If you still don't want to provide the charger and protection circuit and then our Moderation Team will consider to lock this thread for safety reason.

 

And some safety goggles.
Even the professionals get it wrong sometimes. Can you do better than them?
http://abc7news.com/travel/battery-fire-on-plane-points-to-danger-in-the-sky/3020423/
https://www.cbsnews.com/news/lithium-battery-fire-risk-samsung-galaxy-note-7/
e.t.c.

Mind you - My lithium incidents = 0. Various trusted brand manufacturers = gonna need a bigger computer...........

 

How do adapters with a current limiter work?
I've got a current limiter circuit with a mosfet and if the current that flows through it is above 2A, the voltage will decrease which means the current will be limited at 2A. Do the adapters work like this as well? Or are there systems where u can limit the current at 2A and keep the voltage stable.

Oh, please read up on how power supplies work. And basic Ohms law. This question too indicates your lack of understanding.

No, you cannot have a power supply that reaches the 2A current limit and keeps the voltage stable.
Just think, if there is a 10V supply and you put on a 5 ohm load, the current drawn is 2A.
Now add another 5 ohm load so the total load is now 2.5 ohms.
How can the supply limit the current to 2 Amps and the voltage still be 10 Volts?

As the moderator has asked, where is your circuit?
You can tell from many of the comments that we are very concerned with your project as you obviously do not have the understanding to safely build this battery charge controller and have it in your mind to ignore the advice given.

 

I apologize for my stubborness, I just wanted to create something on my own.

I'll take the advice and I've been looking for a charging protection module.
The one i've found is: https://www.amazon.com/SMAKN®-TP5100-Synchronous-anti-reflux-charging/dp/B015DX0422
Does this one seem alright?


Adapter voltage = 12V
TP5100 Vin = 8.8V~17V (2 cell in series)
TP5100 Vout (max charge) = 8.4V (2 cell in series)
TP5100 Iout (max) = 1A-2A (variable by replacing a resistor)

Is this good enough?