My understanding (and it is pretty rudimentary) is that traditional lead-acid batteries are considered fully charged at an open-circuit terminal voltage of 12.6V and that they are floated at 13.8V.

A 20A charging current for a 200AH battery is a nominal charge time of 10H (also known as a C/10 charge rate where C is the current that would nominally charge/discharge the battery in one hour), which is pretty common.

Here's a site that might fill in some gaps in your basic knowledge of the process.

http://batteryuniversity.com/learn/article/charging_the_lead_acid_battery

So you are wanting a constant-current, voltage limited charger. Let's tackle that in two pieces. First, starting with a 28V supply (that's what you have, correct?), let's see how we could deliver a constant 20A to a battery that has a terminal voltage that is somewhere between, say, 11V and 15V (to give us some room to work with, we can narrow it if we need to).

Consider the following circuit:



What is the relationship between the voltage Vc and the current Ic?

 

Vc=Ic R1 ?

 

Vc=Ic R1 ?

Nope.

To find Ic we need to know the value of Re and the voltage across it.

Ic = Ve/Re

where Ve is shown below.


What do you know about the relationship between Vc, Ve, and Vbe?

 

@Claudia94: You mentioned that you've been doing a lot of online research, which is good. Have you tried actually testing out a charging circuit? Sometimes I've learned things just by experimenting with a circuit on a breadboard. As long as you don't wire the batteries in reverse or leave them charging for way too long a period of time, there shouldn't be any danger of them exploding.

 

But I might recommend doing something on a much smaller scale so that you aren't dealing with the kind of power and heat that you will need to deal with eventually.

Instead of 20A, use 20mA. Then use a resistor in place of your battery that will produce about 13V when it has 20mA going through it. To simulate the battery charging up, put a variable resistor in series with it that will let you increase the load resistance until you get 13.8V at something less than 1mA. That will let you get all of your control circuits all figured out and then you only have to boost the output capacity, which won't be too hard to do.

 

thank you @WBahn i will see what i can do and post again. just tell me if i have to re-do a whole new circuit , or the last circuit i posted is enough, and just needs some adjustment.

and about the relation between Vc, Vbe and Ve . i don't know anything about it. i'm pretty sure i did not study any of that, just easier examples which i already almost forgot

 

One of the fundamental concepts of E&M is that the voltage between any two points is independent of the path taken between them -- that's true as long as we are working with "conservative electric fields" and, while not all electric fields are conservative, 99.99+% of the ones we will ever work with (from a design and analysis standpoint) are. That's comparable to saying that the height between the top floor and the bottom floor of a building is independent of whether we measure it up the outside of the building, up the stairwell, or up the elevator shaft. So if we go from the base of the transistor up through the source to the top rail we get a voltage difference of Vc. If we instead go from the base of the transistor and up through the emitter and then through Re we get a voltage difference of Vbe + Ve. So, with that in mind, what is Ve in terms of the other two voltages and then what is an expression for Ic?

As for that last circuit you posted, I wouldn't even know where to begin in trying to salvage that.

 

i couldn't answer. asking me these question is really pointless, 99% chance i won't know the answer. also it took me to read your post few times to understand it what you wrote. sorry to disappoint you.

 

Well, I'm not going to just do the design for you. I'll walk you through it as slowly as needed, but you are going to have to put in a lot of effort and learn how all the pieces work along the way.

 

i couldn't answer. asking me these question is really pointless, 99% chance i won't know the answer. also it took me to read your post few times to understand it what you wrote. sorry to disappoint you.

No pain,no gain!