Did you know that many digital cameras are sensitive to near infrared?
The more recent cameras employ special "hot mirrors", or infrared-blocking filters, to remove the infrared from the image and produce more realistic colors. Older digital cameras did not have this filter, and can be used to "see in the dark".
I've got a Sony Mavica FD-51, several years old now. While photographing a wireless video camera, I noticed that the LEDs that surrounded the lens of the camera appeared white in the viewfinder of the Mavica, but were dark to the unaided eye.
Infrared illuminator LEDs on a wireless camera.
The obvious thing to do next would be to build an "illuminator", a kind of infrared spotlight, to use with the camera to poke around at night. Especially now that Halloween is coming...
There are IR illuminator kits on eBay for under $30 - but never being one to go just halfway, I immediately sat down to design my own. I started with a hexagonal layout.
Having seen several lots of 100 IR LEDs go on eBay, I figured I'd use that as a target - 100 LEDs. Turns out that 90 LEDs make a nice hexagon six LEDs on a side, with the one in the center missing. Why is the one in the center missing? Because LEDs need series resistors, of course.
I figured that rather than use 90 resistors, I could group the LEDs into chains and use one resistor for each chain. I knew that whatever number of LEDs I arranged in a hexagon would be divisible by six, since a hex has six sides, IF I didn't use the spot in the middle. That would make the total number of LEDs odd - and not divisible by six.
After a great deal of screwing around, this is what I came up with -
Component layout for 90 LED illuminator.
Fifteen groups of six LEDs and one series resistor each, all crammed into a circular PC board 3.5 inches across. Note that the traces shown (in blue) are not the right width. Power attaches at the extreme left and right.
Assuming the LEDs I use are Siemens SFH485 (Specs , PDF), with a voltage drop across each LED of 1.5 volts and a current of 100 mA, the supply voltage required would be at 12 volts (actually 9 volts, but more is preferred to give some overhead for the resistors to regulate the current), and the current draw would be 1.5 amps.
Of course, this illuminator design could be used with visible light LEDs. Imagine the light thrown by ninety 10,000 mcd "ultra-bright" white LEDs? That's a lot of light! However, white LEDs have a higher voltage drop but don't draw as much current - typically 20 mA at 3.6 volts - making a 24 volt 300 mA supply appropriate.
IR LEDs come in various wavelengths, most commonly 940 nm, 880 nm (like the aforementioned Siemens LEDs), and 840 nm. Which would be best for this application?
The 840 nm LEDs tend to be barely visible, since their spectral half-width is wide enough to overlap the extreme red portion of the visible spectrum. 880 nm and 940 nm wouldn't be visible, but they might be too far into the infrared for the camera - I don't know how fast IR sensitivity drops off. Obviously, some testing is in order.
If the 880 or 940 nm LEDs are just as bright as the 840 nm LEDs, I think I'd tend to prefer them. What is the point of a night vision if you're just going to use a visible (albeit dim) flashlight?
Okay, time to nitpick.
"Illuminator" is the wrong word. Illumination is the effect of luminous flux striking a surface. Luminous flux, however, is radiant flux (aka total radiant power) adjusted for visibility - since infrared is invisible, there is NO luminous flux, and thus no illumination. So it's not an illuminator.
There is radiant flux, and thus irradiation, so I guess it's an irradiator. Hmm... perhaps I should call it a "flashdark"? Works for me!
© 2003 W. E. Johns