Electron and Hole Concentrations

The Fermi distribution function can be used to calculate the concentrations of electrons and holes in
a semiconductor, if the densities of available states in the valence and conduction bands are known.

n and p can be given in terms of D(E) and f(E)

Number of electrons top of conduction band From Boltzmann approximation
per cm3 in the entire n f ( E ) Dc ( E )dE
conduction band Ec

8mn 2mn 
  E  E F  kT We can choose the integration
n
h3 Ec
E  Ec e dE limit from 0 to ꝏ.

We can write  8mn 2mn  Ec  EF  kT    E  Ec  kT

(E – EF) = (EC – EF + E – EC) 
h 3
e 0
E  E c e d ( E  Ec )

Introducing a new variable 𝑥 = (𝐸 − 𝐸 )/𝑘𝑇 The above integral becomes Gamma function

𝑥𝑒 𝑑𝑥 = 𝜋/2
 Electron and Hole Concentrations
32 Nc is called the effective
n  Nce  ( Ec  E F ) / kT  2mn kT  density of states (of the
where N c  2  2 
 h  conduction band).

A larger n implies smaller (EC – EF). What happens when (EC – EF) is negative?

Similarly, we can have the number of holes per cm3 in the entire valence band

32 Nv is called the effective

 ( E F  Ev ) / kT  2m p kT 
p  Nve where N v  2  2  density of states of the
 h  valence band.

For Si, NC = 2.8 X 1019 cm-3 and NV = 1.04 X 1019 cm-3.

The closer EF moves up to EC , the larger n is; the closer EF moves down to EV, the
larger p is. This shows how strong n or p type the semiconductor is!
 What do we mean by these?
n  N c e  ( Ec  EF ) / kT p  N v e  ( E F  Ev ) / kT

• If we know the position of the Fermi level, we can find out electron and hole concentration.

• Conversely, if we know the electron/hole concentration, we can estimate the position of the
Fermi level with respect to the band edges (Ec or Ev).

• (Ec – EF ) negative means that Fermi level will lie above Ec. This implies a degenerately n-doped
semiconductor: Instead of the impurity level, there will be impurity band.

• This heavily doped semiconductor, will behave like metal, and even at 0 K, electrons will be
available in the conduction band.

• Likewise (EF - Ev) negative means that Fermi level will lie below Ev. This implies a degenerately p-
doped semiconductor: Instead of the impurity level, there will be impurity band, and even at 0 K,
holes will be available in the valence band.
 Electron and Hole Pictorially
Conduction band
Concentrations
• DC (E) is proportional to E1/2, so 𝑫𝑪 𝑬 𝒇 𝑬 = 𝒏(𝑬) 𝒇 𝑬
the density of states in the DC(E)
DC(E)
conduction band increases with
n
electron energy. On the other
hand, the Fermi function becomes
extremely small for large energies. 𝒇 𝑬
• The result is that the product Undoped semiconductor
f(E) DC (E) decreases rapidly above
EC , and very few electrons occupy 𝑫𝑽 𝑬 [𝟏 − 𝒇 𝑬 ] = 𝒑(𝑬)
energy states far above the [𝟏 − 𝒇 𝑬 ]
conduction band edge. DV(E)
DV(E)
• Similarly, the probability of
finding an empty state (hole) p
in the valence band [1 - f(E)]
decreases rapidly below EV ,
and most holes occupy states Valence band
near the top of valence band. 𝒇 𝑬 = 𝟎 𝒇 𝑬 =1
 n-type p-type
Electron and Hole
Concentrations:
After doping
DC(E) DC(E)

n
n
EFi – Fermi level of intrinsic
or undoped semiconductor 𝒇 𝑬

𝒇 𝑬
EF – Fermi level of either
n-type or p-type doped
semiconductor
DV(E) DV(E)
p
p

𝒇 𝑬 =𝟎 𝒇 𝑬 =𝟏 𝒇 𝑬 =𝟎 𝒇 𝑬 =𝟏
What decides the Fermi level position?
 The Fermi Level and Carrier Concentrations
Where will be EF for (a) n = 1017 cm-3? And for (b) p = 1014 cm-3?
For Si (at room temperature), NC = 2.8 x 1019 cm-3 and NV = 1.04 x 1019 cm-3

(a) n  N c e  ( Ec  EF ) / kT At room temperature, kT = 26 meV

Ec  EF  kT ln N c n   0.026 ln 2.8 1019 / 1017   0.146 eV

(b) For p = 1014cm-3, p  N v e  ( EF  Ev ) / kT

EF  Ev  kT ln N v p   0.026 ln 1.04 1019 / 1014   0.310 eV
0.146 eV
Ec Ec
EF

EF
0.310 eV
Ev Ev
(a) (b)
What else decides the Fermi level position?
 Electron and Hole Concentrations (T = 0K to T = 300 K)

(a) (b) (a) (b)

Energy band diagram at T = 0K for (a) n-type and The energy-band diagram showing (a) the discrete donor
(b) p-type semiconductors energy state and (b) the effect of a donor state being ionized.

