Actually this could be used to charge any sort of battery where you want a constant current and a constant voltage.

In this instructable I will take you through the whole process to producing a final boxed system.

It will take an input from any AC/DC adapter with a jack. You simply have to make sure that the adapter is rated for the voltage and current you want to create. This system will allow up to 36V and 2Amps.

Step 1: The Components & Tools Needed

Components needed are :

Project Box,
220nF capacitor,
100nF capacitor,
selection of resistors between 1and 5 Ohms,
5K/10K potentiometer,
820 Ohm resistor
Wiring - some jump leads ( suitable for connecting parts of circuit board together), some cable with two power carrying leads inside ( pos + neg )
Crocodile/Spade clips
2.1mm or 2.5mm input jack ( depending on your power source )
Copper stripboard

The whole circuit revolves around the L200C current/voltage regulator ( the circuit diagram we will be sticking with is shown below ). You can download the datasheet from HERE

Tools needed are

Soldering Iron
Screwdriver ( Philips ) and a very small flatbladed screwdriver
<p>Is there a way to connect the charger in parallel to the battery and load? to charge and use 'simultaneously'?</p>
Yes, just connect your load in parallel with the battery. Just make sure the load doesn't draw more current than the charging current or else you won't be charging the battery but draining it.
Great work! But i just want to confirm that am i supposed to connect positive out to positve terminal of battery and negative out to negative terminal of battery
<p>I'm looking for a charger to charge a 48v battery bank. Ideally it would be compatible with input from a solar panel or two. Does anybody know if there's a variant of this circuit that could be pushed up to 48+v (whatever the charged voltage level is). </p><p>Thanks in advance!</p>
Very impressive! I'm scratching my head trying to build a 2.37 volt battery charger with 15 amp out put, I work on chargers go a living but having trouble building one. Trying to charge 2vdc battery cells <br>Any ideas that might help?
<p>Hi. Though I can buy one cheaper I'd like to make one for the experience. What size is the project box and do you remember where you got it? Thank you.</p>
<p>Does it have an advantage over a regular battery charger? </p>
<p>Nice job, did it and work's great!</p>
<p>Does this charger switch off after the battery is full?</p>
<p>How can I modify this to charge a 24v lead acid battery?</p>
<p>NVM I read it it haha</p>
<p>I have a smaller 6v 4Ah battery. Will this circuit charge my battery too? or it needs some modification? thanks.</p>
you can make same circuit with two LM317 <br>first you put LM317 to limit current and then 2nd to regulate voltage
question: you say &quot;It will take an input from any AC/DC adapter with a jack. You simply have to make sure that the adapter is rated for the voltage and current you want to create.&quot; <br>So does this mean if I want to charge a 12v battery that I need a 12V input?
12v battery needs 13,8v when its full so you would need 14-15v adapter and step down voltage to 13,8-14v using l200c and resistors
would i be able to use diffrent caps with this circuit, i have a bag of them that i bought but none in the nf range all of them are uf and the srcap boards only have 1nf and 10nf caps on them...
Yep, 100% American Made.
The issue I was recommending caution with is not the circuit diagram as there is nothing wrong with it. <br><br>I was commenting on the fact that, for the resistance R3, the author used different value resistors in parallel to achieve a resistance equivalent to a given value (step 5). He then calculates a power dissipation as thought all the parallel resistors are sharing the current equally. <br><br>This is in fact false as, while in current limiting mode, the L200C will always hold pin 5 at 0.45V higher than pin 2 (in order to give you your current control). Hence the power dissipated by any of the parallel resistances in R3 is a function of it's own resistance (0.45^2/R_parallel).<br><br>Lets explain this in terms of what the author did:<br>In step 5;<br> To get a resistance of 0.625 ohm(for R3), he used a 1ohm, a 2.5ohm and a 5ohm.<br> So R3 = 1 / ( 1/1 + 1/2.5 + 1/5 ) = 0.625 Ohms<br> All is well so far...but then he calculates the total current to figure out power dissipation. However, having unbalanced parallel resistors, they wont equally share the current. In other words, if you have low value resistors, you have to calculate power dissipation individually.<br><br> For example, his three resistor will be drawing different currents totaling the value he calculated:<br> I3_1 = 0.45/1 = 0.45 Amps<br> I3_2 = 0.45/2.5 = 0.18 Amps<br> I3_3 = 0.45/5 = 0.09 Amps<br> I3 = I3_1+I3_2+I3_3 = 0.72 Amps<br><br> So although overall he is dissipating 0.72*.45 = 0.324 Watts as he stated, the first resistance is taking most of it at 0.45*0.45 = 0.2025 Watts (getting pretty close to 1/4 Watts resistor's limit). My word of caution is to not use too low of a value for any of the parallel resistors for R3 (since that low value will take most of the current).
so what do we have to change in the schematics for it to be ok? can you expain i dont know much of electronics thanks you :)
In essence, nothing. The schematics are A-ok. I was commenting on the fact that in step 5, the author replaces a single resistor in the schematic with a few resistors to give an equivalent resistance of the needed value. I then urged caution in the selection of these many resistors as pairing a very low value with a very high value will result in the low value taking most of the current.
