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I use a probe for converting optical power to an electrical current. I need to convert Volt to candela/m2. The sensor, a photodiode, measure the brightness of the monitor leaning the probe on the glass (so there is not distance between the monitor and the sensor). The sensor provides the voltage (0.4V). Load resistance: 50 omega, Responsivity (0.725 A/W), diameter sensor (0.009 m). I expect that the conversion will give me 350cd/m2, in fact it is the maximum value of brightness of my monitor.

Sensor: http://www.thorlabs.de/thorcat/TTN/SM1PD1B-SpecSheet.pdf Vout = P * Responsivity * RL = current * RL

RobertAalto
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    Volts measure an electric potential, candela per square meter the density of luminous intensity. There is no possible "conversion" between quantities of different nature. Now, your sensor has a responsivity of $0.725 \mathrm{A.W^{-1}}$. What does it mean to you ? – Tom-Tom Nov 24 '15 at 08:40
  • From the forumulas (9) and (10) from http://www.nist.gov/calibrations/upload/sp250-37.pdf it means something if you consider illuminance responsivity Rv,i (A/lx) = Responsivity * Area and luminous intensity Iv (cd) = dist^2 * current / Rv,i but I have not the distance, so I do know how to make the conversion – RobertAalto Nov 24 '15 at 13:11
  • You haven't defined your source configuration, so you can't calculate the optical power density at your receiver. And it's clear you don't understand the difference between photopic output (candela) and spectral power output. You need to get some local tutoring on this. – Carl Witthoft Nov 24 '15 at 13:27
  • So, is it impossible to measure the cd/m2 from a random monitor with a photodiode? – RobertAalto Nov 24 '15 at 13:57

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