The number of silicon atoms per m3 is 5 × 1028. This is doped simultaneously with 5 × 1022 atoms per m3 of Arsenic and 5 × 1020 per m3 atoms of Indium. Calculate the number of electrons and holes. Given that ni = 1.5 × 1016 m–3. Is the material n-type or p-type?

To determine the number of electrons and holes in the silicon sample, we first need to understand the effects of doping with both Arsenic (n-type dopant) and Indium (p-type dopant). Let's break this down step by step.

Understanding Doping in Semiconductors

Doping is the process of adding impurities to a semiconductor to change its electrical properties. In this case, Arsenic introduces extra electrons (n-type), while Indium creates holes (p-type). The concentrations of these dopants will help us figure out the overall carrier concentration in the silicon.

Calculating Carrier Concentrations

We start with the intrinsic carrier concentration of silicon, denoted as ni, which is given as 1.5 × 10¹⁶ m^–3.

• Silicon atoms: 5 × 10²⁸ m^–3
• Arsenic (n-type): 5 × 10²² m^–3
• Indium (p-type): 5 × 10²⁰ m^–3

Finding the Number of Electrons

In an n-type semiconductor, the number of electrons can be approximated by the concentration of the n-type dopant (Arsenic) minus the concentration of the p-type dopant (Indium). Thus, we calculate:

n (electrons) = n_Arsenic - n_Indium

n = 5 × 10²² m^–3 - 5 × 10²⁰ m^–3 = 4.95 × 10²² m^–3

Finding the Number of Holes

To find the number of holes, we can use the mass action law, which states that the product of the electron concentration and hole concentration in intrinsic semiconductors is constant:

n × p = n_i²

Rearranging gives us:

p = n_i² / n

Substituting the values:

p = (1.5 × 10¹⁶)² / (4.95 × 10²²)

p = 2.25 × 10³² / 4.95 × 10²² ≈ 4.55 × 10⁹ m^–3

Determining the Type of Semiconductor

Now that we have the concentrations of electrons and holes, we can classify the type of semiconductor. In this case:

• Electrons (n) = 4.95 × 10²² m^–3
• Holes (p) = 4.55 × 10⁹ m^–3