Question E5D05:
What parasitic characteristic causes electrolytic capacitors to be unsuitable for use at RF?  

Choices:  
A: Skin effect  
B: Shunt capacitance  
C: Inductance  
D: Dielectric leakage    

Correct Answer is  C: Inductance  

Correct answer: C — Inductance

Why it’s correct:
- Electrolytic capacitors have significant equivalent series inductance (ESL) from their leads and their “jelly-roll” (wound foil) construction. At RF, this inductance dominates.
- The capacitor’s impedance is effectively:
  $$Z(\omega) \approx ESR + j\omega L - \frac{j}{\omega C}$$
- At the self-resonant frequency (SRF),
  $$f_{SRF} = \frac{1}{2\pi\sqrt{LC}}$$
  Above $f_{SRF}$ the part looks inductive, so it will not provide RF bypass or coupling as intended.

Example: For a typical electrolytic with $C=10\,\mu\text{F}$ and $L=30\,\text{nH}$,
$$f_{SRF} \approx \frac{1}{2\pi\sqrt{(10\times10^{-6})(30\times10^{-9})}} \approx 290\,\text{kHz}$$
That’s well below most RF, so the part behaves like an inductor at RF.

Why the other choices are incorrect:
- A: Skin effect — While skin effect increases conductor resistance at RF, it is not the main reason electrolytics fail at RF. ESL is the dominant limitation.
- B: Shunt capacitance — The component’s purpose is capacitance; “shunt capacitance” is not the parasitic that makes it unsuitable at RF.
- D: Dielectric leakage — Mainly a DC issue (leakage current). It affects efficiency and biasing, not the RF suitability like ESL does.

Useful formulas and rules of thumb:
- Capacitive reactance: $$X_C = \frac{1}{2\pi f C}$$
- Inductive reactance: $$X_L = 2\pi f L$$
- Self-resonance when $X_L = X_C$: $$f_{SRF} = \frac{1}{2\pi\sqrt{LC}}$$
- Rule of thumb: Use electrolytics for bulk DC filtering and add a small ceramic (e.g., 0.01–0.1 µF NP0/C0G or X7R) with very short leads for RF bypassing.