EE 211B

Introduction to Physical Electronics

Chapter 4:
The Semiconductor in Equilibrium

School of Electrical Engineering

KAIST
Introduction (1)

§ Objective:
• Learn about concentration of electrons and holes in a
semiconductor
Derive them in thermal equilibrium

How do we change them?

Relation to Fermi energy level?

Introduction (2)

§ Motivation
• To eventually derive current that flows in
semiconductor!
Current is the rate at which “charge” flows. Charge
carriers in a semiconductor are “electrons and

holes”.
!"
Current (rate at which charge flows): 𝐼 = !#

Current density (current per unit area)

Outline

1. Charge carriers in Semiconductors

2. Intrinsic Semiconductor

3. Dopant Atoms and Energy Levels

4. Extrinsic Semiconductor

5. Statistics of Donors and Acceptors

6. Charge Neutrality

7. Position of Fermi Energy Level

Distribution of Electrons and Holes (1)
# of electrons per unit energy per unit volume
Density of states in the conduction band
Fermi-Dirac probability function

(1) n ( E ) = gc ( E ) f F ( E )
(2) p ( E ) = gv ( E ) éë1 - f F ( E )ùû

Density of states in the valence band

# of holes per unit energy per unit volume

• Number of electrons per unit volume is n0 = ò n ( E ) dE (3)

• Number of holes per unit volume is p0 = ò p ( E ) dE (4)

Distribution of Electrons and Holes (2)
• .n(E) = gc (E) × fF (E)

• Density of allowed electronic • Fermi-Dirac probability function:

energy states (E increases, Probability a certain state will be
density increases) occupied

• EF: Fermi level when p=½ (statistical

number so does not have to correspond
to an allowed energy level)
Distribution of Electrons and Holes (3)

• Electron
#
𝑛! = # 𝑔$ 𝐸 𝑓% (𝐸)𝑑𝐸 (1)
"#

• Hole
#
𝑝! = # 𝑔& 𝐸 (1 − 𝑓% (𝐸))𝑑𝐸 (2)
"#
Thermal-Equilibrium Concentration (1)
• Thermal-equilibrium electron concentration (𝒏𝟎)
At T > 0
𝐸

𝑔$ (𝐸)

𝐸#

DOS 𝑓! (𝐸)
Thermal-Equilibrium Concentration (2)
• Thermal-equilibrium electron concentration (𝒏𝟎)

(1) 𝑛% = ∫ 𝑔$ 𝐸 𝑓# 𝐸 𝑑𝐸
1
𝑓% 𝐸 = ≈ 𝑒 "(("(! )/+" , (2)
1+ 𝑒 (("(! )/+" ,

4𝜋 2𝑚-∗ //0
(3)
𝑔$ = 𝐸 − 𝐸$
ℎ/
$& #$'
# %
(4)
𝑛! = 𝑁" 𝑒 (&
'/&
2𝜋𝑚#∗ 𝑘% 𝑇
𝑁" = 2 (5)
ℎ&
∗
𝑚#∗ = 𝑚)*+,# (= 6&/' 𝑚-∗& 𝑚.∗ //'
for Si) (6)
Thermal-Equilibrium Concentration (3)
• Thermal-equilibrium hole concentration (𝒑𝟎 )
At T > 0
𝐸

𝐸#

DOS 1 − 𝑓! (𝐸) (1)

𝑔) (𝐸)
Thermal-Equilibrium Concentration (4)
• Thermal-equilibrium hole concentration (𝒑𝟎 )

𝑝% = ∫ 𝑔) 𝐸 (1 − 𝑓# 𝐸 )𝑑𝐸
1
1 − 𝑓% 𝐸 = ≈ 𝑒 (("(! )/+" ,
1+ 𝑒 "(("(! )/+" ,
//0
4𝜋 2𝑚1∗
𝑔$ = 𝐸 − 𝐸$
ℎ/
$' #$*
# % &
(1)
𝑝! = 𝑁' 𝑒 (
'/&
2𝜋𝑚2∗ 𝑘% 𝑇
𝑁1 = 2
ℎ&
∗ '/& &/'
𝑚+∗ = 𝑚-./,+ = 𝑚33 '/&
+ 𝑚.3
Outline

