Question

Which from following combinations is an example for construction of n-type semiconductor?

• A. Si doped with B
• B. Si doped with P
• C. Si doped with Ga
• D. Si doped with In

Hint:

For n-type semiconductors, dope a Group 14 element (like Si or Ge) with a Group 15 element (like P, As, Sb). For p-type semiconductors, dope a Group 14 element with a Group 13 element (like B, Al, Ga, In).

Answer 1

The Correct Option is D

Concept:
An n-type semiconductor is formed when a pure semiconductor such as silicon is doped with a pentavalent impurity. Pentavalent atoms have $5$ valence electrons. When such an atom replaces a silicon atom, four electrons form covalent bonds and the fifth electron becomes free. Thus: \[ \text{Majority charge carriers = electrons} \]
Step 1: Base semiconductor
Silicon belongs to Group 14 and has: \[ 4 \text{ valence electrons} \] It forms four covalent bonds in crystal structure.
Step 2: Check each dopant

• Boron (B): Group 13, $3$ valence electrons
• Phosphorus (P): Group 15, $5$ valence electrons
• Gallium (Ga): Group 13, $3$ valence electrons
• Indium (In): Group 13, $3$ valence electrons

Step 3: Identify correct dopant for n-type
Only phosphorus has $5$ valence electrons. So when phosphorus is added to silicon: \[ \text{One extra electron is donated} \] Hence electrons become majority carriers.
Step 4: Final Answer
The combination that forms an n-type semiconductor is: \[ \boxed{\text{Si doped with P \] Quick Tip:
Remember:

• Group 15 impurity $\Rightarrow$ n-type semiconductor
• Group 13 impurity $\Rightarrow$ p-type semiconductor