Semiconductor Laser: Understanding the Power

[Beer-monster]

I seem to be missing some information in my notes or something because this question seems to have come straight out of the blue.

The optical power generated by a semi-conductor laser is given by the epression

$$P = A(J - J_{th}) \frac{n_ih\nu}{e}$$

where A is the junction area, J is the current density and Jth is the threshold current density for lasing. Given that ni for GaAs is 0.8 and the refractive index for GaAs is 3.6, and the cavity mirrors are formed by a GaAs/air boundary. Estimate the efficiency of the laser.

Anyone know how I could get started with this, or know of a good source of info on semiconductor lasers and other optics that might help?

 

[quantumdude]

I think you can reason this out by looking at the units. $(J-J_{th})$ seems to be the net current density. Multiply that by $A$ and you have the net current (assuming that $J-J_{th}$ is parallel to $A$). That gives you the number of Coulombs per second. Divide by $e$ and you have the number of electrons per second. $h\nu$ is the number of Joules per photon, and $n_i$ is the number of photons per electron. Multiply all that together and you get Joules per second, or Watts, which is a measure of power.

 

[berkeman]

Sometimes google (and the Internet it serves) amaze me. I googled the following:

gaas semiconductor laser efficiency

and one of the first few interesting hits was this:

http://britneyspears.ac/physics/vcsels/vcsels.htm "Vertical Cavity Surface Emitting Lasers (VCSELs)"

and at the same site, "The Britney's Guide to Semiconductor Physics" http://britneyspears.ac/lasers.htm

Now that's one site I just bookmarked. I'll have to go back sometime and figure out what the heck Britney's doing there, but whatever...