Measuring Current recieved from a Photodiode

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crush

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Hi all,
I cant figure out how to calculate the current from a photodiode!
I am using a photodiode in reverse bias to detect and infrared signal incident on it from a transmitter.
Have built the recieving circuit by incorporating a gyrator circuit- as a constant current source to the LED.
when a signal of the frequency of interest in incident on the photodiode, the time constant of the photodiode is such that it cannot react fast enough, hence producing a low signal at the end of the photodiode. This low signal is then rectified and amplified by a transimpedance amplifier.

(For circuit diagra, - refer: http://generous.boy.googlepages.com/photodioderecievingcircuit)
Have the circuit going. I was wanting to test the accuracy of my circuit, hence:

How do i calculate the amount of current that is top be produced by the photodiode? (for experimental pursposes keeping the height of the transmitter constant at 5 cm and moving the transmitter away from the reciever to 1 cm, 2 cm. 3 cm and so on)

the photodiode i am using is a BP104 (https://www.electro-tech-online.com/custompdfs/2006/11/ZD-1947.pdf)
the transmitter generic 5mm (**broken link removed**)

I am driving the LED for short bursts at 1A.

any ideas?
 
Look at the spectral sensitivity specification for your part. This spec says that given an amount of incident optical power falling on the surface, a certain amount of current will be generated. The tough part is figuring out how much optical power is falling on surface. It is a geometry & unit conversion nightmare. But you can probably find some help on the web. Have you googled lately?
 
place a resistor of known value in parallel with the diode, and meausre the Vdrop across the resistor ... it'll need to be a LARGE value resistor otherwise it'll just kill the photovoltaic effect - there is very very little current
 
The photodiode is used in the reverse-biased mode, not in the photo-voltaic mode. So it "leaks" current when light shines on it.
To measure its current just measure the voltage drop across the resistor that is feeding it the reverse bias.
 
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