BQ24600: Estimating power dissipation and Tj

Hi Chris,

   BQ24600 is a controller, so the IC itself does not consume much current, however the junction temperature depends on choice of external components (RDSon of FET etc) as well as test conditions (which contribute to switching losses) and how close in proximity on the layout the switching FETs are to the IC. It varies layout to layout and test condition to test condition so it is hard to estimate junction temperature. You can look at obtaining an EVM to test and see how much the temperature rise is.

Thanks, but I do believe that the bq24600's gate drive circuitry for the two external FETs is responsible for most of the power being dissipated in the chip.  Is that not so?  I will do some research on gate drives. 

I have measured case temperature on my board, but I'm a little skeptical about applying the junction-to-top characterization parameter (ψjt = 0.6 °C/W) because it seems unreal that junction temperature is little different than case temperature for this chip, which surely is dissipating only a fraction of a watt.  Nonetheless, I would like to figure out how much power is being burned in the chip, versus in the FETs and inductor.

Chris

Hi Chris,

 Yes, the gate driver will contribute to the losses as well, so you will see the heat spots localized more on the switching FETs, and the IC due to gate drive losses. This depends a lot on the Qg (selection of external switching FETs), and as BQ24600 has higher switching frequency (1.2MHz) this will contribute to larger gate drive loss as well.