If the base resistor is very small, the transistor will operate in the

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Examoogle · Accepted Answer

Saturation region

Explanation: When the base resistor ($R_B$) is very small, the base current ($I_B$) increases significantly. This pushes the BJT into the saturation region, where the collector current reaches its maximum limit and the collector-emitter voltage ($V_{CE}$) drops to a very low value.

Working Principle: The relationship is governed by the base input loop $V_{CC} = I_B R_B + V_{BE}$. As $R_B$ approaches zero, $I_B$ is limited primarily by the voltage source and internal resistances, causing $I_B > I_{B,sat} = \frac{I_{C,sat}}{\beta}$. In this state, both the base-emitter and base-collector junctions become forward-biased, effectively acting as a closed switch.

Key Formulas: $I_{B} = \frac{V_{CC} - V_{BE}}{R_B}$ — base current calculation; $I_{C(sat)} \approx \frac{V_{CC} - V_{CE(sat)}}{R_C}$ — saturation collector current

Key Points: In saturation, the transistor behaves as a closed switch ($V_{CE} \approx 0.2V$ for Si).; Forward biasing the Collector-Base junction is the hallmark of saturation.; A small base resistor results in high $I_B$, ensuring $I_C$ is limited by the external circuit components ($R_C$) rather than current gain $\beta$.

Advantages: Used in digital logic design; Provides high switching efficiency

Disadvantages: Excessive charge storage limits switching speed; Power dissipation is non-zero due to $V_{CE(sat)}$

Applications: Digital switching circuits; Relay drivers; LED indicators

Additional Info: If $R_B$ is very large, the transistor would be driven into the Cutoff region.. The Active region is achieved when $I_C = \beta I_B$.

Answer: C is correct — A very small base resistor leads to a high base current, driving the BJT into the saturation region where it acts as a short circuit.

Exam Tip: Always verify if the condition $I_B > I_{C}/\beta$ is met to determine if a transistor is in the saturation region.

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