1. Equilibrium Carrier Statistics
Chapter 4
Equilibrium Carrier Statistics

2. DENSITY OF STATES
The density of state function per unit volume, g(E)dE, gives the number of available quantum states in the energy interval between E and E + dE.
What do we need to know?
i) E – k relationship (dispersion relation)
𝐸= ℏ 2 𝑘 2 2 𝑚 𝑒 ∗ = ℏ 𝑘 𝑥 2 𝑚 𝑥𝑥 + 𝑘 𝑦 2 𝑚 𝑦𝑦 + 𝑘 𝑧 2 𝑚 𝑧𝑧
If 𝑚 𝑥𝑥 = 𝑚 𝑦𝑦 = 𝑚 𝑧𝑧 : isotropic effective mass
If not : anisotropic effective mass
ii) Proceed with constant energy surface.
isotropic effective mass → spherical constant energy surface
anisotropic effective mass → ellipsoidal constant energy surface

3. General Derivation = 𝑆 𝑑 𝑆 𝑑 𝑘 𝑛
E + dE E
constant energy surface
𝑑 𝑘 = 𝑑 𝑘 𝑛 + 𝑑 𝑘 𝑙
𝑑 𝑆 :
differential surface area for constant energy surface
Total volume in k-space between E and E + dE, 𝑥 𝑦 𝑧
𝑘 𝑥
𝑘 𝑦
𝑘 𝑧
2𝜋/𝑎
2𝜋/𝑏
2𝜋/𝑐
𝑉=(𝑎𝑏𝑐)
= 𝑆 𝑑 𝑆 𝑑 𝑘 𝑛
constant energy surface
Unit volume occupied by each value of 𝑘
= 2𝜋 3 𝑉
: minimum volume in 𝑘 -space
Number of states/each unit volume = 1
= 2, including spin

4. Total number of states between E and (E + dE)lunit volume
spin
𝑔 𝐸 𝑑𝐸= 2 𝑆 𝑑 𝑆 𝑑 𝑘 𝑛 (2𝜋) 3 /𝑉 /𝑉= 2 (2𝜋) 3 𝑆 𝑑 𝑆 𝑑 𝑘 𝑛
The differential energy,
𝑑𝐸= 𝛻 𝑘 𝐸∙𝑑 𝑘 = 𝛻 𝑘 𝐸 𝑑 𝑘 𝑛
𝑔 𝐸 𝑑𝐸= 2 (2𝜋) 3 𝑆 𝑑 𝑆 𝛻 𝑘 𝐸 𝑑𝐸
𝑔 𝐸 = 2 (2𝜋) 3 𝑆 𝑑 𝑆 𝛻 𝑘 𝐸
Then,
general expression for density of states
𝑆 𝑑 𝑆 𝛻 𝑘 𝐸 =
𝑆 𝑑 𝑆 𝑑 𝑘 𝑛 𝑑𝐸 = (𝑑 𝑘 ) 𝑑𝐸
or equivalently,
volume element
𝑔 𝐸 = 2 (2𝜋) (𝑑 𝑘 ) 𝑑𝐸
𝑔 𝐸 𝑑𝐸 =2 (𝑑 𝑘 ) (2𝜋) 3
Then,
𝑔 𝐸 𝑑𝐸 = 𝑠𝑝𝑖𝑛 𝑚𝑖𝑛𝑖𝑚𝑎 𝑏𝑎𝑛𝑑 (𝑑 𝑘 ) (2𝜋) 𝑁
volume element
dimension
= 1 for 1-D
2 for 2-D
3 for 3-D
= 2
= 1 for Γ
3 for X
4 for L
heavy hole
light hole
split-off

5. Density of States for Free Particle in 3 - D
𝐸= ℏ 2 𝑘 2 2𝑚 = ℏ 2 2𝑚 𝑘 𝑥 2 + 𝑘 𝑦 2 + 𝑘 𝑧 2
𝑔 𝐸 = 2 (2𝜋) 3 𝑆 𝑑 𝑆 𝛻 𝑘 𝐸
spherical constant energy surface
parabolic E-k relationship
𝑆 𝑑 𝑆 =4𝜋 𝑘 2
𝛻 𝑘 𝐸 = 𝑎 𝑥 ℏ 2 𝑚 𝑘 𝑥 + 𝑎 𝑦 ℏ 2 𝑚 𝑘 𝑦 + 𝑎 𝑧 ℏ 2 𝑚 𝑘 𝑧 = ℏ 2 𝑚 𝑘
𝑔 𝐸 = 2 (2𝜋) 3 𝑆 𝑑 𝑆 𝛻 𝑘 𝐸 = 2 (2𝜋) 3 ∙ ℏ 2 𝑚 𝑘∙4𝜋 𝑘 2 = 𝑚𝑘 𝜋 2 ℏ 2 = 𝑚 3/2 𝜋 2 ℏ 3 𝐸 1/2
𝑔 𝐸 𝑑𝐸 =2 (𝑑 𝑘 ) (2𝜋) 3
𝑑𝑘= 𝑚 ℏ ℏ 2 2𝑚𝐸 𝑑𝐸
𝐸= ℏ 2 𝑘 2 2𝑚
𝑑 𝑘 =4𝜋 𝑘 2 𝑑𝑘,
from
𝑔 𝐸 𝑑𝐸 = 𝑚 3/2 𝜋 2 ℏ 3 𝐸 1/2 𝑑𝐸
𝑔 𝐸 = 𝑚 3/2 𝜋 2 ℏ 3 𝐸 1/2
Density of States for Free Particle in 2 - D and 1 - D
i) 2-D
Neglecting energy minima and band, the volume element (𝑑 𝑘 ) becomes area element.

