In series battery discharge test question: how do buck converters draw current? - Page 1
Forum section: electronics / beginners

author: Barkuti
author_rank: Contributor
timestamp: February 01, 2018, 06:42:23 pm
content: Hello fellows!Received a 6-pack package of Samsung ICR18650-30B cells from an AliExpress seller recently, and I need to do some sort of accurate testing over 'em. No analyzing charger yet, I have aLii-500on the slow boat from china though.Thanks toHKJ's testingI know these cells should deliver ±9.5W of energy when discharged at 0.2A down to 2.8V. For reference graph:The test is fairly old and done on his old station (no 4 terminal holder) but the slow discharge figures should be nicely accurate.I do plan on setting a low voltage light load attached to an XL4015 buck module fed by all 6 cells in series connected by 8×1.5mm neodymium magnets and some copper sheet. My idea is to adjust the buck module for ±4.8W total power and measure the time it takes for its input voltage to drop below 16.8V starting from ±4.2V loaded cells. This means I need to adjust buck module load power to ±4.08W for 85% module efficiency.But, well, as this is it may already be hard enough to attain any meaningful results. Problem is… how is the buck module going to draw power from the cells? Will it draw short bursts (low duty cycle, <20%) of 1+A current? Since I need to toss in the disspated power in the contacts this is of importance, I believe. Thepower factor, or sort of.Thanks and cheers

author: Barkuti
author_rank: Contributor
timestamp: February 01, 2018, 10:41:34 pm
content: Oh well, the high switching frequency plus the decoupling caps, thanks for pointing this outblueskull.Yes, main issue here is to determine if the cells are what they should. Wrapping and markings seem legit, cells' flat tops have 4 wide lobes instead of 3 but I know cells from the big 4 manufacturers are assembled in a variety of genuine plants.With respect to total load power I aim to adjust it in a way that average current draw is close to 0.2A, aiming for ±12 hours of discharge time if all goes well. This is all homecooked, will have to watch out for the input voltage as I don't have a 6S BMS at hand.Thanks and cheers

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 02, 2018, 12:28:05 am
content: FYI, if you're building this from scratch rather than using a battery manager -- battery impedance rises as voltage falls, so there comes a point when you can't maintain output power. This is important for switching supplies, which increase input current as input voltage falls, to maintain a regulated output.That's apparent from the discharge curves, though it may not be obvious or explicit. Simply, the curves drop away that much faster, at higher currents, whether you've been drawing a high current or not. It's a state-of-charge thing.I built a flashlight from scratch, usingdiscretecircuitry, and added a low-input-voltage throttle for this reason. Indeed, in my case it was not only nice to the battery but a required function, as the control circuit is also powered from the single cell, and if the control browns out, the whole circuit is... not very happy!Tim

author: Barkuti
author_rank: Contributor
timestamp: February 02, 2018, 04:10:30 pm
content: Battery impedance rises as voltage falls… yes I knowT3sl4co1l. However, this mostly happens near the end of discharge point/voltage.Speaking of state of charge, my testing setup is now at the 12 hour mark, input voltage is still above 18V. Adjusted output voltage to the (led) load is 3.23V, current 1.26A… 4.0698W / n = 4.677931034483W total load power for efficiency n = 0.87.12h delivered energy: E = 4.677931034483W × 12h = 56.135172413796Wh, looks really good (goal is 57Wh).My observed voltage profile has not matched theaforepostedcurve(s), but the output energy does.Cheers

author: Barkuti
author_rank: Contributor
timestamp: February 02, 2018, 05:03:57 pm
content: 6S input voltage fell below 16.7V (on my +0.5%meter) briefly after hitting 12h 39m on the clock, at such point input current was close to 300mA.Guesstimated delivered energy = 4.677931034483W × 12.65h =59.17582758621WhAs far as my wisdom goes, looks great and perfectly within error margin.Cheers

author: Barkuti
author_rank: Contributor
timestamp: February 07, 2018, 03:16:40 am
content: Quote from: Barkuti on February 02, 2018, 04:10:30 pmAnd I knowwhy. These cells are fake, found 'em destroyed after charging them up to 4.35V and letting them rest for an hour.Related:Seems like I got a six-pack of counterfeit/fake Samsung ICR18650-30B cells :-/Cheers my dears