Semiconductor Physics Notes FE Sem 1 Applied Physics | MU

Every diode, transistor, LED and solar cell is made from a semiconductor, and its behaviour depends on how electrons fill the energy bands. This module distinguishes direct and indirect band gap semiconductors, explains electrical conductivity in terms of drift velocity and mobility for both conductors and semiconductors, and introduces the Fermi-Dirac distribution function. It then locates the Fermi level in intrinsic, n-type and p-type semiconductors and shows how it shifts with temperature and doping concentration. It maps to course outcome CO6, and both conductivity numericals and Fermi level diagrams are standard exam questions.

Summary

Direct and indirect band gap semiconductors: In a semiconductor the valence band and conduction band are separated by a band gap Eg. Plotting energy against crystal momentum (wave vector k), if the minimum of the conduction band and the maximum of the valence band occur at the same k, the material has a direct band gap; an electron can fall from the conduction band to the valence band and emit a photon of energy about Eg without any change in momentum, so recombination is radiative and fast. Examples: GaAs (1.42 eV), InP, GaN. If the conduction band minimum and the valence band maximum lie at different k, the gap is indirect; recombination needs a phonon (lattice vibration) to supply the momentum difference, so it is slow and the energy is released mostly as heat. Examples: Si (1.12 eV), Ge (0.67 eV). Hence LEDs and laser diodes use direct gap materials, while Si and Ge are used for transistors, diodes and ICs.

Drift velocity and mobility in conductors: In a metal the free electrons move randomly with high random speeds (about \(10^5\) to \(10^6\) m/s) but zero average velocity. When an electric field E is applied they acquire a small average velocity opposite to the field, the drift velocity \(v_d\), typically about \(10^{-4}\) m/s, limited by collisions with vibrating ions. The drift velocity is proportional to the field, \(v_d = \mu E\), where the mobility \(\mu\) (m²/V s) is the drift velocity per unit field; in terms of the mean time between collisions \(\tau\), \(\mu = e\tau/m\). The current density is \(J = nev_d\).

Conductivity in conductors: Combining \(J = nev_d\) with \(v_d = \mu E\) gives \(J = ne\mu E = \sigma E\), which is Ohm's law in point form, with conductivity \(\sigma = ne\mu = ne^2\tau/m\) and resistivity \(\rho = 1/\sigma\). In a metal n is fixed (about \(10^{28}\) to \(10^{29}\) per m³), and as temperature rises the lattice vibrations increase, \(\tau\) and \(\mu\) fall, so the conductivity decreases and the resistance increases with temperature.

Electrical conductivity of semiconductors: A semiconductor has two kinds of charge carriers, electrons in the conduction band and holes in the valence band, which drift in opposite directions but both contribute current in the same direction. So \(\sigma = e(n\mu_e + p\mu_h)\). In an intrinsic (pure) semiconductor \(n = p = n_i\), so \(\sigma_i = n_ie(\mu_e + \mu_h)\). In an n-type semiconductor with donor concentration \(N_D\) (all ionised) \(n \approx N_D\), \(p = n_i^2/N_D\) is negligible and \(\sigma_n \approx N_De\mu_e\); similarly for p-type \(\sigma_p \approx N_Ae\mu_h\). The intrinsic carrier concentration rises exponentially with temperature, \(n_i \propto T^{3/2}e^{-E_g/2kT}\), so unlike metals the conductivity of a semiconductor increases with temperature (negative temperature coefficient of resistance). Electron mobility exceeds hole mobility (for Si about 0.135 and 0.048 m²/V s).

Fermi-Dirac distribution function: Electrons obey Pauli's exclusion principle, so their distribution over energy states is given by the Fermi-Dirac function \(f(E) = 1/(1 + e^{(E - E_F)/kT})\), the probability that a state of energy E is occupied at temperature T. At T = 0 K, f = 1 for all states below the Fermi energy \(E_F\) and f = 0 above it, so the Fermi level is the highest filled level at absolute zero. At any T > 0, \(f(E_F) = 1/2\), so the Fermi level is the energy at which the probability of occupation is 50 percent; the function becomes smeared over a range of a few kT around \(E_F\). The probability that a state is empty (a hole) is \(1 - f(E)\).

