Electronic – Photodiode Circuit for measuring light intensity

The responsively number given in the table is specific to your device and is exactly what you need (but only partly - see below). There is no "single" parameter for all sensors as it changes from manufacturer to manufacturer. This is primarily determined by QE (Quantum Efficiency) both internal and external QE that is all bundled up in the one number of responsivity.

What you need is a mapping from Lux to Watts, and then the responsivity maps from watts to current.

All detectors will need a passivation layer on top of them to protect the underlying detector material (Here it's Si) so you'll have layers of SiO2 and other material on top. This is important as the External -QE is concerned with getting the light into the Si. This is explained using fresnel equations, but is best understood by the need to match the index of refraction in air (~ 1.0) to that of Si (~ 3.8), the use of AR (Anti-reflection) coatings, and the interaction of light with the passivation layers greatly affects the external QE of the sensor. Once the light gets into the sensor, internal QE is now the concerning factor. As the light penetrates the Si, it leaves a trail of E/H pairs (electron/hole) which are then swept up in E-fileds in the Si substrate. While the E/H generation is understood the E-fields are what determine which electrons/holes get collected. If you generate a E/H pair but it doesn't get collected then you lose internal QE. The electric fields are in turn created through the distribution of dopants and the applied voltages to the device.

In short, even though the Si absorption characteristics are well understood, individual diodes can vary wildly with design. The good news is that this can determined with the appropriate experimental setup. For example the QE of image sensors (say in the green) can vary between manufacturers from as low as 20% up to 98%. In teh NIR (say around 850 nm) these values diverge even more from 1% to 40%.

Radiometry is the measurement of light in quantitative units, Lux is the same curves with the human photopic response laid over top. Consider that mapping as a dimensionless attenuation factor that is dependant upon wavelength.

Ideally what you have is the illumination vs wavelength spectra, the photopic curve again vs. wavelength (which is easily found on-line) and the sensor response vs. wavelength and from those you'd calculate the amount of current flowing.

You have two deficiencies though. One is that you have not identified your illumination spectra and two, the sensor is only defined at 3 points.

A short hand way of calculating is to use the simple estimate (and it will be only an estimate) of 1 lux =\$\frac{1}{683} \frac{watts}{m^2}\$ @ 556 nm (green). Basically this is saying that if you have a green laser at \$ 1 \frac{w}{m^2} \$ then it will appear as 683 Lumens to the human eye.

You will need to understand the difference between luminance and illuminance. So this means you will need to also say what the imaging/collections system is and in particular it's F/#.

Knowing the relationship between wavelength and energy for light \$ E = \frac{hc}{\lambda}\$ where h = planck's constant, C = speed of light. Will allow you to determine the photon flux. And from that you can come up with the shot noise of the system.

Once you can provide the illuminant wavelength dependance, the collection optics f/# and various other parts I'll come back and fill in the details. Or if you want to use the pointers here to answer the question I can check out the answer for you.

The peak spectral response of your photodiode is actually about 580 nm. So get rid of the blue/white LED, and go with a yellow-green ultra-high-brightness LED. Then get a narrowband filter at about the wavelength of your LED. Thorlabs, for instance, has a good selection for not too much money http://www.edmundoptics.com/optics/optical-filters/bandpass-filters/visible-bandpass-interference-filters/3429/. Don't worry too much about perfect matching of your filter and LED, LEDs usually have a pretty wide (10's of nm) spectrum. The use of a narrow-band filter will be critical in getting rid of unwanted sunlight.

Once you've done that, get a better op amp. If you're trying for cheap, even something like a TL081 will be better than an LM358. You'll have to use a split supply, something like +/- 12 or +/- 15 volts. Get over it. You need a split supply anyways in order to bias your photodiode. If you really don't like providing a higher supply voltage, and LF356 will work fine at +/- 5 volts. Also be aware that you probably should put a small capacitor across your feedback resistor for stability.

This may or may not fix your problem - it will help, but it may not be enough. There are two ways to work around this. The first is to get a second photodiode, and use a different filter on the input, with (let's say) a 100 nm difference in filter wavelength. Just as an example, let's say you use a 580 nm and a 680 nm filter. Both will receive about the same amount of power from sunlight, but only the 580 will get LED power. So you would detect a return only if a) the 580 signal level is high enough, and b) the 680 signal is distinctly lower, like less than 1/3 of the 580 signal.

If this doesn't work, you would need to modulate your LED, and look for the right frequency in the 580 return. The simplest (and least effective) approach is simply to put a bandpass filter on the 580 signal, so only the LED variation gets through. For tougher cases, you need a synchronous demodulator, which sounds scary but can be as simple as an op amp and a FET. See http://www.analog.com/media/en/technical-documentation/technical-articles/Use-Synchronous-Detection-to-Make-Precision-Low-Level-Measurements-MS-2698.pdf, figure 4. This is also called a lock-in amplifier. If you're willing to learn how to use one, you can do amazing things pulling a signal out of noise and background.