Semiconductor Physics: Mastery Quiz 5
Semiconductors form the backbone of modern electronics, from diodes and transistors to integrated circuits and solar cells. This quiz explores the physics that governs charge carriers in semiconductor materials — energy bands, doping, carrier concentration, drift and diffusion, and the behaviour of p–n junctions.
You will apply concepts such as the mass-action law, Fermi–Dirac statistics, and the built-in potential of a junction. Expect medium-level questions that reward careful reasoning rather than pure recall.
Before you begin
- Use ##…## for inline math and ###…### for display math.
- Read each option fully — distractors are designed to be plausible.
- Only one option is correct per question.
Tip: Keep the intrinsic carrier concentration ##n_i## and the mass-action law ##np = n_i^2## in mind throughout.
Q1. At absolute zero temperature, the valence band of a pure semiconductor is:
At ##T = 0## K, all electrons occupy the lowest available states, so the valence band is completely filled and the conduction band is empty. This is why an intrinsic semiconductor behaves as an insulator at absolute zero.
Q2. The energy gap of a semiconductor is typically in the range:
Typical semiconductor band gaps lie roughly between ##1## eV and ##3## eV (e.g., Si ##\approx 1.1## eV, GaAs ##\approx 1.42## eV). Conductors have negligible gaps and insulators have gaps well above ##3## eV.
Q3. In an intrinsic semiconductor at thermal equilibrium, the number of electrons in the conduction band equals:
In a pure (intrinsic) semiconductor, every electron excited to the conduction band leaves behind one hole in the valence band, so ##n = p = n_i##.
Q4. The mass-action law for a semiconductor at thermal equilibrium states that:
At equilibrium the product of electron and hole concentrations is constant: ##np = n_i^2##. This holds for both intrinsic and doped material at a given temperature.
Q5. Adding a pentavalent impurity (such as phosphorus) to silicon produces:
Pentavalent atoms donate an extra electron, creating donor levels just below the conduction band. The material becomes n-type with electrons as majority carriers.
Q6. In a p-type semiconductor, the majority carriers are:
Trivalent acceptor impurities create holes in the valence band, so holes are the majority carriers in p-type material, while electrons are minority carriers.
Q7. The Fermi level in an n-type semiconductor at thermal equilibrium lies:
Donor doping raises the Fermi level toward the conduction band. In a p-type material it lies closer to the valence band.
Q8. The drift velocity of charge carriers in a semiconductor under an applied electric field ##E## is given by:
Drift velocity is proportional to the electric field: ##v_d = \mu E##, where ##\mu## is the mobility of the carrier. Mobility depends on scattering mechanisms and temperature.
Q9. The conductivity of an n-type semiconductor is best expressed as:
The general expression is ##\sigma = e(n\mu_e + p\mu_h)##. In strongly n-type material the hole term is negligible, but the full form is always correct.
Q10. As temperature increases in an extrinsic n-type semiconductor (before intrinsic behaviour dominates), the electron concentration:
In the extrinsic (freeze-out to saturation) range, increasing temperature ionises more donors, slightly increasing ##n##. At much higher temperatures intrinsic generation dominates.
Q11. The built-in potential of a p–n junction is given by:
The contact potential is ##V_0 = \dfrac{kT}{e}\ln\left(\dfrac{N_A N_D}{n_i^2}\right)##. It increases with doping and decreases with rising temperature.
Q12. In a forward-biased p–n junction, the width of the depletion region:
Forward bias reduces the potential barrier, so the depletion width decreases. Reverse bias widens the depletion region.
Q13. The ideal diode equation is:
The Shockley diode equation is ##I = I_0\left(e^{eV/kT} - 1\right)##, where ##I_0## is the reverse saturation current. For large forward bias the ##-1## term is negligible.
Q14. The Hall coefficient of an n-type semiconductor is:
For electrons the Hall coefficient ##R_H = -\dfrac{1}{ne}## is negative. A positive Hall coefficient indicates hole-dominated (p-type) conduction.
Q15. In a p–n junction under reverse bias, the current is mainly due to:
Reverse bias raises the barrier, blocking majority-carrier diffusion. The small reverse current comes from thermally generated minority carriers drifting across the junction.
Q16. The diffusion current density of electrons is proportional to:
Diffusion current arises from a concentration gradient. The electron diffusion current density has magnitude proportional to ##D_n\left|\dfrac{dn}{dx}\right|##; the sign depends on the chosen current-direction convention. Drift current, by contrast, is driven by the electric field.
Q17. The Einstein relation connecting diffusion constant and mobility is:
The Einstein relation is ##\dfrac{D}{\mu} = \dfrac{kT}{e}##, so ##D = \mu kT/e##. It links diffusion and drift at thermal equilibrium.
Q18. A semiconductor has ##n_i = 1.5 \times 10^{10}\ \text{cm}^{-3}## and is doped with donors so that ##n = 1.5 \times 10^{15}\ \text{cm}^{-3}##. The hole concentration ##p## is approximately:
Using ##np = n_i^2##: ##p = \dfrac{n_i^2}{n} = \dfrac{(1.5\times10^{10})^2}{1.5\times10^{15}} = 1.5\times10^{5}\ \text{cm}^{-3}##. This illustrates how heavy doping suppresses minority carriers.
Q19. The depletion capacitance of a p–n junction under reverse bias:
The junction behaves like a parallel-plate capacitor with ##C \propto 1/\sqrt{V_0 + V_R}##. Increasing reverse bias widens the depletion region and lowers the capacitance.
Q20. Which statement about the Fermi–Dirac distribution is correct for electrons in a semiconductor?
The Fermi–Dirac function ##f(E) = \dfrac{1}{1 + e^{(E-E_F)/kT}}## gives the occupation probability of an electron state. For ##E - E_F \gg kT## it reduces to the Maxwell–Boltzmann approximation.