CSD16415Q5: How to calculate the power loss of a CSD16415Q5？

Hello Hanbo,

Thanks for your interest in TI FETs. The power loss in the FET is dependent on the application and operating conditions. The CSD16415Q5 is optimized for hot swap and OR'ing applications and most of the power dissipation is due to conduction loss, I² x Rds(on). If you know the DC or rms current thru the FET, then this can be used to calculate the conduction loss. One thing to keep in mind, the on resistance, Rds(on) varies with VGS and temperature. It is specified in the datasheet at VGS = 4.5V & 10V. I use the max value specified in the datasheet and then use Figure 8 to estimate the increase due to elevated temperature. For example, at VGS = 10V, max Rds(on) = 1.15mΩ at 25°. If the operating temperature is say 100°C then the multiplier from Figure 8 is 1.25: Max Rds(on) at T = 100°C is 1.15mΩ x 1.25 = 1.44mΩ. You can use the formula in your post to estimate the junction temperature rise but the value for RθJA specified in the datasheet is based on minimum and 1in² Cu pad sizes. Your actual results will depend on the PCB layout and stackup and may actually be better than the value in the datasheet. Please see the link below for more information on how TI tests and specs thermal resistance for our FETs.

http://e2e.ti.com/blogs_/b/powerhouse/archive/2016/06/10/understanding-mosfet-data-sheets-part-6-thermal-impedance

Let me know if you have any questions.

Best Regards,

John Wallace

TI FET Applications