6. 𝑔 𝐸 𝑑𝐸 =2 2𝑑𝑘 (2𝜋) 1 = 2𝑚 𝜋ℏ 𝐸 𝑑𝐸 𝑔1𝐷 𝐸 = 2𝑚 𝜋ℏ 𝐸
𝑔 𝐸 𝑑𝐸 =2 (𝑑 𝑘 ) (2𝜋) 𝑁 ky
area element
(𝑑 𝑘 )=2𝜋𝑘𝑑𝑘
dimension
= 2 for 2-D 𝑘
𝑔 𝐸 𝑑𝐸 =2 2𝜋𝑘𝑑𝑘 (2𝜋) 2 = 𝜋 ℏ 2𝑚𝐸 𝑚 ℏ 𝑚𝐸 ℏ 2 𝑑𝐸 = 𝑚 𝜋 ℏ 2 𝑑𝐸 kx
𝑑𝑘= 𝑚 ℏ ℏ 2 2𝑚𝐸 𝑑𝐸,
𝐸= ℏ 2 𝑘 2 2𝑚
𝑔2𝐷 𝐸 = 𝑚 𝜋 ℏ 2
: constant
ii) 1-D
The volume element (𝑑 𝑘 ) becomes line element and N = 1.
(𝑑 𝑘 )=2𝑑𝑘
𝑔 𝐸 𝑑𝐸 =2 2𝑑𝑘 (2𝜋) 1 = 𝑚 𝜋ℏ 𝐸 𝑑𝐸
𝑑 𝑘 kx 𝑘
g(E)
𝑔1𝐷 𝐸 = 2𝑚 𝜋ℏ 𝐸
𝑔1𝐷
𝑔3𝐷
𝑔2𝐷 E

7. Specific Materials Conduction Band – GaAs
isotropic effective mass and parabolic E-k relationship near 𝛤 band minimum (k = 0)
𝐸− 𝐸 𝐶 ≅ ℏ 2 𝑘 2 2 𝑚 𝑒 ∗ = ℏ 2 2 𝑚 𝑛 ∗ 𝑘 𝑥 2 + 𝑘 𝑦 2 + 𝑘 𝑧 2
: spherical constant
energy surface
same as free particle except that
𝑚→ 𝑚 𝑛 ∗
𝐸→(𝐸− 𝐸 𝐶 )
𝑔 𝐶 𝐸 = 𝑁 𝑆 𝑚 𝑛 ∗ 𝜋 2 ℏ 3 (𝐸 − 𝐸 𝐶 ) 1/2
for 𝐸> 𝐸 𝐶
where 𝑁 𝑆 (number of equivalent minima) = 1 for 𝛤 band
Conduction Band – Si, Ge
( 1 2 ×8) equivalent band minima
6 equivalent band minima
almost parabolic E-k relationship near L or X band minimum, but anisotropic effective mass
𝐸− 𝐸 𝐶 ≅ ℏ 2 2 𝑚 𝑙 ∗ 𝑘 𝑥 2 + ℏ 2 2 𝑚 𝑡 ∗ 𝑘 𝑦 2 + 𝑘 𝑧 2
: ellipsoidal constant
energy surface
Let
𝑘 𝑥 2 𝑚 𝑙 ∗ = 𝑘 𝑥 ′ 2 , 𝑘 𝑦 2 𝑚 𝑡 ∗ = 𝑘 𝑦 ′ 2 , 𝑘 𝑧 2 𝑚 𝑡 ∗ = 𝑘 𝑧 ′ 2
then,
𝑑 𝑘 𝑥 = 𝑚 𝑙 ∗ 𝑑 𝑘 𝑥 ′ , 𝑑 𝑘 𝑦 = 𝑚 𝑡 ∗ 𝑑 𝑘 𝑦 ′ ,𝑑 𝑘 𝑧 = 𝑚 𝑡 ∗ 𝑑 𝑘 𝑧 ′

8. 𝐸− 𝐸 𝐶 = ℏ 2 2 𝑘 ′ 2 = ℏ 2 2 𝑘 𝑥 ′ 2 + 𝑘 𝑦 ′ 2 + 𝑘 𝑧 ′ 2
𝐸− 𝐸 𝐶 = ℏ 𝑘 ′ 2 = ℏ 𝑘 𝑥 ′ 2 + 𝑘 𝑦 ′ 2 + 𝑘 𝑧 ′ 2
: spherical constant energy surface in k’-space
where (𝑑 𝑘 ) = 𝑑 𝑘 𝑥 𝑑 𝑘 𝑦 𝑑𝑧= 𝑚 𝑙 ∗ 𝑚 𝑡 ∗ /2 𝑑 𝑘 𝑥 ′ 𝑑 𝑘 𝑦 ′ 𝑑 𝑘 𝑧 ′
= 𝑚 𝑙 ∗ 𝑚 𝑡 ∗ /2 (𝑑 𝑘 ′ )
𝑔 𝐶 𝐸 𝑑𝐸 =2 (𝑑 𝑘 ) (2𝜋) 3 ,
(𝑑 𝑘 ′ )= 4𝜋 𝑘 ′ 2 𝑑 𝑘 ′ =4𝜋 2(𝐸 − 𝐸 𝐶 ) ℏ ℏ 2(𝐸 − 𝐸 𝐶 ) 𝑑𝐸
and
𝑔 𝐶 𝐸 𝑑𝐸 = 𝑚 𝑙 ∗ 𝑚 𝑡 ∗ 𝜋 3 𝑑 𝑘 ′ = 2 𝑚 𝑙 ∗ 𝑚 𝑡 ∗ /2 4𝜋 2(𝐸 − 𝐸 𝐶 ) ℏ ℏ 2(𝐸 − 𝐸 𝐶 ) 𝑑𝐸
Then,
= 𝑁 𝑒𝑙 𝜋 2 ℏ 𝑚 𝑙 ∗ 𝑚 𝑡 ∗ 𝐸− 𝐸 𝐶 𝑑𝐸
equivalent band minima
2 𝜋 2 ℏ 𝑚 𝑛 ∗ 𝐸− 𝐸 𝐶 𝑑𝐸
Nel = 6 for Si (not at X-band edge)
Nel = ×8=4 for Ge (L-band edge)
6 2/3 𝑚 𝑙 ∗ 𝑚 𝑡 ∗
for Si
𝑚 𝑛 ∗ = 𝑁 𝑒𝑙 𝑚 𝑙 ∗ 𝑚 𝑡 ∗ ∴
𝑚 𝑛 ∗ =
4 2/3 𝑚 𝑙 ∗ 𝑚 𝑡 ∗
for Ge

