Question E6B05:
What characteristic of a PIN diode makes it useful as an RF switch?  

Choices:  
A: Extremely high reverse breakdown voltage  
B: Ability to dissipate large amounts of power  
C: Reverse bias controls its forward voltage drop  
D: Low junction capacitance    

Correct Answer is  D: Low junction capacitance  

To be a good RF switch, a device must pass RF with low loss when ON and block RF with high isolation when OFF. A PIN diode does this because its OFF-state junction capacitance is very small.

Why D (Low junction capacitance) is correct:
- In the OFF (reverse-biased) state, a PIN diode behaves like a very small capacitor. Small C means large capacitive reactance:
  $$X_C = \frac{1}{2\pi f C}$$
  Example: If $C = 0.2\ \text{pF}$ at $f = 100\ \text{MHz}$,
  $$X_C \approx \frac{1}{2\pi (100\times10^6)(0.2\times10^{-12})} \approx 8\ \text{k}\Omega$$
  This large impedance gives high isolation; for a 50 Ω system, series OFF isolation is roughly:
  $$\text{Isolation} \approx 20\log_{10}\left(\frac{X_C}{Z_0}\right) \approx 20\log_{10}\left(\frac{8000}{50}\right) \approx 44\ \text{dB}$$

Why the other choices are incorrect:
- A: Extremely high reverse breakdown voltage
  - Not the defining feature for RF switching. Many diodes can have high breakdown, but RF isolation hinges on low capacitance, not breakdown rating.
- B: Ability to dissipate large amounts of power
  - Power handling is package- and current-dependent. The key RF-switch advantage of a PIN diode is its RF behavior (low C when OFF, current-controlled low resistance when ON), not exceptional power dissipation.
- C: Reverse bias controls its forward voltage drop
  - Incorrect. Forward voltage is mainly set by material and forward current. In PIN diodes, the RF ON resistance is controlled by forward bias current; reverse bias does not control forward drop.

Useful rules of thumb:
- OFF state: small $C_j$ gives high $X_C$ and good isolation.
  $$X_C = \frac{1}{2\pi f C}$$
- ON state: a PIN diode acts like a current-controlled resistor at RF.
  $$R_{\text{ON}} \approx \frac{K}{I_F}$$
  where $K$ is a device-dependent constant; higher forward current lowers $R_{\text{ON}}$ and insertion loss.

Bottom line: Low junction capacitance in the OFF state is what makes the PIN diode an effective RF switch.