1. Charge carriers in Semiconductors ✓

2. Intrinsic Semiconductor

3. Dopant Atoms and Energy Levels

4. Extrinsic Semiconductor

5. Statistics of Donors and Acceptors

6. Charge Neutrality

7. Position of Fermi Energy Level

Intrinsic Carrier Concentration (1)
𝐸
( "(
" #+ ,!$
𝑛! = 𝑁$ 𝑒 "

𝑔$ (𝐸)
( "(%
" !$
+ ,
𝑝! = 𝑁& 𝑒 "

electrons
For intrinsic semiconductors,
electrons and holes are
𝐸%2 created in pairs, so

𝑛! = 𝑝! = 𝑛"
holes

6%
5
𝑔& (𝐸) 𝑛4& = 𝑁" 𝑁1 𝑒 7&8 (1)

The intrinsic Fermi level 𝐸%2 is near the midgap

Intrinsic Carrier Concentration (2)
?9
>
𝑛<= = 𝑁$ 𝑁) 𝑒 @: A

• Note 1: Practically, intrinsic silicon is a very good insulator

𝑛' for silicon at 300 K ~1.5×10() 𝑐𝑚*+
The volume density of silicon atoms ~5×10,, 𝑐𝑚*+ (i.e. there are
about 1 electron per 10(, atoms)

• Note 2: From the formula, 𝑛4 exponentially depends on the band-gap

𝐸- of good insulators are 3.5~6 eV; 𝑛' is very small

• Note 3: Nc, Nv
𝑁. and 𝑁/ are called the effective density of the states in the CB and VB,
respectively. At 300 K, they are ~10(0 𝑐𝑚*+
Intrinsic Carrier Concentration (3)

Notice that 𝑁. ’s are quite different, due to the fact that 𝑚1∗ ’s are different.
In comparison, 𝑁/ ’s are not that different.
Recall that conduction bands of different semiconductors can differ a lot,
while valence bands look quite the same.

Orders are VERY different !

This is mainly due to the
difference in the band-gap
Intrinsic Fermi-Level Position
• We make use of the fact that, for intrinsic semiconductors, electrons
and holes are created in pairs

(1) 𝑛! = 𝑝! = 𝑛2

( "( ( "(%
(2) " #+ ,!$ " !$
+ ,
𝑁$ 𝑒 " = 𝑁& 𝑒 "

The intrinsic Fermi level

3 𝑚1∗ must shift away from the
(3) 𝐸%2 = 𝐸324561 + 𝑘7 𝑇 𝑙𝑛 band with the larger density
4 𝑚-∗ of states in order to maintain
equal numbers of electrons
( and holes
where 𝐸#"$%&' = (𝐸* + 𝐸+ )
)

• If 𝑚1∗ = 𝑚3∗ , 𝐸4' = 𝐸5'6-73

• If 𝑚1∗ > 𝑚3∗ , 𝐸4' is below 𝐸5'6-73 (curvature of the CB becomes greater)
• If 𝑚1∗ < 𝑚3∗ , 𝐸4' is above 𝐸5'6-73 (curvature of the VB becomes greater)
Temperature Dependency
• ni is very temperature-sensitive

• For Silicon
When T = 300 → 330 K (10%
increase), ni = 1010 → 1011 cm-3 (10x
increase!)
Equation Summary
• Thermal-equilibrium electron concentration (# per volume)
n= ∫ g (E) × f (E)dE
c F

p = ∫ g (E) × [1− f (E)]dE

v F
Note: Exp. depend
• Two useful equation sets for n, p ence on T

n = N c e(EF −EC )/kT n0 = p0 = ni Rewrite

n = ni e(EF −Ei )/kT
Nc = f(ni)
p = N v e(EV −EF )/kT EF = Ei p = ni e(Ei −EF )/kT
intrinsic case

• Equation for ni • np product

ni = N c NV e−EG /2kT np = ni2 "#e(EF −Ei )/kT e(Ei −EF )/kT $% = ni2
*NC, NV = Effective density of states * Under most cases – equilibrium and non-degenerate
(~effective mass, temperature dependent)