9. Valence Band – Si, Ge, GaAs
degenerated hh and lh band at k = 0 k
heavy hole
light hole
split-off
Neglecting split-off band,
𝑔 𝑉 𝐸 = 𝑔 ℎℎ 𝐸 + 𝑔 𝑙ℎ 𝐸 = 𝑚 ℎℎ ∗ 𝜋 2 ℏ 3 ( 𝐸 𝑉 −𝐸) 1/ 𝑚 𝑙ℎ ∗ 𝜋 2 ℏ 3 ( 𝐸 𝑉 −𝐸) 1/2
= 𝑚 𝑝 ∗ 𝜋 2 ℏ 3 ( 𝐸 𝑉 −𝐸) 1/2
𝑚 𝑝 ∗ = 𝑚 ℎℎ ∗ 𝑚 𝑙ℎ ∗
𝑚 𝑝 ∗ = 𝑚 ℎℎ ∗ 𝑚 ℎℎ ∗ ∴

10. 𝑔1𝐷 𝐸 2 = 2𝑚 𝜋ℏ 𝐸 Density of States of Low-Dimensional Semiconductors
𝑔2𝐷 𝐸 3 = 𝑚 ∗ 𝜋 ℏ 2
i) 2-D semiconductor (quantum well)
Each subband has constant density states.
𝑔2𝐷 𝐸 = 𝑚 ∗ 𝜋 ℏ 2 E4 E4 E3 E3 E2 E2 E1 E1
ii) 1-D semiconductor (quantum wire)
𝑔1𝐷 𝐸 2 = 2𝑚 𝜋ℏ 𝐸 E1 E2 E3

11. FERMI FUNCTION Maxwell-Boltzmann : classical Bose-Einstein
Fermi-Dirac
: classical
Energy distribution functions:
quantum mechanical
compose of the product of two terms:
1) the number of energy states with the energy interval
2) the probability that a particle occupies the states
Fermi-Dirac distribution function is applied for the particle system which obeys the Pauli Exclusion Principle.
No two electrons can have identical quantum states.
Bose-Einstein distribution function is applied for the particle system which does not obey the Pauli Exclusion Principle.

12. Fermi-Dirac Statistics
Derivation Proper
Each allowed state can accommodate one and only one electron.
Two macroscopic constraints:
𝑖 𝑁 𝑖 = 𝑁
: total number of particles
constants
𝑖 𝑁 𝑖 𝐸 𝑖 = 𝐸 𝑇
: total energy of particle system
What is the most probable distribution subject to these constraints?
How many ways can we arrange Ni particles among the Si states in a given interval?

13. 𝑊 𝑖 = 𝑆 𝑖 ! 𝑆 𝑖 − 𝑁 𝑖 ! 𝑁 𝑖 ! 𝑊= 𝑖 𝑊 𝑖 = 𝑖 𝑆 𝑖 ! 𝑆 𝑖 − 𝑁 𝑖 ! 𝑁 𝑖 !
Like putting Ni objects into Si boxes, but the boxes will hold only 1 particle.
Level electrons we can put first in Si boxes, second into Si -1 boxes………last into Si – Ni + 1.
Total number of different ways of putting labeled electrons into the boxes:.
𝑁 𝑖 𝑆 𝑖 − 𝑁 𝑖 +1 = 𝑆 𝑖 𝑆 𝑖 −1 𝑆 𝑖 −2 ….. 𝑆 𝑖 − 𝑁 𝑖 +1 = 𝑆 𝑖 ! 𝑆 𝑖 − 𝑁 𝑖 !
However, electrons are indistinguishable and there are Ni! ways of labeling the electron.
∴ The number of physically different ways of putting Ni electrons among Si states
in the Ei energy level is
𝑊 𝑖 = 𝑆 𝑖 ! 𝑆 𝑖 − 𝑁 𝑖 ! 𝑁 𝑖 !
for any Ei level
The total number of different ways in which N electrons can be arranged in the multilevel system (i.e., number of ways putting N1 electrons into S1 states, N2 electrons into S2 states………Ni electrons onto Si states),
𝑊= 𝑖 𝑊 𝑖 = 𝑖 𝑆 𝑖 ! 𝑆 𝑖 − 𝑁 𝑖 ! 𝑁 𝑖 !

