I have an RC circuit connected to a switch node as a snubber. I am facing difficulties determining the required power rating for the resistor. I have looked at this and this, however, it remains unclear to me.

The switching has a frequency of 115kHz, duty cycle 15%, and amplitude of 90V.

The RC circuit has a 10ohm and a 330pF.

To my understanding after reading the other posts, I believe these statements are true:

1. Since the time constant is very small, 3.3ns, the capacitor will be fully charged and discharged once per cycle. 15% on-time of 115kHz lasts for 1.3us, which is much greater than 16.5ns (5*time constant).
2. The energy dissipated by the resistor is the energy stored in the capacitor, which is (C*V^2)/2 or 1.34uJ. For each cycle, the capacitor is charged once and discharged once, resulting in 2x 1.34uJ = 2.68uJ.
3. The average power dissipated by the resistor is 2.68uJ x 115kHz = 0.3W.

Therefore I need to pick a resistor with at least 0.3W rating. However, right now I am using a 0603 resistor rated for 0.1W and everything runs fine, no magic smoke or burned resistor. Am I missing something here?

I cannot tell if the resistor is hot because other components add heat to the PCB; I am not sure if the resistor is hot because of its dissipated power or because of the surrounding components.

I have an RC circuit connected to a switch node as a snubber. I am facing difficulties determining the required power rating for the resistor. I have looked at this and this, however, it remains unclear to me.

The switching has a frequency of 115 kHz, duty cycle 15%, and amplitude of 90 V.

The RC circuit has a 10 Ω and a 330 pF.

To my understanding after reading the other posts, I believe these statements are true:

1. Since the time constant is very small, 3.3 ns, the capacitor will be fully charged and discharged once per cycle. 15% on-time of 115 kHz lasts for 1.3 us, which is much greater than 16.5 ns (5*time constant).
2. The energy dissipated by the resistor is the energy stored in the capacitor, which is (C*V^2)/2 or 1.34 uJ. For each cycle, the capacitor is charged once and discharged once, resulting in 2x 1.34 uJ = 2.68 uJ.
3. The average power dissipated by the resistor is 2.68 uJ x 115 kHz = 0.3 W.

Therefore I need to pick a resistor with at least 0.3 W rating. However, right now I am using a 0603 resistor rated for 0.1 W and everything runs fine, no magic smoke or burned resistor. Am I missing something here?

I cannot tell if the resistor is hot because other components add heat to the PCB; I am not sure if the resistor is hot because of its dissipated power or because of the surrounding components.