When there is no base current in a transistor switch, the output voltage from the transistor is
A
Low
B
High
C
Unchanged
D
Unknown

Question

When there is no base current in a transistor switch, the output voltage from the transistor is

Examoogle · Accepted Answer

High

Explanation: When the base current ($I_B$) of an NPN transistor is zero, the transistor operates in the cutoff region. In this state, the collector-emitter junction acts as an open circuit, preventing collector current ($I_C$) from flowing, causing the output voltage at the collector to pull up to the supply voltage ($V_{CC}$).

Working Principle: In the cutoff region, the base-emitter junction is reverse-biased or at zero potential, resulting in $I_B = 0$. Since $I_C = \beta I_B$, the collector current becomes zero. According to Kirchhoff's Voltage Law at the output loop: $V_{CE} = V_{CC} - I_C R_C$. With $I_C = 0$, $V_{CE} = V_{CC}$, which represents a 'High' logic state.

Key Formulas: $I_C = \beta I_B$ — Collector current relation; $V_{CE} = V_{CC} - I_C R_C$ — Output voltage equation

Key Points: A transistor switch acts as an open circuit when $I_B = 0$.; Cutoff region is defined by $I_B = 0$ and $I_C = 0$.; In digital logic, this corresponds to the 'OFF' state of the transistor, resulting in a 'High' signal at the collector terminal.; The collector-emitter resistance ($R_{CE}$) is effectively infinite in the cutoff region.

Advantages: Low power consumption during cutoff; Clear separation between digital logic states

Disadvantages: Not suitable for linear amplification in this mode; Susceptible to noise if not properly biased

Applications: Digital logic gates (NOT, NAND, NOR); Electronic switching circuits; Microcontroller interface drivers

Additional Info: In saturation mode, the transistor acts as a closed switch, pulling the output voltage to a 'Low' level (approximately $V_{CE(sat)} \approx 0.2V$).. Option A (Low) describes the saturated state, not the cutoff state.

Answer: B is correct — When $I_B = 0$, the transistor enters the cutoff region, acting as an open switch and causing the output voltage to rise to the supply voltage level ($V_{CC}$).

Exam Tip: Always remember the two fundamental states of a BJT switch: $I_B = 0$ implies Cutoff ($V_{out} = V_{CC}$), and $I_B \ge I_{B(sat)}$ implies Saturation ($V_{out} \approx 0V$).

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