(a) (b)
(a) (b)
Energy band diagram showing complete ionization of The energy-band diagram showing (a) the discrete acceptor
(a) donor states and (b) acceptor states. What about EF? energy state and (b) the effect of an acceptor state being ionized.
The Fermi Level: Effects of Carrier Concentrations & Temperature
Low Moderate Heavy doping

n  N c e  ( Ec  EF ) / kT
E F  Ec  kT ln  N c n 
EC  E F  kT ln  N c n 

As n increases, NC /n decreases  ln(NC /n) decreases  EC - EF is smaller

 EF increases (as EC is fixed for a semiconductor). Similarly for a p-type
semiconductor.

For higher temperatures, thermally generated electron-hole pair will move the Fermi
level slightly down for n-type and will move slightly up for p-type semiconductor.
 The np Product and the Intrinsic Carrier Concentration
 ( E  E ) / kT
Multiply n  N c e c F and p  N v e  ( EF  Ev ) / kT
 E g / kT
np  N c N v e  ( Ec  Ev ) / kT  N c N v e
2
np  ni What does it mean?

Doping a semiconductor with equal amount of n and p type dopants is equivalent to intrinsic semiconductor.

 E g / 2 kT Intrinsic concentration of a semiconductor is

ni  N c N v e strongly dependent on the band gap and the
temperature.
In an intrinsic (undoped) semiconductor, n = p = ni .
ni is the intrinsic carrier concentration = 1.5 x 1010 cm-3 for Si (300 K).
• ni for GaAs = 2 × 106/cm3(300 K)! Why? Because the band gap is 1.46 eV (larger than Si (1.12 eV).
• ni for Ge = 2 × 1013/cm3(300 K)! Why? Because the band gap is 0.67 eV (smaller than Si (1.12 eV).
 The Intrinsic Fermi Level
• In an intrinsic semiconductor, n = p. Therefore Ec – EF ≈ EF – Ev and the Fermi level is nearly at the
middle of the band gap, i.e., EF ≈ Ec – Eg/2. We know that this level is called the intrinsic Fermi level,
Ei or EFi.
 E g / 2 kT
ni  N c N v e

n  N c e  ( Ec  EF ) / kT
• For intrinsic semiconductor  ( Ec  E Fi ) / kT
ni  N c e

• EFi would be at the mid-gap, Ec – Eg/2, if Nc = Nv. For silicon, EFi is very close to the mid-gap.
 Additional learning on carrier concentration
N 
 ( Ec  E F ) / kT Ec  E F  kT ln C 
n  Nce  n 
Given the donor concentration, we can find out how far away the EC
Fermi level is from Ec EF
EFi
The above principle can be extended for intrinsic semiconductor
EV
 ( Ec  E Fi ) / kT
ni  N c e
n  ni e ( EF  EFi ) / kT
Given the donor concentration,
we can find out how far away the
EC
Fermi level is from the intrinsic Ed
Fermi level (assuming Nd = n) EF
EFi
Given the acceptor concentration,
we can also find out how far away EV
the Fermi level is from the intrinsic
 ( E F  E Fi ) / kT
Fermi level (assuming Na = p) p  ni e
 Additional learning on carrier concentration
N 
 ( Ec  E F ) / kT Ec  E F  kT ln C 
n  Nce  n 
Given the donor concentration, we can find out how far away the EC
Fermi level is from Ec EF
EFi
The above principle can be extended for intrinsic semiconductor
EV
 ( Ec  E Fi ) / kT
ni  N c e
n  ni e ( EF  EFi ) / kT
Given the donor concentration,
we can find out how far away the
EC
Fermi level is from the intrinsic Ed
Fermi level (assuming Nd = n) EF
EFi
Given the acceptor concentration,
we can also find out how far away EV
the Fermi level is from the intrinsic
 ( E F  E Fi ) / kT
Fermi level (assuming Na = p) p  ni e
More on donors, acceptors and intrinsic carrier concentration
Donor/Acceptor and intrinsic carrier concentration
N-type semiconductor: ND is the concentration
of the donor atoms:
• If ND >> ni (intrinsic carrier concentration)
Thermal agitation then the concentration of electrons in n-type
semiconductor:

of
of
electrons

P-type semiconductor: NA is the concentration

of the acceptor atoms:
• If NA >> ni then the concentration of holes in
p-type semiconductor:

How to find the concentration

Thermal agitation of both the electrons and the of
of
holes in a n-type or a p-type
semiconductor?
Donor/Acceptor and intrinsic carrier concentration

If NA >> ni or ND >> ni
is not true then what
If ND >> ni  nn ≈ ND happens?
ND

We’ll come back to

this after an example.

If NA >> ni  pp ≈ NA
NA
 Example: Carrier Concentrations
• What is the hole concentration in an N-type semiconductor with 1015 cm-3
of donors?
2
ni 10 20 cm -6
ND = nn = 1015 cm-3. pn   15 3  105 cm -3
nn 10 cm
• After increasing the temperature by 60 C, n remains the same at 1015
 E g / kT
cm-3 while p increases by about a factor of 2300 because ni 2  e
• As ni increases, p increases but n does not increase in an n-type
semiconductor because 1015 + 103 = 1015.
• What happens to the Fermi level after raising the temperature by 60 C?
• The Fermi level will move down to accommodate the fact that there are
certain population of holes in the valence band.
• What is np if pp = 1017cm-3 in a P-type silicon wafer?
2
ni 10 20 cm -3
np   17 3  103 cm -3
p p 10 cm
What happens if we have a host semiconductor
doped with both n and p type material?
 General Theory of n and p
A semiconductor at thermal equilibrium,
n
doped with both n and p type dopants.

