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 cell a 3 \dfrac{\text{\# of atoms per unit cell}}{a^3} a 3 # of atoms per unit cell
Unit: 1 Angstrom = 1e-10 m
Miller Index: ( 1 x -intersect , 1 y -intersect , 1 z -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) ( x -intersect 1 , y -intersect 1 , z -intersect 1 ) × 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 × 1 0 − 34 J-s h=6.625\times10^{-34}~\text{J-s} h = 6 . 6 2 5 × 1 0 − 3 4 J-s
Photon’s energy E = h ν = ℏ ω E=h\nu=\hbar\omega E = h ν = ℏ ω
Photon’s momentum p = h λ p=\frac{h}{\lambda} p = λ h (de Broglie wavelength )
Schrodinger’s Equation
Total wave function: Ψ ( x , t ) = ψ ( x ) ϕ ( t ) = ψ ( x ) e − j ( E / ℏ ) t \Psi(x,t)=\psi(x)\phi(t)=\psi(x)e^{-j(E/\hbar)t} Ψ ( x , t ) = ψ ( x ) ϕ ( t ) = ψ ( x ) e − j ( E / ℏ ) t .
Time-Independent Wave Function
∂ 2 ψ ( x ) ∂ x 2 + 2 m ℏ 2 ( E − V ( x ) ) ψ ( x ) = 0. \frac{\partial^2\psi(x)}{\partial x^2}+\frac{2m}{\hbar^2}(E-V(x))\psi(x)=0.
∂ x 2 ∂ 2 ψ ( x ) + ℏ 2 2 m ( E − V ( x ) ) ψ ( x ) = 0 .
∣ ψ ( x ) ∣ 2 |\psi(x)|^2 ∣ ψ ( x ) ∣ 2 is the probability density function. ∫ − ∞ ∞ ∣ ψ ( x ) ∣ 2 d x = 1 \int_{-\infty}^{\infty}|\psi(x)|^2\mathrm{d}x=1 ∫ − ∞ ∞ ∣ ψ ( x ) ∣ 2 d x = 1 .
Boundary conditions for solving ψ ( x ) \psi(x) ψ ( x ) :
ψ ( x ) \psi(x) ψ ( x ) must be finite, continuos, single-valued.
∂ ψ / ∂ x \partial\psi/\partial x ∂ ψ / ∂ x must be finite, continuos, single-valued.
k k k is wave number. p = ℏ k p=\hbar k p = ℏ 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 ) = − e 2 4 π ϵ 0 r , ∇ 2 ψ ( r , θ , ϕ ) + 2 m 0 ℏ 2 ( E − V ( r ) ) ψ ( r , θ , ϕ ) = 0 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
V ( r ) = 4 π ϵ 0 r − e 2 , ∇ 2 ψ ( r , θ , ϕ ) + ℏ 2 2 m 0 ( E − V ( r ) ) ψ ( r , θ , ϕ ) = 0
Solution: E n = − m 0 e 4 ( 4 π ϵ 0 ) 2 2 ℏ 2 n 2 E_n=\dfrac{-m_0 e^4}{(4\pi\epsilon_0)^22\hbar^2n^2} E n = ( 4 π ϵ 0 ) 2 2 ℏ 2 n 2 − m 0 e 4 , n n n is the principal quantum number .
Chapter 3: Introduction to the Quantum Theory of Solids
Silicon: 1 s 2 2 s 2 2 p 6 3 s 2 3 p 2 1s^22s^22p^63s^23p^2 1 s 2 2 s 2 2 p 6 3 s 2 3 p 2 . Energy levels split and formed two large energy bands: valence band & conduction band.
E E E vs. k k k curve for free particle: E = k 2 ℏ 2 2 m E=\dfrac{k^2\hbar^2}{2m} E = 2 m k 2 ℏ 2 . (parabolic relation)
E E E vs. k k k diagram in reduced zone representation (for Si crystal, derived from Kronig-Penney model ).
Electrical Conduction in Solids
Drift Current:
J = q N v d J=qNv_d J = q N v d ( A / cm 2 ) (\text{A}/\text{cm}^2) ( A / cm 2 ) , N N N being the volume density.
If considering individual ion velocities, J = q ∑ i = 1 N v i J=q\sum_{i=1}^{N}v_i J = q ∑ i = 1 N v i .
