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Calculating Switching Losses in BQ24650

Other Parts Discussed in Thread: BQ24650

Hi All-

I'm creating a circuit to handle an 8A charge current with a BQ24650 chip, and as part of design verification, I want to be sure I'm calculating the switching losses of the FET correctly.  To be sure I have it correctly, I've calculated the switching losses here for the 'typical' switching circuit used in TI's EVAL board.  

Are these calculations correct?

Thanks!

Parameters from the design, MOSFET, and BQ24650:

Design Parameters Value MOSFET Parameters Value BQ24650 Parameters Value
Output Inductance 10uH Rdson 27 mOhms Vregn 6V

Output Capacitance

15uF Qg 4.9nC Ron 6 Ohms
Switching Freq, Fs 600kHz Qgd 1.5nC Roff 1.4 Ohms
Vin 21V Qgs 1.4nC
Vout 12.6V Vplt 2.5V
Duty (D) 0.6

From equation 17 on page 22 of the BQ24650 Datasheet, I can find that the conduction losses of the top MOSFET are:

D * Ichrg * Ichrg * Rdson

And the switching losses of the top MOSFET are:

0.5 * Vin * Ichrg * (Ton - Toff)*Fs

To calculate Ton and Toff, I need to calculate Qsw (equation 19), Ion (equation 20), and Ioff (equation 20):

Qsw = Qgd + 0.5 * Qgs   Qsw = 2.2nC

Ion = (Vregn - Vplt) / Ron  and Ioff = Vplt / Roff     Ion = 0.58A, Ioff = 1.78A

Ton = Qsw / Ion and Toff = Qsw / Ioff.    Ton = 3.7nSec and Toff = 1.2nSec

Using the equation for the power loss from consuction and switching, I find conduction losses of the top FET at 0.064W and switching losses of the top FET at 0.063W.  

Equation 21 on page 22 shows that the bottom FET only has conduction losses:

Pbottom = (1-D) * Ichg * Ichg * Rdson.  Pbottom = 0.04W.

Are these calculations correct?  Have I missed anything?  Also, why are there no switching losses for the bottom FET?

Thanks a ton!