newbie question, early stages, resistors for battery testing - Page 1
Forum section: electronics / beginners

author: carpetpaul2129
author_rank: Contributor
timestamp: February 17, 2022, 08:00:31 pm
content: hi guys, wow, what a subject this is ! I recently completed a project where I dismantled a cordless tyre inflator and replaced its 3 cell 18650 pack with some new cells (on the fifth attempt of soldering and reassemblyreally happy with the result, so saved the faulty cells and started to wonder what else I could repair. My 2015 Dyson cordless vacuum was selected as it keeps cutting out when used on full power, again, disassembled and cells removed, tested for voltage when charged and look fine. Capacity, tested with an Imax B6 charger/discharger at maximum rate of (asked for 1.0amp) but got 0.4amp, was about two third of stated spec. I wondered how else I might test these cells ? there are 6 cells in the Dyson vacuum and it states 21.6 volts and 350 watts using 6 cells ( it was a guess when I divided watts by volts, if indeed it's correct, to get16.2 Amps.Now, I don't have an easy local training course available, but I've been through a lot of the Internet and YouTube (I'm a bit of a slow learner) and yet I'm still struggling to unravel even the basics. I'd like to be able to put each of the potentially faulty old cells from the Dyson through a test to see them fail in the way they do when they're in use in the vacuum, and to be able to measure this on a volt meter and possibly see a different result under a heavy load than I have on my 0.4amp discharge. I have some trouble getting my head around this stuff, I'm very keen to learn, but it's hard when you're not used to it. After countless videos on what Volts, Amps, and Ohms are, I'm listening carefully, but not able to apply it yet. could I make the assumption that each cell takes its portion of 350 watts ? (they're in series of course) so, to test one at a time, it need to see a roughly 58 watt load ? (3.6v x 16.3amp = 58.3 watt) how would I replicate this ? I have currently one 20w / 1ohm resister. I contemplated ordering a couple more 20w/1ohm resistors and experimenting putting them in series in the vague hope that those 20watts would add up. I'm so new to this that I haven't even yet used my multimeter to measure Amps, not even once. I'm keen, not unintelligent, but these concepts are foreign if it's not your thing, any advice ?

author: tunk
author_rank: Super Contributor
timestamp: February 17, 2022, 08:51:28 pm
content: Those 20W is the maximum power dissipation, the actual wattagedepends on the current. If you connect one of them in series witha fully charged cell, then the current is 4.2A (I=U/R=4.2V/1ohm)and wattage 17.6W (W=U*I=4.2*4.2). In order to increase the currentyou have to put the resistors in parallell, not series. If you puttwo in series the resistance is 2ohm and the current 2.1A. Puttingtwo in paralell will give you 0.5ohm, 8.4A and ~35W.Is the resistor one of those with an aluminium casing? In that caseit should be put on a heatsink if you will run the test for more than ahandful of seconds.

author: george.b
author_rank: Frequent Contributor
timestamp: February 17, 2022, 08:57:08 pm
content: Quote from: carpetpaul2129 on February 17, 2022, 08:00:31 pmTo make the assumption that each cell takes its equal portion of 350 W, you have to assume they're identical, which is a reasonable first-order approximation.QuoteAgain, as a first-order approximation, that is reasonable. A better approximation would take the actual (as opposed to nominal) cell voltage into consideration.QuoteFor real world battery testing, you'd use a programmable load. As a first-order approximation, associating resistors is reasonable, with caveats - in practice, voltage isn't constant (depends on battery state of charge, discharge rate, internal resistance) and neither is resistance (for a regular resistor, it normally increases with temperature).Quote3.6 volts over a 1-ohm resistor yields a current of 3.6 A (V=IR). Associating resistors in series will reduce this current. You need an equivalent resistance of 0.22 ohms (3.6/16.3), or, approximately, 4 1-ohm resistors in parallel, which would result in an equivalent resistance of 0.25 ohms (1 ohm divided by 4) and yield a current of 14.4 A at 3.6 V (3.6/0.25), or a power of approximately 52 W. In practice, this is all approximate (see caveats above).Do be careful, though - lithium batteries don't like being stressed. Cells not made for high discharge rates will overheat and possibly fail catastrophically.

author: m3vuv
author_rank: Super Contributor
timestamp: February 17, 2022, 09:00:54 pm
content: I imagine some cells are hi IR and voltage sags under load and the bms cuts the load from the battery,if in doubt replace with new cells rather than piss about with used /old cells!.