Electron effective mass :
d E d k 2 = ℏ 2 m \dfrac{\mathrm{d}^E}{\mathrm{d}k^2}=\dfrac{\hbar^2}{m} d k 2 d E = m ℏ 2 .
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 N N quantum wells, each well of length a a a .
Number of states of the whole crystal within ( 0 , π / a ) (0,\pi/a) ( 0 , π / a ) : N π / a \dfrac{N}{\pi/a} π / a N
Number of states of the whole crystal within Δ k \Delta k Δ k : N π / a × Δ k \dfrac{N}{\pi/a}\times\Delta k π / a N × Δ k
Number of states per unit volume within Δ k \Delta k Δ k : N π / a × Δ k 1 N a = Δ k π \dfrac{N}{\pi/a}\times\Delta k\dfrac{1}{Na}=\dfrac{\Delta k}{\pi} π / a N × Δ k N a 1 = π Δ k
For E E E vs. k k k relation:
E = E c + ℏ 2 2 m n ∗ k 2 , k = ± 2 m n ∗ ( E − E c ) ℏ E=E_c+\frac{\hbar^2}{2m^*_n}k^2,\qquad k=\pm\frac{\sqrt{2m^*_n(E-E_c)}}{\hbar}
E = E c + 2 m n ∗ ℏ 2 k 2 , k = ± ℏ 2 m n ∗ ( E − E c )
Extend to 3-D sphere:
g ( E ) = 1 8 d ( 4 π / 3 × ( k / π ) 3 ) d E g(E)=\frac{1}{8}\frac{\mathrm{d}(4\pi/3\times(k/\pi)^3)}{\mathrm{d}E}
g ( E ) = 8 1 d E d ( 4 π / 3 × ( k / π ) 3 )
Final conclusion:
Conduction band: g c ( E ) = 4 π ( 2 m n ∗ ) 3 / 2 h 3 E − E c \text{Conduction band: }g_c(E)=\frac{4\pi(2m^*_n)^{3/2}}{h^3}\sqrt{E-E_c}
Conduction band: g c ( E ) = h 3 4 π ( 2 m n ∗ ) 3 / 2 E − E c
Valence band: g v ( E ) = 4 π ( 2 m p ∗ ) 3 / 2 h 3 E v − E \text{Valence band: }g_v(E)=\frac{4\pi(2m^*_p)^{3/2}}{h^3}\sqrt{E_v-E}
Valence band: g v ( E ) = h 3 4 π ( 2 m p ∗ ) 3 / 2 E v − E
m n ∗ m^*_n m n ∗ is the density of states effective mass for electrons.
m p ∗ m^*_p m p ∗ is the density of states effective mass for holes.
E c E_c E c is the bottom edge of conduction band.
E v E_v E v is the top edge of valence band.
Statistical Mechanics
Fermi-Dirac distribution:
f F ( E ) = 1 1 + exp ( E − E F k T ) f_F(E)=\frac{1}{1+\exp\left(\dfrac{E-E_F}{kT}\right)}
f F ( E ) = 1 + exp ( k T E − E F ) 1
represents the probability of a quantum state is occupied by an electron.
k k k is the Boltzmann Constant. k = 8.62e-5 eV/K
.
When E − E F ≫ k T E-E_F\gg kT E − E F ≫ k T , 1 at the denominator is ignored. Fermi-Dirac distribution becomes Maxwell-Boltzmann distribution.
f F ( E ) = 1 exp ( E − E F k T ) = exp ( − E − E F k T ) f_F(E)=\frac{1}{\exp\left(\dfrac{E-E_F}{kT}\right)}=\exp\left(-\frac{E-E_F}{kT}\right)
f F ( E ) = exp ( k T E − E F ) 1 = exp ( − k T E − E F )
Chapter 4: Semiconductor in Equilibrium
Thermal-Equilibrium Carrier Concentration
Electron (use Maxwell-Boltzmann approximation ):
n 0 = ∫ g c ( E ) f F ( E ) d E ⇒ n 0 = 2 ( 2 π m n ∗ k T h 2 ) 3 / 2 exp [ − E c − E F k T ] 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]
n 0 = ∫ g c ( E ) f F ( E ) d E ⇒ n 0 = 2 ( h 2 2 π m n ∗ k T ) 3 / 2 exp [ − k T E c − E F ]
For simplicity,
n 0 = N c exp [ − E c − E F k T ] n_0=N_c\exp\left[-\frac{E_c-E_F}{kT}\right]
n 0 = N c exp [ − k T E c − E F ]
N c N_c N c is called effective density of states function in the conduction band.
