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 = $\dfrac{\text{number of atoms per unit cell}}{a^3}$
    • Unit: 1 Angstrom = 1e-10 m
    • Miller Index: $(\frac{1}{x\text{-intersect}},\frac{1}{y\text{-intersect}},\frac{1}{z\text{-intersect}})\times\text{lcm}(x,y,z)$
      • (100) plane, (110) plane, (111) plane
      • [100] direction, [110] direction, [111] direction

Chapter 2: Introduction to Quantum Mechanics

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

Schrodinger’s Equation

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

Time-Independent Wave Function

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

  • $|\psi(x)|^2$ is the probability density function. $\int_{-\infty}^{\infty}|\psi(x)|^2\mathrm{d}x=1$.
  • Boundary conditions for solving $\psi(x)$:
    • $\psi(x)$ must be finite, continuos, single-valued.
    • $\partial\psi/\partial x$ must be finite, continuos, single-valued.
  • $k$ is wave number. $p=\hbar k$.
  • 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

$$ V(r)=\frac{-e^2}{4\pi\epsilon_0 r},\quad \nabla^2\psi(r,\theta,\phi)+\frac{2m_0}{\hbar^2}(E-V(r))\psi(r,\theta,\phi)=0 $$

  • Solution: $E_n=\dfrac{-m_0 e^4}{(4\pi\epsilon_0)^22\hbar^2n^2}$, $n$ is the principal quantum number.

Chapter 3: Introduction to the Quantum Theory of Solids

Formation of Energy Band

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

Electrical Conduction in Solids

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

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 $N$ quantum wells, each well of length $a$.

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

For $E$ vs. $k$ relation: $$ E=E_c+\frac{\hbar^2}{2m^_n}k^2,\qquad k=\pm\frac{\sqrt{2m^_n(E-E_c)}}{\hbar} $$

Extend to 3-D sphere: $$ g(E)=\frac{1}{8}\frac{\mathrm{d}(4\pi/3\times(k/\pi)^3)}{\mathrm{d}E} $$

Final conclusion: $$ \text{Conduction band: }g_c(E)=\frac{4\pi(2m^*_n)^{3/2}}{h^3}\sqrt{E-E_c} $$

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

  • $m^*_n$ is the density of states effective mass for electrons.
  • $m^*_p$ is the density of states effective mass for holes.
  • $E_c$ is the bottom edge of conduction band.
  • $E_v$ is the top edge of valence band.

Statistical Mechanics

Fermi-Dirac distribution: $$ f_F(E)=\frac{1}{1+\exp\left(\dfrac{E-E_F}{kT}\right)} $$

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

When $E-E_F\gg kT$, 1 at the denominator is ignored. Fermi-Dirac distribution becomes Maxwell-Boltzmann distribution. $$ f_F(E)=\frac{1}{\exp\left(\dfrac{E-E_F}{kT}\right)}=\exp\left(-\frac{E-E_F}{kT}\right) $$

Chapter 4: Semiconductor in Equilibrium

Thermal-Equilibrium Carrier Concentration

Electron (use Maxwell-Boltzmann approximation): $$ n_0=\int g_c(E)f_F(E)\mathrm{d}E\quad\Rightarrow\quad n_0=2\left(\frac{2\pi m^*_n kT}{h^2}\right)^{3/2}\exp\left[-\frac{E_c-E_F}{kT}\right] $$

For simplicity, $$ n_0=N_c\exp\left[-\frac{E_c-E_F}{kT}\right] $$

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

Hole (use Maxwell-Boltzmann approximation): $$ p_0=2\left(\frac{2\pi m^*_p kT}{h^2}\right)^{3/2}\exp\left[-\frac{E_F-E_v}{kT}\right] $$

For simplicity, $$ p_0=N_v\exp\left[-\frac{E_F-E_v}{kT}\right] $$

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

Intrinsic Carrier Concentration

$n_i=n_0=p_0$ for intrinsic semiconductors. Therefore, $$ n_i=\sqrt{n_0p_0}=\sqrt{N_c N_v}\exp\left(-\frac{E_g}{2kT}\right) $$

Intrinsic Fermi Level Position

For intrinsic semiconductors, $n_0=p_0$, $E_F=E_{Fi}$. Therefore, equating the previous $n_0$ and $p_0$ expressions, we get $$ E_{Fi}-E_\text{midgap}=\frac{3}{4}kT\ln\left(\frac{m^*_p}{m^*n}\right),\quad\text{where}\quad \frac{1}{2}(E_c+E_v)=E\text{midgap} $$

Dopant Atoms and Energy Levels

Introduce $E_d$ in $n$-type semiconductor, which is the discrete donor energy state. Therefore, $E_c-E_d$ is the ionization energy.

