In a common emitter amplifier circuit using an n-p-n transistor, the output voltage is inverted with respect to the input voltage. This means when the input signal increases, the output signal decreases, and vice versa, resulting in a phase shift of 180° between them.

Here's why this happens step-by-step:

Step 1: When the input voltage (applied to the base) increases, the base current increases.

Step 2: This increase in base current causes a larger collector current to flow (since I_C = β I_B).

Step 3: The larger collector current leads to a larger voltage drop across the collector resistor (R_C). According to Ohm's law, V_RC = I_C R_C.

Step 4: The output voltage is taken across the collector resistor. Since the supply voltage V_CC is fixed, the output voltage V_out = V_CC - I_C R_C. Therefore, when I_C increases, V_out decreases.

Final Answer: Thus, an increase in input voltage causes a decrease in output voltage, and a decrease in input voltage causes an increase in output voltage. This inversion corresponds to a phase difference of $180°$.

Related Topics

Transistors: A transistor is a semiconductor device used to amplify or switch electronic signals. The common emitter configuration is widely used for amplification due to its high voltage and current gain.

Amplifier Circuits: Amplifiers are circuits that increase the amplitude of a signal. The common emitter amplifier is known for inverting the input signal.

Formulae

Collector Current: $I_C=\beta I_B$

Output Voltage: $V_{out}=V_{CC}-I_C{R}_C$