Mini Battery Charger Module with LM317

Yes, I did not create a BOM file, because I do not know how to do it in the KiCAD program I use. In addition, since all of the elements appear on the card quite legibly, the hobbyists who will make this circuit will not have difficulty in this regard. The characteristics of the diodes are not very important. Because the main task of diodes is to ensure a sharp transition from one LED to another. So I avoided naming them. But you can choose 1N4148, it will work. Other components are pretty straightforward. If you have difficulty with SMD components, you can practice more and examine the datasheet files.

Please use our plugin and download the file. https://www.pcbway.com/blog/News/PCBWay_Plug_In_for_KiCad_3ea6219c.html

Thank you for your reply. Understand, I'm barely OK to understand the circuit, but consider if someone which cannot read the circuit and without explain, it will be difficult for them and I think more detail can help more people just buy the component and solder them without understanding the circuit. Again, thank you for sharing with us.

Is it possible to have the BOM auto-extracted and list on the page? That can help makers to save time.

Hi Lam Chun. I created and uploaded a BOM file for you: https://disk.yandex.com.tr/d/mCfX6RCIDZtsRA Regards.

We are making a plugin for this,now you could download the file on the quotation page,it include all files

Thank you very much!

Sorry, one more question. What can I use for the "Rs" resistor value? Thanks.

Rs means "Rsense". It is the current limiting resistor for the charging current. The value of this resistor should be determined according to the characteristics of the battery you will use and your fast/normal charging preference. Its formula is: Rs (ohm) = 1.25V / I (A). This will allow you to calculate how many ohm resistors you should use. I (A) is the current you need to draw. This current should be one-tenth of the battery capacity. For example, if the capacity of the battery is 1000mA, you should set the charging current as 100mA. 1.25V / 0.1A = 12.5Ohm. And the strength of the resistor (watts) is: P = I^2 * R. You have to multiply the resistance value by the square of the current.

Sorry, small update. The resistor Rs is on the minus line, not at the output of the LM317. When this happens, the formula changes and becomes: Rs = Vbat / Icharge

Thank you very much for your information, your explanations help a lot.