Re: VCSEL Radiance in Watts-per-steradian

Michelle Hauer wrote:
Okay, what do you think the total output power of the VCSEL is over it's
full 13-degree angular profile (if it is 0.2 mW at the peak)?
Basically, I still need to get an idea of how to specify the source
power in Watts/steradian (just good enough to be believable for an
estimate).

I'm only after a first-order power budget analysis of a short-distance
free-space optical data link (taking into account that the power
diverges as it propagates away from the VCSEL), and I'm stuck on this
Watts-per-steradian detail.

Michelle

-----Original Message-----
From: Phil Hobbs [mailto:pcdh@xxxxxxxxxxxxxxxxxxxxxxxxxxxx]
Posted At: Thursday, June 28, 2007 12:17 PM
Posted To: sci.optics
Conversation: VCSEL Radiance in Watts-per-steradian
Subject: Re: VCSEL Radiance in Watts-per-steradian

Michelle wrote:
For those who don't want to read the detailed version below, my
basic
question is how to specify the radiance of a multimode VCSEL
(modeled
as a point source with a Gaussian angular profile with a 1/e^2 full
width of 13 degrees) in Watts-per-steradian if the L-I curve on a
data
*** shows that it emits 0.2 mW at a bias point of 2 mA.

Details:
------------
I am using the Illumination feature in Code V to estimate the power
(total flux) I would receive from a VCSEL if I place a detector of a
given size at different points in space beyond the VCSEL (there are
also some optical elements between the VCSEL and the detector). The
problem I am having is how to specify the source power of the VCSEL
in
Watts/steradian (I'm modeling it as a point source with a Gaussian
angular profile, just to get a first order answer).

I've looked up some typical VCSEL data from ULM Photonics:
http://www.ulm-photonics.de/docs/technical-
infos/technical_infos_%20welcome.htm
I am assuming:
Bias point = 2 mA (just above threshold of 1.5 mA, with slope eff.
of
0.4 W/A)
Optical Power = 0.2 mW (from L-I curve)

But I need to specify the power in Watts/steradian in my model.

The angular profile (Figure 9) shows a 1/e^2 full width of about 13
degrees at a 2 mA bias. Is it correct to assume that there is
roughly
0.2 mW within this 13 degree width, which is about 0.04 steradians?
(I
used the equation for the solid angle of a cone to compute this:
2*pi*[1-cos(half-angle)] ). This works out to about 5 mW/sr.

If I use this value (5 mW/sr) for a point source in Code V with a a
Gaussian angular profile with a 13-degree 1/e^2 full width, then I
only get a total flux of about 0.1 mW over a flat, square detector
spanning 13 degrees at the output of the VCSEL. This is half the
0.2
mW that I roughly expected. I understand that some of the power
will
be in the tails of the Gaussian so I won't get the whole 0.2 mW
source
power, but the 1/e^2 points (13 degrees) correspond to the two-sigma
width, which should contain 95% of the total power, not half. Am I
calculating the solid angle wrong? I also thought that perhaps I
was
getting less power because of the obliquity factor (cosine falloff)
since the detector is flat, but I spherically-curved the detector
and
I still only get about 0.1 mW total flux. By trial and error, I
found
I have to increase the source power to about 11 mW/sr to get 0.2 mW
flux on a detector spanning 13 degrees (full width and height).

Another thing that bothers me is that the angullar profile plot
(Figure 9 on the website above) shows a peak power of nearly 0.2 mW
at
0 degrees. They don't mention the solid angle of the detector(s)
used
to take this profile data (it seems to me that the y-axis of this
plot
should really be in mW/sr to be accurate; if it was then I could
integrate the curve to get the total power within the 13 degree
width). I don't believe that if I operate this VCSEL at 2 mA, I
will
measure 0.2 mW with a tiny detector at 0 degrees (as their plot
would
indicate).

Any help would be much appreciated!

Thanks,
Michelle

They're probably specifying the peak value of the Gaussian, whereas
you're using the average. Those VCSELs are sufficiently non-Gaussian
that you'll be lucky to get within a factor of 2 anyway.

Cheers,

Phil Hobbs


If you want to use a Gaussian model, then you need to normalize it correctly. The math isn't difficult once you see the problem, which is that you're giving it the average elevation of the Himalayas when it's expecting the peak elevation of Everest.

Cheers,

Phil Hobbs
.