Fermi level in an intrinsic semiconductor: Since n = p, the Fermi level lies in the middle of the band gap: \(E_F = (E_c + E_v)/2 + (3kT/4)\ln(m_h^*/m_e^*)\). If the effective masses of electrons and holes are equal, \(E_F = (E_c + E_v)/2\) exactly, at the centre of the gap, and it stays there at all temperatures. With \(m_h^* > m_e^*\) it rises slightly with temperature, but the shift is very small, so for practical purposes the Fermi level of an intrinsic semiconductor is at mid gap.

Fermi level in extrinsic semiconductors and its variation: In an n-type semiconductor donor levels lie just below the conduction band. At 0 K the Fermi level lies midway between the donor level and the conduction band bottom, \(E_F = (E_c + E_d)/2\). As temperature rises the donors ionise and then intrinsic carriers become important, so \(E_F\) moves down towards the middle of the gap; at high temperature the material behaves as intrinsic. Increasing the donor concentration moves \(E_F\) up towards the conduction band (and for very heavy doping into it). In a p-type semiconductor acceptor levels lie just above the valence band; at 0 K \(E_F = (E_v + E_a)/2\); with rising temperature \(E_F\) moves up towards mid gap, and with increasing acceptor concentration it moves down towards the valence band. These shifts explain why devices lose their n or p character at high temperatures.

Mind map

Deriving conductivity of a conductor from drift velocity

Important definitions

Formulas

Key points to remember

• Direct gap (GaAs, InP, GaN): efficient light emission, used in LEDs and laser diodes.
• Indirect gap (Si, Ge): recombination needs a phonon, gives heat; used in electronic devices.
• Drift velocity is tiny (about 10^-4 m/s) compared with random thermal or Fermi speeds (about 10^5 to 10^6 m/s).
• σ = neμ for metals; σ = e(nμe + pμh) for semiconductors.
• Metals: σ decreases with temperature (μ falls). Semiconductors: σ increases with temperature (ni rises sharply).
• Electron mobility is greater than hole mobility.
• f(EF) = 1/2 at any temperature above 0 K; at 0 K f(E) is a step function.
• Intrinsic EF is at (or very near) the middle of the band gap.
• n-type EF lies near the conduction band; p-type EF lies near the valence band.
• Rising temperature pushes EF of extrinsic semiconductors towards mid gap; rising doping pushes it towards the respective band.

Solved examples

Example 1. Find the conductivity and resistivity of intrinsic silicon at 300 K if \(n_i = 1.5\times10^{16}\) m^-3, \(\mu_e = 0.135\) and \(\mu_h = 0.048\) m²/V s.

\(\sigma_i = n_ie(\mu_e + \mu_h) = 1.5\times10^{16}\times1.6\times10^{-19}\times(0.135 + 0.048) = 2.4\times10^{-3}\times0.183 = 4.39\times10^{-4}\) S/m. \(\rho = 1/\sigma = 2.28\times10^{3}\) Ω m.

Example 2. Intrinsic germanium has \(n_i = 2.5\times10^{19}\) m^-3, \(\mu_e = 0.39\) and \(\mu_h = 0.19\) m²/V s. Find its resistivity.

\(\sigma = 2.5\times10^{19}\times1.6\times10^{-19}\times(0.39 + 0.19) = 4\times0.58 = 2.32\) S/m. \(\rho = 1/2.32 = 0.431\) Ω m.

Example 3. Silicon is doped with \(10^{21}\) donor atoms per m³. Find its conductivity and the hole concentration (\(n_i = 1.5\times10^{16}\) m^-3, \(\mu_e = 0.135\) m²/V s).

