Question E6E10:
What advantage does surface-mount technology offer at RF compared to using through-hole components?  

Choices:  
A: Smaller circuit area  
B: Shorter circuit board traces  
C: Components have less parasitic inductance and capacitance  
D: All these choices are correct    

Correct Answer is  D: All these choices are correct  

Correct answer: D — All these choices are correct.

Why D is correct:
- Surface-mount parts are physically smaller and mount directly on the copper, so they enable:
  - Smaller circuit area (A).
  - Shorter traces between parts (B).
  - Much lower parasitic inductance and capacitance than long through-hole leads (C).
- All three benefits improve RF performance, so “All these choices” is the best answer.

Why the other choices are “incorrect” by themselves:
- A, B, and C are each true statements, but each names only one advantage. The exam asks for the best overall answer, which is D because it includes all of them.

Useful rules of thumb and formulas:
- Lead/trace inductance: about 1 nH per mm (≈25 nH per inch).
- Inductive reactance:
  $$X_L = 2\pi f L$$
  Examples:
  - 1 nH at 100 MHz: $X_L \approx 0.63\ \Omega$
  - 1 nH at 1 GHz: $X_L \approx 6.3\ \Omega$
  - A 10 mm total lead length (≈10 nH) at 500 MHz: $X_L \approx 31\ \Omega$ (significant!)
- Parasitic capacitance matters at RF:
  $$X_C = \frac{1}{2\pi f C}$$
  Examples:
  - 1 pF at 1 GHz: $X_C \approx 159\ \Omega$
  - 3 pF at 100 MHz: $X_C \approx 530\ \Omega$
- SMT reduces lead length and pad-to-pad spacing, minimizing both $L$ and $C$.
- Shorter traces reduce transmission-line effects and coupling. Rule of thumb: keep trace lengths well below a small fraction of a wavelength:
  $$\lambda \approx \frac{300}{f\ (\mathrm{MHz})}\ \text{meters},\quad \text{use } \ell \lesssim \frac{\lambda}{20}$$

Bottom line: SMT’s compact size directly yields shorter paths and lower parasitics, which is why all listed advantages apply at RF.