Install an amp regulator on a lead acid battery charger
A current limiter can be installed, but the amount of current to be limited must be known.
The most appropriate way to control the current for a charger is by thyristor powered by pulsating current.
To assemble this regulator you need to have knowledge in electronics.
I don't know if you have that requirement. So well it would be necessary to know what the charger is like.
Hello, to electronically regulate 15 Amp, if you actually deliver 13.8V you will not be able to, because the consumption of the circuit will lower your voltage, on the other hand I think the charger has to deliver more voltage, between 15V and 16V, if it is That is a charger.
Hello friend route, there I leave you the most readable circuit, if you lowered the other, you will notice some minor modifications, they are due to changing the 2n2222 for the tip31, if you are going to use the 2n2222, leave the 10k resistor of the 2222 base, and the 47 Ohms base of the mosfet, I would use the tip, the resistance of 100k based on the tip is because if you put a much smaller you can pass the voltage 555 (not current, eye to data, voltage), Depending on the voltage of the charger, there are chargers that deliver up to 18V, that would break the 555 no matter the current. In practice it is like that.
I would also leave the tip, it will never break, the first one that happened to him I did it with 2n2222, but since I do not know the amount of use that is going to be given, it is better that over and not missing, the Mosfet is also something excessive, it is 130 A, but it is for the same reason.
With that current and assuming that the device has already started, it is in a state of very poor regulation, this without considering the current taken by the base of the transistor.
If we see that the "hfe" of the TIP31 is in the order of 50 and he will take all the current flowing through R1 it would give us an Ic = 1.5 [mA].
You have to correct this.
Something very very basic:
Ic = hfe Ib
If Ic = 3 [A] and hfe = 10
Ib = 3/10 = 0.3 [A]
The merit of the transistor as an amplifier is to have a large gain factor in cc, that is to say a hfe >> (greater major). I took 50 to favor the proposed circuit. I insist: the current in base and in the zener is extremely low so that the configuration behaves properly.
Regarding what you think about my theoretical knowledge and especially as far as Zeners is concerned, it is rushed and naive.
You know that, I get tired of you, if it seems to you that the Zener connect anodo a positive, do it like this, and if you want to give an answer to the question also do it, but stop humping me, please, I don't have to hold on You already gave your opinion in my answer, now propose a solution.
¡Aprende a reparar con estos cursos YoReparo!
Curso Localización de Fallas en Refrigeración en Heladeras/Neveras On/Off
