Questions: Semiconductor Physics (Doping and p-n Junctions)

Question 1

In n-type silicon doped with phosphorus, the Fermi level shifts toward the conduction band. Why doesn't it shift all the way into the conduction band?

• A. The crystal structure prevents the Fermi level from entering a band
• B. At finite temperature, the Fermi level must balance electron occupation: it rises high enough that the donor states are mostly ionized and the conduction band has enough electrons to match the donor density, but thermal broadening keeps it below the band edge except at extreme doping
• C. The Fermi level is fixed at the middle of the gap by definition
• D. Phosphorus atoms repel the Fermi level away from the conduction band

Answer: (B)

The Fermi level position is determined self-consistently by charge neutrality: the number of ionized donors (plus holes) must equal the number of conduction electrons. At moderate doping (~10^15-10^18 cm^-3), E_F sits between the donor level and the conduction band edge. At very high doping (>~10^19 cm^-3 in Si), E_F can actually enter the conduction band, creating a 'degenerate' semiconductor that behaves metallically. Temperature also matters: at very low T, carriers freeze out onto donors and E_F drops; at very high T, intrinsic carriers dominate and E_F returns to mid-gap.

Question 2

The law of mass action states that np = n_i^2 in a semiconductor at thermal equilibrium, regardless of doping. What is the physical origin of this constraint?

• A. It follows from conservation of charge in the crystal
• B. It results from the product of the electron and hole Fermi-Dirac distributions: n depends on (E_F - E_c) and p depends on (E_v - E_F), so their product np is independent of E_F and depends only on the gap and temperature
• C. It is an empirical observation with no theoretical derivation
• D. It holds only for intrinsic semiconductors

Answer: (B)

n = N_c exp(-(E_c - E_F)/k_BT) and p = N_v exp(-(E_F - E_v)/k_BT). The product np = N_c N_v exp(-E_g/k_BT) = n_i^2, where the Fermi level cancels. This means that in equilibrium, increasing the electron concentration (by n-type doping) necessarily decreases the hole concentration, and vice versa. The constraint is purely thermodynamic and holds for any equilibrium doping — it breaks down only under non-equilibrium conditions (illumination, injection).

Question 3

A p-n junction in equilibrium has a built-in potential but produces no current. Explain why, despite the electric field in the depletion region.

Model answer: In equilibrium, the Fermi level is uniform throughout the junction. The built-in electric field in the depletion region creates a drift current (minority carriers swept across by the field), but this is exactly balanced by a diffusion current (majority carriers diffusing against the field from the high-concentration side). The two currents cancel at every point, giving zero net current. This must be the case: a p-n junction in thermal equilibrium is a closed system at uniform temperature, and any net current would violate the second law of thermodynamics (you could extract work from a system at uniform temperature).

Forward bias reduces the built-in potential, allowing diffusion current to exceed drift current — net current flows. Reverse bias increases the barrier, suppressing diffusion and leaving a small reverse saturation current from thermally generated minority carriers.

Question 4

Why does the depletion region of a p-n junction widen under reverse bias and narrow under forward bias?

Model answer: Under reverse bias, the external voltage adds to the built-in potential, increasing the electric field across the junction. This stronger field sweeps more mobile carriers away from the junction, uncovering more fixed ionized dopant charges on both sides — the depletion region widens. Under forward bias, the external voltage opposes the built-in potential, reducing the field. Majority carriers can now diffuse further into the junction, neutralizing some of the fixed charges, and the depletion region narrows. At strong enough forward bias, the depletion region nearly vanishes and large currents flow.

The depletion width scales as W ∝ √(V_bi - V_applied) for an abrupt junction, where V_bi is the built-in potential. Reverse bias makes V_applied negative, increasing W. This voltage-dependent width is the basis of varactor diodes (voltage-tunable capacitors).