The small currents in reverse biased condition of p-n diode are due to

Options

A. Electrons
B. Holes
C. Majority charge carriers
D. Thermal agitation of minority charge carriers.

Answer

To solve the question regarding the small currents in the reverse-biased condition of a p-n diode, we can break it down step by step.

### Step 1: Understand the p-n Junction
A p-n junction diode consists of two types of semiconductor materials: p-type (which has an abundance of holes) and n-type (which has an abundance of electrons). When these two materials are joined, they form a junction.

**Hint:** Remember that p-type has holes as majority carriers and n-type has electrons as majority carriers.

### Step 2: Reverse Bias Condition
In reverse bias, the p-type side is connected to the negative terminal of a voltage source, and the n-type side is connected to the positive terminal. This connection increases the potential barrier at the junction.

**Hint:** Visualize how the voltage source affects the depletion region of the diode.

### Step 3: Effect of Increased Potential Barrier
The increased potential barrier in reverse bias prevents the majority charge carriers (holes from the p-side and electrons from the n-side) from crossing the junction. As a result, there is minimal current flow due to these majority carriers.

**Hint:** Recall that majority carriers do not contribute to current in reverse bias.

### Step 4: Minority Carriers
However, there are minority carriers present in both regions. In the p-type region, the minority carriers are electrons, and in the n-type region, the minority carriers are holes. These minority carriers can still move across the junction, albeit in very small numbers.

**Hint:** Think about how thermal energy affects the movement of minority carriers.

### Step 5: Thermal Agitation
The small current that flows in reverse bias is due to the thermal agitation of these minority carriers. Thermal energy can excite some of the minority carriers enough to cross the junction, leading to a small reverse current.

**Hint:** Consider how temperature influences the behavior of charge carriers in semiconductors.

### Conclusion
Thus, the small currents in the reverse-biased condition of a p-n diode are primarily due to the thermal agitation of minority charge carriers. Therefore, the correct answer is option D.

**Final Answer:** The small currents in reverse-biased condition of a p-n diode are due to the thermal agitation of minority charge carriers.