- #1

kelly0303

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I would like to convert the maximum voltage in these peaks, call it ##V_{max}## to the laser power corresponding to that (basically the laser power output when my laser frequency is perfectly on resonance with my optical cavity), call it ##P_{max}##. From the manual from that link I have the formula ##R(\lambda) = \frac{I_{PD}}{P}##, where in my case, for ##\lambda = 1064## nm I get ##I_{PD} = 0.45P##.

Also if I assume the resistance to be ##R = 50## Ohm (is this right?) and ignore the dark current, the measured voltage would be ##V_{max}=RI_{max}=RI_{PD}=0.45RP## so in my case I would get ##P = \frac{V_{max}}{0.45R}=0.1/(50\times0.45)=4.5## mW. Is this right or am I oversimplifying it (I don't need a super accurate result, but just a close estimate of the power)? Thank you!