Charge neutrality n  Na  p  Nd
requires that: and 2
np  ni
The above two equations
can be solved for n & p.
1/ 2
N d  N a  N d  N a  
2
2
n     ni 
2  2  
1/ 2
p
N a  N d  N a  N d  
2
2
p     ni  Decided by
2  2   the dominant
doping type &
Let us talk about some special conditions… Where will be the Fermi level?
temperature!
 General Theory of n and p
Degenerately doped
I. N d  N a  ni (i.e., N-type) n  Nd  Na Dopant Elements
semiconductors
2
p  ni n n-type P, As, Sb
p-type B, Ga, In, Al
2
If N d  N a , n  Nd and p  ni N d

Type of doping Ratio

II. N a  N d  ni (i.e., P-type) p  Na  Nd
2 Heavily doped 1:103
n  ni p Moderately doped 1:106
Low doped 1:1010
2
If N a  N d , p  Na and n  ni N a

• The impurity level broadens into an impurity band that merges with the
conduction band in heavily doped semiconductor (i.e., when donors or
acceptors are close to one another).
• This happens for the same reason the energy levels broaden into Heavily doped n or p
bands when atoms are brought close to one another to form a crystal. type semiconductor
 Example: Dopant Compensation
What are n and p in Si with (a) Nd = 61016 cm-3 and Na = 21016
cm-3 and (b) additional 61016 cm-3 of Na?
Since Nd – Na >> ni
(a) n  N d  N a  4 1016 cm 3
2
p  ni / n  10 20 / 4  1016  2.5 103 cm 3

2 x 1016 donors have compensated for 2 x 1016 acceptors.

Electrons are majority carriers, and holes are minority carriers.

(b) Na = 21016 + 61016 = 81016 cm-3 > Nd

Since Na – Nd >> ni
p  N a  N d  8  1016  6  1016  2  1016 cm 3
2
n  ni / p  10 20 / 2 1016  5 103 cm 3
6 x 1016 donors have compensated for 6 x 1016 acceptors.
Holes are majority carriers, and electrons are minority carriers.
What is the effect of temperature on the carrier concentration?
Temperature Dependence of Carrier
Concentrations
The variation of carrier concentration with temperature is given by

 E g / 2 kT
ni (T )  N c N v e
/
∗ ∗ / /

• ni has a strong temperature dependence. The exponential

temperature dependence dominates ni(T). Variations due to the
T3/2 dependence of the density-of-states function is neglected.

• Si doped with Nd = 1015 cm -3, n >> ni at T = 400 K. So Si would be

good n-type.

• Ge doped with 1015 donors per cm3 is not good, if it has to operate Semi log graph
at 400 K because it has ni > n. So Ge would require more donors for
useful n-type functioning at 400 K. Let us understand this more…
 Increase in temperature
Temperature Dependence of Carrier n  p  ni  N c N v e
 E g / 2 kT

Concentrations (n-type doping)

1/ 2
N N 
n   c d  e ( Ec  Ed ) / 2 kT
 2 

• Si is doped n-type, with Nd of 1015 cm-3.

n
• At very low temperatures (large 1/T), negligible intrinsic Semi log scale
electron-hole pairs exist.
• At about T=100 K (1000/T = 10) all the donor atoms are
High Cryogenic
ionized. This temperature range is called the ionization temp Room temp
temp
region. Carrier concentration, n ≃ Nd = 1015 cm-3.
• At this point n is virtually constant with temperature
until the concentration of intrinsic carriers ni becomes
 Increase in temperature
comparable to the extrinsic concentration Nd.
• At even higher temperatures, ni is much greater than Nd , and the intrinsic carriers dominate.
• In most devices, it is desirable to control the carrier concentration by doping rather than by thermal
electron-hole pair generation.
• Therefore one usually dopes the material such that the extrinsic range extends beyond the highest temperature
at which the device is to be used. Device functioning should not vary with T.
Numerical Problems
Generation and Recombination
 Excess carrier concentration
• Let the equilibrium carrier concentrations are denoted by n0 and p0.

• The total electron and hole concentrations can be different from n0

and p0, when light shines on the sample and generates electrons and
holes.

• The differences are called the excess carrier concentrations n’ and p’.

n  n0  n'
p  p0  p '
This situation is not the same as the thermal agitation. Here the light is
shined on the semiconductor.
 Charge Neutrality n’

• If n’ and p’ are created by light or any other

means, they will be equal (as they are created in
pair) because of charge neutrality.
p’
Creation of excess electron and hole
• When a non-zero n’ is present, an equal p’ may be densities by photons
assumed to be present to maintain charge
equality and vice-versa.

n'  p'

n0 also contains thermally generated electrons. Recombination of excess carrier

re-establishing thermal equilibrium
 Electron-Hole Recombination
Recombination Lifetime

• Assuming that the light generates n’ and p’ carriers. If the light is

turned off, n’ and p’ decay with time until they become zero.

• The process of decay is called recombination.

• The time constant of decay is the recombination time or carrier

lifetime,  .

• Recombination is nature’s way of restoring equilibrium (n’= p’= 0).

 Recombination Lifetime
• Relaxation time   ranges from 1 ns to 1 ms in Si and depends on
the density of metal impurities such as Au and Pt (deep traps).

