I have been looking around for any cheap (less than $7) alternative to the official Rasberry Pi camera module. Is there any extremely cheap camera module/feature/extension available for Rasberry Pi?

The goal is to have an array of 25+ camera modules that will asynchronously capture pictures. The quality of the camera does not have to be up to par as the official Rasberry Pi module, even a low quality image would be ok.


  • The goal is to have.....? You are hoping to find a camera that is less than 1/3rd the price of the Pi Cam with more features than the Pi cam they may not be a realistic expectation. – Steve Robillard Apr 17 '16 at 23:40
  • oops sorry I think the rest got cut off, I will make it clearer – redlime9 Apr 18 '16 at 1:27
  • you want to connect 25 cameras to one pi? – edo1 Apr 18 '16 at 8:18
  • The cheapest usable cameras would be some kind of really bad USB webcam. Ebay is likely to be a good source. The only application for 25 simultaneous cameras I can think of is a bullet time rig - you'll never pull that off like this. Worth a look at this post. – goobering Apr 18 '16 at 8:39
  • @goobering thank you for that post, yes thats the kind of idea I have but the cameras are to take photos asynchronously of a certain chemical reaction in different wells. Each well will require one camera. I hope though that I would not need 48 rasberry pi's as used in the bullet time rig – redlime9 Apr 19 '16 at 23:46

You can get one on Aliexpress for under $7, right here:


| improve this answer | |

Currently, eBay has this OV5647 5MP Camera OV5647 Camera Module for Raspberry Pi A/B+/2 Model B W/ Cable (1080p) US $3.97 + Free S/H and this New USB Camera for Raspberry Pi 2 Model B/B+/A+ Not Require Drivers HOT (640x480 video resolution) US $3.99 + Free S/H. According to both specs, the former has a much better video resolution, at least 4x the later, AFAICT. So, I certainly will get the former if I need one.

| improve this answer | |

Not the answer you're looking for? Browse other questions tagged or ask your own question.