The Light-Emitting Diode (LED) — Question 1

Question 1

The recombination energy selects the LED wavelength

In an ideal direct-gap LED, radiative recombination produces photons with energy approximately equal to $E_g=2.20\,\mathrm{eV}$. Use $hc=1240\,\mathrm{eV\,nm}$. The gap temperature coefficient near the operating point is $dE_g/dT=-4.00\times10^{-4}\,\mathrm{eV/K}$.

1. Find the peak wavelength.

2. Derive the wavelength temperature coefficient and estimate the wavelength shift for a 20.0 K rise using linearization.

Question 1 - Solution

The recombination energy selects the LED wavelength

(a) Method and calculation

$$E_\gamma=hc/\lambda\simeq E_g\Rightarrow\boxed{\lambda=hc/E_g}.$$$$\boxed{\lambda=1240/2.20=563.6\,\mathrm{nm}}.$$

(b) Interpretation and further calculation

$$\boxed{\frac{d\lambda}{dT}=-\frac{hc}{E_g^2}\frac{dE_g}{dT}=+.1025\,\mathrm{nm/K}}.$$$$\boxed{\Delta\lambda\simeq(d\lambda/dT)\Delta T=+2.05\,\mathrm{nm}}.$$

Physical check

A decreasing gap shifts emission to longer wavelength. Carrier kinetic energies and spectral broadening can shift the actual peak from this band-edge estimate.