# Distance of IR radiation [closed]

I have recently been looking at IR LEDs but having trouble in terms of finding the distance (in meters, cm or mm) the IR radiation travels through a medium such as air using the below diagram. Can someone help me to find how far this radiation can travel before its intensity is almost 0? • Joey, you'll need to define 'negligible'. It will work pretty much the same as any light with slight difference in absorption by air and moisture. At what distance does the light of a visible LED become negligible? – Transistor Oct 5 '18 at 21:04
• I don't know that it ever reaches zero. The odd photon might make it out as far as interstellar space and keep going forever. That's why you need to set a minimum value of interest. See nichia.co.jp/specification/products/led/… for some further info. – Transistor Oct 5 '18 at 21:13
• That information is not provided in your chart, nor answerable from it. Practically speaking without good lenses or a laser, in clear air your issue is likely to be spread more than attenuation. To actually figure attenuation you'd need to know the absorbtion of air (and anything in it) at a particular wavelength, the original (absolute, not relative) power, and to make a decision as to what constitutes "so little left it might as well be zero" – Chris Stratton Oct 5 '18 at 21:29
• How far do you want the signal to travel? If you are worrying about attenuation over a few millimetres then perhaps a different transmission medium, like a fibre optic cable, would be more useful. Or maybe you want to measure the attenuation for some purpose and need to know a usable distance between the transmitter and receiver, but then we would need more information in the question. – Andrew Morton Oct 5 '18 at 22:09
• For what value of X does 1/(X^2) = 0? – BeB00 Oct 6 '18 at 6:58

[noise floor analysis, to predict minimum energy and thus distance, is at end of this answer]

As part of robot-positioning research, about 30 years ago I was detecting IR LED energy at 30 foot ranges (the distance in my backyard), with detection indicated by an LM567 logic-output enabling a red LED to glow.

My collaborator and I moved around the transmitter, an IR LED of 10 mA diode current (about 15 mW electrical power, so about 1-5 mW optical) emitting in a +- 18 degree beam width, with the 10 mA coming from an NE555 oscillating with some duty cycle at 30 kHz. We moved the transmitter until aligned with the RX sensor both in boresight and angle-of-deflection; at 30 feet, outside in the sunlight, the RED LED would change state, indicating the receiver was "seeing" the transmitter energy.

The receiver has a DC nulling circuit wrapped around a phototransistor, for sunlight attenuation. There were various DC blocking filters between the phototransistor and the first stage of gain, thus sunlight attenuation was not required to be complete, rather the response to DC (sunlight) needed to be adequate to keep the sensor output away from the rails (that is, still able to respond "linearly").

The first stage of gain was discrete, crafted to be "low noise" and to achieve some logarithmic/compression gain curve.

The detector was the LM567 tone-decoder, potentiometer-set to free-run at/very near the 30 kHz not-really-square wave of the transmitter. simulate this circuit – Schematic created using CircuitLab

Now you need to compute the noise floor of the phototransistor, using 1 Mohm base resistors. And compute the LM567 noise-bandwidth and signal-noise ratio needed for reliable "logic output".

Given these numbers, we can begin to define "Almost zero".

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Here is my noise analysis. Assume the 1 Mohm in base of the phototransistor sets the noise floor; 1 kohm is 4 nV/rtHz bandwidth, and 100 Hz bandwidth produces 40 nV RMS noise. For 1 Mohm, the noise voltage rises by sqrt(1Meg / 1K) = sqrt(1,000) or 31.5X, to 1.2 uV RMS. Into a short (perhaps the Rin of the transistor's base is quite low), the noise CURRENT is 1.2 uV/1 Mohm or 1.2 pA RMS.

Assume the tone-decoder PhaseLockLoop bandwidth is 100 Hertz. Thus the noise floor is 40 nV RMS at the base of the phototransistor. And assume the PLL works down to 0 dB SNR, or about 40 nV of signal (ignoring RMS/peakpeak factors).

Photodiodes have about 1 : 0.5 conversion efficiency from optical power in watts, to output current in amps. If this also works for a phototransistor (the collector-base junction), and the Rin of the phototransistor for Beta=100 and reac (1/gm) is 52 ohms at 0.5 mA Collector currenet (for Vce ~~ 4 volts), the Rin is 5,200 ohms and we need 40 nV across that Rin. The signal current must be

40 nV / 5,200 ohms which we'll make 40nV/4kohm

thus Isignal is 4e-8 / 4e+3 = 10 pA signal current, or optical power being twice that at 20 pW optical power (energy).

[I'm not bothering to be exact about how the 100 kohm Rbase noise voltage interacts with the device's Rin]

Now we need to take 1mW optical energy from the LED, give a +-18 degree beamwidth, and determine how much of that optical energy is captured by a 4 mm diameter phototransistor lense at 30 feet.

Will we find this energy to be about 20 pW?

The effective radiated power is boosted greatly, because of the LED lense; we'll assume (360/36)^2 or 100X, to 100 mW optical, radiated uniformly.

We need to compute the surface area of a sphere of radius 10 meters, then compute the energy intercepting the Receiver Sensor lense of size 4mm*4mm.

Area of 10 meter sphere is 4 * PI * Radius^2, or 12 * 10meter * 10meter = 1,200 square meters. How many square mm? Scale up by 1,000 * 1,000, to 1,200,000,000 square mm.

Our receiver sensor lense area is 16 square mm. Thus the fraction of emitter energy is 16 / 1,200,000,000 or approximately 1/100,000,000 (no need to be silly about precision, here).

With 1 mW LED power (could be higher, of course), the received power is 0.100 / 100,000,000 or 1/1,000,000,000 or a billionth of a watt. Or 1,000 pW.

And 5 paragraphs back we computed 20 pW.

Anyhow, we now have 2 numbers that define "almost zero". 20 pW, or 1,000 pW.

• Great answer, a proper optical link budget and signal to noise analysis. One niggle: the gain of a lens or antenna is 42000/(beamwidth^2) because there are only about 360x180 square degrees out there. So the gain would be 32 for a 36 degree beamwidth. – tomnexus Oct 6 '18 at 9:49
• Thanks for the compliment. I was using full hemisphere model, thus 32*2 would be the gain? And note my caveat: I need to review this for any errors....that is not yet done. – analogsystemsrf Oct 6 '18 at 14:00
• A really good answer, but does not address the question asked. The question was, "what was the attenuation of the IR beam with distance in air". You addressed quite the opposite ...though very well. – Jack Creasey Oct 6 '18 at 15:49

Air is not really a definition of anything and is a poor place to start gaining any understanding.

The transmission characteristics and attenuation of the atmosphere is complex since it's made up of multiple gasses, its pressure varies and it has water and pollutants in it.

If you want to know absolutes, then use the HITRAN database.

These lecture notes might get you started.