Question E6B07:
What causes a junction diode to fail from excessive current?  

Choices:  
A: Excessive inverse voltage  
B: Excessive junction temperature  
C: Insufficient forward voltage  
D: Charge carrier depletion    

Correct Answer is  B: Excessive junction temperature  

To fail from excessive current, a diode overheats internally. The actual failure mechanism is excessive junction temperature.

Why B is correct (Excessive junction temperature):
- Forward current causes power dissipation:
  - $$P_{\text{forward}} \approx I_F \cdot V_F \quad\text{or}\quad P \approx I_F^2 \cdot R_D$$
- That power raises the junction temperature:
  - $$T_J = T_A + P \cdot \theta_{JA}$$
  - where $T_J$ is junction temp, $T_A$ is ambient, and $\theta_{JA}$ is thermal resistance.
- If $T_J$ exceeds the diode’s maximum (often 125–175°C), materials degrade, bond wires/fuse can open, or the junction shorts. Also, as temperature rises, $V_F$ of a silicon diode drops by about $-2\ \text{mV}/^\circ\text{C}$, promoting more current at a given drive (thermal runaway).

Why the other choices are incorrect:
- A: Excessive inverse voltage
  - That’s an overvoltage failure (avalanche/Zener breakdown) mechanism, not what “excessive current” directly causes. It’s about $V_R$ exceeding the diode’s $V_{RRM}$, not current heating.
- C: Insufficient forward voltage
  - Too little $V_F$ just means the diode does not conduct significantly; it won’t fail from that.
- D: Charge carrier depletion
  - “Depletion region” is a normal property under reverse bias; “depleting carriers” is not the failure mechanism from too much current.

Useful formulas/rules of thumb:
- Forward dissipation: $$P \approx I_F \cdot V_F$$
- Junction temperature rise: $$\Delta T \approx P \cdot \theta_{JA}$$
- Keep $T_J < T_{J(\max)}$ (check the datasheet; commonly 150°C) with margin.
- Observe rated average forward current $I_F(\text{avg})$ and surge current $I_{FSM}$.
- Heatsinking lowers $\theta_{JA}$; derate power/current as ambient temperature rises.