel explanation: The main charge line consists of the LM317/pass transistor connected to the line positive voltage, the battery, and a 1 ohm sense resistor which is connected to ground. The sense circuitry consists of two opamps measure the voltage across the resistor and across the battery terminals (dividing by 4 to reduce the up to 20 V terminal voltage down to 5V). Both of these opamps are connected to the A/D inputs of the 16F88. Control consists of buffering the low pass filtered PWM output of the 16F88 with an opamp (multiplied by 4) and then using that output to drive the ADJ pin of the LM317. So here's how it works: the PIC gets the battery voltage and C from the user. Then charging starts: 1) Bulk: The LM317 is driven upwards until a voltage of C/4 is drawn across the sense resistor. That state is maintained until the terminal voltages reaches 2.4V/cell. 2) Absorption: Simply drive the LM317 so that 2.4V/cell is maintained. Watch the voltage across the sense resistor until it drops to C/50. 3) Float: Set the LM317 to 2.3V/cell and maintain indefinitely. > If it is connected after the 12V regulator, Not after, before. The battery comes first in the circuit. So the line voltage charges the battery, then the battery feeds the regulators. > with a diode, how could I > obtain 5V when the system is running on battery ? The battery feeds both regulators. Which is why one of them will need to be a switcher. > The 5V regulator > should be located in the schematic after the 12V one No. The 5V regulator's input should also be the battery. They are parallel circuits, not series ones. ASCII Art: Line Voltage -> Battery -> 12V regulator -> 12V regulated. | |-----> 5V regulator -> 5V regulated. > and the battery connection point. But in this case, the 12V > regulator should be up to 6A > (3A + 3A), isn't it ? Asked and answered. Both are 3A parallel circuits connected directly to the battery. > I don't know if I made myself >clear, because my english is not so good, It's fine. I took 5 years of French and I couldn't hope to write it as well as you write English. > but I hope someone of you will understand what I tried to explain. Quite clear. > I think > this is am usual problem to deal when using dual voltage back-up power > supplies. The usual problem is that while it's possible to match input and output voltages pretty well for a single linear supply, it's problematic for multiple linear supplies. And since you want battery backup, conserving power is an issue. BAJ > > Lucian -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads