Series resistor for LDO reg calculations - Page 1
Forum section: electronics / projects

author: jmoschetti45
author_rank: Contributor
timestamp: June 05, 2019, 12:57:27 pm
content: I'm using a pair of lithium batteries in series as a backup source for a raspberry pi zero w. Charging them isn't too complicated, (for now) I have a cheap ebay special bms handling them, and am using an MIC29xxx series LDO reg to drop my 9V rail to 8.4V with an LM317 to current limit at 250mA. Doesn't seem to get too hot to worry me there, a small heat sink is doing fine. There is probably a better way to do this, but that's what was readily available in my parts bins. Under normal circumstances, the 5V for the pi comes from a buck converter straight from the input, but on power failure, it'll have to come from the batteries.Now, to get from the 8.4 down to 5v, I'm using another MIC29xxx series regulator. Which gets incredibly hot under any sort of load. From logging on my bench supply, the pi draws about 295mA idling, to about 725mA running at full capacity. My biggest problem with using a series resistor for this is the changing input voltage. It's not going to stay at 8.4V long. The two 14500 cells have 720mAh rated capacity (actual tests show 729 and 723). The bms claims it'll cut out at 2.5-3V/cell, but we'll see what reality brings there. I may be making a custom bms for it need be, I'd like it to cut out above 2.75V/cell. The datasheet for the MIC29xxx includes the recommended formula for adding the series resistor as this:RMAX= VIN MIN- (VOUT MAX+ VDO) / (IOUT PEAK- IGND)This gives me RMAX= 5.5 - (5.1 + 0.2) / (0.725 + 0.015) = 0.27ohmThe LDO reg is dissipating about 2.5w, which of course makes it toasty, but only when the batteries are completely full. When the batteries are near flat, it's only about 0.6w.The resistor isn't doing me much good here.... (0.725 + 0.015)2* 0.27 = 0.15w. Reducing heat in the LDO reg by 0.15 watts isn't even worth the effort.I've thought about alternatives, such as a buck reg, but there isn't enough voltage difference with the batteries low.Also, due to design constraints, I'm limited to 2 batteries (they are molded into the enclosure, I can't change that on my end). 3 would have been simple. It's also crossed my mind to parallel the cells and use a boost converter, but there isn't much room in the enclosure for that, and I see it as extra inefficiency.I'm open to suggestions...

author: SiliconWizard
author_rank: Super Contributor
timestamp: June 06, 2019, 04:19:18 pm
content: First point, I'm assuming you're talking about LiPo batteries? I really recommend against charging them the way you do, and I suggest using a true charger IC. Your way may shorten the batteries' life significantly at best, or could even be unsafe. Barely limiting the current to 250mA is not enough protection IMO, and the constant voltage of 8.4V must be pretty accurate for LiPo batteries, not sure you'll get enough accuracy with your LDO (that said, lower will be safer than higher).As to the rest of your circuit, as you also suggested, I really recommend using a buck converter, otherwise you're not only generating a lot of heat but also wasting a significant part of the batteries' capacity. Not sure about your comment regarding the "voltage difference". A fully discharged LiPo cell will be around 3V, so 6V for two cells in series (which seems your case here). 6V min is more than enough for many decent buck converters to output a steady 5V. And since you'll be wasting much less energy, you may even be able to decrease the capacity (and thus size) of the required batteries quite a bit.