• The deep traps (due to impurities) capture electrons and holes to

facilitate recombination and are called recombination centers.
How deep the traps
Direct are located?
Recombination
For Si band gap of 1.12
is unfavorable eV, a deep defect is 200
in silicon meV to 600 meV away
from band edges.

An electron–hole pair recombines when an electron drops from the

conduction band into the valence band either directly or through traps.
Recombination Lifetime: Direct and Indirect Band Gap

Radiative Trap Mostly

recombination non-radiative
recombination

Direct band gap Indirect band gap

Example: GaAs Example: Si
Direct recombination is Indirect recombination is rare as
efficient as k conservation k conservation is not satisfied.
is satisfied.
 Rate of recombination (s-1cm-3)

The recombination rate (per cubic centimeter per second) is proportional to n’ and p’

dn n

dt 
Recombination rate increases with increase in the carrier
concentration, and decrease in relaxation time.

n  p

dn n p dp
  
dt   dt
 Example

• A bar of Si is doped with boron at 1015cm-3. It is exposed to light such that electron-
hole pairs are generated throughout the volume of the bar at the rate of 1020/s·cm3.
The recombination lifetime is 10s. What are (a) p0 (b) n0 (c) p’ (d) n’ (e) p (f) n
and (g) the np product?

(a) p0 = Na = 1015 cm-3

(b) n0 = ni2/p0 = 105 cm-3

(c) In steady-state, the rate of generation is equal to
the rate of recombination.
1020/s-cm3 = p’/
 p’= 1020/s-cm3 · 10-5s = 1015 cm-3
 Example

(d) n’= p’= 1015 cm-3

(e) p = p0 + p’= 1015 cm-3 + 1015 cm-3 = 2×1015 cm-3

(f) n = n0 + n’= 105 cm-3 + 1015 cm-3 ~ 1015 cm-3 since n0 << n’

(g) np ~ 21015 cm-3 ·1015 cm-3 = 21030 cm-6 >> ni2 = 1020 cm-6.

The np product can be very different from ni2.

When such a situation occurs, the system is said to be out

of equilibrium.
 Quasi-equilibrium and Quasi-Fermi Levels
• Whenever n’ = p’  0, np  ni2, the semiconductor is not at equilibrium.
Nonetheless, we would like to preserve and use the simple relations:

 ( Ec  E f ) / kT  ( E f  Ev ) / kT
n  Nce p  Nve
• These equations lead to np = ni2.
• When np  ni2, we introduce two quasi-Fermi levels Efn and Efp such that:

 ( Ec  E fn ) / kT  ( E fp  Ev ) / kT
n  Nce p  Nve

• Efn and Efp are the electron and hole quasi-Fermi levels.
• When electrons and holes are at equilibrium, i.e., when np = ni2, Efn and
Efp coincide and this is known as Ef. Otherwise, Efn ≠ Efp.
 Example: Quasi-Fermi Levels
• Consider a Si sample with Nd = 1017 cm–3.
1. Find the location of Ef.
2. Find the location of Efn and Efp when excess carriers are introduced such that n’ = p’ = 1015 cm–3.
( )/
𝑛 = 𝑁 = 10 𝑐𝑚 =𝑁𝑒

𝑁 2.8 × 10
𝐸 − 𝐸 = 𝑘𝑇 ln = 0.026 × ln = 0.15 𝑒𝑉
𝑛 10

Ef is below EC by 0.15 eV

𝑛 = 𝑛 + 𝑛 = 𝑁 + 𝑛 = 1.0 × 10 + 0.01 × 10 = 1.01 × 10 𝑐𝑚

1.01 × 10 𝑐𝑚 = 𝑁 𝑒 ( )/

𝑁 2.8 × 10
𝐸 −𝐸 = 𝑘𝑇 ln = 0.026 × ln = 0.15 𝑒𝑉
1.01 × 10 1.01 × 10

Efn is nearly identical to Ef because n  n0 .

 Example: Quasi-Fermi Levels
𝑛
𝑝=𝑝 +𝑝 = + 𝑝 = 10 + 10 = 10 𝑐𝑚
𝑁
10 𝑐𝑚 = 𝑁 𝑒 ( )/

𝑁 1.04 × 10
𝐸 − 𝐸 = 𝑘𝑇 ln = 0.026 × ln = 0.24 𝑒𝑉
10 10

0.15 eV 0.15 eV
𝐸 𝐸
𝐸 𝐸 𝐸

Non-equilibrium
Thermal Equilibrium Condition
𝐸 0.24 eV

𝐸 𝐸

Thermal equilibrium energy band diagram Quasi Fermi levels for electrons and holes
for Nd = 1017 per cm3 if 1015 per cm3 excess carriers are present
 Discussion
• Since the majority carrier electron concentration does not change, the quasi-Fermi
level for electrons is not different from the thermal-equilibrium Fermi level.

• To be very precise, the electron concentration has increased by a very small amount,
the quasi-Fermi level for electrons has moved slightly closer to the conduction band.

• The quasi-Fermi energy level for the minority carrier holes is significantly different
from the Fermi level and illustrates the fact that we have deviated from thermal
equilibrium significantly.

• Because the hole concentration has increased significantly so that the quasi-Fermi
level for holes has moved much closer to the valence band.

