Electronic – Thin wall to block LED light from hitting photodiode

First comment is that, based on Figure 4, this setup is simply measuring the temperature of the diode, not the temperature of a blackbody radiator based on IR radiation. (Figure 4 appears to show the heat pack in contact with the diode.) Further, blackbody radiation at the temperatures referenced in the paper peak at much longer wavelengths than the cutoff of silicon diodes (about 1100 nm.)

Second comment regarding the gain: The simple model for a photodiode is an ideal current source in parallel with a capacitor. For a transimpedance amplifier as shown, neglect the diode capacitance since is is shorted to virtual ground. The feedback network is 10 meg in parallel with 0.51 nF, which gives a pole in the transfer function at \$\dfrac{\pi \times R \times C}{2}\$ or roughly 49 kHz. This really says nothing about the sampling rate, it's just how the current from the diode is going to be translated to a voltage by the transimpedance circuit. The sampling rate is set by the A/D converter.

Third comment: these amplifiers have high enough gain-bandwidth to be considered ideal for all practical purposes for this circuit. You can do an exact transfer function but a 10% capacitor tolerance would swamp out the non-ideal effects of the amplifier.

Bottom line though, is that this setup is simply measuring the temperature of the diode. It is not measuring temperature via IR response. You can get a much more accurate tempeature sensor from TI, Analog Devices, National, etc. that is integrated on a single chip. Or you can just measure the leakage current of a diode to do the same thing.

There are chips that already do what you want. There is really no need to build your own thing out of color filters and diodes.

Take a look at the TCS3103 or TCS3104 from AMS for example. They output three voltages which are proportional to the incoming red, green and blue light components.