Hello!
Any ideas on how to relate the voltage that I measure in Vout and the incident radiation on the foto transitor (Watt per square meter)?
I'm sending you the fototransistor's datasheet and the my schematics is shown below

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Cheers

David Mendes

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There is a graph of Photocurrent, I-sub-PCE versus E-sub-e. The devices are graded with a dash number from 2 to 6. The top end of each line is at 0.5 mW/cm^2. At that input level the Photocurrent ranges from just over 1 mA to almost 8 mA. If we pick the -4 device we get 3 mA @ 0.5 mW/cm^2

1K Ohms times 3 mA gives a drop of 3 Volts so Vout will be (5 - 3) = 2 VDC

For the same -4 device we get 600 uA @ 0.1 mW/cm^2

1K Ohms times .6 ma gives a drop of 0.6 Volts so Vout will be (5 -.6) or 4.4 VDC

Looks like you might need the -6 device to get closer to ground with 0.5 mW/cm^2

 

Hi,

Generally, radiation like sunlight gets measured in terms of watts/sq. meter with a radiometer. It is not a phototranistor. You can use a phototransistor for such purpose at a specific illumination frequency, but the spectral response is such that its output is way too "lumpy" to resolve over any significant range of wavelengths. For instance, using sunlight as the example, you cannot measure that part of the phototransistor's output that results from 564 nm light and that coming from, say, 380 nm. You would have to use filters, and still miss a significant part of the total radiation.