• We will consider the quasi-Fermi energy levels again when we discuss forward-biased
pn junctions.
 Current Flow Mechanism
(Transport in Semiconductors)
 Motion of Electrons and Holes
Thermal Motion
Even without an applied electric field, carriers are not at rest but possess finite kinetic energy due to
the heat exchange from surrounding in a material system.
Three degrees of freedom,
Average electron or hole kinetic each having energy (1/2)kT
energy (equipartition theorem)
k – Boltzmann constant
T - temperature
vth = thermal velocity

meff - Conductivity effective mass

de Broglie wavelength of thermal electron 

• de Broglie wavelength of thermal electron is a yardstick for quantum confinement in nanostructures.

• As an example, for Si it is 12 nm, implies that when the size of Si crystal becomes comparable to or
smaller than 12 nm, the electrons will be quantum confined.
 Thermal Motion
– • Zig-zag motion is due to collisions or scattering
with ions and/or imperfections in the crystal.
Net vth = 0
• Imperfections could be grain boundaries in crystals
– (geometrical defects), and atomic impurities
(point defects).
• Net thermal velocity is zero.
• Mean free time between collisions is m ~ 0.1ps.

Also called relaxation time (τ)

Example:
• The distance travelled between collisions (mean electrostatic discharge
free path) is a few tens of nm or a few hundred
angstroms. For Si, it is approximately 23 nm.

• However, the picture changes when the charge

carriers are under the influence of the electric field!
 Drift
Electron and Hole Mobilities
• The average velocity of the carriers is no longer zero when an electric field is applied to
the semiconductor. This nonzero velocity is called the drift velocity.

The zig-zag motion

+ is due to scattering
In this case there is a net
drift velocity of the +ve
+ charge in the direction of
the electric field.

E
• Drift is the motion caused by an electric field. Why is it important?

• A faster carrier velocity is desirable, as it allows a semiconductor device or circuit to

operate at a higher speed.
 Electron and Hole Mobilities
m p v = qE  mp τmp – mean free time for hole
The drift momentum gained
τmn – mean free time for electron
between collisions is equal to
the force times the mean free qE  mp
time (assuming for holes). v= mp – hole effective mass
mp mn – electron effective mass

 v is the drift velocity

Force = rate of change of momentum v =  pE
 E is the applied electric field
F = dp/dt = m dv/dt  m dv = F dt q mp • p is the hole mobility and
m v = qE t p = • n is the electron mobility
mp
Mobility is a measure of how quickly the
The negative sign means that the
electrons drift in a direction opposite
v = -  nE charge carriers respond to the applied
electric field (as µ = v /E ).
to the E-field. They do so because the qmn
electrons are negatively charged. n = • Smaller the carrier effective mass,
mn larger the mobility, and vice versa.
• Larger the mean free time, larger the
Engineering !dea: effective mass, mobility. mobility and vice versa.
 Electron and Hole Mobilities
 cm/s cm 2 
v = E ;  has the dimensions of v/E  = .
 V/cm V  s 

Electron and hole mobilities of selected semiconductors

Why electron mobility

Si Ge GaAs InAs
Why InAs electron
is much higher than  n (cm2/V∙s) 1400 3900 8500 30000
the hole mobility? mobility is very high?
 p (cm2/V∙s) 470 1900 400 500
Engineering !dea
Based on the above table alone, which semiconductor and which carriers
(electrons or holes) are attractive for applications in high-speed devices?

• GaAs has a much higher µn than Si (due to a smaller mn). Thus, high-speed transistors can be made with GaAs,
which are typically used in communication systems.

• InAs has an even higher µn due to much smaller mn, but the technology of fabricating InAs devices has not yet
been fully developed.
 Drift Velocity, Mean Free Time, Mean Free Path

Example: Given p = 470 cm2/V·s for Si, what is the hole drift velocity at E = 103 V/cm?
What is mp and what is the distance traveled between collisions (called the mean free path)?
n = p E = 470 cm2/V·s  103 V/cm = 4.7 105 cm/s
mp = pmp/q =470 cm2/V ·s  0.39  9.110-31 kg / 1.610-19 C
= 0.047 m2/V ·s  2.210-12 kg/C = 110-13s = 0.1 ps
mean free path = mpnth ~ 1 10-13 s  2.3107 cm/s
= 2.310-6 cm = 230 Å = 23 nm
This is larger than the typical dimensions of devices manufactured today!
Do we expect ballistic transport?
Nope, because of many other sources of scattering (many other τ’s).
 Mechanisms of Carrier Scattering
What are the root causes of carrier collisions or scattering in a crystal lattice?

There are two main causes of carrier scattering:

1. Phonon (Lattice Vibration) Scattering

2. Ionized-Impurity (Coulombic) Scattering

Phonon scattering - mobility decreases when temperature rises:

1 1
 phonon   phonon    T -3 / 2

phonon density  carrier thermal velocity T  T 1 / 2

 = q /meff
T
vth  T1/2
What about the impurity scattering?
 Impurity (Dopant)-Ion Scattering or Coulombic Scattering
• The dopant ions are fixed charge in the semiconductor crystal. They can make electrons and holes
change the direction of motion through the Coulombic force.
• An electron can be scattered by either a donor (positive) ion or an acceptor (negative) ion:
Electron Boron
_ Ion Electron
– –
+

Attractive Coulomb
Repulsive Coulomb Arsenic Ion scattering
scattering

• The mobility due to impurity scattering is inversely

proportional to the sum of the donor and acceptor ion
vth3 T 3/ 2
impurity   concentrations. Why?
• More impurities  More scattering  Less mobility.
Na  Nd Na  Nd • It is also proportional to T3/2. Why is the mobility
higher, i.e., the scattering weaker, at a higher T?
There is less change in the direction of travel if • At a higher temperature, the electron has a higher
the electron zips by the ion at a higher speed! thermal velocity and flies by the ion in a shorter time,
and its direction of motion is less affected by the ion.
 Current Flow Mechanism
(Transport in Semiconductors)
 Mechanisms of Carrier Scattering
What are the root causes of carrier collisions or scattering in a crystal lattice?

