Tool/software:
Hello,
In one of our project for Aircraft we need to pas strict radiated EMI limits. We found that we have some ringing at switching point of buck designed with LM70880-Q1. VIN = 28V, Vout = 12.3V, Iout(max)= 4A, fsw = 500kHz We added 3,3R in series with bootstrap capacitor, it improved ringing (see yellow line in the following screenshot from scope. But still some ringing remain.
So we added also RC snubber - R = 3,6R / 0603, C = 1nF / 0603 / 50V
QUESTION:
My question is. How to calculate estimated power rating for snubber resistor. According document from Philip C. Todd - Snubber Circuits Theory, Design and Application, 1993
Should be:
So when I have fsw = 500kHz, C = 1nF and Vmax at switching point is: Vmax = 30,0013V
Then P = 500kHz * 1nF * (30,0013)^2 = 450.050 mW.
Does that mean that we need at least 500mW resistor like e.g. 1206 from Panasonic ERJ8BQF3R6V ?
Could You please confirm if my calculation is correct. I'm asking because I used 0603 resistor in snubber on the prototype and it looks from thermocamera that power rating is enough. I don't want use 1206 due to space limitations.
Thank You, Martin
