Question E6A05:
How does DC input impedance at the gate of a field-effect transistor (FET) compare with that of a bipolar transistor?  

Choices:  
A: They are both low impedance  
B: An FET has lower input impedance  
C: An FET has higher input impedance  
D: They are both high impedance    

Correct Answer is  C: An FET has higher input impedance  

Correct answer: C — An FET has higher input impedance.

Why C is correct:
- At DC, the FET gate draws essentially no current:
  - MOSFET: gate is insulated by an oxide, so input resistance is extremely high (limited by leakage), often $>10^9$ to $10^{12}\ \Omega$.
  - JFET: gate is a reverse-biased PN junction with only tiny leakage current, so DC input resistance is also very high.
- Since $R_{\text{in}} \approx V/I$, and $I_{\text{gate}} \approx 0$ at DC, FET input impedance is very large.

Why the others are incorrect:
- A: “They are both low impedance” — Incorrect. A BJT base-emitter junction conducts DC current like a diode, giving relatively low input impedance, while an FET gate does not conduct DC.
- B: “An FET has lower input impedance” — Incorrect. It’s the opposite; FETs have higher DC input impedance.
- D: “They are both high impedance” — Incorrect. The BJT base requires base current, so its DC input impedance is not high.

Useful formulas and rules of thumb:
- BJT base draws current: $I_B \approx I_C/\beta$. The small-signal input resistance at the base is
  $$r_\pi \approx \frac{\beta}{g_m} \approx \frac{\beta \cdot 26\ \text{mV}}{I_C}$$
  Often this is only a few kΩ to tens of kΩ at typical bias currents, i.e., “low” compared to FET gates.
- FET gate DC input resistance:
  $$R_{\text{in, FET}} \approx \frac{1}{I_{\text{leak}}} \to \text{very large (MΩ–GΩ+)}$$
- Rule of thumb:
  - MOSFET/JFET gate: essentially no DC current → very high input impedance.
  - BJT base: forward-biased junction needs current → relatively low input impedance.