There are two main causes of carrier scattering:

1. Phonon (Lattice Vibration) Scattering

2. Ionized-Impurity (Coulombic) Scattering

Phonon scattering - mobility decreases when temperature rises:

1 1
 phonon   phonon    T 3 / 2

phonon density  carrier thermal velocity T  T 1 / 2

 = q /meff
T
vth  T1/2
What about the impurity scattering?
 Impurity (Dopant)-Ion Scattering or Coulombic Scattering
• The dopant ions are fixed charge in the semiconductor crystal. They can make electrons and holes
change the direction of motion through the Coulombic force.
• An electron can be scattered by either a donor (positive) ion or an acceptor (negative) ion:
Electron Boron
_ Ion Electron
– –
+

Attractive Coulomb
Repulsive Coulomb Arsenic Ion scattering
scattering

• The mobility due to impurity scattering is inversely

proportional to the sum of the donor and acceptor ion
vth3 T 3/ 2
impurity   concentrations. Why?
• More impurities  More scattering  Less mobility.
Na  Nd Na  Nd • It is also proportional to T3/2. Why is the mobility
higher, i.e., the scattering weaker, at a higher T?
There is less change in the direction of travel if • At a higher temperature, the electron has a higher
the electron zips by the ion at a higher speed! thermal velocity and flies by the ion in a shorter time,
and its direction of motion is less affected by the ion.
 Total Mobility Matthiessen's rule 1 1 1
 
  phonon impurity
1 1 1
 
Electrons   phonon  impurity

• Silicon hole mobility is about one-third of the

electron mobility. Part of this difference in
mobility can be explained by the difference in
the effective mass. The rest is attributable to
Holes the difference in the scattering mean free time.

• The inverse proportionality to dopant density is

not followed at the limit of very large Na or Nd.

The electron and hole mobilities of silicon at 300 K. At low • The reason is free-carrier screening. When the
dopant concentration, the electron mobility is dominated carrier concentration is large, the carriers can
by phonon scattering; at high dopant concentration, it is distribute themselves to partially screen out the
dominated by impurity ion scattering. Coulombic field of the dopant ions.
 Temperature Effect on Mobility
Lattice scattering - phonon
• At small dopant concentrations, μ decreases with increasing
T, indicative of the dominance of phonon scattering.

10 1 5 • At very high dopant concentration and low temperature,

where impurity scattering is expected to dominate, μ
increases with increasing T. Beyond a certain T, µ decreases.

• With increase in the dopant concentration, mobility

decreases.

• It is not easy to measure mobility accurately and that we

should presume the existence of uncertainties in
experimental data in general.

• Also, it is not easy to model mobility very accurately as there

Schematic plot of the temperature are several different scattering mechanisms working together,
dependence of μn. and it is not possible to separate them out.
 Velocity Saturation
For a microelectronic device of say 10 micron, application of say 5 V on the device would mean an
electric field = 5 V/10 µm = 5 V/10 x 10-4 cm  5 x 103 V/cm  5000 V/cm!

• In small devices, the electric field can easily reach

105 V/cm.
• If the electron mobility is 103 cm2/V·s, the drift
velocity, according to v = μE, should be 108 cm/s.
• In reality, electron and hole velocities saturate at
around 107cm/s and do not increase beyond that, Saturation of electron
no matter how large E is. drift velocity at high
electric fields for Si.
• When the kinetic energy of a carrier exceeds the
optical phonon energy, Eopt, it generates an optical
phonon and loses the kinetic energy.
• Therefore, the velocity does not rise above
saturation velocity, vsat which is 107 cm/s.

• Velocity saturation has a damaging effect on

device speed.
 Drift Current and Conductivity
p-type semiconductor bar
When an electric field is applied in a semiconductor,
E Jp electrons and holes acquire drift velocity, and the
consequent motion results in electric current, which
unit is drift current.
+ area
A is the cross sectional area
normal to the current flow
+n
More useful way of quantifying v – drift velocity
current in semiconductor devices p – hole density per cm3
q – charge
Hole current density Jp = qpv A/cm2 or C/cm2·sec

Example: If p = 1015cm-3 and v = 104 cm/s, then

Jp= 1.610-19C  1015cm-3  104cm/s
= 1.6 C/s  cm 2  1.6 A/cm2
 Drift Current and Conductivity
Hole current density in n – electron concentration
terms of mobility Jp,drift = qpv = qppE p – hole concentration
q – electronic charge
Electron current density in µ - mobility
terms of mobility Jn,drift = –qnv = – qn (–nE ) σ – conductivity
ρ – resistivity
Jn,drift = qnnE J – current density
E – electric field
The total drift
current density is
Jdrift = Jn,drift + Jp,drift = (qnn+qpp)E =  E

Conductivity of a semiconductor:  = (qnn + qpp) 1/ohm-cm or siemens/cm

1/ = ρ  is resistivity (ohm-cm)

Usually, in the conductivity, only one of the components is significant because

of the large ratio between the majority and minority carrier densities.
Relationship between Resistivity
and Dopant Density For silicon at room temperature
As the dopant density increases, conductivity increases.
Therefore resistivity decreases. Do you see an ambiguity  = 1/
here?