14. Under thermal equilibrium, the most probable distribution or arranged of electrons is the one that is most disordered.
That is, the distribution of electrons which can occur in the largest number of ways is the most probable one.
∴ The most probable distribution occurs for maximum W subject to the constant constraints of
𝑖 𝑁 𝑖 = 𝑁,
𝑖 𝑁 𝑖 𝐸 𝑖 = 𝐸 𝑇
To easily maximize, treat with 𝑙𝑛𝑊
𝑙𝑛𝑊=𝑙𝑛⁡( 𝑖 𝑊 𝑖 )= 𝑖 𝑙𝑛𝑊 𝑖 = 𝑖 𝑙𝑛𝑆 𝑖 !− ln⁡(𝑆 𝑖 − 𝑁 𝑖 !) − 𝑙𝑛𝑁 𝑖 !
Maximize W with respect to Ni methods of “ Largrangian undetermined multipliers”
𝛼 − 𝑖 𝑁 𝑖 + 𝑁 = 0,
𝛽 − 𝑖 𝐸 𝑖 𝑁 𝑖 + 𝐸 𝑇 = 0
𝜕 𝜕 𝑁 𝑖 𝑙𝑛𝑊+𝛼 − 𝑖 𝑁 𝑖 + 𝑁 +𝛽 − 𝑖 𝐸 𝑖 𝑁 𝑖 + 𝐸 𝑇 = 0

15. 𝜕 𝜕 𝑁 𝑖 𝑙𝑛𝑊 −𝛼−𝛽 𝐸 𝑖 = 0
Using Stirling’s approximation,
𝑙𝑛𝑥!≅𝑥 𝑙𝑛𝑥 - 𝑥
𝑙𝑛𝑊= 𝑖 𝑆 𝑖 𝑙𝑛𝑆 𝑖 − 𝑆 𝑖 − (𝑆 𝑖 − 𝑁 𝑖 ) ln⁡(𝑆 𝑖 − 𝑁 𝑖 )+( 𝑆 𝑖 − 𝑁 𝑖 ) −𝑁 𝑖 𝑙𝑛 𝑁 𝑖 + 𝑁 𝑖
= 𝑖 𝑆 𝑖 𝑙𝑛 𝑆 𝑖 𝑆 𝑖 − 𝑁 𝑖 − 𝑁 𝑖 𝑙𝑛 𝑁 𝑖 𝑆 𝑖 − 𝑁 𝑖
…..(𝑎)
𝜕 𝜕 𝑁 𝑖 𝑙𝑛𝑊 = 𝜕 𝜕 𝑁 𝑖 𝑆 𝑖 𝑙𝑛 𝑆 𝑖 𝑆 𝑖 − 𝑁 𝑖 − 𝑁 𝑖 𝑙𝑛 𝑁 𝑖 𝑆 𝑖 − 𝑁 𝑖 =𝑙𝑛 𝑆 𝑖 − 𝑁 𝑖 𝑁 𝑖
𝑆 𝑖 − 𝑁 𝑖 𝑁 𝑖 = 𝑒 𝛼+𝛽 𝐸 𝑖
∴𝑙𝑛 𝑆 𝑖 − 𝑁 𝑖 𝑁 𝑖 =𝛼+𝛽 𝐸 𝑖
𝑓 𝐸 𝑖 = 𝑁 𝑖 𝑆 𝑖 = 𝑒 𝛼+𝛽 𝐸 𝑖
called “Fermi-Dirac distribution function”
If we define a Fermi energy,
𝐸 𝐹 ≡ −𝛼 𝛽
𝑓 𝐸 𝑖 = 𝑒 𝛽( 𝐸 𝑖 − 𝐸 𝐹 )
…..(𝑏)
𝛽 can be evaluated by thermodynamics.

16. We can define the Helmholtz function,
𝐹 ≡ 𝐸 𝑇 −𝑇𝑆
…..(𝑐)
where 𝐸 𝑇 = internal energy of a system
𝑇 = temperature
S = entropy
P = pressure
V = volume
and the Gibbs function as
𝐺 ≡ 𝐸 𝑇 −𝑇𝑆 +𝑃𝑉
…..(𝑑)
F and G are minimum at thermal equilibrium.
𝑓 𝐸 𝑖 = 𝑒 𝛽( 𝐸 𝑖 − 𝐸 𝐹 )
…..(𝑏)
Boltzmann definition of entropy at thermal equilibrium:
𝑆 = 𝑘𝑙𝑛𝑊
…..(𝑒)
: most probable arrangement of particles in a crystal.
Substituting (b) into (a),
𝑙𝑛𝑊 = 𝑖 𝑆 𝑖 𝑙𝑛 𝑆 𝑖 𝑆 𝑖 − 𝑁 𝑖 − 𝑁 𝑖 𝑙𝑛 𝑁 𝑖 𝑆 𝑖 − 𝑁 𝑖 = 𝑖 𝑆 𝑖 𝑙𝑛 1+ 𝑒 𝛽( 𝐸 𝐹 − 𝐸 𝑖 ) + 𝑁 𝑖 −1 𝛽( 𝐸 𝑖 − 𝐸 𝐹 ) =𝛽 𝑖 𝑁 𝑖 𝐸 𝑖 − 𝑖 𝑁 𝑖 𝐸 𝐹 + 𝑖 𝑆 𝑖 𝑙𝑛 1+ 𝑒 𝛽( 𝐸 𝐹 − 𝐸 𝑖 ) =𝛽 𝐸 𝑇 − 𝐸 𝐹 𝑁 + 𝑖 𝑆 𝑖 𝑙𝑛 1+ 𝑒 𝛽( 𝐸 𝐹 − 𝐸 𝑖 )
≈ 𝑁 𝑖
𝜕 𝜕 𝐸 𝑇 𝑙𝑛𝑊 =𝛽

