Semiconductor physics — current components in an n-type bar under an electric field Consider an n-type semiconductor bar placed in an applied electric field. Electrons and holes drift in opposite directions because of their opposite charges. Evaluate the following statements: 1) The net current is contributed by both electrons and holes, with electrons as the majority carriers. 2) The net current equals the sum of the electron and hole current components. 3) The net current equals the difference between the electron and hole currents. Which statements are correct?

Introduction / Context:
The drift current in a semiconductor arises from the motion of both types of charge carriers under an electric field. In an n-type bar, electrons are majority carriers while holes are minority carriers. Understanding how these contributions combine is fundamental to device analysis.

Given Data / Assumptions:

• Uniform n-type semiconductor bar.
• Steady-state drift under an applied electric field; diffusion is not the focus here.
• Electrons carry negative charge and drift against the field; holes carry positive charge and drift with the field.

Concept / Approach:
Current density is the algebraic sum of carrier contributions: J = J_n + J_p, where J_n = q n μ_n E and J_p = q p μ_p E (taking proper signs into account within the drift velocities). Even though electrons and holes move in opposite physical directions, their charge signs result in current contributions that add to produce the net current along the field direction. In an n-type material, electrons dominate the magnitude because n » p.

Step-by-Step Solution:
Identify majority carriers: electrons (n-type).Write drift components: J_n and J_p as above.Net current density: J = J_n + J_p → sum of components, not their difference.Therefore, statements (1) and (2) are correct; (3) is incorrect.

Verification / Alternative check:
Diode and transistor equations derive total current as the sum of individual carrier currents. Sign conventions ensure that contributions are additive in the chosen reference direction.

Why Other Options Are Wrong:
Options claiming only (2) or only (1) ignore the other true statement; “difference” is incorrect because the algebraic combination is a sum once carrier charge signs are treated properly.

Common Pitfalls:

• Confusing physical drift directions with current directions and thus incorrectly subtracting currents.
• Ignoring minority-carrier contribution; while small in n-type, it is not zero.

Final Answer:
1 and 2