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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 )
Grommett
Crocodile/Spade clips
2.1mm or 2.5mm input jack ( depending on your power source )
Copper stripboard
L200C
Heatsink

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
Drill

Step 2: The Box

The project box is made from ABS plastic, if you plan to use the chip to its full potential you may need a metal box. This will be explained a little later. It should be of suffient size to allow insertion of your copper stripboard and also have some headroom for the L200C chip - this chip can generate some heat and unless the box is metal you dont want it pressing against the box.

You can see that a hole has been drilled into the box to accomodate the DC Input jack. If you look at the DC input you will see that it has 3 tabs. The one attached to the centre is the positive, the next one out is the negative - these are the only two we are interested in.

Please be aware that jack plugs have polarity too - usually the polarity is as shown in the 2nd picture - always check. ( I even ringed the important info around in red )

Step 3: First Things First

Check that the copper stripboard fits into your box, you may need to trim it - I have designed the circuit so it will fit a board with 23 holes and 9 strips. One hole either end is not used to allow it to slide into the slots provided by the project box. Best to make sure of a fit now before you start any soldering.

You will also need to drill a 2nd hole in the other end of the box. The black wire containing your main two output power lines should fit through the plastic grommit. Drill the hole, install the grommit and check out the cable runs through - it should be a tightish fit so your cable wont pull out and strain the curcuit board.

Step 4: What Voltage/Current Should I Use?

You should charge your lead acid battery according to the specification of the manufacturer. Below you can see the one I was charging - 6.5volts at .7Amps. Build you circuit around the typical batteries you need to charge.

Step 5: The Circuit

I include two versions of the circuit board, You have the traditional circuit diagram and a graphical representation of the copper stripboard.

C1 is a 220nF capacitor
C2 is a 100nF capacitor

The two capacitors help smooth and filter the input and the output voltages.

R2 is a 820 Ohm Resisitor.

W1 through to W6 are all jumper wires of various lengths. Most electronics shops have them available.

The X marks you see on the tracks are breaks in the copper strips. You can break them using a stripboard track breaking tool - a supplier I use for them can be found at Electronic Projects Online

R1 is the 5K or 10K potentiometer.

The 3 x R3 resistors make up the value of Ohms you need to supply the correct current. Notice that they are set up in parallel. This is using 0.25W capable resistors making a total of 0.75W. The current passes directly through these resistors so it need to be rated correctly. We will talk about the equations for calculating correct values shortly.

Finally you can see the L200C. It has the pins numbered which you can match up from the datasheet. You will have to do a small amount of gentle bending to get the pins lines up as I have them - sadly the pins are just a little too close together to fit perfectly into the strip board.

Pin 1 accepts accepts the positive lead from the powersupply. Pin 3 is ground ( negative ). Pin 5 is the output. Pin 2 and Pin 4 are used to determine the correct voltage and current.

Equations!

R3 = 0.45 / Amps

So in my case I wanted it to limit the current to 700mA
R3 = 0.45 / 0.7 = 0.64 Ohms

In my case I used 3 different resistors to get close to that value - 1,2.5 and 5 Ohms. The way to calculate resistors in parallel is

1 / (( 1/R1)+(1/R2)+(1/R3))

in my case that is

1 / (( 1/1) + (1/2.5) + (1/5))
= 1 / ( 1 + 0.4 + 0.2 ) = 1 / 1.6 = 0.625 Ohms

Which is close enough! To work out the current you get from a set Ohm value you can go backwards - its useful to find out how your approximations with resistors gets you.

Current = 0.45 / 0.625 Ohms = 0.72Amps

The power going through R3 is 0.45*0.45 / R3 in Ohms

In my case this is 0.45*0.45 / 0.625 = 0.324W, considering the 3 resistors allow a total of 0.75W we are well within the tolerance.

Working out the value of R1 is easy.

R1 = (Vout/2.77 - 1) * R2

We know what R2 is 820 Ohms and we know what we want out VOut to be so ( in my case )

R1 = ((6.5V/2.77) - 1) * 820 = 1104 Ohms

The simplest way is to attach your multimeter to Vout and then adjust the potentimeter.

IMPORTANT POINTS
1) your Volts IN needs to be about 2Volts higher than your required Volts out.
2) The chip burns of the excess voltage/current as heat. To keep the heat down try not to have VIN much greater than VOut - taking into account point 1.

To work out the Watts being dissipated by the chip you need to do (Vin-Vout) * current selected. Mine version is 12V-6.5V * 0.7 = 3.85W. I have also clipped a heatsink to my chip and the box DOES get quite warm - though it seems quite capable of dealing with it. Things might get very tricky if Vin was 24V and Vout was 6V and you were at the full 2A current.... pretty hot at 36W .. FAN PLEASE lol

Step 6: Building the Circuit - Step One

Make sure you have your soldering area setup and your components near to hand. I use a sponge to help keep my components in the board when I turn it over to solder... hmmm it just occurred to me.. would blue-tack or some kind of putty help to hold them in place... I will try that next and and let you know..

Print out the strip board diagram and have it where you can see it. Remember that as you set your components onto the board you need to leave that one hole border left and right so you can slide it into the box.

If you have had little experience soldering - do not worry - there are plenty of links on the internet and a strip board is one of the easiest ways to get some practice in.

Step 7: Building the Circuit - Step Two

Once you have built the circuit minus the final power leads, its a good idea just to tie on some temporary leads ( so that they touch the correct copper row ) so you can test the circuit. First measure the current with your multi-meter and then the voltage. Adjust the potentiometer until you get the required voltage. Then you can solder in the final power leads and then insert the circuit.

You will then need to attach the input power leads to the DC input jack ( shown in picture 3 and 4 ). You should also add the headsink to the L200C - you can see it in picture 4. You can see that the spades/crocodile clips have been connected too in Picture 4.

One final tip - if you circuit board is loose fitting, you can add a few dabs of glue where the board is slotted into the box, ie on the runners. This will stop the board moving up and down. You can also see from the images that I have the board situated so that the chip is as close to the centre as possible - as far away from the plastic as I could manage. Saying that, in the configuration I choose the box doesnt get hot.

Step 8: Finishing Up

The first picture shows the box with all the connections made. The 2nd with the lid on and 3rd and 4th charging the battery.

If anyone is interested in purchasing a kit to build yourself I have a few for sale in my ebay shop

http://stores.ebay.co.uk/Electronic-Widgets-Inc

There are actually two kits, a basic and an advanced kit. The basic kit provides you with a much more detailed explanation that found here but with pretty much the same outcome. It gives you all the components you need to build it apart from the tools. The advanced kit comes with two knobs and larger potentiometers so you can adjust both the current and the voltage. There are also metal box versions.
<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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