17. 𝑓 𝐸 = 1 1+ 𝑒 𝐸− 𝐸 𝐹 /𝑘𝑇 𝜕 𝜕 𝐸 𝑇 𝑙𝑛𝑊 = 1 𝑘 𝜕𝑆 𝜕 𝐸 𝑇 = 1 𝑘𝑇 𝑆 = 𝑘𝑙𝑛𝑊
𝜕 𝜕 𝐸 𝑇 𝑙𝑛𝑊 = 1 𝑘 𝜕𝑆 𝜕 𝐸 𝑇 = 1 𝑘𝑇
𝑆 = 𝑘𝑙𝑛𝑊
𝜕𝐹 𝜕 𝐸 𝑇 ≡1−𝑇 𝜕𝑆 𝜕 𝐸 𝑇 =0
𝐹 ≡ 𝐸 𝑇 −𝑇𝑆
∴𝛽= 1 𝑘𝑇
α=− 𝐸 𝐹 𝑘𝑇
and
𝑓 𝐸 𝑖 = 1 1+ 𝑒 𝐸 𝑖 − 𝐸 𝐹 /𝑘𝑇
Replacing Ei with continuous variable E in energy band,
𝑓 𝐸 = 1 1+ 𝑒 𝐸− 𝐸 𝐹 /𝑘𝑇
𝐸− 𝐸 𝐹 (eV)

18. SUPPLEMENTAL INFORMATION
Equilibrium Distribution of Carriers
𝑛 𝐸 =𝑔 𝐶 𝐸 𝑓(𝐸)
𝑔 𝐶 𝐸 = 𝑁 𝑆 𝑚 𝑛 ∗ 𝜋 2 ℏ 3 (𝐸 − 𝐸 𝐶 ) 1/2
𝑔 𝑉 𝐸 = 𝑚 𝑝 ∗ 𝜋 2 ℏ 3 ( 𝐸 𝑉 −𝐸) 1/2
𝑝 𝐸 =𝑔 𝑉 𝐸 [1−𝑓 𝐸 ]
𝑓 𝐸 = 1 1+ 𝑒 𝐸− 𝐸 𝐹 /𝑘𝑇

19. The Energy Band Diagram
ℰ (𝑥,𝑦,𝑧)=−𝛻𝑉(𝑥,𝑦,𝑧)
ℰ 𝑥 =− 𝑑𝑉(𝑥) 𝑑𝑥
ℰ 𝑥 = 1 𝑞 𝑑 𝐸 𝐶 (𝑥) 𝑑𝑥 = 1 𝑞 𝑑 𝐸 𝑉 (𝑥) 𝑑𝑥
𝑑ℰ 𝑥 𝑑𝑥 = 𝜌(𝑥) 𝜖
𝑃.𝐸. 𝑥 = 𝐸 𝐶 (x) − 𝐸 𝑟𝑒𝑓
𝑃.𝐸. 𝑥 =−𝑞𝑉(𝑥)
𝑉 𝑥 =− 1 𝑞 𝐸 𝐶 𝑥 − 𝐸 𝑟𝑒𝑓

20. Donors, Acceptors, Band Gap Centers
Intrinsic material, n = p = ni
Extrinsic material

22. Ionization(Binding) Energy of Donor and Acceptor
Energy required for electron in solid to make a transition from the donor level to the
conduction band and become (quasi) free. r3
e, m0 r2
n=1
n=2
n=3
e, m* r3 r2 r1
n=1
n=3
n=2 Si
= 11.7 for Si
hydrogen atom
in vacuum
donor atom
in Si r1
n =  , E = 0
n = 3, E3
n = 2, E2
n = 1, E1
hydrogen atom
n =  , Ed= Ec
n = 3, Ed3
n = 2, Ed2
n = 1, Ed1
donor atom
Ec= Ed
~6meV
n=1, Ed1= Ed
, n=1, 2, 3,

23. EQUILIBRIUM CONCENTRATION RELATIONSHIPS
Formulas for n and p
Electron concentration in the conduction band
≈ ∞ if 𝐸 𝑡𝑜𝑝 - EF >> kT
due to exponential dependence of f(E)
𝑛= 𝐸 𝐶 𝐸 𝑡𝑜𝑝 𝑛 𝐸 𝑑𝐸= 𝐸 𝐶 𝐸 𝑡𝑜𝑝 𝑔 𝐶 𝐸 𝑓 𝐸 𝑑𝐸
𝑔 𝐶 𝐸 = 𝑚 𝑛 ∗ 𝜋 2 ℏ 3 (𝐸 − 𝐸 𝐶 ) 1/2
Hole concentration in the valence band
𝑔 𝑉 𝐸 = 𝑚 𝑝 ∗ 𝜋 2 ℏ 3 ( 𝐸 𝑉 −𝐸) 1/2
𝑝= 𝐸 𝑏𝑜𝑡𝑡𝑜𝑚 𝐸 𝑉 𝑝 𝐸 𝑑𝐸= 𝐸 𝑏𝑜𝑡𝑡𝑜𝑚 𝐸 𝑉 𝑔 𝑉 𝐸 [1−𝑓 𝐸 ]𝑑𝐸
𝑓 𝐸 = 1 1+ 𝑒 𝐸− 𝐸 𝐹 /𝑘𝑇
≈− ∞
𝑛= 𝐸 𝐶 ∞ 𝑔 𝐶 𝐸 𝑓 𝐸 𝑑𝐸= 𝑚 𝑛 ∗ 𝜋 2 ℏ 3 𝐸 𝐶 ∞ (𝐸 − 𝐸 𝐶 ) 1/2 1+ 𝑒 𝐸− 𝐸 𝐹 /𝑘𝑇 𝑑𝐸
using change of variable,
ξ = 𝐸− 𝐸 𝐶 𝑘𝑇 ,
η = 𝐸 𝐹 − 𝐸 𝐶 𝑘𝑇
𝑑ξ = 1 𝑘𝑇 𝑑𝐸, (𝐸 − 𝐸 𝐶 ) 1/2 = ξ 1/2 (𝑘𝑇) 1/2 ,𝐸 − 𝐸 𝐹 =ξ - η
𝑛= 𝑚 𝑛 ∗ (𝑘𝑇) 𝜋 2 ℏ ∞ ξ 1/2 1+ 𝑒 ξ − η 𝑑ξ = 𝑁 𝐶 ℱ 1/2 η 𝐶
then
= 𝜋 𝑁 𝐶 𝐹 1/2 η 𝐶
Similarly, for holes in valence band
𝑝= 𝑁 𝑉 ℱ 1/2 η 𝑉
= 𝜋 𝑁 𝑉 𝐹 1/2 η 𝑉

