Introduction to Semiconductor Devices

This post contains my lecture notes of taking the course VE320: Introduction to Semiconductor Devices at UM-SJTU Joint Institute.

Chapter 1: The Crystal Structure of Solids

  • Unit cell vs. primitive cell
    • primitive cell: smallest unit cell possible
  • Three lattice types
    • Simple cubic
    • Body-centered cubic
    • Face-centered cubic
  • Volume density = # of atoms per unit cella3\dfrac{\text{\# of atoms per unit cell}}{a^3}
    • Unit: 1 Angstrom = 1e-10 m
    • Miller Index: (1x-intersect,1y-intersect,1z-intersect)×lcm(x,y,z)(\frac{1}{x\text{-intersect}},\frac{1}{y\text{-intersect}},\frac{1}{z\text{-intersect}})\times\text{lcm}(x,y,z)
      • (100) plane, (110) plane, (111) plane
      • [100] direction, [110] direction, [111] direction

Chapter 2: Introduction to Quantum Mechanics

  • Plank’s constant: h=6.625×1034 J-sh=6.625\times10^{-34}~\text{J-s}
    • Photon’s energy E=hν=ωE=h\nu=\hbar\omega
    • Photon’s momentum p=hλp=\frac{h}{\lambda} (de Broglie wavelength)

Schrodinger’s Equation

Total wave function: Ψ(x,t)=ψ(x)ϕ(t)=ψ(x)ej(E/)t\Psi(x,t)=\psi(x)\phi(t)=\psi(x)e^{-j(E/\hbar)t}.

Time-Independent Wave Function

2ψ(x)x2+2m2(EV(x))ψ(x)=0.\frac{\partial^2\psi(x)}{\partial x^2}+\frac{2m}{\hbar^2}(E-V(x))\psi(x)=0.

  • ψ(x)2|\psi(x)|^2 is the probability density function. ψ(x)2dx=1\int_{-\infty}^{\infty}|\psi(x)|^2\mathrm{d}x=1.
  • Boundary conditions for solving ψ(x)\psi(x):
    • ψ(x)\psi(x) must be finite, continuos, single-valued.
    • ψ/x\partial\psi/\partial x must be finite, continuos, single-valued.
  • kk is wave number. p=kp=\hbar k.
  • Steps of solving wave function:
    • Consider Schrodinger’s equation in each region. Solve separately.
    • Apply boundary conditions to determine coefficients.

Electron Energy in Single Atom

V(r)=e24πϵ0r,2ψ(r,θ,ϕ)+2m02(EV(r))ψ(r,θ,ϕ)=0V(r)=\frac{-e^2}{4\pi\epsilon_0 r},\quad \nabla^2\psi(r,\theta,\phi)+\frac{2m_0}{\hbar^2}(E-V(r))\psi(r,\theta,\phi)=0

  • Solution: En=m0e4(4πϵ0)222n2E_n=\dfrac{-m_0 e^4}{(4\pi\epsilon_0)^22\hbar^2n^2}, nn is the principal quantum number.

Chapter 3: Introduction to the Quantum Theory of Solids

Formation of Energy Band

  • Silicon: 1s22s22p63s23p21s^22s^22p^63s^23p^2. Energy levels split and formed two large energy bands: valence band & conduction band.
  • EE vs. kk curve for free particle: E=k222mE=\dfrac{k^2\hbar^2}{2m}. (parabolic relation)
  • EE vs. kk diagram in reduced zone representation (for Si crystal, derived from Kronig-Penney model).

Electrical Conduction in Solids

  • Drift Current:
    • J=qNvdJ=qNv_d (A/cm2)(\text{A}/\text{cm}^2), NN being the volume density.
    • If considering individual ion velocities, J=qi=1NviJ=q\sum_{i=1}^{N}v_i.
  • Electron effective mass:
    • dEdk2=2m\dfrac{\mathrm{d}^E}{\mathrm{d}k^2}=\dfrac{\hbar^2}{m}.

Metals, Insulators, Semiconductors

  • Metals:
    • Has a partially filled energy band.
  • Insulators:
    • Has an either completely filled or completely empty energy band.

Density of States Function

Consider 1-D crystal with NN quantum wells, each well of length aa.

