How to lower voltage in battery charger by 1V - Page 1
Forum section: electronics / beginners

author: mamkody
author_rank: Newbie
timestamp: October 25, 2024, 03:46:23 pm
content: HiI've bought cheap battery charger that supposed to be 21V 2A. Unfortunately the output voltage is 21.5V which is too high and overcharge the battery (5s no BMS). How I can reduce the output voltage in easy way to max 20.5V?

author: VinzC
author_rank: Frequent Contributor
timestamp: October 25, 2024, 04:31:45 pm
content: Have you measured the output voltage at 2A load?

author: tunk
author_rank: Super Contributor
timestamp: October 25, 2024, 04:40:21 pm
content: 2-3 diodes in series with the output?The transformer could look a bit on the small side for 42W.

author: IanB
author_rank: Super Contributor
timestamp: October 25, 2024, 04:48:59 pm
content: Quote from: mamkody on October 25, 2024, 03:46:23 pmHave you tested this by charging a battery and measuring the end of charge voltage on the battery?If it is 4.3 V per cell, then you would be correct, but if it happens to be 4.2 V per cell then there is nothing to worry about.Also, charging to 4.3 V might not be so bad, it depends on the cell. At the worst it will shorten the working life of the cells a bit. It's up to you how valuable your cells are, and if you want to get a bit more out of them at the cost of wearing them out a bit faster.

author: tooki
author_rank: Super Contributor
timestamp: October 25, 2024, 05:25:23 pm
content: Ummm… that thing is just a power supply, I think. It doesn’t have a Li-ion charger as such, as best I can tell. I suspect it is intended to be used with batteries with built-in BMS and charging circuit, giving it just enough headroom to regulate down to the required voltages.Lithium ion batteries can’t just be connected to a voltage and left alone. They need a two-stage process that starts with a constant-current phase, then changes to a constant-voltage phase, in each case ending with accurate measurement of voltage and current.I would not use this thing.P.S. Please delete the [attach=x] tags from your post — that way the pictures can expand without a page reload.

author: mamkody
author_rank: Newbie
timestamp: October 25, 2024, 05:55:39 pm
content: Quote from: tooki on October 25, 2024, 05:25:23 pmThat's why I want to reduce the end voltage to a maximum of 20–20.5V to leave some headroom in case the cells become unbalanced over time. I don't use a BMS, as I balance the cells manually from time to time.

author: tooki
author_rank: Super Contributor
timestamp: October 25, 2024, 06:00:53 pm
content: You still need an actual charging circuit. Do you not understand how lithium ion batteries need to be charged?

author: mamkody
author_rank: Newbie
timestamp: October 25, 2024, 06:22:56 pm
content: Ive already used this charger to charge the battery and it works fine but it stop charging when battery reach 21.5v which is too high. It should stop charging at 20.5v

author: Zero999
author_rank: Super Contributor
timestamp: October 25, 2024, 06:41:10 pm
content: What'd U1's part number?

author: tooki
author_rank: Super Contributor
timestamp: October 25, 2024, 06:53:25 pm
content: Quote from: mamkody on October 25, 2024, 06:22:56 pmThat is not how you charge lithium ion batteries!!!

author: showman
author_rank: Contributor
timestamp: October 25, 2024, 08:28:10 pm
content: Quote from: tooki on October 25, 2024, 06:53:25 pmYou don'tusuallycharge it like that, but in principle there is no problem. The constant current phase does not really have to be constant current, unless you want the optimal (fastest possible) charging. The current just has to be limited to some appropriate value. Given the puny transformer and the integrated switcher, I'd be surprised if this thing can output 2 A even into empty batteries and if the cells are large enough, whatever this thing outputs is ok current-wise. The final voltage is too high of course, but this might be adjustable with one of the resistors on the board. Also given manual balancing, there's no issue with that either.

author: mamkody
author_rank: Newbie
timestamp: October 25, 2024, 09:14:47 pm
content: Quote from: Zero999 on October 25, 2024, 06:41:10 pmGG1034A06B2404Quote from: showman on October 25, 2024, 08:28:10 pmYou mean that one, as this is the only resistor on low voltage side. If yes what value should i use?

