Question E6B03:
What property of an LED's semiconductor material determines its forward voltage drop?  

Choices:  
A: Intrinsic resistance  
B: Band gap  
C: Junction capacitance  
D: Junction depth    

Correct Answer is  B: Band gap  

Correct answer: B — Band gap

Why this is correct:
- An LED conducts and emits light when the carrier energy overcomes the semiconductor band gap, releasing photons by recombination.
- The forward voltage is set primarily by the band gap energy $E_g$:
  $$V_f \approx \frac{E_g}{q}$$
  Since 1 eV corresponds to 1 V per electron charge, this is often quoted as:
  $$V_f \approx E_g(\text{eV}) \text{ volts}$$
- Because photon energy is tied to color, shorter-wavelength (bluer) LEDs have larger $E_g$ and thus higher $V_f$.

Useful relations and rules of thumb:
- Photon/band-gap energy vs wavelength:
  $$E_g \approx \frac{hc}{\lambda} \quad\Rightarrow\quad E_g(\text{eV}) \approx \frac{1240}{\lambda(\text{nm})}$$
  Example: Red 630 nm → $E_g \approx 1240/630 \approx 1.97$ eV → $V_f \approx 2.0$ V. Blue 470 nm → $E_g \approx 2.64$ eV → $V_f \approx 2.6$–3.4 V (materials and series resistance add some variation).
- Diode I–V (for context):
  $$I = I_S\left(e^{V_D/(nV_T)} - 1\right)$$
  The “knee” for LEDs appears near $V_f \approx E_g/q$.
- Temperature: $V_f$ decreases with temperature (roughly a few mV/°C).

Why the other choices are incorrect:
- A: Intrinsic resistance
  - Series/intrinsic resistance affects the slope of the I–V curve at higher currents (additional drop), but it does not set the fundamental forward knee voltage.
- C: Junction capacitance
  - Influences switching speed and AC behavior, not the DC forward voltage drop.
- D: Junction depth
  - Affects capacitance, breakdown, and speed; it is not the primary determinant of the LED’s forward voltage threshold.