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View Full Version : pic: IR captured in photo (4 IR LEDs @.1A DC ea)


CD47-Bot
11-02-2004, 21:41
[cdm-description=photo]16829[/cdm-description]

Rickertsen2
11-02-2004, 21:42
Thread created automatically to discuss this image in the Picture Gallery.

http://www.chiefdelphi.com/pics/bin/107655041239.jpg (http://www.chiefdelphi.com/forums/pictures.php?&action=single&picid=6136)
Yup make yourself an array of them and you have a flashlight that only your camera can see. = really bootleg night vision

Damian Manda
12-02-2004, 14:02
Ehat camera did you use for this picture? My camera can hardly see anything when pointed at the beacon.

KenWittlief
12-02-2004, 16:14
any old camcorder will see IR remotes and LEDs

try turning the lights off, or hold the LED closer to the camera - BTW this is a good way to check and see if they are working

Joe Ross
12-02-2004, 22:22
Ehat camera did you use for this picture? My camera can hardly see anything when pointed at the beacon.

Somebody else posted that newer cameras have better filters and block more of the IR

Dale(294engr]
13-02-2004, 16:38
Ehat camera did you use for this picture? My camera can hardly see anything when pointed at the beacon.

Camera used: Olympus 3040, digital, 3 megapixel, flash off, macro focus on

This photo was taken with an IR LED powered from a
5 VDC voltage source drawing .11A (33 ohm kit current limit resistor)
i.e. 100% ON duty cycle....
to make it easy for the digital camera to see.

See my other IR Beacon photos in the gallery
showing the much dimmer (but visible) LED illumination when pulsed/burst
<1 mS burst of 40 KHz pulses, & low duty 100Hz rep>

emitter 0:
pulsed at 40 kHz (50-50 duty square wave)
burst for 40 cycles = 1mS (total ON time = 500uS)
and 100 Hz repetition rate (total Off time = 9500uS)
500*100/(9500+500) = 5% ON duty time

use emitter 1 (to improve cameras ability to see IR a bit)
80 cycles = 2mS (total ON time = (.5*2mS) =1000uS)
and 100 Hz repetition rate (total Off time = 9000uS)
1000*100/(9000+10000) = 10% ON duty time