4500PP06 - Power Received by Detector Differential area on...

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Unformatted text preview: Power Received by Detector Differential area on surface of sphere d A = (r d)(r sin d) Differential solid angle d = dA = sin d d r2 /2 2 Total power emitted by LED Pt = =0 =0 /2 2 IRo cos sin d d sin2 2 0 0 = IRo = IRo Cone of light received by the detector has a half apex angle of = t where sin t = a r Power received by the detector is t 2 PD = =0 =0 IRo cos sin d d sin2 2 2 t 2 = IRo 0 0 = IRo sin t = IRo = IRo = Pt a2 r2 a2 a2 + d2 a2 a2 + d2 For a = 1.00 mm, d = 10.0 mm, and Pt = 1.00 milliwatt PD = 9.90099 W ...
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This note was uploaded on 04/29/2008 for the course ECE 4500 taught by Professor Gaylord during the Spring '08 term at Georgia Institute of Technology.

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4500PP06 - Power Received by Detector Differential area on...

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