24. where
ℱ 1/2 η = 𝜋 𝐹 1/2 η
𝐹 1/2 η ≡ 0 ∞ ξ 1/2 𝑑ξ 1+ 𝑒 ξ −η
: Fermi-Dirac integral of order 1/2
𝑁 𝐶 =2 𝑚 𝑛 ∗ 𝑘𝑇 2π ℏ /2
: effective density of states at conduction band edge
𝑁 𝑉 =2 𝑚 𝑝 ∗ 𝑘𝑇 2π ℏ /2
: effective density of states at valence band edge

25. For nondegenerated semiconductors,If
𝐸 𝐹 ≤ 𝐸 𝐶 −3𝑘𝑇 ( η 𝐶 ≤ − 3),
1+ 𝑒 ξ − η −1 ≈ 𝑒 −(ξ − η)
𝑛= 𝑁 𝐶 ℱ 1/2 η 𝐶 ≈ 𝑁 𝐶 2 𝜋 0 ∞ ξ 𝑒 − 𝜉 − 𝜂 𝑑ξ
≈ 𝑁 𝐶 2 𝜋 𝑒 η 𝐶 0 ∞ ξ 1/2 𝑒 −ξ 𝑑ξ
then
=𝑁 𝐶 𝑒 η 𝐶 =𝑁 𝐶 𝑒 −( 𝐸 𝐶 − 𝐸 𝐹 )
0 ∞ 𝑦 1/2 𝑒 −𝑎𝑦 𝑑𝑦 = 𝜋 2𝑎 𝑎
𝛤- function
likewise If
𝐸 𝐹 ≥ 𝐸 𝑉 +3𝑘𝑇,
𝑝 =𝑁 𝑉 𝑒 −( 𝐸 𝐹 − 𝐸 𝑉 )
𝑛𝑝 =𝑁 𝐶 𝑁 𝑉 𝑒 − 𝐸 𝐺 𝑘𝑇 = 𝑛 𝑖 2
Mass-action law
𝑛 𝑖 = 𝑁 𝐶 𝑁 𝑉 𝑒 −𝐸 𝐺 2𝑘𝑇

26. Physical Meaning of “Effective Density of States”
𝑔 𝐶 (𝐸)= 𝑁 𝐶 δ(𝐸 − 𝐸 𝐶 )
𝑛= 0 ∞ 𝑔 𝐶 𝐸 𝑓 𝐸 𝑑𝐸
=𝑁 𝐶 𝑒 −( 𝐸 𝐶 − 𝐸 𝐹 )
= 𝑁 𝐶 𝑓 𝐸
𝑛= 0 ∞ 𝑔 𝐶 𝐸 𝑓 𝐸 𝑑𝐸
= 0 ∞ 𝑁 𝐶 δ(𝐸 − 𝐸 𝐶 ) 𝑓 𝐸 𝑑𝐸
= 𝑁 𝐶 𝑓 𝐸
for nondegenerate
Charge Neutrality Relationship
From Poisson’s equation
𝛻∙ ℰ = 𝜌 𝐾 𝑆 𝜀 0 ,
𝜌= 𝑞(𝑝 −𝑛+ 𝑁 𝐷 + − 𝑁 𝐴 − )
For uniformly doped semiconductor at equilibrium, ℰ = 0 and charge neutrality requires that
𝑝 −𝑛+ 𝑁 𝐷 + − 𝑁 𝐴 − =0
𝑝 −𝑛+ 𝑁 𝐷 − 𝑁 𝐴 =0
dopant sites totally ionized

27. Relationships for ND+ and NA-
Donor and Acceptor Statistics
: ʓ(𝜇, 𝑇)= 𝑎𝑙𝑙 𝑠𝑡𝑎𝑡𝑒𝑠 𝑎𝑙𝑙 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑒 𝑁𝜇−𝐸 /𝑘𝑇
Gibbs grand sum
chemical potential
= 𝐺𝑖𝑏𝑏𝑠(𝑠𝑡𝑎𝑡𝑒𝑠) ʓ(𝜇, 𝑇)
Probability in a state
Fermi-Dirac distribution
Occupancy Energy N E
ʓ(𝜇, 𝑇)= 𝑒 𝑁𝜇−𝐸 /𝑘𝑇 =1+ 𝑒 𝜇−𝐸 /𝑘𝑇
f(𝐸)= 𝐺𝑖𝑏𝑏𝑠(𝑠𝑡𝑎𝑡𝑒𝑠) ʓ(𝜇, 𝑇) = 1∙ 𝑒 𝜇−𝐸 /𝑘𝑇 1+ 𝑒 𝜇−𝐸 /𝑘𝑇 = 1 1+ 𝑒 𝐸− 𝐸 𝐹 /𝑘𝑇
Probability of occupancy,
replacing 𝜇
with EF.

