Question 1

A semiconductor has an intrinsic carrier concentration of ni = 1.5 × 10¹⁰ cm⁻³ at 300 K. If the semiconductor is doped with donor impurities to make it n-type with carrier concentration n = 1.5 × 10¹⁵ cm⁻³, what is the hole concentration p in this n-type semiconductor?

Accepted Answer

In thermal equilibrium, the mass action law states n·p = ni², therefore p = ni²/n = (1.5 × 10¹⁰)²/(1.5 × 10¹⁵) = 1.5 × 10⁵ cm⁻³.

Question 2

Which of the following statements correctly describes the energy band structure of a semiconductor at 0 K?

Accepted Answer

At absolute zero temperature, all electrons occupy the lowest available energy states; the valence band is completely filled and the conduction band is empty in an intrinsic semiconductor.

Question 3

A p-n junction diode is forward biased with an applied voltage of 0.6 V. If the thermal voltage kT/e at room temperature is 0.026 V, what is the approximate diode current relative to its reverse saturation current I₀ using the ideal diode equation I = I₀(e^(eV/kT) − 1)?

Accepted Answer

With V = 0.6 V and kT/e = 0.026 V, eV/kT ≈ 23.08, so e^(23.08) ≈ 11 × 10⁹, which gives I ≈ 11 × 10⁹ × I₀ or approximately 23000 × I₀ (order of magnitude).

Question 4

In a semiconductor material, the forbidden energy gap (band gap) is defined as:

Accepted Answer

The band gap Eg is the minimum energy difference between the valence band and conduction band, which must be overcome to create a free electron-hole pair.

Question 5

A silicon semiconductor is doped with acceptor impurities to create a p-type semiconductor. Which of the following is the primary charge carrier in this p-type material?

Accepted Answer

In a p-type semiconductor, acceptor atoms create holes in the valence band, and these holes become the majority carriers responsible for electrical conduction.

Question 6

Which of the following statements about the depletion width of a p-n junction is correct?

Accepted Answer

Reverse bias increases the depletion width by expanding the region depleted of charge carriers, whereas forward bias narrows it.

Question 7

In a zener diode, the breakdown occurs due to which mechanism when the reverse bias voltage exceeds the zener voltage?

Accepted Answer

Zener breakdown is caused by quantum tunneling of electrons through the narrow depletion region potential barrier under high reverse bias.

Question 8

A solar cell made of silicon with band gap 1.1 eV is illuminated with photons of energy 2.5 eV. What is the maximum kinetic energy acquired by an electron excited to the conduction band?

Accepted Answer

Excess kinetic energy = photon energy - band gap = 2.5 - 1.1 = 1.4 eV.

Question 9

In semiconductor physics, the term 'hole' represents a vacant state in the valence band. How does hole conduction compare to electron conduction in terms of effective charge and direction of motion?

Accepted Answer

Holes are treated as positive charge carriers that move in the opposite direction to electron flow, with effectively positive charge +e.

Question 10

A silicon semiconductor has a band gap of 1.1 eV at room temperature. What is the minimum energy required to create an electron-hole pair?

Accepted Answer

The band gap energy directly equals the minimum energy required to excite an electron from valence band to conduction band.

Question 11

In an n-type semiconductor, the majority charge carriers are electrons. Which element is typically used as a dopant to create n-type silicon?

Accepted Answer

Phosphorus is a pentavalent element with 5 valence electrons, providing free electrons to create n-type conductivity in silicon.

Question 12

The resistivity of a p-type semiconductor is 0.5 Ω·cm. If holes have a mobility of 200 cm²/(V·s) and charge e = 1.6 × 10⁻¹⁹ C, what is the hole concentration? (Assume ρ = 1/(e·μ·N_h))

Accepted Answer

Using N_h = 1/(e·ρ·μ) = 1/(1.6 × 10⁻¹⁹ × 0.5 × 200) ≈ 6.25 × 10¹⁵ cm⁻³.

Question 13

In a p-n junction at thermal equilibrium, what is the direction of the electric field within the depletion region?

Accepted Answer

The electric field in the depletion region points from the positively charged n-region to the negatively charged p-region.

Question 14

A germanium p-n junction has a built-in potential of 0.3 V. If the reverse saturation current is 10 μA and thermal voltage V_T = 26 mV at 300 K, what is the forward current when 0.15 V forward bias is applied? (I = I_s(e^(V/V_T) - 1), assume e^(V/V_T) >> 1)

Accepted Answer

I ≈ I_s·e^(V/V_T) = 10 × e^(0.15/0.026) = 10 × e^5.77 ≈ 10 × 10 = 100 μA (approximately).

Question 15

A silicon p-n junction diode has a built-in potential of 0.7 V at room temperature. When forward biased with an external voltage of 0.5 V, what is the net potential barrier across the junction?

Accepted Answer

Net barrier = Built-in potential − Forward bias voltage = 0.7 − 0.5 = 0.2 V.

Question 16

In a semiconductor, the width of the depletion region in a p-n junction depends on which of the following factors?

Accepted Answer

Depletion width W ∝ √(V_built-in + V_reverse) / √(doping concentration), so it depends on both doping level and applied reverse voltage.

Question 17

A germanium p-n junction has a reverse saturation current of 10 μA at 25°C. Assuming the saturation current doubles for every 10°C rise in temperature, what will be the reverse saturation current at 45°C?

Accepted Answer

Temperature increase = 45 − 25 = 20°C, which is 2 intervals of 10°C; I_s(45°C) = 10 × 2² = 40 μA.

Question 18

In a p-n junction, if the doping concentration of the p-side is N_A = 10¹⁸ cm⁻³ and the n-side is N_D = 10¹⁵ cm⁻³, the depletion width extends more into which region and why?

Accepted Answer

Depletion width extends more into the lightly doped side; W_n/W_p = N_A/N_D = 10³, so the n-region (less doped) extends much farther into the depletion region.

NDA Semiconductor Electronics

Semiconductor Electronics — Practice Questions