Hole (use Maxwell-Boltzmann approximation ):
p 0 = 2 ( 2 π m p ∗ k T h 2 ) 3 / 2 exp [ − E F − E v k T ] p_0=2\left(\frac{2\pi m^*_p kT}{h^2}\right)^{3/2}\exp\left[-\frac{E_F-E_v}{kT}\right]
p 0 = 2 ( h 2 2 π m p ∗ k T ) 3 / 2 exp [ − k T E F − E v ]
For simplicity,
p 0 = N v exp [ − E F − E v k T ] p_0=N_v\exp\left[-\frac{E_F-E_v}{kT}\right]
p 0 = N v exp [ − k T E F − E v ]
N v N_v N v is called effective density of states function in the valence band.
Intrinsic Carrier Concentration
n i = n 0 = p 0 n_i=n_0=p_0 n i = n 0 = p 0 for intrinsic semiconductors. Therefore,
n i = n 0 p 0 = N c N v exp ( − E g 2 k T ) n_i=\sqrt{n_0p_0}=\sqrt{N_c N_v}\exp\left(-\frac{E_g}{2kT}\right)
n i = n 0 p 0 = N c N v exp ( − 2 k T E g )
Intrinsic Fermi Level Position
For intrinsic semiconductors, n 0 = p 0 n_0=p_0 n 0 = p 0 , E F = E F i E_F=E_{Fi} E F = E F i . Therefore, equating the previous n 0 n_0 n 0 and p 0 p_0 p 0 expressions, we get
E F i − E midgap = 3 4 k T ln ( m p ∗ m n ∗ ) , where 1 2 ( E c + E v ) = E midgap 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}
E F i − E midgap = 4 3 k T ln ( m n ∗ m p ∗ ) , where 2 1 ( E c + E v ) = E midgap
Dopant Atoms and Energy Levels
Introduce E d E_d E d in n n n -type semiconductor, which is the discrete donor energy state. Therefore, E c − E d E_c-E_d E c − E d is the ionization energy.
Introduce E a E_a E a in p p p -type semiconductor, which is the discrete acceptor energy state. Therefore, E a − E v E_a-E_v 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 [ E F − E F i k T ] n_0=n_i\exp\left[\frac{E_F-E_{Fi}}{kT}\right]
n 0 = n i exp [ k T E F − E F i ]
p 0 = n i exp [ − E F − E F i k T ] p_0=n_i\exp\left[-\frac{E_F-E_{Fi}}{kT}\right]
p 0 = n i exp [ − k T E F − E F i ]
n 0 p 0 = n i 2 n_0p_0=n_i^2 n 0 p 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 n n -type semiconductor .
When the acceptor concentration is high enough to split the discrete acceptor energy state, the semiconductor is called a degenerate p p p -type semiconductor .
Statistics of Donors and Acceptors
f D ( E ) = 1 1 + 1 2 exp ( E d − E F k T ) , n d = N d − N d + . 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^+.
f D ( E ) = 1 + 2 1 exp ( k T E d − E F ) 1 , n d = N d − N d + .
Reason for 1 / 2 1/2 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 = N d 1 + 1 2 exp ( E d − E F k T ) , n d = N d − N d + . 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 = 1 + 2 1 exp ( k T E d − E F ) N d , n d = N d − N d + .
n d n_d n d represents the density of electrons occupying the donor state .
n d = N d × f D ( E d ) n_d=N_d\times f_D(E_d) n d = N d × f D ( E d ) .
p a = N a 1 + 1 g exp ( E F − E a k T ) , p a = N a − N a + . 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 = 1 + g 1 exp ( k T E F − E a ) N a , p a = N a − N a + .
p a p_a p a represents the probability of holes occupying the acceptor state .
g g 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 n n -type compensated: N d > N a N_d>N_a N d > N a
p p p -type compensated: N a > N d N_a>N_d N a > N d
Equilibrium Electron and Hole Concentrations
At equilibrium, overall charge is neutral.
n 0 + N a − = p 0 + N d + ⇔ n 0 + ( N a − p a ) = p 0 + ( N d − n d ) n_0+N_a^-=p_0+N_d^+\quad\Leftrightarrow\quad n_0+(N_a-p_a)=p_0+(N_d-n_d)
n 0 + N a − = p 0 + N d + ⇔ n 0 + ( N a − p a ) = p 0 + ( N d − n d )
Assume complete ionization :
n 0 + N a = n i 2 n 0 + N d ⇒ n 0 = ( N d − N a ) 2 + ( N d − N a 2 ) 2 + n i 2 . 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}.
n 0 + N a = n 0 n i 2 + N d ⇒ n 0 = 2 ( N d − N a ) + ( 2 N d − N a ) 2 + n i 2 .
