The Light-Emitting Diode (LED) — Question 9

Question 9

Optical extraction reduces the heat retained by an LED

At a fixed operating current .100 A, an LED voltage is 3.00 V and externally emitted optical power is .0900 W. All remaining electrical power becomes heat in the package. Its steady thermal resistance to ambient is $R_\theta=80.0\,\mathrm{K/W}$, with ambient temperature $25.0^\circ\mathrm C$.

1. Find dissipated heat and junction temperature.

2. If improved extraction raises output to .120 W at unchanged electrical input and thermal resistance, find the new junction temperature and change.

Question 9 - Solution

Optical extraction reduces the heat retained by an LED

(a) Method and calculation

$$P_{\mathrm{el}}=IV=.300\,\mathrm W\Rightarrow\boxed{P_h=P_{\mathrm{el}}-P_{\mathrm{opt}}=.210\,\mathrm W}.$$$$\boxed{T_j=T_a+R_\theta P_h=25.0+16.8=41.8^\circ\mathrm C}.$$

(b) Interpretation and further calculation

$$\boxed{P_{h,\mathrm{new}}=.180\,\mathrm W,\quad T_{j,\mathrm{new}}=39.4^\circ\mathrm C}.$$$$\boxed{\Delta T_j=-R_\theta\Delta P_{\mathrm{opt}}=-2.40\,\mathrm K}.$$

Physical check

The power balance counts only photons that leave the package as optical output. Internally reabsorbed light ultimately contributes to heat under the stated assumptions.