author: coppercone2
author_rank: Super Contributor
timestamp: February 17, 2022, 09:02:03 pm
content: that dyson likes to cut off when its clogged, be sure to change the hepa filter on it if it has one..

author: carpetpaul2129
author_rank: Contributor
timestamp: February 17, 2022, 09:46:15 pm
content: thank you for the great replies :Tunk : thanks, the (cheap) resistor that I got is a ceramic one, seemed to survive several minutes, but point taken re heat sink, thanks. Man, I'm so confused as to how to put a load on the cells, I'm sorry to be dumb, this stuff must be obvious to you experienced guys, but wow.... I'm not there yet, lolGeorge B : thanks (and to Tunk) for pointing out that its series rather than parallel, I'm learning slowly, and I did also get close to the 0.22 ohms calculation, although I've totally forgot how I got there nowm3vuv : yep, replacing with new cells, which I already have, but I'm looking to learn about why the faulty cells failed, and also do the same test to those new cells and see how they handle it before assembling back into a (hopefully) working packcoppercone2 : agreed, but already disassembled as far as it will go and cleaned all the insides out, which weren't clogged, I think the cells have had it, I always used on high power and it's 7 years oldall the advice is really great, thank you all so much, and sorry, but I'll probably be back with more daft questionscheersPaul

author: carpetpaul2129
author_rank: Contributor
timestamp: February 17, 2022, 09:56:25 pm
content: lol, next daft question happened already....I watched a GCSE schools YouTube video about adding resistors into a circuit as series vs parallel, the teacher described resisters in series as 'adding up' increasing the resistance, and he showed the resistance value as having doubled, 50 ohm single resistance became 100 ohm total, is this not twice the resistance, twice the work for the cell ? then he showed the same resistors in parallel, describing a separate path for current and a halving of the current to (close to) 25 ohmsI'm simply not clear on the basic concepts, a strong resistor with a high value 'holds up' the current, heats up like crazy and stresses the cell with a certain amount of load (I guess I can work out the wattage knowing the voltage of the cell at that point and the resistance value of the resistor ?)if theres a chance you can give me a couple of examples I can work through, would be most appreciatedmany thanksPaul

author: george.b
author_rank: Frequent Contributor
timestamp: February 17, 2022, 11:59:39 pm
content: Quote from: carpetpaul2129 on February 17, 2022, 09:56:25 pmNope, it's actually half the "work" (the correct term here would be "power", as "work" is actually a technical thing and basically means energy - power is work over time) for the cell. I assume you're familiar with Ohm's Law at this point, yeah? So, voltage equals current times resistance: V = I*R. Plug in two values and you get the third.If you connect a 50 ohm resistance to a 50 volt source, you get a current of 1 A: 50 = 50*I, the current, "I", can only be 1 A. If you connect a 100 ohm resistance instead, you get 50 = 100*I, I = 0.5 A...Quote...at which point you can use whichever equation you prefer: P = R*I², P = V*I, P = V²/R to work out the power. For the first case, you're draining 1 A from a 50 V source, and that's a power of 50 W. Alternatively, that's a 50 ohm resistance connected to a 50 V source, and that's the same 50²/50 = 50 W, you get the idea.For the second case, you're draining 0.5 A from a 50V source, so that's a power of 25 W. With greater resistance you get a lesser current, so the battery is giving out less power. Consider free air, with a resistance that you could call infinite for these purposes: infinite resistance, zero current, therefore zero power. If greater resistance meant greater power/stress/"work" from the battery, then batteries wouldn't hold a charge, just from standing there, since there would be infinite resistance from the air between their terminals, which would cause them to "work infinitely hard", so to speak, which in turn would immediately cause them to be depleted.It might help a bit to think of a sometimes criticized, yet close enough for argument's sake, analogy: a battery is a water reservoir, the wires are pipes, voltage is the pressure, current is the water flow, resistance is a restriction to the water flow. The reservoir doesn't "work harder" when there's a restriction, there's just less water flowing through the pipes.

author: SmallCog
author_rank: Frequent Contributor
timestamp: February 18, 2022, 04:32:40 am
content: Give some thought to having a load to test with, this can be DIY'dOr bought from eBay/Banggood/Aliexpress etc:https://www.ebay.com.au/itm/-/144028730454(example not a specific recommendation)Or there are proper lab items available that depending on your local area may be available for reasonable second hand pricesAs for the resistors remember that they resist the flow of current. The higher the resistance, the lower the current flow, and the lower the current flow the lower the heat/power (all for a steady voltage)