Question E6B02:
Which characteristic of a Schottky diode makes it a better choice than a silicon junction diode for use as a power supply rectifier?  

Choices:  
A: Much higher reverse voltage breakdown  
B: More constant reverse avalanche voltage  
C: Longer carrier retention time  
D: Lower forward voltage drop    

Correct Answer is  D: Lower forward voltage drop  

Correct answer: D — Lower forward voltage drop

Why D is correct
- A Schottky diode uses a metal–semiconductor junction, giving a much lower forward voltage drop than a silicon PN junction.
- Typical values:
  - Silicon PN rectifier: $V_f \approx 0.7$ to $1.0\ \text{V}$ at moderate currents
  - Schottky rectifier: $V_f \approx 0.2$ to $0.45\ \text{V}$
- Lower $V_f$ means less heat and higher efficiency in rectifiers:
  $$P_{\text{loss}} \approx I \times V_f$$
  Rule of thumb: Each $0.1\ \text{V}$ reduction in $V_f$ saves about $0.1 \times I$ watts.
- In a bridge rectifier, two diodes conduct at once, so the drop is about $2V_f$; Schottky diodes can save close to a volt total.

Why the other choices are incorrect
- A: Much higher reverse voltage breakdown  
  Schottky diodes usually have lower reverse voltage ratings and higher reverse leakage than silicon PN diodes. So this is false.

- B: More constant reverse avalanche voltage  
  Schottky diodes are not used for avalanche/zener behavior; their reverse characteristics are leakier and less well-defined for voltage regulation. This is not an advantage of Schottkys.

- C: Longer carrier retention time  
  Schottkys are majority-carrier devices with very little charge storage (short reverse recovery). They do not have longer retention; they have shorter—another advantage for high-frequency switching, but not the reason asked here.

Useful formulas/rules
- Conduction loss per diode: $$P \approx I_{\text{avg}} \times V_f$$
- For bridge rectifier under load: $$P_{\text{bridge}} \approx 2 \times I_{\text{avg}} \times V_f$$
- Efficiency improvement from Schottky ≈ proportional to the reduction in $V_f$ at the operating current.