EE362 | Spring 2019 18

Outline

1. Charge carriers in Semiconductors ✓

2. Intrinsic Semiconductor ✓

3. Dopant Atoms and Energy Levels

4. Extrinsic Semiconductor

5. Statistics of Donors and Acceptors

6. Charge Neutrality

7. Position of Fermi Energy Level

 Dopants / Impurity atoms
• Adding small, controlled amounts of dopants (impurity atoms)

• Increase number of carriers (electrons or holes)

• Alter the conductivity of semiconductor

• Elements in Group III and V can be

used as the dopants to Si.
Donors (1)
Donor (atom)

Intrinsic Dopant added Electron released

from donor

• In intrinsic Si, all 4 valence electrons contribute to covalent bonding.

• In Si doped with P as a substitutional impurity
4 valence electrons of P contribute to covalent bonding
1 electrons loosely bound to P (donor electron)
• Donor electron can easily break the bond with P and freely moves (free
electron) à jump to CB of Si
Donors (2)

• Energy to elevate the donor electron into CB is less than that for the electron
involved in covalent bonding
• So, the energy state of the donor electron (𝐸6 ) should be located near 𝐸. .
• With small energy added (e.g., thermal energy), donor electron is elevated to
CB, leaving behind positively charge P ion (impurity ion)
• Now, number of electrons is greater than number of holes, so majority carrier
is electron à n-type semiconductor
Acceptor (1)
Acceptor (atom)

Hole created due to

Dopant added acceptor

• In intrinsic Si, all 4 valence electrons contribute to covalent bonding.

• In Si doped with B as a substitutional impurity
3 valence electrons of B contribute to covalent bonding
1 covalent bonding is empty
Acceptor (2)

• With small energy added (e.g., thermal energy), an electron in a Si atom can
occupy the empty covalent bonding, leaving behind an empty position in the
Si atom à Hole is created (acceptor hole)
• The energy state of the acceptor hole (𝐸𝑎) should be located near 𝐸𝑣.
• Now, number of holes is greater than number of electrons, so majority carrier
is hole à p-type semiconductor
Ionization Energy (1)
• Energy required for donor electrons to be elevated to CB

𝐸$ − 𝐸4 = Ionization Energy

• Energy required for electrons in VB to be elevated to acceptor level =

Energy required for acceptor holes to move to VB

𝐸6 − 𝐸8 = Ionization Energy
Ionization Energy (2)
- Review: The Bohr Model -
• Quantization of Angular Momentum
Using the quantization of the angular momentum,

𝐿 = 𝑚𝑣𝑟# = 𝑛ℏ

N. Bohr could calculate the Rydberg constant,

which agreed excellently with the experimental
value:
§ electrostatic force = centripetal force
;8 @1 8
= (classical)
<=>9 ? 8 ?

FA K) A
è = (quantum)
GHIB JCA JC
Ionization Energy (3)
• We try to estimate the ionization energy using the Bohr’s model.

• From the Bohr’s model, we get the Bohr’s radius, which coincides
with the most probable distance of an electron from the nucleus of
an atom.

• Roughly speaking, we ‘imagine’ that the outermost valence electron

(which is loosely bound to the nucleus) of the dopant atom orbits
around the nucleus at the ‘most probable’ distance from the nucleus,
which can be determined by using the Bohr’s model.

• Let’s determine the distance in the followings.

Ionization Energy (4)
• Using the Bohr’s model,

FA KB ) A
(1) = (Hydrogen Atom)
GHIB JCA J
∗ A
FA K DE )
è = (Inside the Silicon) (2)
GHIDE JCA JC
∗
where 𝑚," is the conductivity effective mass of Si
(here the material is assumed to be Si)

• Why conductivity effective mass?

In this case, we regard the electron as a ‘classical’ particle orbiting
around the nucleus.
𝐹 = 𝑞𝐸 = 𝑚∗ 𝑎, where 𝑚∗ = 𝑚"∗ (See Appendix F.3)
 Ionization Energy (5)
.!"
(1) • Put n = 1 à 𝑟( = ∗ 𝑎! (this is the most probable distance we were looking for)
#!"

