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I have an IR emitter (EMIRS200_AT01T_BR080_Series) with the datasheet showing the Optical output power versus the opening angle as below.

Graphic

Now, I would like to understand how interpretation the opening angle (integral rotation of a cone). Does 90° mean the perpendicular direction from the emitter and lower angles until 0° that means to be parallel to the floor?

I'm using a Thin Film Pyroelectric Dual Channel Sensor as a receiver

NicoCaldo
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2 Answers2

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I think this graphics explains the terminology well

enter image description here

Source

G. B.
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What the graph is telling is the way that the emitter is transmitting light in terms of angles. So, if you had a really wide angle photodiode to catch all of the emitted beam, you would catch all of the power emitted. However, if you used some form of collimator that restricted the angle of reception to about 8 degrees, you would only catch half of the emitted light power.

Andy aka
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  • I'm using an IR gas sensor located directly in front of the emitter at around 10 cm from the emitter. Is this position considered as 90° on that table? – NicoCaldo Jul 30 '19 at 09:46
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    @NicoCaldo no not at all. The sensor has an active area of 1 square mm and, using the graph in your question, the angle of beam divergence is about 20 degrees. If your sensor was up real close to the emitter, all of the radiated light might just about fall onto all of the 1 sq mm of detector and you would get full power transfer but, your receiver is 10 cm away and that point source of light has spread to an area that is hundreds (or thousands) of times bigger AND the power has thinned out by the same amount hence, that light hitting the PD 1 sq mm is much, much smaller. Do the math! – Andy aka Jul 30 '19 at 09:53