  • Number of states of the whole crystal within (0,π/a)(0,\pi/a): Nπ/a\dfrac{N}{\pi/a}
  • Number of states of the whole crystal within Δk\Delta k: Nπ/a×Δk\dfrac{N}{\pi/a}\times\Delta k
  • Number of states per unit volume within Δk\Delta k: Nπ/a×Δk1Na=Δkπ\dfrac{N}{\pi/a}\times\Delta k\dfrac{1}{Na}=\dfrac{\Delta k}{\pi}

For EE vs. kk relation:

E=Ec+22mnk2,k=±2mn(EEc)E=E_c+\frac{\hbar^2}{2m^*_n}k^2,\qquad k=\pm\frac{\sqrt{2m^*_n(E-E_c)}}{\hbar}

Extend to 3-D sphere:


Final conclusion:

Conduction band: gc(E)=4π(2mn)3/2h3EEc\text{Conduction band: }g_c(E)=\frac{4\pi(2m^*_n)^{3/2}}{h^3}\sqrt{E-E_c}

Valence band: gv(E)=4π(2mp)3/2h3EvE\text{Valence band: }g_v(E)=\frac{4\pi(2m^*_p)^{3/2}}{h^3}\sqrt{E_v-E}

  • mnm^*_n is the density of states effective mass for electrons.
  • mpm^*_p is the density of states effective mass for holes.
  • EcE_c is the bottom edge of conduction band.
  • EvE_v is the top edge of valence band.

Statistical Mechanics

Fermi-Dirac distribution:


  • represents the probability of a quantum state is occupied by an electron.
  • kk is the Boltzmann Constant. k = 8.62e-5 eV/K.

When EEFkTE-E_F\gg kT, 1 at the denominator is ignored. Fermi-Dirac distribution becomes Maxwell-Boltzmann distribution.


Chapter 4: Semiconductor in Equilibrium

Thermal-Equilibrium Carrier Concentration

Electron (use Maxwell-Boltzmann approximation):

n0=gc(E)fF(E)dEn0=2(2πmnkTh2)3/2exp[EcEFkT]n_0=\int g_c(E)f_F(E)\mathrm{d}E\quad\Rightarrow\quad n_0=2\left(\frac{2\pi m^*_n kT}{h^2}\right)^{3/2}\exp\left[-\frac{E_c-E_F}{kT}\right]

For simplicity,


  • NcN_c is called effective density of states function in the conduction band.

Hole (use Maxwell-Boltzmann approximation):

p0=2(2πmpkTh2)3/2exp[EFEvkT]p_0=2\left(\frac{2\pi m^*_p kT}{h^2}\right)^{3/2}\exp\left[-\frac{E_F-E_v}{kT}\right]

For simplicity,


  • NvN_v is called effective density of states function in the valence band.

Intrinsic Carrier Concentration

ni=n0=p0n_i=n_0=p_0 for intrinsic semiconductors. Therefore,

ni=n0p0=NcNvexp(Eg2kT)n_i=\sqrt{n_0p_0}=\sqrt{N_c N_v}\exp\left(-\frac{E_g}{2kT}\right)

Intrinsic Fermi Level Position

For intrinsic semiconductors, n0=p0n_0=p_0, EF=EFiE_F=E_{Fi}. Therefore, equating the previous n0n_0 and p0p_0 expressions, we get

EFiEmidgap=34kTln(mpmn),where12(Ec+Ev)=EmidgapE_{Fi}-E_\text{midgap}=\frac{3}{4}kT\ln\left(\frac{m^*_p}{m^*_n}\right),\quad\text{where}\quad \frac{1}{2}(E_c+E_v)=E_\text{midgap}

Dopant Atoms and Energy Levels

Introduce EdE_d in nn-type semiconductor, which is the discrete donor energy state. Therefore, EcEdE_c-E_d is the ionization energy.

Introduce EaE_a in pp-type semiconductor, which is the discrete acceptor energy state. Therefore, EaEvE_a-E_v is the ionization energy.

The Extrinsic Semiconductor

An extrinsic semiconductor is defined as a semiconductor in which controlled amounts of specific dopant or impurity atoms have been added so that the thermal-equilibrium electron and hole concentrations are different from the intrinsic carrier concentration.

Equilibrium Distribution of Electrons and Holes



  • n0p0=ni2n_0p_0=n_i^2 still holds.

Degenerate and Non-degenerate Semiconductor

When the donor concentration is high enough to split the discrete donor energy state, the semiconductor is called a degenerate nn-type semiconductor.

When the acceptor concentration is high enough to split the discrete acceptor energy state, the semiconductor is called a degenerate pp-type semiconductor.