• The total energy, using 𝑟( , is given by, (refer to the textbook for details)
∗
𝑚,"
𝐸 = 𝑇 + 𝑉 ⇒ 𝐼𝑜𝑛𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑒𝑛𝑒𝑟𝑔𝑦 = 𝐸( = ) ×𝐸( 𝐻𝑦𝑑𝑟𝑜𝑔𝑒𝑛 (2)
𝜖,"
• Note:
1. This approach just gives the order of magnitude for ionization energy.
2. The actual ionization energy depends on specific dopants/acceptors used.
(See the tables below)
Outline

1. Charge carriers in Semiconductors ✓

2. Intrinsic Semiconductor ✓

3. Dopant Atoms and Energy Levels ✓

4. Extrinsic Semiconductor

5. Statistics of Donors and Acceptors

6. Charge Neutrality

7. Position of Fermi Energy Level

Effect of doping (1)
• Fermi energy is shifted

(# "(!
" + (1)
𝑛! = 𝑁$ 𝑒 ",
& # '& !$ (& !$ '& !
" (2)
= 𝑁$ 𝑒 )" *
(! "(!$
(3)
= 𝑛2 𝑒 +" ,

(𝐸% > 𝐸%2 ) (𝑛! > 𝑛2 )

(! "(%
" +
𝑝! = 𝑁& 𝑒 ",

( "(
" !+ ,!$ (4)
= 𝑛2 𝑒 "

(* note: 𝑝! < 𝑛" .

n-type This will be explained later)

Effect of doping (2)
• Fermi energy is shifted

(# "(!
" +
𝑛! = 𝑁$ 𝑒 ",

(! "(!$
= 𝑛2 𝑒 +" ,

(* note: 𝑛! is smaller than 𝑛" )

(! "(%
" +
𝑝! = 𝑁& 𝑒 ",

( "(
" !+ ,!$
= 𝑛2 𝑒 "

(𝑝! > 𝑛2 )
p-type
Effect of doping (3)

n-type p-type
 𝑛0 𝑝0 product
• One of the fundamental principles of semiconductors in equilibrium!

?G >?GE
𝑛% = 𝑛< 𝑒 @: A
(1) 𝑛% 𝑝% = 𝑛<= (2)
? >?
> G GE
𝑝% = 𝑛< 𝑒 @: A

• Notice: The 𝑛F 𝑝F product is valid, only when the Boltzmann

approximation is valid. That is

1
≈ 𝑒 "(("(! )/+" ,
1 + 𝑒 (("(! )/+" ,
Fermi Dirac Integral (1)
• If we use the Fermi-Dirac distribution (not the Boltzmann approximation),
0
𝑛! = < 𝑔* 𝐸 𝑓1 𝐸 𝑑𝐸 4𝜋 2𝑚4∗ 6/7
(1) 𝑔3 = 𝐸 − 𝐸3
/$ ℎ6
1
𝑓+ 𝐸 =
1 + 𝑒 (-.-!)/1"2
#
4𝜋 2𝑚4∗ 6/7
𝐸 − 𝐸#
(2) 𝑛! = # 𝑑𝐸 𝜂 ≡ 𝐸 − 𝐸3 (3)
ℎ6 (# 1 + 𝑒
(-.-! )/1" 2
𝐸+ − 𝐸3
𝜂+ ≡
4/) 0 𝑘8 𝑇
2𝜋𝑚2∗ 𝑘3 𝑇 𝜂(/) 𝑑𝜂
(4) = 4𝜋 <
ℎ) ! 1 + exp 𝜂 − 𝜂1
≡ 𝐹9/0 (𝜂% )
• Then,
4
∗
2𝜋𝑚2 𝑘3 𝑇 ) 𝐸1 − 𝐸*
𝑛! = 4𝜋 𝐹(/)
ℎ) 𝑘3 𝑇
4
∗
2𝜋𝑚' 𝑘3 𝑇 ) 𝐸+ − 𝐸1
𝑝! = 4𝜋 𝐹(/)
ℎ) 𝑘3 𝑇
Fermi Dirac Integral (2)

Comparison of 𝐹//& and

the Boltzmann approximation
Nondegenerate Semiconductors
• Small amount of dopant atoms (impurity atoms)
- No interaction between dopant atoms

