BQ21040: When the charging current of 0.8a is output by BQ21040 and the ambient temperature is 25 degrees, what degree will the IC reach?

With 5V input, an estimate of junction temperature would be (max dropout voltage) * 0.8 A * RθJB  + 25C = (5-2.5V) * .8A * 45.5 C/W  + 25 C = 116 C. This also depends on board layout. The IC temperature is not recommended to be measured with the TS pin, the device has internal thermal regulation when it approaches 125C. The TS pin is intended to measure the temperature of the battery pack.

We do not recommend you use a 100k NTC as this would put the temperature range of the device outside normal range. We suggest a 10K NTC such as SEMITEC 103AT-2. This gives the normal operation of 0 - 45C.

What temperature range are you trying to achieve here?

Best,

Hi there,

I haven’t heard back from you, I’m assuming you were able to resolve your issue. If not, just post a reply below or create a new thread if the thread has locked due to time-out.

Regards,