Statistics of Donors and Acceptors

fD(E)=11+12exp(EdEFkT),nd=NdNd+.f_D(E)=\frac{1}{1+\dfrac{1}{2}\exp\left(\dfrac{E_d-E_F}{kT}\right)},\qquad n_d=N_d-N_d^+.

  • Reason for 1/21/2: in conduction band energy states, each state can be occupied by at most 2 electrons (spin up & spin down); while in donor energy state, each state can only be occupied by 1 electron (either spin up or spin down).

nd=Nd1+12exp(EdEFkT),nd=NdNd+.n_d=\frac{N_d}{1+\dfrac{1}{2}\exp\left(\dfrac{E_d-E_F}{kT}\right)},\qquad n_d=N_d-N_d^+.

  • ndn_d represents the density of electrons occupying the donor state.
  • nd=Nd×fD(Ed)n_d=N_d\times f_D(E_d).

pa=Na1+1gexp(EFEakT),pa=NaNa+.p_a=\frac{N_a}{1+\dfrac{1}{g}\exp\left(\dfrac{E_F-E_a}{kT}\right)},\qquad p_a=N_a-N_a^+.

  • pap_a represents the probability of holes occupying the acceptor state.
  • gg, the degeneracy factor, is normally taken as 4.

Complete ionization: all donor/acceptor atoms have donated an electron/a hole to the conduction band/valence band.

Freeze-out: all donor/acceptor energy states are filled with electrons/holes.

Charge Neutrality

A compensated semiconductor is one that contains both donor and acceptor impurity atoms in the same region.

  • nn-type compensated: Nd>NaN_d>N_a
  • pp-type compensated: Na>NdN_a>N_d

Equilibrium Electron and Hole Concentrations

At equilibrium, overall charge is neutral.

n0+Na=p0+Nd+n0+(Napa)=p0+(Ndnd)n_0+N_a^-=p_0+N_d^+\quad\Leftrightarrow\quad n_0+(N_a-p_a)=p_0+(N_d-n_d)

Assume complete ionization:

n0+Na=ni2n0+Ndn0=(NdNa)2+(NdNa2)2+ni2.n_0+N_a=\frac{n_i^2}{n_0}+N_d\quad\Rightarrow\quad n_0=\frac{(N_d-N_a)}{2}+\sqrt{\left(\frac{N_d-N_a}{2}\right)^2+n_i^2}.

  • Only valid for an nn-type semiconductor!

Assume complete ionization:

ni2p0+Na=p0+Ndp0=(NaNd)2+(NaNd2)2+ni2.\frac{n_i^2}{p_0}+N_a=p_0+N_d\quad\Rightarrow\quad p_0=\frac{(N_a-N_d)}{2}+\sqrt{\left(\frac{N_a-N_d}{2}\right)^2+n_i^2}.

  • Only valid for an pp-type semiconductor!

Incomplete ionization:

  • When TT is very high, niNd+n_i\gg N_d^+. Therefore n0=ni=NcNvexp(Eg2kT)n_0=n_i=\sqrt{N_cN_v}\exp\left(\dfrac{-E_g}{2kT}\right);
  • When TT is not high:






Final conclusion:


Position of Fermi Energy Level

With respect to intrinsic Fermi energy level:

n0=niexp(EFEFikT)EFEFi=kTln(n0ni)n_0=n_i\exp\left(\dfrac{E_F-E_{Fi}}{kT}\right)\quad\Rightarrow\quad E_F-E_{Fi}=kT\ln\left(\frac{n_0}{n_i}\right)

p0=niexp(EFiEFkT)EFiEF=kTln(p0ni)p_0=n_i\exp\left(\dfrac{E_{Fi}-E_F}{kT}\right)\quad\Rightarrow\quad E_{Fi}-E_F=kT\ln\left(\frac{p_0}{n_i}\right)

With respect to valence band energy level:

po=Nvexp(EFEvkT)EFEv=kTln(Nvp0)p_o=N_v\exp\left(\dfrac{E_F-E_v}{kT}\right)\quad\Rightarrow\quad E_F-E_v=kT\ln\left(\frac{N_v}{p_0}\right)

  • if we assume NaniN_a\gg n_i, then EFEv=kTln(NvNa)E_F-E_v=kT\ln\left(\dfrac{N_v}{N_a}\right)

With respect to conduction band energy level:


  • Note. n0n_0 can be derived from complete ionization OR incomplete ionization formulas.

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