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Plan to replace solar cell with lens + photodiode

Powerpoint with initial version of hardware design: CBP_large_lens_calibration_system.pptx

lens: https://www.ebay.com/itm/184986621155?chn=ps&_trkparms=ispr%3D1&amdata=enc%3A1G08XAgSuSmibxPFCG1ZFlQ82&norover=1&mkevt=1&mkrid=711-117182-37290-0&mkcid=2&mkscid=101&itemid=184986621155&targetid=1587268788377&device=c&mktype=pla&googleloc=9002000&poi=&campaignid=19894961968&mkgroupid=148855406073&rlsatarget=pla-1587268788377&abcId=9307911&merchantid=113505756&gclid=CjwKCAjw-IWkBhBTEiwA2exyO38BmKS59kDX7yC1bDVaryyAKyGwXfi_9vPYCjG1mPkSEO2YcfymaxoCHIoQAvD_BwE

Should the focused beam fit on a photodiode?

Theoretical answer: Yes, comfortably. d = 2*w0 = M^2 * k(A, D) * lambda * f/D, where k is ideally 1.27, usually between 1.5 and 2, M is the beam quality, which should be pretty high, f=700 mm, D = 260 mm

So for lambda = 1 um, d = 5.38 um

Then the power through an aperture P(r, z = 0) = P0 (1 - exp(-2r^2/w0^2)), if want P/P0 = 0.999, then r/w0 = sqrt(-1/2 ln(.001)) = 1.86, and P/P0 = 0.0001 has r/w0 = 2.145. This beam diameter is still under 15 um, so it very easily fits within the diameter of the photodiode.

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