#### staticV3

##### Major Contributor

- Joined
- Aug 29, 2019

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sqrt(0.1526W*300Ω)=6.766Vrms

sqrt(1.45W*33Ω)=6.917Vrms

Therefore, at 65Ω, the Amp will clip at around 6.8Vrms as well.

To convert that to Watts, you do:

6.8Vrms²/65Ω=0.711W

If you tell me the exact headphone model that you have, then I can also show you how to calculate the peak SPL that you can achieve with the C200.