AN442

Rev. 0.1

17

Therefore:

and

Since the SF 4650 has a mathematical equation:

ED from the contribution of that surface can be derived by plugging in the angles, reflectance R of the surface, the

area A, and the distances RE and RD.

Note that in Figure 9, the dimension, D, represents the distance of the target to the PCB hosting both the SFH 4650

and the Si1120. As D increases, the

ET and ED angles approach 0. Also, the dimensions, R

So, it is indeed a fourth root function, as expected for small targets.

The analysis can be repeated for as many segments as necessary to cover the entire illuminated surface. As

3.10. Calculating Irradiance from a Specular Surface Reflection

When operating in single-port, the window is shared between the detector and the LED. There are actually two

reflection coefficients, one for each polarity. The reflection coefficients are a function of the material differences at

the junction and the refractive index, n. Air = 1, and, for a transparent polycarbonate material, it is approximately

1.6.

The reflection coefficient is in the neighborhood of 10%, depending on the material. However, whatever R is, the

easiest method of calculating the effect of the LED on the detector is to create an image of the specular reflection

and then evaluate at twice the distance.

In this example, we assume that the LED and the detector are 2 cm away. Assume that the glass is 1 cm away so

that the angles of incidence and reflection are 45°. Using the approximate SFH 4650 mathematical model, the SFH

4650 at 45° has a radiant intensity of 175 mW/sr x 0.137 = 24 mW/sr assuming 400 mA current. The SFH 4650

nominally has a 50 mW/sr at 100 mA, and the derating curve chows that, at 400 mA, there is a 3.5x increase. The

equation below is a derived SFH 4650 model that was used to derive this 24 mW/sr result.

E

D

L



DT

cos



R

D

2

---------------------------------------

if

A

0.5

RD

----------

5

units are in

W

m

2

-------



=

E

D

AR



I

E



ET

cos



ED

cos





R

E

2

R

D

2







--------------------------------------------------------------------------------------

units are in

W

m

2

-------

=

I

E

I

1



ET

90 deg

------------------

–







0.75

+

cos

ET









=

E

D

AR



I

0





D

4







------------------------------

=

I

E

I

1



90 deg

------------------

–







0.75

+

cos









=