- Discrete, non-interacting energy states

- Fermi energy within the bandgap

𝐸#
𝐸#

𝐸! > 𝐸!4 𝐸! < 𝐸!4

Degenerate Semiconductors
• Large amount of dopant atoms (~effective density of states)
- Dopant atoms interact with each other

- Band of dopant states widens and overlap the conduction band (n-type)

- Fermi energy lies within the conduction band (n-type)

Outline

1. Charge carriers in Semiconductors ✓

2. Intrinsic Semiconductor ✓

3. Dopant Atoms and Energy Levels ✓

4. Extrinsic Semiconductor ✓

5. Statistics of Donors and Acceptors

6. Charge Neutrality

7. Position of Fermi Energy Level

Statistics of Donors (1)
• Objective: to calculate the distribution function of donor electrons in the
donor energy states
• Suppose there are 𝑁# dopant atoms. The average number of electrons
occupying the donor state is given by

1
(1) 𝑛$ = 𝑁$
1 𝐸 − 𝐸1 𝐸$
1 + exp $
2 𝑘3 𝑇

We are considering
the donor state at 𝐸 = 𝐸9

• Why 1/2 factor ???

Because there are two states (spin-up and spin-down states) for a single donor state
If one electron occupies spin-up state, the second one cannot occupy spin-down state
and vice versa, because the dopant atom allows only one electron in the dopant energy
state.
 Statistics of Donors (2)
• Example: 𝑁) = 3
down state
up state 3(atoms) x2(states) = 6 states

𝑛4 = 1
….

𝑛4 = 2 These configurations
…. are not allowed!!

𝑛4 = 3
We look for the most probable case (as before,
see Chap. 3, statistical mechanics section), then
…. we can get the distribution function which
contains the degeneracy factor 2.
Statistics of Donors (3)

(1)

(= Density of electrons
remaining in the donor level)
 =
Statistics of Donors (4) neutral
+

dopant atom
• Example:
𝑁# = 𝑛# + 𝑁#$

5 =3 + 2

Before ionization After ionization

Conduction
Band

𝐸$

+ + + + +
𝐸4
 Statistics of Acceptors

(1)

(= Density of holes remaining

in the acceptor level) g = degeneracy factor
e.g. g = 4 for Si
because there are HH (heavy hole) and LH (light hole)
Complete Ionization (1)
Doping Doping
Intrinsic
- before ionization - after ionization

𝑛2 𝑛2 𝑛!

𝐸4 𝑁4 𝐸4 𝑛4
𝐸%2 𝐸%

𝑛2 𝑝!
𝑛2
(𝑝! < 𝑛2 )

dopant atoms are

ionized and donor
electrons jump to CB
Complete Ionization (2)
• Let’s assume 𝐸) − 𝐸! ≫ 𝑘% 𝑇

%(': %'! )
(1) 𝑛# = 2𝑁# exp )" * 𝑛# 1
=
%('# %'!) 𝑛+ + 𝑛# 𝑁 − 𝐸" − 𝐸# (3)
(2) 𝑛+ = 𝑁" exp 1 + " exp
)" * 2𝑁# 𝑘, 𝑇

• Example 4.7: Consider phosphorus doping in silicon, for 𝑇 = 300 𝐾 at

a concentration of 𝑁) = 10/Y 𝑐𝑚5' .

𝑛4 𝑁8 : Table 4.1
(4) = 0.0041
𝑛! + 𝑛4 𝐸8 − 𝐸$ = Ionization energy : Table 4.3

Almost no electrons remain in the donor state

= Almost all dopants are ionized

Q: The formula tells us that, if 𝑁$ ↑, ionization rate ↓. Why???

Complete Ionization (3)
• Acceptor

𝑛- 1
=
𝑛+ + 𝑛- 1 + 𝑁. exp − 𝐸- − 𝐸.
4𝑁- 𝑘, 𝑇

• Question: Consider boron doping in silicon, for 𝑇 = 300 𝐾 at a

concentration of 𝑁Z = 10/Y 𝑐𝑚5' . 𝑛Z /(𝑛F + 𝑛Z ) = ?

