Suppose the base resistor is open. The Q point will be

Question

Suppose the base resistor is open.  The Q point will be

Examoogle · Accepted Answer

At the lower end of the load line

Explanation: If the base resistor of a BJT is open, the base current ($I_B$) drops to zero. Consequently, the collector current ($I_C$) also becomes zero, forcing the transistor into the cut-off region.

Working Principle: In a common-emitter configuration, the collector current is defined as $I_C = \beta I_B$. With an open base, $I_B = 0$, which results in $I_C = 0$. Since the voltage drop across the collector resistor ($I_C R_C$) is zero, the collector-to-emitter voltage becomes $V_{CE} = V_{CC}$. On the load line plot, this point corresponds to the intersection of the $I_C = 0$ axis and the $V_{CE} = V_{CC}$ intercept, which is the lower end of the load line (cut-off).

Key Formulas: $I_B = \frac{V_{BB} - V_{BE}}{R_B}$ — Base current equation; $I_C = \beta I_B + (1+\beta)I_{CEO}$ — Collector current equation; $V_{CE} = V_{CC} - I_C R_C$ — Output loop equation

Key Points: An open base condition implies $I_B = 0$.; The transistor enters the cut-off region when $I_B = 0$.; At cut-off, the collector-emitter voltage $V_{CE}$ is approximately equal to the supply voltage $V_{CC}$.; The Q-point (Quiescent point) represents the DC operating condition of the transistor.

Advantages: Clear definition of transistor operating regions; Predictable behavior in bias analysis

Disadvantages: Open circuit usually indicates a fault or component failure; Transistor becomes unusable as an amplifier in cut-off

Applications: Switching circuit analysis; Fault diagnosis in BJT biasing circuits

Additional Info: When $I_B = 0$, $I_C = I_{CEO}$ (collector leakage current), which is typically negligible, effectively placing the transistor at the bottom of the load line.. Option B refers to the Saturation region, which occurs when base current is sufficiently high to drive $V_{CE} \approx 0$.

Answer: C is correct — An open base resistor results in zero base current, pushing the transistor into the cut-off region located at the lower end of the load line.

Exam Tip: Always remember: Cut-off is the 'OFF' state ($I_C \approx 0, V_{CE} \approx V_{CC}$), while Saturation is the 'ON' state ($I_C \approx I_{C,sat}, V_{CE} \approx 0$).

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