author: showman
author_rank: Contributor
timestamp: October 25, 2024, 09:34:53 pm
content: Quote from: mamkody on October 25, 2024, 09:14:47 pmOnly datasheet I found and it's quite limited. Fairly sure it is R3, but not sure if it needs to be larger or smaller.

author: tooki
author_rank: Super Contributor
timestamp: October 25, 2024, 11:07:32 pm
content: Quote from: tunk on October 25, 2024, 04:40:21 pmGood call!The datasheet suggests the chip is for up to 12W.

author: brucehoult
author_rank: Super Contributor
timestamp: October 26, 2024, 12:07:33 am
content: Quote from: showman on October 25, 2024, 08:28:10 pmRight. For 2000 mAh or above it might be ok as is. For smaller capacity a series resistor of 20000/mAh Ohms (and mAh/50 W rating) would ensure initial charge rate was under 1C. Even for 2000 mAh a 10 Ohm 40W series resistor wouldn't hurt, other than increasing charging time. And one more more diodes to limit the final voltage. Internal resistant of the battery is a fraction of an Ohm, so it's basically limiting short-circuit current.

author: showman
author_rank: Contributor
timestamp: October 26, 2024, 12:18:18 am
content: The only problem with diodes is that they will work for short term, but since there is really no hard cutoff, they will still conduct below threshold and keep charging the batteries. I'm too lazy to calculate how long will it take to go for example above 21 V, but at least after infinite time the battery would still charge to 21.5 V.

author: brucehoult
author_rank: Super Contributor
timestamp: October 26, 2024, 01:11:11 am
content: Quote from: showman on October 26, 2024, 12:18:18 amHmm. I'd think a 2000 mAh Li-ion battery would have self-discharge in the 2-10 mA range, which should be enough to keep a silicon diode near its typical forward voltage drop?Also, presumably the battery is being charged in order to use it, so infinite time won't happen.

author: showman
author_rank: Contributor
timestamp: October 26, 2024, 02:09:59 am
content: I think you might be off by couple of orders of magnitude. Should not be some mA, but uA.And if it is in active use, there's likely no issue, but with something like a backup battery that stays charging until the power goes away, it might become one.

author: brucehoult
author_rank: Super Contributor
timestamp: October 26, 2024, 02:42:53 am
content: Quote from: showman on October 26, 2024, 02:09:59 amHow do you calculate that?I went off sources saying that Li-ion batteries lose 1%-5% of their charge per month. There are 30*24 = 720 hours in a month, so losing 5% of 2000 mAh in 720 hours is 2000*0.05/720 = 0.14 mA.Ok ... guess I did slip two orders of magnitude somehow :-( Still seems like significant current, but I don't know what the voltage drop would be. It's not single digit µA or anything.

author: tunk
author_rank: Super Contributor
timestamp: October 26, 2024, 01:23:14 pm
content: Back to R3 mentioned by showman - maybe you could breakthe track between the rectifier and resistor, and solder in a 22kresistor between them?

author: khs70374
author_rank: Newbie
timestamp: December 25, 2024, 05:05:52 pm
content: I stumbled upon the same charger, which looks nice and compact, but mine had a voltage of even 22.5V, which is way too high for an 18V Makita battery, and better should be around 20.5V.As already mentioned, R3 on FB pin defines the output voltage, with a simple linear relation.My R3 was a 4873 = 487kOhm (measured 485kOhm), the correct value should be 485kOhm*20.5V/22.5V = 442kOhm.So a resistor Rp parallel to R3 will do the job with Rp = 1/(1/442kOhm-1/485kOhm) = 5000kOhm = 5MOhm.I soldered a 5.6MOhm resistor parallel to R3, and now I have an output voltage of 20.6V, which is perfect.

author: Siwastaja
author_rank: Super Contributor
timestamp: December 25, 2024, 05:18:41 pm
content: Quote from: brucehoult on October 26, 2024, 01:11:11 amNo, it does not. You are three orders of magnitude off; 2-10 µA would be a better guess.Quote from: brucehoult on October 26, 2024, 02:42:53 amDo not trust Battery University. That is off by an order of magnitude.For 3%/year which would be typical, 2000mAh * 0.03 / (365*24h) = 6.8µA.