In n-p-n transistor, in CE configuration a) the emitter is heavily doped than the collector b) emitter and collector can be interchanged c) the base region is very thin but is heavily doped d) the conventional current flows from base to emitter

Options

A. a and b are correct
B. a and c are correct
C. a and d are correct
D. b and c are correct

Answer

To solve the question regarding the characteristics of an n-p-n transistor in common emitter (CE) configuration, we will analyze each statement one by one.

### Step-by-Step Solution:

1. **Statement A: The emitter is heavily doped than the collector.**
- In an n-p-n transistor, the emitter is indeed heavily doped with n-type material compared to the collector. This is because the emitter needs to inject a large number of charge carriers (electrons) into the base region to ensure efficient operation. Therefore, this statement is **correct**.

2. **Statement B: Emitter and collector can be interchanged.**
- The emitter and collector cannot be interchanged in an n-p-n transistor. The emitter is designed to inject carriers into the base, while the collector is meant to collect them. The doping levels are different, with the emitter being heavily doped and the collector being lightly doped. Thus, this statement is **incorrect**.

3. **Statement C: The base region is very thin but is heavily doped.**
- The base region in an n-p-n transistor is indeed very thin, but it is not heavily doped. In fact, the base is lightly doped compared to the emitter. Therefore, this statement is **incorrect**.

4. **Statement D: The conventional current flows from base to emitter.**
- In an n-p-n transistor, the conventional current actually flows from the emitter to the base. The arrow in the transistor symbol indicates the direction of conventional current flow, which is from the emitter to the base. Therefore, this statement is **incorrect**.

### Summary of Statements:
- A: Correct
- B: Incorrect
- C: Incorrect
- D: Incorrect

### Final Answer:
The correct statements are A (the emitter is heavily doped than the collector) and D (the conventional current flows from base to emitter is incorrect). Therefore, the correct option is **A only**.