# Photodiode current estimate in an Infrared Proximity sensor

1. Mar 15, 2015

### vst98

I am trying to estimate a current which I can get from the photodiode in a simple proximity sensor application
having a IR LED emitter and a photo-diode detector. The LED and the photodiode are mounted in the same plane, on a PCB board, and separated about 10mm.
If I have a flat Lambertian surface (like a gray card) parallel to the PCB and a LED emitter like VSMY2850 where all
of the emitted power is inside the half angle, then irradiance at the photodiode (detector) can be found from

Ed = Ls * π * sin21/2)

I think Ls can be written as

Ls = M/π = R*Es

where R is the reflectance of the Lambertian surface (gray card) and Es is the irradiance of the disk.
Since all of the radiated power of the VSMY2850 is contained within its half angle

Es= P/A = P/(a2*π)

a - radius of Lambertian disk
P - radiant power emitted on the disk from the LED (~ 55mW for 100mA)

but since tan(θ1/2)=a/d , I have

Ed ~ 1/d2 *cos21/2)

So for θ1/2=10°
Ed ~ 1/d2 * 0.970

and for say θ1/2=3°
Ed ~ 1/d2 * 0.997

This made me a little confused, considering the same power, if I used LED with 3° (typical for highly focused LED) or If I placed a lens (like a PMMA Plano convex) on the VSMY2850 to get more focused beam from it, not much would change in terms of the current I would get from the photodiode.
I'm not sure if this is correct.

2. Mar 16, 2015

### vst98

By using a narrow angle LED (or an additional lens on the LED), the LED beam spot would start to overspill the target at greater distances then when using wider angle LED.
This is where I can see an advantage if I choose a narrow angle LED.
But if the whole LED beam spot is on the target, difference between 3° and 10° half angle LED's are not much, right ?

d is the distance between the sensor plane and the target and the distance under consideration are from about ~0.5m to maximum I can get.
(forgot to write this in my first post)