The Light-Emitting Diode (LED) — Question 4

Question 4

External optical power counts extracted photons per electron

An LED is driven at $I=20.0\,\mathrm{mA}$. Injection efficiency is .900, internal radiative efficiency .800, and photon extraction probability .250. Each extracted photon has energy $2.00\,\mathrm{eV}$. Use $e=1.60\times10^{-19}\,\mathrm C$.

1. Find external quantum efficiency and extracted photon rate.

2. Find optical output power and electrical-to-optical efficiency at forward voltage $V_f=2.40\,\mathrm V$.

Question 4 - Solution

External optical power counts extracted photons per electron

(a) Method and calculation

$$\boxed{\eta_{\mathrm{ext}}=\eta_{\mathrm{inj}}\eta_{\mathrm{int}}\eta_{\mathrm{out}}=.180}.$$$$\boxed{\dot N_\gamma=\eta_{\mathrm{ext}}I/e=2.25\times10^{16}\,\mathrm{s^{-1}}}.$$

(b) Interpretation and further calculation

$$\boxed{P_{\mathrm{opt}}=\dot N_\gamma(2.00e\,\mathrm V)=7.20\,\mathrm{mW}}.$$$$\boxed{\eta_{\mathrm{power}}=P_{\mathrm{opt}}/(IV_f)=.00720/.0480=.150}.$$

Physical check

Quantum efficiency counts photons; power efficiency weights each photon by its energy relative to electrical energy per charge. They are generally unequal.