The same steps as in the case of donors. You will get the result that
borons are almost completely ionized.
Complete Ionization (4)

n-type p-type

Complete ionization:
𝑛$ = 0 (that is, no electrons remaining in the donor states)
9
𝑁$ = 𝑁$ (that is, the number of positive ions (or ionized impurities)
is the same as the number of donor atoms)

Complete ionization:
𝑛& = 0 (that is, no holes remaining in the acceptor states)
:
𝑁& = 𝑁& (that is, the number of negative ions (or ionized impurities)
is the same as the number of acceptor atoms)
Complete Freeze-Out (1)
• Donor
1
𝑛# = 𝑁#
1 𝐸 − 𝐸/
1 + 2 exp #
𝑘, 𝑇

• At 𝑇 = 0
Before using the formula, the physics tells you that there is no thermal
energy to raise the donor electron to the conduction band which means
that 𝑛# = 𝑁# .
So, in the above formula, the relationship 𝐸/ > 𝐸# must hold.

• This case is called ‘complete freeze-out’.

Complete Freeze-Out (2)

n-type p-type

Complete freeze-out:
𝑛$ = 𝑁$ (that is, all electrons remaining in the donor states)
9
𝑁$ = 0 (that is, no atoms are ionized)

Complete freeze-out:
𝑛& = 𝑁& (that is, all holes remaining in the acceptor states)
𝑁&9 = 0 (that is, no atoms are ionized)
Temperature Dependence

Zero or Very Low Intermediate High

Temperature Temperature Temperature

𝐸" 𝐸" 𝐸"

+ + 𝐸! + + + + +
𝐸! 𝐸!

𝑛! = 0
𝑛! = 𝑁!
𝑁!# = 𝑁!
𝑁!# = 0
Partially Completely
Freeze-out
Ionized Ionized
Outline

1. Charge carriers in Semiconductors ✓

2. Intrinsic Semiconductor ✓

3. Dopant Atoms and Energy Levels ✓

4. Extrinsic Semiconductor ✓

5. Statistics of Donors and Acceptors ✓

6. Charge Neutrality

7. Position of Fermi Energy Level

Charge Neutrality (1)
• In thermal equilibrium, the semiconductor crystal is electrically neutral.
Why?
Intrinsic semiconductor is neutral, of course (e-h pair creation).
Extrinsic semiconductor is also neutral because we put neutral dopants.
Ionization creates the same number of positive ions and electrons (for n-
type doping)

• Compensated semiconductor: both donor and acceptor impurity atoms

are present in the same region. Why do we put both types of impurities?
Compensation takes place during the device fabrication
n-type compensated: 𝑁$ > 𝑁&
p-type compensated: 𝑁& > 𝑁$

• Completely compensated semiconductor (same characteristics as

intrinsic semiconductor): 𝑁6 = 𝑁4
Charge Neutrality (2)

𝑛! + 𝑁6" = 𝑝! + 𝑁4: (1)

𝑛! = total electron concentration

in the CB

𝑝! = total hole concentration

in the VB
Charge Neutrality (3)

Compensation
Donors and acceptors
compensate each other

Note: It is energetically much

more favorable for donor
electrons fill the acceptor holes
than they jump to the CB.
Charge Neutrality (4)

Some donor electrons also annihilate the holes in the VB

This is why 𝑝! is smaller

than 𝑛2 (intrinsic hole
concentration)

This also affects the

electron distribution
in the CB (discussed shortly)
n-type Compensated Semiconductor (1)
• Let’s assume complete ionization
𝑁-% = 𝑁-
𝑁#$ = 𝑁#

(1) 𝑛+ + 𝑁-% = 𝑝+ + 𝑁#$ 𝑛+ + 𝑁- = 𝑝+ + 𝑁# (2)

• We use the relationship 𝑛$ 𝑝$ = 𝑛%& 𝑝+ = 𝑛01 /𝑛+ (3)

Majority electron concentration

0
𝑁4 − 𝑁6 𝑁4 − 𝑁6
𝑛! = + + 𝑛20 (4)
2 2

Minority hole concentration (after getting 𝑛! )

𝑝! = 𝑛20 /𝑛! (5)
n-type Compensated Semiconductor (2)
• Note (1): in the previous slide, we could get 𝑝$ by
𝑛+ + 𝑁- = 𝑝+ + 𝑁#

𝑛+ = 𝑛01 /𝑝+

• We can also calculate hole directly ...

