The Light-Emitting Diode (LED) — Question 10

Question 10

A pulsed drive produces a finite optical rise and decay

An ideal LED population follows $dN/dt=I(t)/e-N/\tau$. A rectangular current pulse $I_0$ lasts from t=0 to $t_p=\tau\ln2$, starting from N=0. Radiative lifetime is $\tau_r$; photon rate is $N/\tau_r$.

1. Find N at pulse end and its subsequent decay.

2. Integrate the total number of photons emitted during and after the pulse, and express it using injected charge and internal efficiency $\eta_{\mathrm{int}}=\tau/\tau_r$.

Question 10 - Solution

A pulsed drive produces a finite optical rise and decay

(a) Method and calculation

$$0<t<t_p:\quad N=\frac{I_0\tau}{e}(1-e^{-t/\tau})\Rightarrow\boxed{N(t_p)=I_0\tau/(2e)}.$$$$\boxed{N(t>t_p)=\frac{I_0\tau}{2e}e^{-(t-t_p)/\tau}}.$$

(b) Interpretation and further calculation

$$\int_0^\infty Ndt=\frac{I_0\tau}{e}\left[t_p-\tau(1-e^{-t_p/\tau})\right]+N(t_p)\tau=\frac{I_0\tau t_p}{e}.$$$$\boxed{N_\gamma=\frac1{\tau_r}\int_0^\infty Ndt=\eta_{\mathrm{int}}\frac{I_0t_p}{e}=\eta_{\mathrm{int}}\frac{Q_{\mathrm{inj}}}e}.$$

Physical check

The population returns to zero after the pulse, so every injected carrier is eventually removed by one of the competing recombination paths. The pulse shape changes timing, not the integrated efficiency in this linear model.