CdS Photocells

Thanks, Phil. You probably saved me some time and effort.

Nice book.

Reply to
John S
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This fixture was used to calibrate photoreceivers. It needed to make a uniform, mostly position-independent illumination. We had to bead-blast the inner surface of the tube to minimize the effect of goofy wall reflections.

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John Larkin         Highland Technology, Inc 
picosecond timing   precision measurement  

jlarkin att highlandtechnology dott com 
http://www.highlandtechnology.com
Reply to
John Larkin

Transferring optical power calibrations accurately is hard. It would be ver y nice if there were an optical equivalent to the Fluke 87, but there isn't . 1% absolute accuracy is pretty good going in optics, though of course rel arive accuracies in the 1E-6 to 1E-8 range are quite possible.

Cheers

Phil Hobbs

Reply to
Phil Hobbs

ery nice if there were an optical equivalent to the Fluke 87, but there isn 't. 1% absolute accuracy is pretty good going in optics, though of course r elarive accuracies in the 1E-6 to 1E-8 range are quite possible.

Hmm, well if we use some semi (say Si) and just quote photo current.. ignoring the surface reflection. (if you use an AR coating then you have t ell.) I feel pretty good to the ~10-20% level.... That's close enough for me.

Is there any experiment that needs 1% absolute optical power measurement? Changes in intensity are a whole 'nother matter, as you say.

George H.

Reply to
George Herold

Well, we've got used to doing without, of course. Light bulb and csmera mak ers are probably pretty happy with that state of affairs. ;)

Cheers

Phil Hobbs

Reply to
Phil Hobbs

Net planetary insolation measurement just reached the 1% level in recent years. I was at an awards ceremony for the scientist who received the accolade for achieving it (a life's work!).

They believe that the Earth *system* absorbs about 403 watts/m^2 net in solar radiation. The net emitted (black body?) radiation is believed to be within 1% of that, so the error margins on whether the earth is gaining or losing heat is around the 4w/m^2 mark, or about 2 terawatts globally.

The thermal balance is surprisingly fine, in any case. Better measurement than 1% is needed to detect an imbalance.

Clifford Heath

Reply to
Clifford Heath

Thanks. I think I found the discussion on pages 399-400. As I said this is really a learning exercise. So far I have thrown out everything in the Nuts and Volts article except the Cds sensors. And may throw those out after I try them.

I like your advice on designing optical instruments. Most of the advise applies to all engineering.

Dan

Reply to
dcaster

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