Optics question (Fresnel equations)

You just treat the two polarizations independently and add up the photocurrents when you're done.

LEDs are pretty well unpolarized when you look at them from a distance.

There are polarization effects with angle, due to the Fresnel reflections from the top surface. If the LED has a flat top facet, p-polarized light escapes better, so there's a tendency for the light to be somewhat radially-polarized. Textured surfaces and lensed packages smear that out pretty well, though, so to leading order your LED should be unpolarized.

Thus, it's a good guess to assume the LED light has equal amounts of s- and p-polarized light. These don't interfere, so the total photocurrent is just the sum of the s and p photocurrents.

Cheers

Phil Hobbs

OK, thanks, makes sense now.

The LED is an Osram SFH4050, the top surface is slightly frosted so as you say, hopefully I can just treat it as 50:50 split between s and p polarization.

One piece of odd behaviour I did see with this LED - I assumed the output power would be roughly linear with current, and lose efficiency and tail off as the die heated up. But going up in 50 uA steps to about 5 mA (max is 100) there's a noticeable upward curve. At first I thought I'd somehow screwed up the photodiode amp, but I tested it on an Ophir Nova II and got the same results. I don't remember seeing this before with other LEDs.

So long as the output power is long-term stable to within a few dB it won't matter (there isn't space for a monitor photodiode in the design). I will put a few of them on continuously for a few months, just to check.

Sort of, it's a non-contact fluid sensor.

Depends on the device. There's normally a bit of a toe at low currents--in the low tens of microamps for normal display LEDs--but then it's pretty linear.

If your LED is something unusual it might behave differently.

Cheers

Phil Hobbs

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