This is my first time working with lasers, I made a data transmission system with a laser using normal components, and now I would like to change It to a very long range system (~2-3 KM).

My question is, would I be able to use these cheap IR photodiodes: Link

if I use a very powerful laser Such as > 5-watt laser? If not do you have any recommendations for a photodiode for the range I specified?

  • \$\begingroup\$ Is it fiber optics, or over the air? \$\endgroup\$ – Lior Bilia Nov 2 '18 at 10:49
  • \$\begingroup\$ @LiorBilia Over the air \$\endgroup\$ – G. Doe Nov 2 '18 at 10:52
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    \$\begingroup\$ (Laser) light has a certain wavelength. The receiver (photo diode) must be able to "see" the wavelength that the transmitter (laser) emits. You should mention the wavelength of the laser and include a datasheet of the IR photodiode and check that the laser's wavelength matches the diode's sensitivity curve. \$\endgroup\$ – Bimpelrekkie Nov 2 '18 at 11:11
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    \$\begingroup\$ 5W @ 808nm as a long free space path? Not in any sense eye safe, you really don't want to do that. Pick a wavelength that is eye safe and use a beam expander to get the power density at the transmitter down (this also lowers the diffraction limit so you actually get more power density in the far field). What are you using as an optical bandpass at the receiver? \$\endgroup\$ – Dan Mills Nov 2 '18 at 12:05
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    \$\begingroup\$ Caution: unless you have proper optics safety training, I would strongly recommend not using any laser over 5 miliwatts! Pointing a 5 watt laser into open air with no additional safety design is extremely dangerous to bystanders and probably illegal in your jurisdiction. \$\endgroup\$ – pjc50 Nov 2 '18 at 14:34

So I can't find the data sheet for the diode in question but I have approximated with a similar diode

The current though the diode is proportional to the intensity of the light on the diode. in your case we have a 808nm laser, lets round that to 800nm for simplicity. For your requirement there are a lot of variables to take into account such as environmental factors, background radiation etc. solar radiation spectrum

The peak response from the diode is at 940nm where we have an intensity of roughly 1W/m^2 so roughly 0.0001 W/cm^2 or 0.1mW/cm^2 so we can assume that the response from the diode is 0.

Here is the relevant graphs from the data sheet: enter image description here From our data sheet we can see that peak response is at 940nm so your 800nm laser will have a relative intensity of about 0.3 (quite low, I would suggest getting closer to that 940nm mark)

lets say we are converting current to voltage with a resistor and op-amp. set our logic threshold to everything above 1V is logical 1, everything below 1 is logical 0. lets say that our system turns 50uA of current though the diode into 1V output with a simple resistor and op-amp with a gain of 20000. This means that the diode will need to receive 1.1mW/cm^2 of 940nm light, but will need 3.67mW/cm^2 from your 800nm laser.

lets say that you're going to be transmitting in clear conditions over 2km with negligible weather conditions (you're traveling down the rabbit hole of optical comms, SNR and environmental effects) but lets assume that your laser pulse is 50% as intense when it arrives so you're going to need at least a 7.4mW laser. I feel I have approximated too broadly there and the actual required power is going to be possibly orders of magnitude higher, but the premise is the same. you will need to take that 3.7mW received power and work backwards taking into account how the light diffuses and spreads as it travels and what the effects the environment will have on the light pulse. Maybe other commenters can expand on these effects.

I don't know much about optical comms, but I'm assuming that real laser or optical communication systems are finely tuned to specific frequencies so finding a diode with a narrow response and a laser with a tightly matching wavelength will be very important for signal integrity.

  • \$\begingroup\$ Thank you for taking time to write all these details :) very helpful, I will dig deeper into what you explained and so some more researches. \$\endgroup\$ – G. Doe Nov 2 '18 at 20:43
  • \$\begingroup\$ No worries @G.Doe, happy to help, good luck with your project \$\endgroup\$ – B_Gsy Nov 5 '18 at 14:16

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