28. Electrons in a band with spin
Occupancy Energy N
↑ E
↓ E
↑↓ E 2
ʓ(𝜇, 𝑇)= 𝑒 𝑁𝜇−𝐸 /𝑘𝑇 =1+ 𝑒 𝜇−𝐸 /𝑘𝑇 + 𝑒 𝜇−𝐸 𝑘𝑇 + 𝑒 2 𝜇−𝐸 /𝑘𝑇
Probability of occupancy
= 𝐺𝑖𝑏𝑏𝑠(𝑠𝑡𝑎𝑡𝑒𝑠) ʓ(𝜇, 𝑇) = 1∙ 𝑒 𝜇−𝐸 /𝑘𝑇 +1∙ 𝑒 𝜇−𝐸 𝑘𝑇 +2∙ 𝑒 𝜇−𝐸 /𝑘𝑇 ʓ(𝜇, 𝑇) = 2 𝑒 𝜇−𝐸 /𝑘𝑇 (1+ 𝑒 𝜇−𝐸 /𝑘𝑇 ) 𝑒 𝜇−𝐸 /𝑘𝑇 2 =2× 1 1+ 𝑒 𝐸− 𝐸 𝐹 /𝑘𝑇
replacing 𝜇
with EF.
Donors
Occupancy Energy N
↑ ED 1
↓ ED 1
ʓ(𝜇, 𝑇)= 𝑒 𝑁𝜇−𝐸 /𝑘𝑇 =1+ 𝑒 𝜇−𝐸𝐷 /𝑘𝑇 + 𝑒 𝜇−𝐸𝐷 𝑘𝑇 =1+2 𝑒 𝜇−𝐸𝐷 /𝑘𝑇
Probability of occupancy = probability that the donor atoms are unionized (neutral donors)
𝑁 𝐷 0 𝑁 𝐷 = 1∙ 𝑒 𝜇−𝐸𝐷 /𝑘𝑇 +1∙ 𝑒 𝜇−𝐸𝐷 /𝑘𝑇 1+2 𝑒 𝜇−𝐸𝐷 /𝑘𝑇 = 𝑒 𝐸𝐷− 𝐸 𝐹 /𝑘𝑇
where
𝑁 𝐷 = 𝑁 𝐷 0 + 𝑁 𝐷 +
replacing 𝜇
with EF.

29. Probability of being unoccupied = probability that the donor atoms are ionized
𝑁 𝐷 + 𝑁 𝐷 = 𝑒 𝐸 𝐹 −𝐸𝐷 /𝑘𝑇 = 1 1+ 𝑔 𝐷 𝑒 𝐸𝐹− 𝐸 𝐷 /𝑘𝑇
where
𝑔 𝐷 =
degeneracy factor for donors 2
Acceptors
Occupancy Energy N
0 ↑
0 ↓ EA
ʓ(𝜇, 𝑇)= 𝑒 𝑁𝜇−𝐸 /𝑘𝑇 = 𝑒 𝜇−𝐸𝐴 /𝑘𝑇 =4+ 𝑒 𝜇−𝐸𝐴 /𝑘𝑇 hh lh
Probability of occupancy = probability of being ionized acceptors
where
𝑁 𝐴 = 𝑁 𝐴 0 + 𝑁 𝐴 −
𝑁 𝐴 − 𝑁 𝐴 = 1∙ 𝑒 𝜇−𝐸𝐴 /𝑘𝑇 4+ 𝑒 𝜇−𝐸𝐴 /𝑘𝑇 = 𝑒 𝐸𝐴− 𝐸 𝐹 /𝑘𝑇 = 1 1+ 𝑔 𝐴 𝑒 𝐸𝐴− 𝐸 𝐹 /𝑘𝑇
replacing 𝜇
with EF. 4
𝑔 𝐴 =
degeneracy factor for acceptors
Deep level trap centers
𝑁 𝐷 + → 𝑁 𝑇 + , 𝑔 𝐷 → 𝑔 𝑇 , 𝐸 𝐷 → 𝐸 𝑇
for donor-like
𝑁 𝐴 − → 𝑁 𝑇 − , 𝑔 𝐴 → 𝑔 𝑇 , 𝐸 𝐴 → 𝐸 𝑇
for acceptor-like
𝑁 𝑇 + 𝑁 𝑇 = 1 1+ 𝑔 𝑇 𝑒 𝐸 𝐹 −𝐸𝑇 /𝑘𝑇 = 1 1+ 𝑒 𝐸𝐹− 𝐸 𝑇 ′ /𝑘𝑇
where
𝐸 𝑇 ′ = 𝐸 𝑇 - kT ln 𝑔 𝑇

30. CONCENTRATION AND EF CALCULATION

31. Equilibrium Carrier Concentrations
From charge neutrality
𝑝 −𝑛+ 𝑁 𝐷 + − 𝑁 𝐴 − =0
and assuming nondegeneracy
𝑁 𝑉 𝑒 −( 𝐸 𝐹 − 𝐸 𝑉 ) −𝑁 𝐶 𝑒 −( 𝐸 𝐶 − 𝐸 𝐹 ) + 𝑁 𝐷 1+ 𝑔 𝐷 𝑒 𝐸𝐹− 𝐸 𝐷 /𝑘𝑇 − 1 1+ 𝑔 𝐴 𝑒 𝐸𝐴− 𝐸 𝐹 /𝑘𝑇 =0
Freeze-out/extrinsic T (ND >> NA or NA >> ND )
In a donor-doped semiconductor (ND >> NA),
n >> p and 𝑁 𝐷 + >> 𝑁 𝐴 − (except in the extreme T→0)
∴𝑛 ≈ 𝑁 𝐷 +
𝑁 𝐷 + = 𝑁 𝐷 1+ 𝑔 𝐷 𝑒 𝐸𝐹− 𝐸 𝐷 /𝑘𝑇 = 𝑁 𝐷 1+ 𝑔 𝐷 𝑛 𝑁 𝐶 𝑒 𝐸𝐶− 𝐸 𝐷 /𝑘𝑇 = 𝑁 𝐷 1+ 𝑛 𝑁 ξ
𝑛 2 + 𝑁 ξ 𝑛 − 𝑁 ξ 𝑁 𝐷 =0
where
𝑁 ξ ≡ 𝑁 𝐶 𝑔 𝐷 𝑒 − 𝐸𝐶− 𝐸 𝐷 /𝑘𝑇
𝑛=− 𝑁 ξ 𝑁 ξ 𝑁 ξ 𝑁 𝐷 1/2
𝑛= 𝑁 ξ 𝑁𝐷 𝑁 ξ 1/2 −1 ≈ 𝑁 𝐷 or
typically 𝑁 ξ ≫𝑁𝐷 in the extrinsic temperature region
almost fully ionized at room temperature

