SEMICONDUCTORS ONE MARK QUESTIONS

Contents
Introduction
Intrinsic semiconductor
Extrinsic semiconductor
Density of states
Fermi-Dirac distribution function
Problems on Fermi Function
Electrons concentration in CB
Holes concentration in VB
Direct & indirect bandgap semiconductors
Hall effect
Problems on Hall effect
P-N junction diode
Light Emitting Diode (LED)
Problems on LED
Solar Cell
Previous Years' Questions
Quiz Hub
1 Mark Questions
  1. Explain what an intrinsic semiconductor is and give one example.
    An intrinsic semiconductor is a pure semiconductor without impurities, where conduction is due to thermally generated electron-hole pairs. Example: Silicon (Si).
  2. Explain how doping increases the conductivity of a semiconductor.
    Doping introduces donor (n-type) or acceptor (p-type) atoms, providing extra electrons or holes, which increases charge carrier density and conductivity.
  3. Why is the conductivity of an intrinsic semiconductor low at room temperature?
    At room temperature, only a small number of electrons can jump across the band gap, so carrier concentration is very low.
  4. Explain the significance of the Fermi–Dirac function in semiconductors.
    The Fermi–Dirac distribution \( f(E)=\dfrac{1}{1+e^{(E-E_f)/kT}} \) gives the probability that an energy state \( E \) is occupied by an electron.
  5. Show that at \( E=E_f, f(E)=1/2 \) at any finite temperature.
    The Fermi–Dirac distribution \[ f(E)=\dfrac{1}{1+e^{(E-E_f)/kT}} \] Substituting \( E=E_f \) in \(f(E) \): \[ f(E_f)=\dfrac{1}{1+e^0}=\dfrac{1}{2} \].
  6. Define Fermi energy and Fermi level for a semiconductor.
    - Fermi Energy: The energy level at which the probability of finding an electron is 50% at absolute zero.
    - Fermi Level: The energy level in a semiconductor that determines the occupancy of electron states, varying with temperature and doping.
  7. Draw the Fermi level in intrinsic and extrinsic semiconductors.
    black body radiation
    • Intrinsic: \( E_f \) lies at the middle of band gap.
    • n-type: \( E_f \) shifts closer to conduction band.
    • p-type: \(E_f \) shifts closer to valence band.
  8. What is \(f(E)\) at \(T=0\) K?
    • If \(E \le E_f \): \( f(E)=1 \) (states filled).
    • If \(E > E_f \): \( f(E)=0 \) (states empty).
  9. Why are direct bandgap semiconductors preferred in the designing of LED?
    Because electron-hole recombination in direct band gap materials emits photons efficiently without requiring phonon assistance.
  10. Differentiate between n-type and p-type semiconductors.
    • n-type: doped with pentavalent atoms, majority carriers are electrons.
    • p-type: doped with trivalent atoms, majority carriers are holes.
  11. Explain forward bias and reverse bias in a PN junction diode.
    • Forward bias: p-side connected to +ve, reduces barrier, allows current.
    • Reverse bias: p-side to –ve, increases barrier, current is minimal.
  12. Why does the reverse current saturate in a PN junction diode?
    Reverse current is due to minority carriers, which are limited and independent of applied voltage.
  13. Draw and explain the I–V characteristics of a PN junction diode.
    • Forward bias: current increases exponentially after threshold voltage.
    • Reverse bias: very small saturation current until breakdown.
  14. Name one material used for making LEDs.
    Gallium Arsenide (GaAs) or Gallium Nitride (GaN).
  15. Calculate the wavelength of radiation emitted by a LED with band gap energy of 1.5 eV.
    The relationship between energy and wavelength is \( E = \frac{hc}{\lambda} \), where \( h = 6.634 \times 10^{-34} \, \text{J·s} \), \( c = 3 \times 10^8 \, \text{m/s} \) and \( 1eV = 1.602 \times 10^{-19} \text{J} \).
    Rearranging for \( \lambda \): \[ \lambda = \frac{hc}{E} = \frac{(6.634 \times 10^{-34}) \times (3 \times 10^8)}{1.5 \times 1.602×10^{-19}} \approx 8.27 \times 10^{-7} \, \text{m} = 827 \, \text{nm} \] So, the wavelength is approximately 827 nm.
  16. Explain the working principle of an LED.
    In forward bias, electrons recombine with holes at the junction, releasing energy as photons (light emission).
  17. What is the principle of a solar cell and define it?
    A solar cell works on the photovoltaic effect, converting light energy into electrical energy at a PN junction.
  18. Why does a solar cell generate current under illumination but not in the dark?
    Photons generate electron-hole pairs under light; in dark, no carriers are generated, so no current flows.
  19. What is Hall effect? Mention any two applications.
    The Hall effect is the generation of a voltage difference (Hall voltage) across an electrical conductor or semiconductor when a current flows through it perpendicular to an applied magnetic field.
    Applications:
    - Measuring magnetic fields.
    - Determining carrier concentration
    - mobility in semiconductors.
  20. How can the type of semiconductor (n or p) be identified using the Hall effect?
    • n-type: Hall coefficeint negative (electrons as majority carriers).
    • p-type: Hall coefficeint positive (holes as majority carriers).
