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AMU 2026 science PYQ — Which statement about phase shift in a common-emitter amplifier i… | Mathem Solvex

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AMU 2026 science PYQ — Which statement about phase shift in a common-emitter amplifier i… | Mathem Solvex | Mathem Solvex

Explanation

1. Conceptual Mechanism of a Common-Emitter (CE) Amplifier

In a Common-Emitter configuration, the AC input signal is applied between the base and the emitter, and the amplified output signal is taken from the collector with respect to the emitter.

The relationship between the input voltage ( $V_{in}$ ) and the output collector voltage ( $V_{out}$ ) can be analyzed dynamically by tracing the path of the current through the circuit loops:

Positive Half-Cycle of Input: When the input voltage rises in the positive direction, it increases the forward bias voltage across the base-emitter junction. This increase in base-emitter voltage drives a larger base current ( $I_{b}$ ). Since the collector current is directly proportional to the base current ( $I_{c} = β I_{b}$ ), the collector current ( $I_{c}$ ) also rises.
Voltage Drop across the Collector Resistor ( $R_{c}$ ):
The output voltage at the collector terminal is determined by Kirchhoff's Voltage Law (KVL) applied across the collector-emitter loop:
$V_{out} = V_{cc} - I_{c} R_{c}$
As $I_{c}$ increases during the positive half-cycle of the input, the voltage drop across the collector resistor ( $I_{c} R_{c}$ ) increases. Because this larger drop is subtracted from the fixed DC supply voltage ( $V_{cc}$ ), the net output voltage at the collector terminal ( $V_{out}$ ) drops, pulling the output signal into its negative half-cycle.
Negative Half-Cycle of Input:
Conversely, when the input voltage goes negative, the base-emitter forward bias reduces, decreasing both $I_{b}$ and $I_{c}$ . A smaller $I_{c}$ reduces the voltage drop across $R_{c}$ , which causes the net output voltage ( $V_{out} = V_{cc} - I_{c} R_{c}$ ) to rise toward the supply level, driving the output signal into its positive half-cycle.

2. Conclusion on Phase Relationship

Because a positive-going input creates a negative-going output, and a negative-going input creates a positive-going output, the output signal is an exact inverted replica of the input signal. This structural inversion represents a phase inversion or a phase shift of exactly 180°.

Correct Answer

The correct option is (c) Output is $18 0^{\circ}$ out of phase with input.

Explanation

1. Conceptual Mechanism of a Common-Emitter (CE) Amplifier

In a Common-Emitter configuration, the AC input signal is applied between the base and the emitter, and the amplified output signal is taken from the collector with respect to the emitter.

The relationship between the input voltage ( $V_{in}$ ) and the output collector voltage ( $V_{out}$ ) can be analyzed dynamically by tracing the path of the current through the circuit loops:

Positive Half-Cycle of Input: When the input voltage rises in the positive direction, it increases the forward bias voltage across the base-emitter junction. This increase in base-emitter voltage drives a larger base current ( $I_{b}$ ). Since the collector current is directly proportional to the base current ( $I_{c} = β I_{b}$ ), the collector current ( $I_{c}$ ) also rises.
Voltage Drop across the Collector Resistor ( $R_{c}$ ):
The output voltage at the collector terminal is determined by Kirchhoff's Voltage Law (KVL) applied across the collector-emitter loop:
$V_{out} = V_{cc} - I_{c} R_{c}$
As $I_{c}$ increases during the positive half-cycle of the input, the voltage drop across the collector resistor ( $I_{c} R_{c}$ ) increases. Because this larger drop is subtracted from the fixed DC supply voltage ( $V_{cc}$ ), the net output voltage at the collector terminal ( $V_{out}$ ) drops, pulling the output signal into its negative half-cycle.
Negative Half-Cycle of Input:
Conversely, when the input voltage goes negative, the base-emitter forward bias reduces, decreasing both $I_{b}$ and $I_{c}$ . A smaller $I_{c}$ reduces the voltage drop across $R_{c}$ , which causes the net output voltage ( $V_{out} = V_{cc} - I_{c} R_{c}$ ) to rise toward the supply level, driving the output signal into its positive half-cycle.

2. Conclusion on Phase Relationship

Correct Answer

The correct option is (c) Output is $18 0^{\circ}$ out of phase with input.

AMU 2026 — science PYQ

Choose the correct answer:

Explanation

1. Conceptual Mechanism of a Common-Emitter (CE) Amplifier

2. Conclusion on Phase Relationship

Correct Answer

Explanation

1. Conceptual Mechanism of a Common-Emitter (CE) Amplifier

2. Conclusion on Phase Relationship

Correct Answer