32. Extrinsic/Intrinsic T (relatively high T)
𝑁 𝐷 + ≈ 𝑁 𝐷 , 𝑁 𝐴 − ≈ 𝑁 𝐴
Then,
𝑝 −𝑛+ 𝑁 𝐷 − 𝑁 𝐴 =0
: charge neutrality
𝑛𝑝= 𝑛 𝑖 2
: nondegenerated semiconductor
𝑛 2 − 𝑁 𝐷 − 𝑁 𝐴 𝑛 − 𝑛 𝑖 2 =0
∴𝑛= 𝑁 𝐷 − 𝑁 𝐴 𝑁 𝐷 − 𝑁 𝐴 𝑛 𝑖 2 1/2
𝑝= 𝑛 𝑖 2 𝑛 = 𝑁 𝐴 − 𝑁 𝐷 𝑁 𝐴 − 𝑁 𝐷 𝑛 𝑖 2 1/2
For donor-doped semiconductor, extrinsic T (ND >> NA, ND >> ni),
𝑝≈ 𝑛 𝑖 2 𝑁 𝐷
𝑛 ≈ 𝑁 𝐷 ,
For acceptor-doped semiconductor, extrinsic T (NA >> ND, NA >> ni),
𝑛≈ 𝑛 𝑖 2 𝑁 𝐴
𝑝 ≈ 𝑁 𝐴 ,
For intrinsic T,
𝑛 𝑖 ≫ 𝑁 𝐷 − 𝑁 𝐴
𝑛 ≈ 𝑛 𝑖 , 𝑝 ≈ 𝑛 𝑖 ,
For compensation,
𝑁 𝐷 − 𝑁 𝐴 ≈0

33. Determination of EF Exact position of Ei
For intrinsic semiconductor, n = p, NA = ND = 0, EF = Ei
𝑁 𝐶 𝑒 −( 𝐸 𝐶 − 𝐸 𝑖 ) =𝑁 𝑉 𝑒 −( 𝐸 𝑖 − 𝐸 𝑉 )
∴ 𝐸 𝑖 = 𝐸 𝐶 + 𝐸 𝑉 𝑘𝑇 2 𝑙𝑛 𝑁 𝑉 𝑁 𝐶 = 𝐸 𝐶 + 𝐸 𝑉 𝑘𝑇𝑙𝑛 𝑚 𝑝 ∗ 𝑚 𝑛 ∗
Freeze-out/extrinsic T (ND >> NA or NA >> ND )
In a donor-doped semiconductor (ND >> NA, ND >> ni)
𝑛 ≈ 𝑁 𝐷 +
𝑛= 𝑁 𝐶 𝑒 −( 𝐸 𝐶 − 𝐸 𝐹 ) = 𝑁 ξ 𝑁 𝐷 𝑁 ξ 1/2 −1
𝐸 𝐹 = 𝐸 𝐶 +𝑘𝑇𝑙𝑛 𝑁 ξ 2 𝑁 𝐶 𝑁 𝐷 𝑁 ξ 1/2 −1
or equivalently (at low temperature extrinsic region)
𝑁 𝐶 𝑒 −( 𝐸 𝐶 − 𝐸 𝐹 ) = 𝑁 𝐷 1+ 𝑔 𝐷 𝑒 𝐸𝐹− 𝐸 𝐷 /𝑘𝑇 ≈ 𝑁 𝐷 𝑔 𝐷 𝑒 𝐸𝐹− 𝐸 𝐷 /𝑘𝑇
if EF – ED > 0 and as T goes small.
𝐸 𝐹 = 𝐸 𝐶 + 𝐸 𝐷 𝑘𝑇𝑙𝑛 𝑁 𝐷 𝑔 𝐷 𝑁 𝐶
𝐸 𝐶 − 𝐸 𝐹 = 𝐸 𝐶 + 𝐸 𝐷 𝑘𝑇𝑙𝑛 𝑔 𝐷 𝑁 𝐶 𝑁 𝐷 or

34. Extrinsic/Intrinsic T (relatively high T)
𝑛≈ 𝑁 𝐷 , 𝑝≈ 𝑁 𝐴
𝑛 =𝑛 𝑖 𝑒 ( 𝐸 𝐹 − 𝐸 𝑖 )
𝑛 =𝑁 𝐶 𝑒 −( 𝐸 𝐶 − 𝐸 𝐹 )
𝑝=𝑛 𝑖 𝑒 ( 𝐸 𝑖 − 𝐸 𝐹 )
𝑝 =𝑁 𝑉 𝑒 −( 𝐸 𝐹 − 𝐸 𝑉 )
For ND >> NA and ND >> ni
𝐸 𝐹 − 𝐸 𝑖 =𝑘𝑇𝑙𝑛 𝑁 𝐷 𝑛 𝑖 or
𝐸 𝐶 − 𝐸 𝐹 =𝑘𝑇𝑙𝑛 𝑁 𝐶 𝑁 𝐷
For NA >> ND and NA >> ni
𝐸 𝑖 − 𝐸 𝐹 =𝑘𝑇𝑙𝑛 𝑁 𝐴 𝑛 𝑖 or
𝐸 𝐹 − 𝐸 𝑉 =𝑘𝑇𝑙𝑛 𝑁 𝐶 𝑁 𝐴
What happens for partially compensated donor and acceptor with ND > NA?
Read “degenerate semiconductor consideration’