All donors ionised: \(n \approx N_D = 10^{21}\) m^-3. \(\sigma \approx N_De\mu_e = 10^{21}\times1.6\times10^{-19}\times0.135 = 21.6\) S/m, about 50000 times the intrinsic value. Holes: \(p = n_i^2/n = (1.5\times10^{16})^2/10^{21} = 2.25\times10^{11}\) m^-3, negligible.

Example 4. A copper wire of cross-section 1 mm² carries 1 A. If copper has \(8.5\times10^{28}\) free electrons per m³, find the drift velocity.

\(v_d = I/(neA) = 1/(8.5\times10^{28}\times1.6\times10^{-19}\times10^{-6}) = 1/(1.36\times10^{4}) = 7.35\times10^{-5}\) m/s, about 0.07 mm/s.

Example 5. Copper has conductivity \(5.8\times10^{7}\) S/m and \(8.5\times10^{28}\) electrons per m³. Find the electron mobility and relaxation time.

\(\mu = \sigma/(ne) = 5.8\times10^{7}/(8.5\times10^{28}\times1.6\times10^{-19}) = 4.26\times10^{-3}\) m²/V s. \(\tau = \mu m/e = 4.26\times10^{-3}\times9.1\times10^{-31}/1.6\times10^{-19} = 2.43\times10^{-14}\) s.

Example 6. Find the probability that an energy level 0.1 eV above the Fermi level is occupied at 300 K.

\(kT = 8.62\times10^{-5}\times300 = 0.0259\) eV. \((E - E_F)/kT = 0.1/0.0259 = 3.87\). \(f = 1/(1 + e^{3.87}) = 1/(1 + 47.9) = 0.0205\), about 2 percent.

Example 7. At what temperature will a level 0.5 eV above the Fermi level have a 1 percent probability of occupation?

\(1/(1 + e^{x}) = 0.01\) gives \(e^{x} = 99\), \(x = \ln99 = 4.595\). So \(kT = 0.5/4.595 = 0.1088\) eV and \(T = 0.1088\times1.6\times10^{-19}/1.38\times10^{-23} = 1262\) K.

Example 8. GaAs has a direct band gap of 1.42 eV. Find the wavelength of light emitted by band-to-band recombination. Why is silicon (1.12 eV) not used for LEDs?

\(\lambda = hc/E_g = 6.63\times10^{-34}\times3\times10^{8}/(1.42\times1.6\times10^{-19}) = 8.75\times10^{-7}\) m, about 875 nm (near infrared); the short cut 1.24/1.42 μm gives 873 nm, the difference being rounding of constants. Silicon's gap is indirect: recombination needs a phonon, so it is slow and most energy goes into heat rather than light, giving very poor emission efficiency.

Important questions and answers

Q1. Distinguish between direct and indirect band gap semiconductors with E-k diagrams. (5 marks)

Direct: conduction band minimum and valence band maximum at the same k; electron-hole recombination emits a photon of energy Eg directly, no momentum change needed; high probability of radiative recombination; short carrier lifetime; used in LEDs, laser diodes, photodetectors; examples GaAs, InP, GaN, CdS. Indirect: conduction band minimum and valence band maximum at different k; recombination needs a phonon to conserve momentum; mostly non-radiative, energy released as heat; longer carrier lifetime; used in diodes, transistors, ICs and solar cells; examples Si, Ge, GaP. Diagram: two E-k parabolas, one directly above the other (direct) and one shifted sideways (indirect), with vertical photon arrow and slanted phonon-assisted arrow.

