Semiconductor Physics: Quiz 2
Semiconductors form the backbone of modern electronics, from diodes and transistors to integrated circuits. This quiz explores the physics that makes them work: energy bands, charge carriers, doping, p–n junctions, and the behaviour of devices under bias.
You will apply concepts such as the mass-action law, Fermi–Dirac statistics, depletion regions, and rectification. Expect multi-step reasoning rather than simple recall.
- Review band theory and intrinsic carrier concentration.
- Practise doping calculations and carrier relationships.
- Reason through junction behaviour and diode characteristics.
Work carefully with units and signs; several questions hinge on direction of current or bias polarity.
Q1. At absolute zero, the valence band of a pure semiconductor is:
At ##0\,\text{K}## the valence band is fully occupied by electrons and the conduction band is empty, so the material behaves as an insulator.
Q2. The energy gap of a semiconductor is typically of the order of:
Semiconductor band gaps are usually between about ##0.1\,\text{eV}## and ##3\,\text{eV}##, intermediate between conductors and insulators.
Q3. In an intrinsic semiconductor at temperature ##T##, the number of electrons in the conduction band equals:
In a pure (intrinsic) semiconductor every electron excited to the conduction band leaves one hole behind, so ##n = p = n_i##.
Q4. The intrinsic carrier concentration of a semiconductor depends on temperature approximately as:
The standard result is ##n_i \propto T^{3/2}\exp\!\left(-\dfrac{E_g}{2k_BT}\right)##, showing strong exponential dependence on the band gap.
Q5. Adding a pentavalent impurity such as phosphorus to silicon produces:
Pentavalent donors contribute extra electrons, making the material n-type with electrons as majority carriers.
Q6. In an n-type semiconductor, the minority carriers are:
In n-type material electrons are majority carriers and holes are the minority carriers.
Q7. The law of mass action for a doped semiconductor states that:
At thermal equilibrium the product of electron and hole concentrations is constant: ##np = n_i^2##, independent of doping.
Q8. A silicon sample is doped with donors so that ##n = 10^{17}\,\text{cm}^{-3}##. If ##n_i = 1.5\times10^{10}\,\text{cm}^{-3}##, the hole concentration is approximately:
Using ##p = n_i^2/n = (1.5\times10^{10})^2/10^{17} \approx 2.25\times10^{3}\,\text{cm}^{-3}##.
Q9. In a p–n junction at equilibrium, the diffusion of majority carriers across the junction is balanced by:
The built-in field creates a drift current that exactly cancels the diffusion current at equilibrium.
Q10. The depletion region of a p–n junction contains:
The depletion region is depleted of mobile carriers and contains fixed, ionised dopant ions that create the built-in field.
Q11. When a p–n junction is forward biased, the width of the depletion region:
Forward bias reduces the potential barrier, so the depletion width decreases and current flows readily.
Q12. The barrier potential of a silicon p–n junction is approximately:
A silicon diode has a forward barrier of about ##0.7\,\text{V}## (germanium is about ##0.3\,\text{V}##).
Q13. In reverse bias, the current through an ideal p–n junction is:
Reverse bias gives a small, nearly voltage-independent saturation current carried by minority carriers, until breakdown.
Q14. The dynamic resistance of a diode is defined as:
Dynamic (ac) resistance is the reciprocal of the slope of the I–V curve: ##r_d = \dfrac{\Delta V}{\Delta I}##.
Q15. A half-wave rectifier using one diode produces an output that:
A half-wave rectifier passes one half-cycle, so the ripple frequency equals the input frequency.
Q16. A full-wave rectifier's output ripple frequency, for a ##50\,\text{Hz}## input, is:
Full-wave rectification uses both half-cycles, doubling the ripple frequency to ##2\times50 = 100\,\text{Hz}##.
Q17. In a transistor, the base region is made:
A thin, lightly doped base ensures most injected carriers reach the collector, giving good current gain.
Q18. For a transistor in common-emitter configuration, the current gain ##\beta## is related to ##\alpha## by:
Since ##\alpha = \dfrac{\beta}{1+\beta}##, rearranging gives ##\beta = \dfrac{\alpha}{1-\alpha}##.
Q19. The Fermi level in an intrinsic semiconductor lies:
For an intrinsic semiconductor the Fermi level lies approximately at the centre of the forbidden gap.
Q20. Zener breakdown in a heavily doped diode occurs mainly because:
Heavy doping makes the depletion region very narrow, so a strong field enables band-to-band tunnelling (Zener effect).