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<math> P=\frac{V}{(.38 Amps/W) * (10^4 V/Amps)} = 196 \mu W </math>
 
With the .38 Amps/W as the responsivity of the photodiode and the 100000<math> 10^5 </math> V/Amps is the transimpedance gain of the photodiode. The results agree! This is the power with the chopper on, so double it if we want the actual power that could reach the photodiode.
 
 
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