Q2. Define drift velocity and mobility. Derive an expression for the conductivity of a conductor. (5 marks)

Drift velocity: average velocity of free electrons along the direction of the electric force when a field is applied. Mobility: drift velocity per unit field, \(\mu = v_d/E\). Derivation: an electron in field E has acceleration eE/m; over the relaxation time τ it gains \(v_d = eE\tau/m\), so \(\mu = e\tau/m\). In time dt the electrons in a length \(v_d dt\) of a wire of area A cross a section: charge \(dq = neAv_d dt\), so \(I = neAv_d\) and \(J = nev_d = ne\mu E\). By Ohm's law \(J = \sigma E\), so \(\sigma = ne\mu = ne^2\tau/m\) and \(\rho = m/ne^2\tau\).

Q3. Derive the expression for the conductivity of an intrinsic semiconductor and of n-type and p-type semiconductors. (5 marks)

In a semiconductor, current is carried by electrons (concentration n, mobility μe) and holes (p, μh). Electron current density \(J_e = ne\mu_eE\); hole current density \(J_h = pe\mu_hE\). Both add: \(J = e(n\mu_e + p\mu_h)E\), so \(\sigma = e(n\mu_e + p\mu_h)\). Intrinsic: \(n = p = n_i\), \(\sigma_i = n_ie(\mu_e + \mu_h)\). n-type: \(n \approx N_D \gg p\), \(\sigma_n \approx N_De\mu_e\). p-type: \(p \approx N_A \gg n\), \(\sigma_p \approx N_Ae\mu_h\). Minority carriers from \(np = n_i^2\).

Q4. Explain the Fermi-Dirac distribution function and its variation with temperature. (5 marks)

\(f(E) = 1/(1 + e^{(E - E_F)/kT})\) gives the probability that a state of energy E is occupied by an electron. At T = 0 K: for \(E < E_F\), the exponent is minus infinity, f = 1 (all states filled); for \(E > E_F\), f = 0 (all empty). So the graph is a step at \(E_F\), which is the highest occupied level at 0 K. At T > 0 K: for \(E = E_F\), f = 1/2 always; some electrons just below \(E_F\) are excited to states just above it, so the step is rounded over a width of a few kT; at higher T the rounding is wider. The curve is symmetric about \(E_F\): the probability that a state ΔE above \(E_F\) is filled equals the probability that a state ΔE below is empty. Graph: f(E) vs E with curves for 0 K, T1 and T2 > T1, all passing through 0.5 at \(E_F\).

Q5. Show that the Fermi level of an intrinsic semiconductor lies at the middle of the band gap. (5 marks)

Electron concentration in the conduction band: \(n = N_ce^{-(E_c - E_F)/kT}\). Hole concentration in the valence band: \(p = N_ve^{-(E_F - E_v)/kT}\), where \(N_c\) and \(N_v\) are effective densities of states, \(N_c \propto (m_e^*)^{3/2}\) and \(N_v \propto (m_h^*)^{3/2}\). For an intrinsic semiconductor n = p: \(N_ce^{-(E_c - E_F)/kT} = N_ve^{-(E_F - E_v)/kT}\). Taking logs: \(2E_F = E_c + E_v + kT\ln(N_v/N_c)\), so \(E_F = (E_c + E_v)/2 + (3kT/4)\ln(m_h^*/m_e^*)\). If \(m_h^* = m_e^*\), or since the second term is very small, \(E_F = (E_c + E_v)/2\): the Fermi level lies at the middle of the band gap and is practically independent of temperature.

Q6. Explain with diagrams the position of the Fermi level in n-type and p-type semiconductors and its variation with temperature and doping concentration. (6 marks)

n-type: donor level Ed lies about 0.01 to 0.05 eV below Ec. At 0 K all donor electrons are bound and \(E_F = (E_c + E_d)/2\), between Ed and Ec. As T rises, donors ionise and later intrinsic generation becomes significant, so n approaches ni and \(E_F\) falls gradually towards the mid gap; at very high T the material becomes intrinsic and \(E_F = E_i\). Increasing donor concentration raises \(E_F\) towards (and for degenerate doping into) the conduction band. p-type: acceptor level Ea lies slightly above Ev. At 0 K \(E_F = (E_v + E_a)/2\). With rising T, \(E_F\) rises towards mid gap; with increasing acceptor concentration it moves down towards the valence band. Diagrams: band diagrams showing Ec, Ev, Ed or Ea and EF, plus a plot of EF versus T showing EF starting near the band edge and approaching the intrinsic level Ei as T increases, with curves for different doping levels.

