= Light-emitting diode efficiency
Efficiency has several denominators. Specify a steady <electric current> $I>0$, positive terminal <electric potential difference> $V_f$, and a simplified narrow emission <spectrum> with <photon> <energy> $E_\gamma$. Assume at most one radiative <photon> per injected <Electron> and <electron hole> pair. Let $\eta_{\rm inj}$ be the fraction of the electrical pair-supply rate $I/q$ reaching the active layer, $\eta_{\rm rad}$ the fraction of those pairs recombining radiatively, and $\eta_{\rm ext}$ the fraction of internally emitted <photons> that escape. Then counting the successive events gives
$$
\dot N_{\rm active}=\eta_{\rm inj}\frac Iq,\qquad
\dot N_{\rm internal}=\eta_{\rm rad}\dot N_{\rm active},\qquad
\dot N_{\rm out}=\eta_{\rm ext}\dot N_{\rm internal}.
$$
Define <internal quantum efficiency> here conditionally on pairs reaching the active layer: $\eta_{\rm IQE}=\eta_{\rm rad}$. The internal <photon> yield per electrical pair is instead $\eta_{\rm inj}\eta_{\rm IQE}$. Stating this convention matters when another definition folds injection into <internal quantum efficiency>. The <external quantum efficiency> and <wall-plug efficiency> are
$$
\eta_{\rm EQE}=\frac{\dot N_{\rm out}}{I/q}
=\eta_{\rm inj}\eta_{\rm IQE}\eta_{\rm ext},\qquad
\eta_{\rm WPE}=\frac{P_{\rm opt}}{IV_f}
=\eta_{\rm EQE}\frac{E_\gamma}{qV_f},\qquad
P_{\rm opt}=\dot N_{\rm out}E_\gamma.
$$
These are <photon>-count and <energy>-flow identities for the declared model. For a broad <spectrum>, replace $E_\gamma$ by the mean escaping <photon> <energy>, weighted by photon number, when computing optical <power>.
A stipulated example chooses $I=20\,\mathrm{mA}$, $V_f=3.20\,\mathrm V$, $\lambda=450\,\mathrm{nm}$, $\eta_{\rm inj}=0.90$, $\eta_{\rm IQE}=0.80$ and $\eta_{\rm ext}=0.50$. For every 100 electrical pairs, 90 reach the active layer, 72 emit internally, and 36 <photons> escape. Thus $\eta_{\rm EQE}=0.36$, not $0.80$. Using $q=1.602176634\times10^{-19}\,\mathrm C$ gives
$$
\begin{aligned}
I/q&=1.24830\times10^{17}\,\mathrm{s}^{-1},\\
\dot N_{\rm out}&=4.49389\times10^{16}\,\mathrm{s}^{-1},\\
P_{\rm opt}&=19.8375\,\mathrm{mW},\qquad IV_f=64.0\,\mathrm{mW},\\
\eta_{\rm WPE}&=0.309960.
\end{aligned}
$$
The <dimensional analysis> is transparent: $\mathrm{A/C}=\mathrm{s}^{-1}$, $\mathrm{J/s}=\mathrm W$, and $\mathrm{CV}=\mathrm J$. The optical and electrical efficiencies differ because $E_\gamma/q=2.75520\,\mathrm V$ differs from the chosen $V_f$.
\Image[/blue-led-photon-energy-and-efficiency.png]
{title=<Photon> <energy>, <wavelength> and a chosen <light-emitting diode efficiency> budget}
{description=Original calculations, not measurements or specifications of the historical devices. Left: $E_\gamma=hc/\lambda$, with chosen blue and converted-light <wavelengths>. Right: successive <photon> counts per 100 electrical pairs for $\eta_{\rm inj}=0.90$, $\eta_{\rm IQE}=0.80$, $\eta_{\rm ext}=0.50$. The optional converter calculation uses the separately stipulated absorption fraction and <fluorescence quantum yield> below.}
{source=blue-led-photon-energy-and-efficiency.py}
The factors suggest different improvements. Better confinement can reduce injection leakage; suppressing <nonradiative recombination> raises the radiative fraction; improving <light extraction efficiency> helps already-emitted <photons> escape. A <photon> trapped by <total internal reflection> has already passed the radiative step and can still be lost through absorption. Improving extraction therefore need not change the internal radiative fraction.
<Shuji Nakamura>'s lecture cautions that why <indium gallium nitride> emits efficiently despite many defects was not completely settled in that account.https://journals.aps.org/rmp/pdf/10.1103/RevModPhys.87.1139[§IV.B, pp. 1144–1145: high defect density, localization hypothesis and stated uncertainty]{ref}
As an independent consistency check on the 1994 reported values, treating its broad <spectrum> as monochromatic at the $450\,\mathrm{nm}$ peak gives
$$
\eta_{\rm EQE}\approx\frac{(1.5\times10^{-3}\,\mathrm W)q}{(0.020\,\mathrm A)(hc/(450\,\mathrm{nm}))}=0.02722.
$$
This agrees at the quoted precision with the reported $2.7\%$. Its optical/electrical <power> ratio is instead $1.5\,\mathrm{mW}/(20\,\mathrm{mA}\times3.6\,\mathrm V)=2.0833\%$. The first estimate uses a peak in place of a photon-number-weighted mean, so agreement is a consistency check rather than an exact reanalysis of the measured <spectrum>.
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