LM5145: SLVAEQ9 document question

Hi Imi-san,

It seems the question is centered around the rise/fall times of the switch current, which is associated with the Qgs2 interval. My understanding is that it's the average voltage during this time, which is (Vth + Vpl)/2, that is relevant for that calculation.

I will ask the authors of the app note to confirm and refine the expressions if needed.

Regards,

Tim

Hi Imi-san,

The turn-off (tf) calculation looks good. However, the turn-on voltage during the Qgs2 interval should be reviewed. The correct voltage difference is Vdr - (Vpl + Vth)/2, as this is the average voltage applied across the gate resistance during that time interval.

You have Vdr - (Vpl - Vth)/2, which is incorrect. The correct expression is Vdrv - (1/t2)*Int[....], where Int[...] is what you have for the integral. If you calculate that, you should get Vdr - (Vpl + Vth)/2

Regards,

Tim

Thank you, Imi-san.