1
𝑁# − 𝑁- 𝑁# − 𝑁-
(1) 𝑝+ = − + + 𝑛01
2 2

• Theoretically, this should work, but in practice, never do this, because

you usually end up subtracting two big numbers to get a very small
number!

For minority hole concentration (n-type semiconductor)

get 𝑛$ first and then use 𝑝$ = 𝑛%& /𝑛$
 n-type Compensated Semiconductor (3)
• Note (2):
𝑛+ + 𝑁- = 𝑝+ + 𝑁# Doping Doping
- before ionization - after ionization
= 𝑛01 /𝑛+ + 𝑁#

• Let’s assume 𝑁- = 0 for simplicity

(1) 𝑛+ = 𝑛01 /𝑛+ + 𝑁# (CORRECT)

𝑛2 𝑛!
• Looking at the figure to the right,
you may simply think that 𝑁4 𝐸4 𝑛4
𝐸%
(2) 𝑛+ = 𝑛0 + 𝑁# (WRONG)

𝑛2 𝑝!
• But this is NOT true, due to the
fact that the annihilation of holes (𝑝! < 𝑛2 )
Let’s assume
by the donor electrons and the
complete ionization
redistribution of electrons as a
consequence.
n-type Compensated Semiconductor (4)
• Example 4.9: Determine the thermal-equilibrium electron and hole
concentrations in silicon at T = 300 K for given doping concentrations
(a) Let 𝑁! = 10'( 𝑐𝑚)* and 𝑁+ = 0
(b) Let 𝑁! = 5×10', 𝑐𝑚)* and 𝑁+ = 2×10', 𝑐𝑚)*

à Lesson: If 𝑁4 − 𝑁6 ≫ 𝑛2 , 𝑛! ≈ (𝑁4 − 𝑁6 )

• Example 4.10: Determine the thermal-equilibrium electron and hole

concentrations in germanium at T = 300 K for given doping
concentration. Let 𝑁! = 2×10'- 𝑐𝑚)* and 𝑁+ = 0

à Lesson: If the donor impurity concentration is not too

different from the intrinsic concentration, 𝑛! is also influenced
by the intrinsic concentration.
Temperature Dependence (1)

𝑁$ ≫ 𝑛" 𝑁$ ≪ 𝑛"
𝑛! ≈ 𝑁$ 𝑛! ≈ 𝑛"

𝑁! = 5×10'-

𝑛" is a strong function

of temperature

Partial ionization
due to low thermal
energy
Temperature Dependence (2)
p-type Compensated Semiconductor
• Let’s assume complete ionization
𝑁-% = 𝑁-
𝑁#$ = 𝑁#

𝑛+ + 𝑁-% = 𝑝+ + 𝑁#$ 𝑛+ + 𝑁- = 𝑝+ + 𝑁#

• We use the relationship 𝑛$ 𝑝$ = 𝑛%& 𝑛+ = 𝑛01 /𝑝+

Majority hole concentration

1
𝑁- − 𝑁# 𝑁- − 𝑁#
(1) 𝑝+ = + + 𝑛01
2 2

Minority electron concentration (after getting 𝑛! )

𝑛+ = 𝑛01 /𝑝+
Outline

1. Charge carriers in Semiconductors ✓

2. Intrinsic Semiconductor ✓

3. Dopant Atoms and Energy Levels ✓

4. Extrinsic Semiconductor ✓

5. Statistics of Donors and Acceptors ✓

6. Charge Neutrality ✓

7. Position of Fermi Energy Level

Position of Fermi Energy Level (1)
% &%
: $ '
• Assuming the Boltzmann approximation, from the (1) 𝑛! = 𝑁8 𝑒 () *

equilibrium distribution of electrons and holes /' :/'"