Q7. Why does the conductivity of a metal decrease and that of a semiconductor increase with temperature? (3 marks)

Metal: σ = neμ. The number of free electrons n is fixed and very large. Rising temperature increases lattice vibrations, so electrons collide more often, τ and μ decrease, and σ decreases. Semiconductor: σ = e(nμe + pμh). Mobility also decreases with temperature, but the carrier concentration \(n_i \propto T^{3/2}e^{-E_g/2kT}\) increases exponentially as more electrons are excited across the gap. This increase dominates, so σ increases with temperature.

Q8. What is the probability of occupation of the Fermi level at any temperature? Prove it. (3 marks)

Put \(E = E_F\) in \(f(E) = 1/(1 + e^{(E - E_F)/kT})\): exponent is zero, \(e^0 = 1\), so \(f(E_F) = 1/(1 + 1) = 1/2\). Thus at any temperature above 0 K the probability that the Fermi level is occupied is 50 percent, which can be used as the definition of the Fermi level.

Q9. Define mobility. Why is electron mobility greater than hole mobility? (3 marks)

Mobility is the drift velocity per unit electric field, \(\mu = v_d/E\), unit m²/V s. Electrons in the conduction band move freely through the crystal and have a smaller effective mass. Hole movement is really the successive jumping of valence electrons into neighbouring vacancies, which is a more constrained process with larger effective mass, so holes drift more slowly; in Si, μe is about 0.135 and μh about 0.048 m²/V s.

Q10. A silicon sample is doped with \(10^{21}\) donors per m³. Find its conductivity (\(\mu_e = 0.135\) m²/V s). (3 marks)

\(\sigma \approx N_De\mu_e = 10^{21}\times1.6\times10^{-19}\times0.135 = 21.6\) S/m; resistivity \(= 1/21.6 = 0.0463\) Ω m.

Exam tips

• Draw E-k diagrams for direct and indirect band gaps; the sideways shift in the indirect case is the key point examiners look for.
• In conductivity numericals check units: concentrations in m^-3 and mobility in m²/V s give σ in S/m.
• For extrinsic semiconductors ignore minority carriers unless asked; use np = ni² to find them.
• Remember f(EF) = 1/2 and draw the Fermi-Dirac curve at 0 K and higher temperatures on the same axes.
• For Fermi level questions draw band diagrams for intrinsic, n-type and p-type side by side, then a sketch of EF versus T.
• Use kT = 0.0259 eV at 300 K to save time in Fermi-Dirac numericals.
• State clearly: metals have positive and semiconductors negative temperature coefficient of resistance.

FAQs

Is the Fermi level an actual energy level that electrons occupy?

Not necessarily. In a semiconductor it usually lies in the band gap where there are no states. It is a reference energy at which the occupation probability would be one half.

Why do LEDs use GaAs or GaN instead of silicon?

They have direct band gaps, so electron-hole recombination emits photons efficiently. Silicon's indirect gap wastes most recombination energy as heat.

Why is drift velocity so small when current flows almost instantly?

The electric field (signal) travels along the wire at a large fraction of the speed of light, setting all electrons drifting at once; each electron itself moves only a fraction of a millimetre per second.

Does doping change the band gap?

For normal doping levels the band gap is essentially unchanged. Doping adds donor or acceptor levels inside the gap and shifts the Fermi level, which changes the carrier concentrations.

These notes follow the University of Mumbai first-year engineering syllabus (NEP 2020, Applied Physics) and were prepared with AI help and checked.