DOPANT DENSITY cm-3

For n-type donor in majority
 ≈ qnn 1/ = ρ But
P-type

Electrons

N-type

Holes

Nd range 1014 – 1020, µn range 1400 – 100. σ still

increases as n goes up by 106 order whereas µn
decreases by one order. Note Nd = n at T = 300 K. RESISTIVITY (cm)
 Example: Resistance with doping
(a) What is the resistivity () of silicon doped with 1017cm-3 of arsenic?
(b) What is the resistance (R) of a piece of this silicon material 1m long and
0.1 m2 in cross-sectional area?

Difference Between Resistance and Resistivity

• Resistance is macroscopic or an extrinsic property and depends on the geometry of the conductor
or semiconductor, whereas resistivity is microscopic or an intrinsic property of a conductor or
semiconductor.
• This means that given a temperature, the resistivity of a conductor is always the same and is
independent of its length or size.
• Resistivity of a semiconductor changes with the doping, & temperature.
• Resistance and resistivity of a material are related to each other through an equation Resistivity =
resistance x area of cross-section/length. That is, R = ρl/A. Therefore resistance varies with area
of cross section and length of a conductor.
• The same material (say Cu) would have the same resistivity, but the resistance becomes higher
as the cross section area gets smaller (copper wire versus a copper disc).
 Example: Resistance with doping
(a) What is the resistivity () of silicon doped with 1017cm-3 of arsenic?
(b) What is the resistance (R) of a piece of this silicon material 1m long and Increased phonon scattering
0.1 m2 in cross-sectional area?  Decreased mobility
 Decreased conductivity
(a) Using the N-type curve in the previous figure, we find that  = 0.09 -cm.  Increased resistivity
 Increased resistance.
(b) R = L/A = 0.09 -cm  1 m / 0.1 m2
= 0.09 -cm  10-4 cm/ 10-9 cm2
= 9  103 
= 9 KΩ
Example: Resistance with temperature 10 15

• By what factor will R increase or decrease from T = 300 K to T = 400 K?

• The temperature dependent factor in  (and therefore ) is n.

• From the mobility vs. temperature curve for 1017cm-3, we find that n
decreases from 750 at 300K to 400 at 400K. Therefore conductivity
decreases, and so resistivity increases. As a result, R increases by a
factor
• Why R increases?
• Increased scattering due to phonons!
 Current Flow Mechanism
(Transport in Semiconductors)
 Diffusion
• When light is flashed at a small region near the center of a long
thin bar of semiconductor material, electron-hole pairs, in
addition to those available at thermal equilibrium, are generated
at the center because of the light energy absorption by the
semiconductor.

Spreading of a pulse of • These excess carriers will move randomly to the right and to the
electrons by diffusion. left of the generation zone, resulting in an outwards flux of
carriers away from the center.

• These carriers are said to diffuse (like a drop of ink in a glass of

water quickly mixes with water).

Equilibrium is disturbed. Diffusion takes place so that the

distribution is uniform and equilibrium is achieved.
 Diffusion Current
Flux density

Dn(p) - electron(hole)
diffusion constant.

Particles diffuse from a higher-concentration

location to a lower-concentration location.
Flux density

Particles could be electrons, holes or even ions, atoms

Diffusion Current If the electron/hole concentration is not uniform, there will be electron/hole
diffusion current, which is proportional to the gradient of the electron/hole
concentration.

, ,

Concentration gradient
Dn(p) - electron(hole) diffusion constant
Ln(p) - Electron (hole) diffusion length
τn(p) – Electron (hole) lifetime

• Larger the concentration gradient,

more the diffusion current.

• Larger the Dn is, faster the electrons

Electron flow Hole flow diffuse.
Current flow Current flow
 Example

• Assume that, in an n-type gallium arsenide semiconductor at T = 300 K, the electron concentration
varies linearly from 2 x 1016 to 5 x 1015 cm-3 over a distance of 0.20 cm. Calculate the diffusion
current density if the electron diffusion coefficient is Dn = 195 cm2/s.

The diffusion current density is given by

𝑑𝑛 ∆𝑛 2 × 10 − 5 × 10
𝐽 , = 𝑒𝐷 ≈ 𝑒𝐷 = (1.6 × 10 )(195) = 2.34 𝐴/𝑐𝑚2
𝑑𝑥 ∆𝑥 0.20
 Total Current – Review of Four Current Components
dn
Jn = Jn,drift + Jn,diffusion = qnnE + qDn
dx
dp
Jp = Jp,drift + Jp,diffusion = qppE – qD p
dx
These are current density equations

JTOTAL = Jn + Jp
E(x)
Electric field
Jp (diffusion), Jp (Drift)
n(x)
Concentration
variation Jn (diffusion)
p(x)
Jn (Drift)
Hole current = direction of hole motion
Electron current = opposite direction of electron flow.
 Semiconductor in an electric field
1.5 V
Potential drops (gradient) across the Si bar
Potential decreases with
External electric field 𝑑𝑉
− increase in x, so negative
E 𝑑𝑥
1.5 V slope.

