Electronic – 660 Mbits per second photo-diode amplifier on 3.3V rail

First, the Lighthouse puts out a lot more optical power. Instead of a puny little 5 mW laser, they apparently use a string of lasers, and you can easily get 10 times more power from that sort of setup.

Second, you seem to have missed the basic idea about emitters/receivers entirely. If your laser produces a visible line (and it seems to do so), why on earth would you try to detect it with an IR receiver? The BPW34 data sheet www.vishay.com/docs/81521/bpw34.pdf gives it a peak sensitivity at 950 nm, but it also works at visible wavelengths, but at lower sensitivity. You're using a visible laser, so I'd expect lower sensitivity from your setup than is possible with an IR emitter of the same power. And going to a BPW FS will only make matters worse, since the FS incorporates a visible-rejection filter, so it won't see your visible laser at all. This also explains why you could not detect other lasers with an IR receiver, although there are other issues such as modulation which you need to learn about.

What you do need is a narrow bandpass filter at your laser wavelength. You can find them on eBay. If you get something, keep in mind that, for it to be really effective, you'll have to make a light-tight housing for your diode so that only light which has gone through the filter hits the diode. You really need to do something like this anyways. Try this experiment. With the laser off, put a box over your PD and take a reading. Call this your zero point. Now take the box off and take another reading, and call this your ambient level. Finally, turn on your laser and take a third reading - call it your active level. If the ambient is significantly different from zero (and it will be) this represents a laser light level which is unmeasurable, since the laser is being masked by the ambient.

An amplifier is probably a good idea, but this is clearly going to be a learning experience for you.

In this circuit the junction is generating a photocurrent, and in the above circuit the op amp will generate just enough current through the feedback resistor to cancel the photocurrent such that the voltage across the junction is zero volts (in the ideal case.) That's because any difference in voltage between the + and - inputs gets amplified by the very large voltage gain of the op-amp and the feedback serves to servo the voltage back to zero. So the only offset you will have is the input offset voltage of the op-amp.

Note that this is not the fastest response circuit that you can get. The junction capacitance is large with no bias, so if you need response speed you may want to explore a TIA with reverse bias of the photodiode.

Also, it's relatively uncommon for a photodiode to have 3 leads. Are you sure it's a photodiode?