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BQ24195: Charge current issue

Part Number: BQ24195

Hi

m using BQ24195 as charger and first boost-up chip in my design, my battery is Swislight 26650(3.7V,5000mAh).

the boost-up is functional, however the charger aint.

voltage of one of my battery is 3.6V, the VBUS is powered by a programmable DC power supply, set the limitation of 5V*2A.

when i power up the VBUS, the current displays nearly 100mA, and i check the datasheet, which means the VBAT is lower than 2V.

different batteries and chips attempt the same result. 

m wondering discussions and solutions.

THX

  • Hi user,

    It sounds like you may be running into a current limit. There are a number of things that can be causing this.

    Can you provide a schematic for this design? Also, are you using the bq24195 EVM, or is this already on your test board?

    Thanks,
    John
  • hi. John

    thank you for reply, and here is my schematic. i used the bq24195 EVM with another kind of battery and i thought no problem like this.

    please help me this, thx

  • Hi,

    I have noticed two differences in the schematics.

    1. You are showing D+ and D- not connected to anything. These pins are used for input source detection which set the input current limit.

    2. There may be an issue with the way the thermistor network is set up.Right now it looks like the TS pin will be half of VREGN, which is very close to the hot temperature threshold. Please refer to the design procedure on page 18 and 19 (section 8.3.3.3)

    Thanks,
    John
  • Hi, John
    i pull D- to 2.7V and this aint solve the problem
    RT1 is set 5.1kohm, RT2 is set 10kohm, this shall avoid the second cause.
    and
    i saw the follow website
    e2e.ti.com/.../2313866
    does it really work?
    wait ur answer
  • Hi,

    Yes this works.

    I just verified in lab using the bq24195 EVM. With D+ and D- open the input current limit get automatically set to 100mA. With D+ shorted to D- the input current limit increased to the value set by the ILIM resistor.

    Also I would like to note that it is recommended to use an actual thermistor network instead of only using a resistor divider.

    Thanks,
    John