Question E6A03:
Why does a PN-junction diode not conduct current when reverse biased?  

Choices:  
A: Only P-type semiconductor material can conduct current  
B: Only N-type semiconductor material can conduct current  
C: Holes in P-type material and electrons in the N-type material are separated by the applied voltage, widening the depletion region  
D: Excess holes in P-type material combine with the electrons in N-type material, converting the entire diode into an insulator    

Correct Answer is  C: Holes in P-type material and electrons in the N-type material are separated by the applied voltage, widening the depletion region  

To understand a reverse-biased PN junction, remember: the applied voltage pulls majority carriers away from the junction, expanding the depletion region and preventing charge flow across it.

1) Why the correct answer is correct (Choice C)
- In reverse bias, the positive terminal is on the N-side and the negative terminal is on the P-side.
- This pulls electrons in N-type and holes in P-type away from the junction.
- Result: the depletion region widens, the electric field across it strengthens, and it becomes a high-resistance barrier to majority-carrier flow.
- Only a tiny reverse saturation current from minority carriers flows.

2) Why the other choices are incorrect
- A: “Only P-type semiconductor material can conduct current”  
  Incorrect. Both P-type (holes) and N-type (electrons) conduct. The lack of conduction under reverse bias is due to the widened depletion region, not an inability of N-type material to conduct.

- B: “Only N-type semiconductor material can conduct current”  
  Incorrect for the same reason as A. Both types conduct; the junction barrier under reverse bias blocks majority carriers.

- D: “Excess holes … combine … converting the entire diode into an insulator”  
  Incorrect. Reverse bias reduces recombination at the junction; it does not convert the whole diode into an insulator. The bulk remains semiconductive; only the depletion region widens and blocks majority-carrier current. Significant reverse current can flow if breakdown occurs.

3) Useful formulas and rules of thumb
- Diode I–V relationship:
  $$I = I_S\left(e^{\frac{V_D}{n V_T}} - 1\right)$$
  where:
  - $I_S$ = reverse saturation current (very small),
  - $V_D$ = diode voltage (negative in reverse bias),
  - $n$ ≈ 1–2,
  - $V_T = \frac{kT}{q} \approx 25.9\,\text{mV at }300\,\text{K}$.
  In reverse bias ($V_D < 0$ and not in breakdown), $e^{V_D/(nV_T)} \approx 0$, so:
  $$I \approx -I_S$$

- Depletion width increases with reverse bias (conceptual form):
  $$W \propto \sqrt{V_{bi} + V_R}$$
  where $V_{bi}$ is the built-in potential and $V_R$ is the applied reverse voltage. Wider $W$ → higher barrier → negligible majority-carrier conduction.

- Breakdown note: If $V_R$ exceeds the breakdown voltage $V_{BR}$ (Zener or avalanche), the diode will conduct in reverse.

- Temperature rule of thumb: $I_S$ roughly doubles for about every 10 °C rise in temperature.