7
\$\begingroup\$

I have an idea for a project where I use a pc fan to make an oscillator. The oscillation is created by the cycle of light/dark falling on a light sensor as the blades of the fan rotate.

My initial research points to the use of the photodiode in combination with a op-amp to provide a suitable line audio output. My question is: What are the critical parameters for the circuit?

Here is the datasheet for the photodiode i'm prototyping: http://www.jaycar.co.nz/products_uploaded/ZD1948.pdf

Rob

\$\endgroup\$
2
  • 1
    \$\begingroup\$ Gijs Gieskes's work provided inspiration for this project. Check out his site and the videos - some amazing creative stuff. gieskes.nl vimeo.com/gijs \$\endgroup\$
    – Rob
    Jan 18, 2010 at 22:43
  • \$\begingroup\$ There are a lot of simpler ways to make a oscillator. Why such a kludge? \$\endgroup\$
    – Olin Lathrop
    Oct 7, 2013 at 11:38

4 Answers 4

Reset to default
4
\$\begingroup\$

If you just use a 555 timer you won't need to amplify it, you can plug it directly into a line level input (microphone in on computer, aux in on hifi etc). By putting the photocell on the bottom of the fan and a light source above the fan you could get some interesting results. The speed of the fan would create an oscillation, also the 'brightness' (pitch of harmonic content) of the oscillator could be adjusted by changing the intensity of the overhead light source. Amos is right, you could use a very similar circuit to the one in my Posc, I'll draw up an example just give me a mo....

\$\endgroup\$
3
  • 1
    \$\begingroup\$ It's real easy to fine tune the circuit, you can put a resistor in series with the photocell, or you can change the value of the ceramic capacitor (the smaller the value of your cap, the higher the pitch and brightness of your oscillator). For instance - A 0.1uf cap will take the 555 timer to the limits of the human hearing range when you shine a bright light into the photocell, but when you cover the cell the pitch should be lower than the human hearing range and will be audible as pulses not tones. \$\endgroup\$
    – Jim
    Jan 18, 2010 at 14:27
  • 2
    \$\begingroup\$ You can also use a similar trick to listen to the discreet frequencies found in the laser of an optical mouse. Just hold the laser on the photocell of said circuit, and listen to the crazy noises! \$\endgroup\$
    – Jim
    Jan 18, 2010 at 14:36
  • 1
    \$\begingroup\$ To be exact, it's making a square wave oscillator to create a tone, then your using the fan to modulate the pitch of that tone. It's not a sine wave by any means, but you won't get that from a PC fan! I'm positive it would make a pretty great sounding oscillator tho \$\endgroup\$
    – Jim
    Jan 18, 2010 at 14:49
2
\$\begingroup\$

Alex who is a member here has details over on his tinkerlog site of something he calls Synchronizing Fireflies, these may have some useful details in.

But of more use to you is probably Jim's site Sonodrome which has details of his Posc on it. This has two oscillators controlled by light dependent resistors and can output to a stereo. There is a schematic etc in there somewhere.

\$\endgroup\$
2
  • \$\begingroup\$ I like the use of the ping-pong balls to make the glowing orbs and self organising systems are fascinating in the patterns that emerge. \$\endgroup\$
    – Rob
    Jan 18, 2010 at 22:39
  • \$\begingroup\$ Me too, I thought they were fantastic. \$\endgroup\$
    – Amos
    Jan 19, 2010 at 11:01
2
\$\begingroup\$

The critical parameters are the light and dark currents. The dark current has a maximum value of 30nA. The light current has a minimum value of 30uA and a typical value of 40uA. Since the ratio is 1000:1 there will be a large difference between on and off.

If you use an op-amp to output a line level I would set the gain resistor so that a 40uA current gives you 80-90% of the line voltage level.

The downside is that the level may be different for different diodes. You could put a pot in the feedback loop for adjustment.

If you want to get more complicated you could have the photodiode trigger an analog switch (SPDT). The precise line level could be set by adjusting the inputs of the switch.

On the Hamamatsu site there is some good technical information and application hints for photodiodes.

\$\endgroup\$
2
  • 1
    \$\begingroup\$ Do not count on reaching either the dark current and light current unless you actually reach light and dark, a fan blade will probably not reach these levels, but they are a good place to start. Plan to have more gain in your pocket. \$\endgroup\$
    – russ_hensel
    Jan 18, 2010 at 1:17
  • \$\begingroup\$ Yes, good point, i'll take some measurements with the components in-situ to get an idea of the dynamic range of the photodiode in normal conditions. \$\endgroup\$
    – Rob
    Jan 18, 2010 at 21:36
1
\$\begingroup\$

Sorry it's not too clear, but you should get the idea, just ask if you wanna know more.

alt text

\$\endgroup\$
8
  • \$\begingroup\$ You may need to make a cover/lid for the top, so it limits the overall light going into the box, but that depends on what kind of sound you want. \$\endgroup\$
    – Jim
    Jan 18, 2010 at 14:18
  • \$\begingroup\$ If you do make this project, I wanna hear it! ([email protected]) Cheers. \$\endgroup\$
    – Jim
    Jan 18, 2010 at 14:59
  • \$\begingroup\$ Wow, thanks for your detailed help Jim. I'll read up about the 555. Lots of work to do! \$\endgroup\$
    – Rob
    Jan 18, 2010 at 21:52
  • \$\begingroup\$ Hi Jim, i've documented some testing results here and some questions: mintmedia.co.nz/Fan_oscillator \$\endgroup\$
    – Rob
    Jan 19, 2010 at 1:26
  • \$\begingroup\$ Dude! you're totally right, the photodiode is different to the photocell, my bad. It won't work in the same way. The photocell is acting as a variable resistor. Sorry, I'm not too hot on photodiodes \$\endgroup\$
    – Jim
    Jan 19, 2010 at 7:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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