In a common emitter amplifier, the input and output voltage are out of phase by

Options

A. 30°
B. 90°
C. 120°
D. 180°

Answer

To determine the phase difference between the input and output voltage in a common emitter amplifier, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Common Emitter Configuration**:
- A common emitter amplifier uses a transistor (NPN or PNP) where the emitter terminal is common to both the input and output circuits. The input is applied between the base and emitter, while the output is taken between the collector and emitter.

2. **Analyzing the Input Signal**:
- When an AC input signal is applied to the base of the transistor, it causes the transistor to conduct. The input voltage (Vin) is the voltage across the base-emitter junction.

3. **Transistor Operation**:
- For an NPN transistor, when the base voltage increases (positive half-cycle of the input signal), the transistor turns on more, allowing more current to flow from collector to emitter. This results in a drop in the collector voltage (Vout) since the collector is connected to a positive supply voltage.

4. **Phase Relationship**:
- As the input voltage increases, the output voltage decreases. This means that when the input is at its maximum positive value, the output is at its minimum negative value. Conversely, when the input voltage decreases, the output voltage increases.

5. **Conclusion on Phase Difference**:
- Since the output voltage is inversely related to the input voltage, the input and output voltages are out of phase by 180 degrees. This means that when the input signal goes through one complete cycle, the output signal goes through the opposite cycle.

### Final Answer:
In a common emitter amplifier, the input and output voltage are out of phase by **180 degrees**.

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