Difficulty: Easy
Correct Answer: source resistor
Explanation:
Introduction / Context:Self-bias is a simple and robust way to bias JFET amplifiers. It avoids an extra negative supply by exploiting the voltage drop across a source resistor, which raises the source potential relative to the (near-zero) gate potential, thereby creating a negative V_GS for an n-channel JFET.
Given Data / Assumptions:
Concept / Approach:Because gate current is approximately zero, the gate sits near 0 V through a high-value resistor. The channel current flowing through R_S creates V_S = I_D * R_S > 0. Therefore V_GS = V_G − V_S ≈ 0 − (I_D * R_S) = −I_D * R_S, which is negative without needing a dedicated negative supply. This negative V_GS self-adjusts with I_D, providing stabilizing feedback against device parameter variations.
Step-by-Step Solution:
Connect gate to ground via a large resistor (e.g., 1 MΩ).Include a source resistor R_S in series with the source lead.Allow I_D to generate V_S = I_D * R_S → V_GS = −I_D * R_S (negative).Verification / Alternative check:Measure V_G, V_S in a lab circuit. With I_D ≈ 2 mA and R_S = 1.5 kΩ, V_S ≈ 3 V, so V_GS ≈ −3 V, consistent with theory and device transfer curves.
Why Other Options Are Wrong:
Common Pitfalls:Omitting the source bypass capacitor analysis; bypassing R_S increases gain but reduces local feedback—trade-offs must be considered.
Final Answer:source resistor
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