so what do we have to change in the schematics for it to be ok? can you expain i dont know much of electronics thanks you :)
You make the same excellent point. I would not have noticed the concern without your insightful comment. I only responded because your very relevant comment appears under step 1 - whereas the problem you very effectively address is not evident to the reader until they see the text and images attached to step 5. Comments seem to be exclusive to the project steps that they are posted under. <br> <br>If I could make a suggestion - I believe that your important comment would be better understood if you proceeded to step 5 and then made the matter clear in a post there - in context. I feel that your comment would have more meaning to readers if they saw it under step 5. <br> <br>Thanks, <br> <br>-- coiley <br> <br> <br>
My apologies for that one. I see how it could be confusing on the first read of this instructible. I usually try to post under the relevant step but it appears that I neglected that step this time.
mathieulj, i read your comment about a mistake in schematics, what would be whe fixed schematics without that error? sorry i dont know much about electronics :)
You have to be careful with your unbalanced resistors in parallel. When you parallel resistors, the lower valued one will take in more current (and hence dissipate more power). With a 1Ohm resistor, it pulls 0.45V/1Ohm = .45A (0.45V^2/1Ohm = 0.20W) no matter what other resistors are in parallel with it ( since the L200 drives the current up to keep this voltage drop constant). With this particular selection of values, you are nearly overdriving a resistor. I am posting this as a word of caution for those who might have wanted to throw a lower value resistor into the mix.
You are referring to the authors choice of using three parallel-resistors-of-different-values for R3 shown in step 5. Gotcha. <br> <br>I used two &quot;identical&quot; 1 Ohm 1W resistors in parallel. This is equivalent to a 0.5 Ohm 2W resistor for R3 - and the identical-resistors-in-parallel divide the current equally. <br> <br>0.5 Ohm R3 makes for a 900mA max charging current. <br>
LOL gotta love that :P
could a different regulator be used as an alternative for the L200C?<br />
&nbsp;does the voltage of the capacitors matter?
could this be modified in some way to charge a 24v electric scooter battery?<br />
I can sorta follow you in this instructable, to a T atleast. But I lack the&nbsp; understanding to modify this for say my drill. How would I gain sufficient expertise in this, as an electrician or something?<br />
If I wanted to charge some lead acid batteries with a solar panel, could I simply hook up the batteries in sequence and then hook them up in sequence with the solar panel and then add a diode to prevent the batteries from discharging through the solar panel? And if this does work, where would the diode go? I'm a bit confuzzled there. Thank you very much. :)
You'll overcharge them that way and lead acid batteries will sweell and burst or just die. Get a charge controlled rated for the amps or watts that the solar panel will put out. They make controller specifically for that. You can make your own controller, but the parts will cost you just as much as if you bought one from harbor freight. Also, is your solar panel made to output between 12 and 14? If it's higher it may be a 24 volt system or 18 volt system. Be careful. You want to charge lead acid batteries with about 13.5 give or take a little. If you have any questions, send me a message. Plus, if it is the right voltage, you want to hook the batteries up in parallel, not series. Make sure you get that right. If you hook them up in series, they'll only get half the voltage each and that can ruin your batteries.
<p>I have two solar panels each of 60w and two batteries each of 12volts connected to an invereter of 24v microtek .I want to charge these lead acid batt. with solar panels and also with electricity one at a time my problem is solar overchrging to batteries . so please help me to show a way for stopping overcharging batteries.</p>
Heh, that and I love how it states right after the .7 amp max that this is less than 1 amp!
nice one, wandering if u have done for 6v 12ah battery charger with led indicator when fully charge.
where can I buy the L200C chip I cant seem to find it.
<a rel="nofollow" href="http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail?name=497-1382-5-ND">http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail?name=497-1382-5-ND</a><br/>
will this charger stop charging when the battery is done? or is it just a trickle charger?
i will, and is this just a form of a trickle charger?
Does that battery actually say 6.5 Bolts and not volts? lol I have a toaster over that says poewr and not power.
XD XD XD and teachers say I spell bad! XD XD XD
Yep, thats Bolts! :)
Thank you for that great Instructable!!! I will build that battery charger soon, you saved me ££££!
I had built the SLA battery charger, again thank you for this great instructable! The first photo is the battery charger under test (charging a 6v SLA battery at 7v at 0.3A) and the other photo is the battery charger under construction.
Glad you liked it, do you have photos of yours I can see?
? The photo I posted on the comment before got deleted? I don't know that is going on...
will it shut off when the battery is charged pls reply
The L200C will gradually decrease the amount of current going through the circuit until it is only trickle charging.
I was just wondering if there were a reference if the manufacturer is not known? I know the output of the battery, but do not know the charging specifications. 6V 5AH....it's for a bicycle light, if it matters....

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