# IR LED Detectable Range

As a part of a project at work I need to detect an IR LED of my choice (I was thinking about 1~3 W , 730 to 950 nm) from a distance of ~100 m with a camera sensor (was looking at IDS UI-3160CP-M-GL) and I'm trying to figure out how to approximate the detectable range with a given LED and/or a given sensor/camera.

The range of IR led is quadratically proportional (not really but close enough) to the emitter intensity $I_e$. The formula to calculate the transmission range is: $$distance = \sqrt{\frac{I_e}{E_e}}$$ where $I_e$ is emitter intensity and $E_e$ is sensitivity of the receiver. Some sources cite $E_e$ as irradiance whose relationship to sensitivity of the receiver I do not understand and thus a source of confusion.

The terms intensity (could not get a consistent definition) and sensitivity of the receiver are not well defined thus another source of confusion. So if some of the more experienced guys could chime in, that would be welcomed.

• Use laser, really. – Eugene Sh. Oct 10 '17 at 19:58
• Can you provide more details about what you're trying to do? In principle, a one watt LED at a hundred meters should show up extremely bright on an IR-sensitive camera with any reasonable shutter speed, provided the background light is not too bright (although there are ways around this). We need more context to understand exactly what you're trying to achieve. – Peter Oct 10 '17 at 20:12
• @Trevor - yep. Integration time (shutter speed), nighttime vs daylight, detector noise floor, ability to modulate the IR source or not. – Peter Oct 10 '17 at 20:19
• And of course how much image processing you want to do to identify it - just pick the hottest pixel vs. do edge detection on a shape. – Peter Oct 10 '17 at 20:20
• The traffic light override systems (used by gov't vehicles to switch the lights) must contend with sunlight, reflections, etc. To make it work, they use a crystal-controlled, low frequency IR pulses; use optical baffles; use thin film optical filtering; use an extremely narrow-band electronic receive filter (Q=10k bandpass); and selected transmitter and receiver angles; before attempting electronic detection. – jonk Oct 10 '17 at 20:34

It is possible to get a rough idea of the numbers involved.

Start with the LED. Assume that this is a 900 nm LED emitting 1 watt of optical power over a hemisphere (pi steradians). This, of course, isn't true. While an efficiency of 1/3 (3 watts of electrical power giving 1 watt of IR) is reasonable, even a wide-angle illuminator won't emit over a view angle of 180 degrees. So assuming such a radiation pattern will underestimate the on-axis power and grossly overestimate the far off-axis power. But it's a start.

Photon energy can be expressed as $$E(ev) = \frac{1.24}{\lambda}$$ so individual photons at 900 nm (0.9 micron) have an energy of about 1.4 eV. 1 eV is about 1.6 x 10^-19 J, so in round numbers the LED will be emitting 10^19 photons per second.

The camera you specified is quite small, and let's figure a lens diameter of 2 cm. Its area is then about 3 x 10^-4 m^2. At 100 meters, the LED radiation will be spread over pi d^2, or about 3 x 10^4 m^2. The energy (and number of photons) will be equal to the ratio of the two, so the camera will receive about 10^-8 times 10^19, or 10^11 photons per second.

The camera you specified uses this sensor. Note that quantum efficiency (electrons produced per photon) falls off rapidly at longer wavelengths, and is about 10% at 900 nm. So the electrons produced on the sensor will run about 10^10 per second. Since the sensor has a sensitivity of about 10,000 electrons for full scale, this means that any frame rate less than about 1 million fps will cause saturation of the affected sensor cell.

Well. That seems to make detection pretty simple, right? Not so fast. As Peter K pointed out, background is an issue. How big an issue?

Unfortunately, the camera data sheet does not provide a focal length for the lens. Let's go with a 10 mm lens. Then this calculator says that the area imaged will be about 300 feet across - let's call it 100 m. Since the image format is about 2000 pixels across, this gives a pixel width of about 5 cm. Area is about 2.5 x 10^-3 m^2.

Here is a handy-dandy blackbody radiation calculator, and you can use it two ways. First, let's assume that you have a not-terribly-good-but-not-terribly-bad filter on your sensor, which only lets through radiation in the 800 to 1000 nm range. Now, plugging this into the calculator, you can specify thermal radiation from the background using a 300 K temperature. The band power released is much, much less than the LED power, so you can ignore it. This makes sense - if the thermal radiation were significant, the chip could be used as a thermal imager, and the company would be really raking in the bucks.

Now run the calculator using 5500 K, which will simulate a reflection of sunlight. You'll get about 4000 W/m^2/sr. Multiplying by pi (to get a full hemisphere) and by 2.5 x 10^-3 (to compensate for the area actually seen by the camera pixel) gives a bit over 30 watts. This is 30 times greater than a similar measure for the LED, and shows that you really need to pay attention to optical filtering. If you want solar reflections to be 20 dB below the LED (a factor of 100), you need to reduce the bandwidth of your optical filter by a factor of about 300, from 200 nm to about .7 nm. Well, OK, you can relax that by a factor of at least 2, since the atmosphere attenuates sunlight a good deal. But this still means an optical bandwidth of about 2 nm, which is a pretty stiff spec. To make things worse, if you check your LED data sheet, you'll find that it's not a laser, and has an emission bandwidth probably on the order of 10 to 20 nm.

And on the other other hand, this is a real worst-case situation, involving a perfect reflection from a roughly 2 inch by 2 inch area around the LED. It seem likely that you can control this. It also means that you need to be damned sure the sensor never looks at the sun. If your target is moving in an arbitrary manner, this may be harder than you think.

So, the takeaway is that, as Peter K suggested, you should have no trouble detecting a 3 watt LED with even a modest set of optics and sensor at the ranges you are talking about. You'll need to be careful with the construction of your LED housing in order to eliminate specular reflections of incident sunlight. You will also need to be careful about LED construction. As I noted, the LED will need to be pointed approximately toward the sensor, say within 30 degrees. If you can't guarantee this, you'll need to provide an array of LEDs, all pointing in different directions and all being strobed.