(2) 𝑛! = 𝑛" 𝑒 ;) <
• n-type semiconductor p-type semiconductor
æN ö æ Nv ö
(1-a) Ec - E F = kT × ln ç c ÷ (1-b) E F - Ev = kT × ln ç ÷
è n0 ø p
è 0ø
æ n0 ö æ p0 ö
(2-a) EF - EFi = kT × ln ç ÷ (2-b) E Fi - E F = kT × ln ç ÷
è ni ø è ni ø
Variation of EF with Doping Concentration (1)
• n-type
1
𝑛+ 𝑁# − 𝑁- 𝑁# − 𝑁-
𝐸/ = 𝐸/0 + 𝑘, 𝑇𝑙𝑛 𝑛+ = + + 𝑛01
𝑛0 2 2

• p-type
𝑝+ 1
𝑁- − 𝑁# 𝑁- − 𝑁#
𝐸/ = 𝐸/0 −𝑘, 𝑇𝑙𝑛 𝑝+ = + + 𝑛01
𝑛0 2 2

• EF vs. doping

n-doping ↑, 𝐸. → 𝐸"

p-doping ↑, 𝐸. → 𝐸/
Variation of EF with Doping Concentration (2)
• Example: Determine the Fermi energy level and the maximum doping
concentration at which the Boltzmann approximation is still valid.

• Consider p-type silicon, at T = 300K, doped with boron. We may assume that
the limit of the Boltzmann approximation occurs when 𝐸/ − 𝐸. = 3𝑘, 𝑇.
Empirically, the FD distribution can be approximated to MB distribution if
𝐸 − 𝐸% > 3𝑘7 𝑇

1 𝐸 − 𝐸/
≈ exp −
𝐸 − 𝐸/ 𝑘, 𝑇
1 + exp
𝑘, 𝑇

This example shows that if the doping density is greater than certain
value (3×1023 in this example), MB approximation should not be used.
Variation of EF with Doping Concentration (3)
• What happens if we can not use the Maxwell-Boltzmann (MB) approx. ?

MB / :/
: $ '
𝑛! = 𝑁* 𝑒 ;) <
𝑛! = ∫ 𝑔$ 𝐸 𝑓% 𝐸 𝑑𝐸 𝐸1 − 𝐸*
𝑛! = 𝑁* ×2𝜋𝐹(/)
FD 𝑘3 𝑇

low or intermediate doping

MB 𝑁9 𝑁9 7
𝑛< = + + 𝑛;7
𝑝! = 𝑛") /𝑛! 2 2
𝑛! = 𝑝! + 𝑁4 𝑝! = 𝑛") /𝑛!
𝑛! = 𝑁$
let’s assume FD
𝑁? = 0 or
𝑝! will be practically zero
high doping
𝑁9 → (𝑁9 − 𝑁? ) (simply because doping density is very high)

𝑁=
MB 𝐸3 − 𝐸+ = 𝑘8 𝑇𝑙𝑛 The result should give 𝐸3 − 𝐸+ >
𝑁9
Fermi energy 3𝑘 𝑇
𝐸+ − 𝐸3 8
FD 𝑁9 = 𝑁3 ×2𝜋𝐹>/7 Numerical solution
𝑘8 𝑇

• Note: formulas related with 𝑛; , Eq. (4.20) and (4.23) will be always valid.
Variation of EF with Temperature
• For n-type semiconductor

æ Nc ö æ n0 ö
Ec - EF = kT × ln ç ÷ EF - EFi = kT × ln ç ÷
N
è dø è ni ø
(Assuming Nd>>ni) Strongly dependent on temperature

T ↑, 𝑛! ↑, 𝐸" → 𝐸#!$%&'

Extrinsic à Intrinsic

T ≈ 0; freeze-out
𝐸" > 𝐸$ (n-type)
𝐸" < 𝐸& (p-type)
 Fermi Energy at Equilibrium (1)
material A material B
①

ß far apart à

② material A and B in contact:

- before reaching new equilibrium
③ material A and B in contact = single entity
= only one Fermi energy
- after reaching new equilibrium

• Analogy: electron = water; Fermi level = water level New Fermi Level
• In contact, electrons should be exchanged to reach a new equilibrium (otherwise, Fermi levels
(EFA and EFB in this example) do not change, even if two materials are in contact.)
Fermi Energy at Equilibrium (2)

• In thermal equilibrium, the Fermi energy level is a constant

throughout a system

Very important!

Starting point for the further analysis of semiconductor diodes!