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

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

Does it have an advantage over a regular battery charger?

Nice job, did it and work's great!

Does this charger switch off after the battery is full?

How can I modify this to charge a 24v lead acid battery?

NVM I read it it haha

I have a smaller 6v 4Ah battery. Will this circuit charge my battery too? or it needs some modification? thanks.

first you put LM317 to limit current and then 2nd to regulate voltage

So does this mean if I want to charge a 12v battery that I need a 12V input?

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.

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).

Lets explain this in terms of what the author did:

In step 5;

To get a resistance of 0.625 ohm(for R3), he used a 1ohm, a 2.5ohm and a 5ohm.

So R3 = 1 / ( 1/1 + 1/2.5 + 1/5 ) = 0.625 Ohms

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.

For example, his three resistor will be drawing different currents totaling the value he calculated:

I3_1 = 0.45/1 = 0.45 Amps

I3_2 = 0.45/2.5 = 0.18 Amps

I3_3 = 0.45/5 = 0.09 Amps

I3 = I3_1+I3_2+I3_3 = 0.72 Amps

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).

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.

Thanks,

-- coiley

I used two "identical" 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.

0.5 Ohm R3 makes for a 900mA max charging current.

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.