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Forum Index : Electronics : SLA battery charger help...
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Grogster Admin Group  Joined: 31/12/2012 Location: New ZealandPosts: 9307 |
I want to include a SLA battery charger in my latest project. Basically, I want to use a float-charger, but current-limited to protect both the battery and the PSU. The normal way I have done this in the past, is a simple series resistor, and a diode gate:  The switchmode PSU has a rating of 1A, so I would like the resistor to limit the current to about 250mA or so. An 82R 5W resistor across a 13.8v supply gives a current of about 170mA, which would be fine, but when I connect this in series with a 12v 7Ah battery, the charge current is zero. I suspect that the battery internal resistance is something I need to consider when choosing the resistor, but I'm not sure how. When I used this R+D charging arrangement before, it was for 9v NiCd batteries with a very small charge current of 17-20mA or so(so I used 330R for R1), but I need more grunt for an SLA so it does not take forever to recharge. What puzzles me, is that with an 82R 5W resistor in series, the charge current is zero. If I bypass the resistor, the charge current is 3A or so. Can someone suggest what I am doing wrong? Device will be powered 24/7, so a simple float-charger is all I really need, but why isn't it working? It worked with the 9v NiCd batteries....  Smoke makes things work. When the smoke gets out, it stops! |
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| Warpspeed Guru  Joined: 09/08/2007 Location: AustraliaPosts: 4406 |
As the battery voltage rises, the current will fall. Current will reach zero when the battery is the same voltage as the charger. Honestly, this sort of thing is just not worth messing around with, when the Chinese can sell a real ripper fully automatic smart battery charger with microprocessor for less than you can probably scrounge the bits to do a home brew. https://www.ebay.com.au/itm/304201718849?hash=item46d3d5dc41:g:qNsAAOSw-nBhuVgZ Cheers, Tony. |
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CaptainBoing Guru Joined: 07/09/2016 Location: United KingdomPosts: 2075 |
wee trinket to calculate IR, but I don't think it is the issue here (assuming the battery isn't sh*gged) I would only use barrier diodes like above if the battery was not rechargeable (like an RTC coin cell and the barrier would be a BAT54C). In your case, you don't care that the shore-supply and battery get together (it is needed). What is your thinking on the blocking, what am I missing? Quick check, no insult intended - You are using Schottky diodes right? something like 1N5819 would probably be a good fit, 40V 1A. Vf will rise with If, with 1N400x this could already be knocking you down to close to 13V. If you have a good battery it could be that Vcc-Vf is close to Vbat - in which case there would be no current flow. Dunno about the bypassing thing, 3A is way above what you say the mains supply will give... more info? As a quick test, consider taking out D2 and dropping R1 so you get what juice you need to charge the battery under load. The circuit will then be "running from the battery" all the time with the mains supply acting as supplemental/charge. I would only use the below for LA (and did for my aquarium controller years ago, pre-mite days... tiddly little 12V SLA motorbike battery, very cute)  Tony's point about a cheapy module off fleabay is spot-on but I totally get the desire to homebrew and relying on someone elses "black box" (esp. mass-produced cheap chinese stuff) is a yellow flag for me . Edited 2021-12-30 19:35 by CaptainBoing |
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| Warpspeed Guru Joined: 09/08/2007 Location: AustraliaPosts: 4406 |
Oh yes, me too. But the reality is that if you wish to make something that does not look like total crap, you need to buy a box, some knobs/switches, and maybe a digital voltmeter module. Already you are up over $50 assuming you already have a transformer, big electrolytic etc..... Or a flea bay cheap Chinese take away, that comes nicely packaged and usually works. Cheers, Tony. |
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| Grogster Admin Group Joined: 31/12/2012 Location: New ZealandPosts: 9307 |
Hi folks.  @ Warpspeed - thanks for the link, but that thing is way, way too big to fit. I need something I can fit into about 50mm of PCB space, which is why I went to the simple R+D float idea. @ CaptainBoing - thanks, yes, using 1N5819's. The diode gate is so that the battery can be charged with the PSU, but if the mains fails, the battery takes over. Fully automatic. The diodes simply prevent the battery voltage from trying to go back into the output of the PSU if the mains fails. 3A was with a bench PSU when testing, but the PSU I will be using, is current-limited to 1A. Removing D2 is a good idea. I just copied what I had with the 9v setup, but the resistor used there was 330R, so the diode was there to bypass the charging resistor when the mains failed, otherwise there would be too much voltage drop across the 330R resistor to power the rest of the circuit. But with a SLA battery, the charge current can be much higher, so a much lower value resistor such as 82R or so, probably won't affect the operation of the rest of the circuit due to the low value. And thanks for the link to the IR calculator. I will play about some more. Smoke makes things work. When the smoke gets out, it stops! |
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| CaptainBoing Guru Joined: 07/09/2016 Location: United KingdomPosts: 2075 |
Yep, nice gentle charge but loads of grunt courtesy of D2 when needed. Maybe the precise inheritance of the original circuit is moot with LA. Re-arranging how to think about it; if using the battery as primary power, the mains PSU becomes primarily charging - straight onto the battery via R1 with supplemental capability when fully charged. <obvious>Don't forget to fuse the the battery. LA is scary powerful... </obvious> I melted a spanner across the terminals of a car battery (deliberately) as an idiot 18yr-old. "Youth is wasted on the young" - G.B.Shaw (varies) In your boots, I would play on the bench and drop R1 down to around 18R, or maybe even 15R if you trust D1 (bit hard on it though), it'll get you nearer to the 1A capability of the mains PSU, good strong charge and more supplemental power when running "normally". Lowering R1 does mean you'll probably break your 50mm2 space limit though, going from 3W to 15W (ish) - unless R1 is mounted on the chassis for heat dissipation in which case it won't matter. Following with interest. Edited 2021-12-31 18:41 by CaptainBoing |