Semiconductor bar acts like an ohmic resistor

Electron energy Constant takes care of reference
at the CB edge
Ec(x) = constant – qV(x) 

• The slope of Ec and Ev indicates the electric field.

1.5 V
• Bands will bend in the electric field (‘+’ slope) and that Ec
and Ev vary in the opposite direction from the voltage.
• That is, Ec and Ev are higher where the voltage is lower.
• The energy of electron in the CB varies with position because the potential varies with the position.
• The electrons roll downhill like stones in the energy band diagram and the holes float up like bubbles.
 Semiconductor in an electric field
Air Drop Kinetic energy
gained in an
electric field EC
Water Bubble

A B Analogy of a water Superimposition of the

drop and a bubble (E, k) band structure on
in empty and filled the E-versus-position.
EC
bottles respectively EV

• When a voltage is applied across a piece of semiconductor, it

alters the band diagram.
• By definition, a positive voltage raises the potential energy of a
positive charge and lowers the potential energy of a negative EV • Electron energy increases
charge. going up, while hole energy
• It therefore lowers the energy diagrams since the energy diagram increases going down.
plots the energy of an electron (a negative charge). • Similarly, electron and hole
• The energy band diagram is lower (at the left) where the voltage wavevectors point in
is higher. The band diagram is higher where the voltage is lower. opposite directions and they
• Ec and Ev are always separated by a constant Eg. move opposite to each other.
 Einstein Relationship
Consider a piece of non-uniformly doped semiconductor.

• The semiconductor is at equilibrium,  ( Ec  E f ) / kT

and therefore the Fermi level Ef is
N-type semiconductor nN e c
constant.
• The left side is more heavily doped dn N ( E  E ) / kT dEc
 ce c f
than the right side, and so Ec is closer Decreasing donor concentration dx kT dx
to Ef on the left side.
n dEc
• Because Ec is not a constant, there is Ec (x)  
an electric field equal to (1/q) dEc /dx kT dx
[= (1/q) dEv /dx ]. Ef n  dV 
• This field is internally created and is     q 
kT  dx 
as real as a field created by an
external voltage.
Ev (x)
n
• Because the semiconductor is at Internal electric field   qE
equilibrium, there cannot be any kT
Jn or Jp.
 Einstein Relationship between D and 
dn n
 qE
From the drift-diffusion dx kT
equation for electron
dn
J n  qn n E + qDn  0 at equilibrium.
dx
qD
0  qn n E  qn n E
kT
Given the mobility Very important
kT kT
one can readily find Dn   n Similarly, Dp  p relationship: µ is
diffusion constant q q drift component
and vice versa. and D is diffusion
These are known as the Einstein relationship. component.

Consequences:
(1) The electron drift and diffusion currents will perfectly cancel each other out for an arbitrary doping
profile, if and only if the above equation for electron is satisfied. The same is true for the hole.
(2) All the scattering mechanisms i.e. phonon and impurity scatterings that impede electron drift
would also impede electron diffusion.
 Example: Diffusion Constant

Calculate the hole diffusion constant in a piece of

silicon with p = 410 cm2 V-1s-1 ?

kT/q = 26 mV at room temperature.

 kT 
D p     p  (26 mV )  410 cm 2 V 1s 1  11 cm 2 /s
 q 
Transport Equations
The transport equations are a set of five equations that govern the behavior of
semiconductor materials and devices. The first two transport equations are the
drift-diffusion equations.

Drift-Diffusion Equations:
E

In one dimensional problem

E
Poisson Equation: Using the expression for Displacement current and from Maxwell’s equation
and Gauss’s law:
Using the relationship between the electric field and the potential
Poisson equation tells us:
given the charge density,
what is the potential and
The Poisson equation vice versa.

where ϵs is the permittivity of the semiconductor and ρ is the local charge density in Coulomb/cm3
in the semiconductor.

If all the doping atoms are ionized, which is the case at room temperature,

In short In one dimension

Continuity Equation
For one-dimensional current flow in the x-direction, consider a differential volume of semiconductor
material. The cross-sectional area of the volume under consideration is A with length dx. An electron
current density Jn(x) (unit: Amps/cm2) enters the volume and a current density Jn(x + dx) flows out of it.

The variation of the number of free electrons in the volume A.dx as

a function of time is given by the number of electrons entering the
volume, minus the number of electrons flowing out of the volume,
plus the number of electrons generated minus the number of
electrons recombined:

U – Intrinsic generation rate

G – Extrinsic generation rate
 Continuity Equation

Continuity Equation for

electrons in one dimension

Continuity Equation for

holes in one dimension
If we neglect the effects of
In three dimensions generation and recombination

Rate of change of carrier

concentration is equal to 𝐧 𝐧
divergence of J plus the
net carrier generation rate.
𝐩 𝐩

The set of equations composed of the drift-diffusion equations, the Poisson equation, and the continuity
equations is called the “transport equations”. The transport equations allows one to derive most properties
of semiconductor devices. These equations are solved numerically and are backbones of device simulators.