The Light-Emitting Diode (LED) — Question 8

Question 8

A series resistor sets LED current and absorbs excess voltage

A 5.00 V supply drives an LED and series resistor. Approximate LED forward voltage by $V_f=2.00\,\mathrm V$ at desired current 15.0 mA. Over a 25.0 K rise, take $dV_f/dT=-2.00\,\mathrm{mV/K}$ and a temperature-independent resistor.

1. Find the required resistance and initial resistor power.

2. Find the new current after the temperature rise and the fractional increase.

Question 8 - Solution

A series resistor sets LED current and absorbs excess voltage

(a) Method and calculation

$$V_s=IR+V_f\Rightarrow\boxed{R=(5.00-2.00)/.0150=200\,\Omega}.$$$$\boxed{P_R=I^2R=.0450\,\mathrm W}.$$

(b) Interpretation and further calculation

$$\Delta V_f=(-.00200)(25.0)=-.0500\,\mathrm V\Rightarrow\boxed{I_{\mathrm{hot}}=(5.00-1.95)/200=15.25\,\mathrm{mA}}.$$$$\boxed{\Delta I/I=.250/15.0=.01667=1.667\%}.$$

Physical check

The resistor supplies negative feedback against changes in diode voltage. The fixed-drop approximation neglects the LED differential resistance and self-heating feedback.