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| Grogster Admin Group Joined: 31/12/2012 Location: New ZealandPosts: 9307 |
The SLA I plan to use is either a 2.3Ah or a 1.2Ah. I had a 2.3Ah one sitting on the bench, so I played with that one. I discharged it to about 8v by putting a 6A load on the battery in the form of a radiator fan. I am aware that you never allow SLA's terminal voltage to fall below about 7v. With the battery now at least partly discharged, the resistor I ended up with was 22R 5W, to give a charging current of about 150mA float/trickle, which I think will work fine. It does not need to be a fast-charge, and a gentle current-limited float charge is probably better for the battery long term, then fast-charging. I will do a few more tests on the bench before I put it on the PCB layout. Yes, I will in-line fuse the SLA. Yes, a short-circuit LA can pass some SERIOUS current, so we don't want any kind of fault condition setting fire to the box!  You'd be surprised how much short-circuit current you can get out of even the small SLA batteries. It usually kills the battery, but while in that state, those small SLA's can still pass serious current in a short-circuit condition. I've also seen those SLA's swell up and bulge in a very unhealthy looking way, if you don't at least current-limit them while charging. One battery I replaced in a panel, was just wired DIRECTLY across the output of a power-supply capable of 5A or so, and I was called in cos the staff said there was a horrible smell coming from the panel. Opened it up, battery so swollen and bulging I'm surprised it had not cracked open. Was hot enough to fry an egg on, and if you put your ear to it, you could hear it sizzling internally. That is NOT how you charge a SLA, but I digress!  Smoke makes things work. When the smoke gets out, it stops! |
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| Warpspeed Guru Joined: 09/08/2007 Location: AustraliaPosts: 4406 |
Seven volts ? I would think 11v would be about the point I would start to really worry about removing all load and recharging. Any gassing at all will lead to water loss and the cell drying out, and then its stone dead. Maybe a trickle charge at the twenty hour rate ? 1.2Ah is pretty small, 120mA at a ten hour charge rate, and 60mA at a twenty hour charge rate. How much battery life are you anticipating ? Three weeks, three months, three years ? The harder you flog it, the sooner it will die. Cheers, Tony. |
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| Grogster Admin Group Joined: 31/12/2012 Location: New ZealandPosts: 9307 |
11v? Really? I'm sure I read it was 7v or so. I could well be wrong, and a man of your inverter knowledge would certainly know his batteries. I'll keep that in mind, and change to around 11v. .....I'm sure I read about it being 7v or so though....(again, I could well be wrong) EDIT: Wanting as much life out of the battery as possible. Most of its life, it is only sitting there as a backup battery as the mains PSU will run things most of the time, so the load on the battery will be very light and intermittent. Edited 2022-01-01 14:06 by Grogster Smoke makes things work. When the smoke gets out, it stops! |
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| Warpspeed Guru Joined: 09/08/2007 Location: AustraliaPosts: 4406 |
Right from Yuasa the Japanese king of sealed lead acid batteries. https://newbedev.com/how-bad-is-it-to-undervoltage-a-12-volt-lead-acid-battery A fully charged 2.3 Ah battery discharged at 6 amps is a discharge rate of 6A/2.3Ah = 2.6 From the above curve, a discharge rate of C = 2.6 the battery will go from full charge down to 11v in about 5 minutes, and it will drop to 8v in about 8 minutes. That is when its brand new and in perfect condition. Its all down hill from there. Will that do the job ? Edited 2022-01-01 15:14 by Warpspeed Cheers, Tony. |
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| Grogster Admin Group Joined: 31/12/2012 Location: New ZealandPosts: 9307 |
Thanks for the link. PLEASE REMEMBER - the 6A load I put on the 2.3Ah SLA was for the SOLE PURPOSE of discharging the battery, so I could then experiment with charge currents. At no time, will the ACTUAL load be anything even close to 6A. Standard load would be in the area of 200mA or so, with occasional peaks to the 800mA or so level. Smoke makes things work. When the smoke gets out, it stops! |
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| Warpspeed Guru Joined: 09/08/2007 Location: AustraliaPosts: 4406 |
Oh very sorry, I misunderstood. O/k so for 2.3Ah battery, discharge rate 0.2A/2.3Ah = .087 At C =.09 it should be good for about ten to eleven hours down to 11v. The usual way of rating lead acid batteries is for a ten hour test discharge. So 2.3Ah should give 230mA for 10 Hours. 6.0Ah should give 600Ma for ten hours, and so on. If you discharge it faster than over the full ten hours, your actual recoverable capacity will be less than its marked Ah rating. At a lower discharge rate it should be possible to get slightly higher than its marked rating. At least in theory when the battery is brand new. For battery sizing, you might work it all out and then fit a battery with about double the required amp hours. That way it can be 50% stuffed and still just meet some minimum requirement. Cheers, Tony. |
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| Grogster Admin Group Joined: 31/12/2012 Location: New ZealandPosts: 9307 |
Cheers. I read that article you linked to - a very interesting read.  Smoke makes things work. When the smoke gets out, it stops! |
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| Warpspeed Guru Joined: 09/08/2007 Location: AustraliaPosts: 4406 |
If you liked that, check out the Battery University website. Its an absolute goldmine of reliable information on every battery type. https://batteryuniversity.com/ Cheers, Tony. |
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