Battery Protection Circuit IC help

[kingofkeys79]

I am designing a PCB (Protection Circuit Board) for a single cell Li-ion battery.

I know I can buy these PCBs, but I have a limited space to work with and since I have worked with printed circuit boards and ICs before I assumed this task would be trivial. Here is what I have so far:

IC for over voltage, under voltage and over current protection of single cell Li-Ion:
http://datasheet.sii-ic.com/en/battery_protection/S8241_E.pdf

I understand how everything works, except one thing: how does the over current protection work? In other words, how do I set my current limit to 500mA or 1A, or 2A. I was looking for some cap and resistor values to control the current limit value, but that doesn't seem to be the case here. I read through the entire datasheet a few times but I cant seem to figure it out!

Someone please help me!

 

[kingofkeys79]

Anyone guys?

 

[HKJ]

The on resistance of the power fets controls the current limit. Remember that this resistance is temperature depend!

 

[kingofkeys79]

Ooo! That makes total sense. So the ideal current limit = (voltage listed on the IC datasheet)/(on-resistance of the two mosfets) :)
Thank you HKJ

 

[kingofkeys79]

I have a follow up question that is confusing me a little.

So if I want a current limit of 0.5A, then according to the equation above (with voltage listed on the IC = 0.1V), then the on-resistance of the two mosfet has to be 180 milliOhms. Would not building such a design give me a lot of power loss? What are my options?

 

[HKJ]

I have a follow up question that is confusing me a little.

So if I want a current limit of 0.5A, then according to the equation above (with voltage listed on the IC = 0.1V), then the on-resistance of the two mosfet has to be 180 milliOhms. Would not building such a design give me a lot of power loss? What are my options?

It will give some power loss, but not that much, remember that it is 0.1 volt from 3.6 volt.
I believe that this voltage drop is the reason some manufacturers uses limits of 10-12A on the LiIon batteries, instead of the more reasonable 6-8A
You can get chips with lower voltage threshold.

Remember that resistance does also include loss in mosfet connections and PCB traces, you can even add a resistor, to get the lower limit.

 

[kingofkeys79]

In that case, if you look on page 27 of the datasheet, it shows a resistor R2 = 1kOhm. Should I add this into my calculation of the current limit. ahhhhh I am loosing my basic circuit analysis skills

Edit: If I do include the R2 into my calculations, then the equation becomes Current Limit = VM/ (R2 + on-resistance)... and that equation doesnt make sense because R2 is HUGE! What am I missing here?

*VM = Voltage threshold for over current detection listed in the IC datasheet (0.08V)

 

[HKJ]

In that case, if you look on page 27 of the datasheet, it shows a resistor R2 = 1kOhm. Should I add this into my calculation of the current limit. ahhhhh I am loosing my basic circuit analysis skills

No, it is for reverse charge protection.
A resistor would have to be added between FET2 and the junction with R2.

 

[kingofkeys79]

Thanks for all your help :) Doing the PCB layout now.