Consider a p-n junction as a capacitor, formed with p and n - materials acting as thin metal electrodes and depletion layer width acting as separation between them. Basing on this, assume that a n-p-n transistor is working as an amplifier in CE configuration. If `C_1 and C_2` are the baseemitter and collector-base junction capacities then

Options

A. A) `C_1 gt C_2 `
B. B) `C_2 lt C_2 `
C. C) `C_1 =C_2 `
D. D) `C_1 =C_2 =0`

Answer

To solve the problem, let's analyze the p-n junctions in the context of an n-p-n transistor working as an amplifier in common emitter (CE) configuration.

### Step-by-Step Solution:

1. **Understanding the p-n Junctions**:
- In an n-p-n transistor, there are two p-n junctions: the base-emitter junction (n-p) and the collector-base junction (p-n).
- The base-emitter junction is forward biased when the transistor is in active mode, while the collector-base junction is reverse biased.

2. **Capacitance of p-n Junctions**:
- A p-n junction can be modeled as a capacitor where the p-type and n-type materials act as the plates, and the depletion region acts as the dielectric.
- The capacitance of a p-n junction is influenced by the width of the depletion region, which varies with the applied voltage.

3. **Depletion Region Width**:
- When the base-emitter junction is forward biased, the depletion region width decreases, leading to an increase in capacitance (C1).
- Conversely, when the collector-base junction is reverse biased, the depletion region width increases, leading to a decrease in capacitance (C2).

4. **Comparing Capacitances**:
- Since C1 (base-emitter capacitance) is influenced by a forward bias (which reduces the depletion width), it will be larger than C2 (collector-base capacitance), which is influenced by a reverse bias (which increases the depletion width).
- Therefore, we can conclude that \( C_1 > C_2 \).

5. **Final Conclusion**:
- Based on the analysis, the correct answer is that \( C_1 \) is greater than \( C_2 \).