Dear all,
Our customer is using CSD18510KCS.
The loss power RRG of the gate resistor at that time is as follows. Is it correct?
◎Condition
・ Vg = 10V
・ Qgtotal = 118nC (typ.)
・ Fsw = 80kHz
PRG = Qgtotal x Vg x fsw= 118nC × 10V × 80kHz
Best Regards,
Y.Ottey
Hi Y.Ottey,
Thanks for promoting TI FETs at your customers. The power calculation you included in your post is the total dissipation to turn the FET on and off. This will be divided between the driver IC, external gate resistor and internal gate resistance of the FET based on the values for the internal resistance of the gate driver, external resistor and internal gate resistance of the FET. For the CSD18510KCS, the internal gate resistance is specified as 0.9ohms (typical) and 1.8ohms (max). See equations below.
