Internal quantum efficiency (IQE)

[whatisreality]

Homework Statement

The IQE is defined as (photons emitted from active region $s^{-1}$) / electrons injected across the junction.
Derive an expression for this quantity, then find the injected electron concentration $n$ which maximises it.

Homework Equations

The Attempt at a Solution

We've been taught that radiative recombination is given by $\sim Bn^2$, SRH recombination by $\sim An$ and Auger by $\sim Cn^3$. So then is the expression for IQE just
$\eta = \frac{Bn^2}{An + Bn^2 + Cn^2}$
It doesn't seem like much of a derivation, so I'm concerned that this is wrong or that I've missed something that needs to also be considered. Thanks for any help!

 

[mark726]

Hi there, great question! You are on the right track with your expression for the IQE. To derive it, we can start with the definition given in the post:
$\eta = \frac{photons emitted from active region}{electrons injected across the junction}$

We can then express the number of photons emitted from the active region in terms of the radiative recombination rate, which is given by $Bn^2$. Similarly, the number of electrons injected across the junction can be expressed in terms of the total electron concentration, $N$, and the recombination rates for SRH and Auger processes, $An$ and $Cn^3$, respectively. This gives us the following expression for the IQE:

$\eta = \frac{Bn^2}{An + Bn^2 + Cn^3}$

To find the injected electron concentration that maximizes this quantity, we can take the derivative of the IQE with respect to $n$ and set it equal to zero:

$\frac{d\eta}{dn} = \frac{2Bn(An + Bn^2 + Cn^3) - Bn^2(A + 3Cn^2)}{(An + Bn^2 + Cn^3)^2} = 0$

Solving for $n$, we get:

$n = \sqrt{\frac{A}{3C}}$

So the injected electron concentration that maximizes the IQE is given by the square root of the ratio between the SRH recombination rate and three times the Auger recombination rate.

I hope this helps! Let me know if you have any further questions.