  21. Define the terms drift and diffusion current.
    - Drift Current: The movement of charge carriers (electrons or holes) in a semiconductor due to an applied electric field. It is proportional to the electric field strength and carrier mobility.
    - Diffusion Current: The flow of charge carriers due to a concentration gradient, where carriers move from regions of high concentration to regions of low concentration.
  22. What are solar cells?
    Solar cells are semiconductor devices that convert sunlight directly into electricity through the photovoltaic effect. They typically consist of a p-n junction, where photons excite electrons, generating a voltage and current.
  23. The \( R_H \) of a specimen is \( 3.66 \times 10^{-4} \, \text{m}^3 \text{C}^{-1} \). Its resistivity is \( 8.93 \times 10^{-3} \, \Omega\text{-m} \). Find mobility and charge carrier concentration.
    - Charge carrier concentration (\( n \)): \( R_H = \frac{1}{nq} \), where \( q = 1.6 \times 10^{-19} \, \text{C} \). So, \( n = \frac{1}{R_H q} = \frac{1}{(3.66 \times 10^{-4}) \times (1.6 \times 10^{-19})} \approx 1.71 \times 10^{22} \, \text{m}^{-3} \). - Mobility (\( \mu \)): \( \mu = \frac{1}{n q \rho} \), where \( \rho = 8.93 \times 10^{-3} \, \Omega\text{-m} \). So, \( \mu = \frac{1}{(1.71 \times 10^{22}) \times (1.6 \times 10^{-19}) \times (8.93 \times 10^{-3})} \approx 4.11 \times 10^{-2} \, \text{m}^2\text{V}^{-1}\text{s}^{-1} \).
  24. Distinguish intrinsic and extrinsic semiconductor.
    - Intrinsic Semiconductor: Pure material, conductivity due to thermal excitation (e.g., Si, Ge).
    - Extrinsic Semiconductor: Doped with impurities (e.g., P or B), conductivity enhanced by added charge carriers (n-type or p-type).
  25. Write any two differences between direct and indirect band gap semiconductors. Give two examples for each.
    Differences:
    1. Direct bandgap allows direct electron-hole recombination with photon emission; indirect requires phonon assistance.
    2. Direct bandgap materials are more efficient for LEDs/lasers; indirect are better for transistors.
    - Examples: - Direct bandgap: GaAs, InP. - Indirect bandgap: Si, Ge.
  26. Calculate the wavelength of light emitted by LED with the band gap of energy 1.8 eV.
    The relationship between energy and wavelength is \( E = \frac{hc}{\lambda} \), where \( h = 6.634 \times 10^{-34} \, \text{J·s} \), \( c = 3 \times 10^8 \, \text{m/s} \) and \( 1eV = 1.602 \times 10^{-19} \text{J} \).
    Rearranging for \( \lambda \): Using \[ E = \frac{hc}{\lambda} \Rightarrow \lambda = \frac{hc}{E} \]: \[ \lambda = \frac{(6.634 \times 10^{-34}) \times (3 \times 10^8)}{1.8\times 1.602×10^{-19}} \approx 6.89 \times 10^{-7} \, \text{m} = 689 \, \text{nm} \] So, the wavelength is approximately 689 nm.
  27. The Hall coefficient and conductivity of an n-type specimen are \( -1.25 \times 10^{-3} \, \text{m}^3/\text{C} \) and \( 112 \, /\Omega\text{-m} \). Find the mobility of electrons.
    Mobility \( \mu = \frac{\sigma |R_H|}{n} \), but since \( R_H = \frac{1}{nq} \) and \( \sigma = nq\mu \), \( \mu = \frac{|R_H|}{\rho} \), where \( \rho = \frac{1}{\sigma} = \frac{1}{112} \approx 8.93 \times 10^{-3} \, \Omega\text{-m} \). So: \[ \mu = \frac{1.25 \times 10^{-3}}{8.93 \times 10^{-3}} \approx 0.14 \, \text{m}^2\text{V}^{-1}\text{s}^{-1} \]
  28. What is the direction of current flow in a forward biased pn junction diode?
    a) From n-type to p-type      b) From p-type to n-type      c) No current flow      d) None of the above
    Answer: b) From p-type to n-type
  29. Find the wavelength associated with an electron with energy 2000 eV.
    Wavelength \( \lambda = \frac{h}{p} \), where \( p = \sqrt{2mE} \), \( m = 9.11 \times 10^{-31} \, \text{kg} \), \( E = 2000 \, \text{eV} = 3.2 \times 10^{-16} \, \text{J} \). \[ p = \sqrt{2 \times (9.11 \times 10^{-31}) \times (3.2 \times 10^{-16})} \approx 2.42 \times 10^{-23} \, \text{kg·m/s} \] \[ \lambda = \frac{6.626 \times 10^{-34}}{2.42 \times 10^{-23}} \approx 2.74 \times 10^{-11} \, \text{m} = 0.0274 \, \text{nm} \] So, the wavelength is approximately 0.0274 nm.
  30. The semiconductor material NOT used in LED is
    a) Silicon carbide      b) GaAsP     c) GaAs      d) Si
    31. Answer: d) Si
  31. On increase of temperature, the Fermi level shifts upwards in
    a) p-type semiconductor      b) n-type semiconductor      c) intrinsic semiconductor      d)none of these
    Answer: c) intrinsic semiconductor