Introduce $E_a$ in $p$-type semiconductor, which is the discrete acceptor energy state. Therefore, $E_a-E_v$ 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

$$ n_0=n_i\exp\left[\frac{E_F-E_{Fi}}{kT}\right] $$

$$ p_0=n_i\exp\left[-\frac{E_F-E_{Fi}}{kT}\right] $$

  • $n_0p_0=n_i^2$ 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 $n$-type semiconductor.

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

Statistics of Donors and Acceptors

$$ f_D(E)=\frac{1}{1+\dfrac{1}{2}\exp\left(\dfrac{E_d-E_F}{kT}\right)},\qquad n_d=N_d-N_d^+. $$

  • Reason for $1/2$: 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).

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

  • $n_d$ represents the density of electrons occupying the donor state.
  • $n_d=N_d\times f_D(E_d)$.

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

  • $p_a$ represents the probability of holes occupying the acceptor state.
  • $g$, 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.

  • $n$-type compensated: $N_d>N_a$
  • $p$-type compensated: $N_a>N_d$

Equilibrium Electron and Hole Concentrations

At equilibrium, overall charge is neutral. $$ n_0+N_a^-=p_0+N_d^+\quad\Leftrightarrow\quad n_0+(N_a-p_a)=p_0+(N_d-n_d) $$

Assume complete ionization: $$ n_0+N_a=\frac{n_i^2}{n_0}+N_d\quad\Rightarrow\quad n_0=\frac{(N_d-N_a)}{2}+\sqrt{\left(\frac{N_d-N_a}{2}\right)^2+n_i^2}. $$

  • Only valid for an $n$-type semiconductor!

Assume complete ionization: $$ \frac{n_i^2}{p_0}+N_a=p_0+N_d\quad\Rightarrow\quad p_0=\frac{(N_a-N_d)}{2}+\sqrt{\left(\frac{N_a-N_d}{2}\right)^2+n_i^2}. $$

  • Only valid for an $p$-type semiconductor!

Incomplete ionization:

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

$$ n_0=N_d^+=N_d\left(1-\frac{1}{1+\dfrac{1}{2}\exp\left(\dfrac{E_d-E_F}{kT}\right)}\right) $$

$$ n_0=\frac{N_d}{1+2\exp\left(-\dfrac{E_d-E_F}{kT}\right)} $$

$$ n_0=\frac{N_d}{1+2\exp\left(\dfrac{E_c-E_d}{kT}\right)\exp\left(\dfrac{E_F-E_c}{kT}\right)} $$

$$ n_0=\frac{N_d}{1+2\exp\left(\dfrac{E_c-E_d}{kT}\right)\dfrac{n_0}{N_c}} $$

$$ 2\exp\left(\frac{E_c-E_d}{kT}\right)n_0^2+N_cn_0-N_dN_c=0 $$

Final conclusion: $$ n_0=N_c\times\frac{-1+\sqrt{1+\dfrac{8N_d}{N_c}\exp\left(\dfrac{E_c-E_d}{kT}\right)}}{4\exp\left(\dfrac{E_c-E_d}{kT}\right)}. $$

Position of Fermi Energy Level

With respect to intrinsic Fermi energy level: $$ n_0=n_i\exp\left(\dfrac{E_F-E_{Fi}}{kT}\right)\quad\Rightarrow\quad E_F-E_{Fi}=kT\ln\left(\frac{n_0}{n_i}\right) $$ $$ p_0=n_i\exp\left(\dfrac{E_{Fi}-E_F}{kT}\right)\quad\Rightarrow\quad E_{Fi}-E_F=kT\ln\left(\frac{p_0}{n_i}\right) $$

With respect to valence band energy level: $$ p_o=N_v\exp\left(\dfrac{E_F-E_v}{kT}\right)\quad\Rightarrow\quad E_F-E_v=kT\ln\left(\frac{N_v}{p_0}\right) $$

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

With respect to conduction band energy level: $$ E_F-E_c=kT\ln\left(\frac{n_0}{N_c}\right). $$

  • Note. $n_0$ can be derived from complete ionization OR incomplete ionization formulas.