Only valid for an n n n -type semiconductor !
Assume complete ionization :
n i 2 p 0 + N a = p 0 + N d ⇒ p 0 = ( N a − N d ) 2 + ( N a − N d 2 ) 2 + n i 2 . \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}.
p 0 n i 2 + N a = p 0 + N d ⇒ p 0 = 2 ( N a − N d ) + ( 2 N a − N d ) 2 + n i 2 .
Only valid for an p p p -type semiconductor !
Incomplete ionization:
When T T T is very high, n i ≫ N d + n_i\gg N_d^+ n i ≫ N d + . Therefore n 0 = n i = N c N v exp ( − E g 2 k T ) n_0=n_i=\sqrt{N_cN_v}\exp\left(\dfrac{-E_g}{2kT}\right) n 0 = n i = N c N v exp ( 2 k T − E g ) ;
When T T T is not high:
n 0 = N d + = N d ( 1 − 1 1 + 1 2 exp ( E d − E F k T ) ) 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 = N d + = N d ⎝ ⎜ ⎜ ⎛ 1 − 1 + 2 1 exp ( k T E d − E F ) 1 ⎠ ⎟ ⎟ ⎞
n 0 = N d 1 + 2 exp ( − E d − E F k T ) n_0=\frac{N_d}{1+2\exp\left(-\dfrac{E_d-E_F}{kT}\right)}
n 0 = 1 + 2 exp ( − k T E d − E F ) N d
n 0 = N d 1 + 2 exp ( E c − E d k T ) exp ( E F − E c k T ) 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 = 1 + 2 exp ( k T E c − E d ) exp ( k T E F − E c ) N d
n 0 = N d 1 + 2 exp ( E c − E d k T ) n 0 N c n_0=\frac{N_d}{1+2\exp\left(\dfrac{E_c-E_d}{kT}\right)\dfrac{n_0}{N_c}}
n 0 = 1 + 2 exp ( k T E c − E d ) N c n 0 N d
2 exp ( E c − E d k T ) n 0 2 + N c n 0 − N d N c = 0 2\exp\left(\frac{E_c-E_d}{kT}\right)n_0^2+N_cn_0-N_dN_c=0
2 exp ( k T E c − E d ) n 0 2 + N c n 0 − N d N c = 0
Final conclusion:
n 0 = N c × − 1 + 1 + 8 N d N c exp ( E c − E d k T ) 4 exp ( E c − E d k T ) . 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)}.
n 0 = N c × 4 exp ( k T E c − E d ) − 1 + 1 + N c 8 N d exp ( k T E c − E d ) .
Position of Fermi Energy Level
With respect to intrinsic Fermi energy level:
n 0 = n i exp ( E F − E F i k T ) ⇒ E F − E F i = k T ln ( n 0 n i ) 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)
n 0 = n i exp ( k T E F − E F i ) ⇒ E F − E F i = k T ln ( n i n 0 )
p 0 = n i exp ( E F i − E F k T ) ⇒ E F i − E F = k T ln ( p 0 n i ) 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)
p 0 = n i exp ( k T E F i − E F ) ⇒ E F i − E F = k T ln ( n i p 0 )
With respect to valence band energy level:
p o = N v exp ( E F − E v k T ) ⇒ E F − E v = k T ln ( N v p 0 ) 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)
p o = N v exp ( k T E F − E v ) ⇒ E F − E v = k T ln ( p 0 N v )
if we assume N a ≫ n i N_a\gg n_i N a ≫ n i , then E F − E v = k T ln ( N v N a ) E_F-E_v=kT\ln\left(\dfrac{N_v}{N_a}\right) E F − E v = k T ln ( N a N v )
With respect to conduction band energy level:
E F − E c = k T ln ( n 0 N c ) . E_F-E_c=kT\ln\left(\frac{n_0}{N_c}\right).
E F − E c = k T ln ( N c n 0 ) .
Note. n 0 n_0 n 0 can be derived from complete ionization OR incomplete ionization formulas.