I designed a battery charging, protection, and boost circuit for a project I am working on. However, I made a complete noob mistake and forgot to find the overcurrent value for the battery protection IC I am using, and it turns out the overcurrent value is too low.

The manufacturer makes multiple protection ICs in the same package size, so I can likely just swap it out with a new IC with a higher overcurrent value. However, overcurrent is specified as overcurrent voltage, and I am not sure how to convert this into an actual current draw.

I need the protection IC to allow charging current of at least 1A (no more than 2A), and a discharge current of at least ~1.5A (and no more than 2A). The battery is a 2200mAh li-po cell so this should be perfectly within spec.

This is the datasheet for the IC (I am using the AP9101CAK6-ANTRG1 variant): https://www.diodes.com/assets/Datasheets/AP9101C.pdf?src-supplier=Digi-Key

Here is what I have tried:

  1. I changed the value of R2 (from the example circuit, R5 in my circuits) to 1k from 2.7k . I noticed no difference in the allowed current draw.
  2. I shorted P- to Battery - to bypass the MOSFETS, this allowed for higher current draw, which confirmed that it is a problem with the protection IC and not the boost converter.

I also am having trouble with getting the battery to charge, I tried connecting the battery directly to the charge IC and charging worked normally, so again there is a problem with the protection circuit.

If it helps at all, I have attached my circuit schematics below

Charge Controller Schematic

Protection and Boost Converter Schematic

Any help would be greatly appreciated as I have struggled with this for several hours, but I think I lack the background knowledge to figure this out.

  • $\begingroup$ Looks like U2 is AP9101C. Can you please share the part number for U1 and U3. I suggest updating the body the question with this information. $\endgroup$ Commented Dec 23, 2019 at 15:16
  • $\begingroup$ have you solved the problem? I am using the same IC (AP9101C) in a very similar circuit and I am having similar problems. Would appreciate, if you could share some solution. $\endgroup$
    – emafors
    Commented Mar 14, 2021 at 12:06
  • $\begingroup$ @emafors, can you describe the issues you are encountering? $\endgroup$ Commented Mar 14, 2021 at 16:29

3 Answers 3


don't know whether you resolved this. Answer: Your part has one of the lowest over current thresholds (60mV in table), pick a higher voltage. Unfortunately I can't see how the voltage translates to current in the datasheet so you will have to base it on the measured cutoff current from the original part. Also did you ever perform a battery discharge test and then discharge it a bit more than the cut off point of this part? By doing this I have managed to get it into an unrecoverable state that can only be rectified by shorting the charge FET, not ideal in an end user application! I'm waiting on support from Diodes Inc. but just interested in other users experience.


The FETs will have an resistance when they are turned on. V=IR, or I=V/R. So:

  • [current limit] = [overcurrent voltage of AP9101C] / [total Rdson of the two MOSFETs]

Or if you've already chosen your MOSFETs and current limit and want to find which AP9101C variant you need:

  • [overcurrent voltage of AP9101C] = [current limit] * [total Rdson of the two MOSFETs]

Disclaimer: Not tried this myself yet. About to try the above, but it seems to make sense to me.


The current limit formulated in Volts is total Rdson x Idc.

My mistake was I used normal FETs 2N7002 and even a small current of 40 mA caused overcurrent events.

After replacing them with two IRML2502 (with 50 mOhm Rdson) FETs, the device works properly. My AP9101CAUTRG has a 0.100 V overcharged limit, which for dual IRML2502 makes the total current of 1 Amp.


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