EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
Lecture 5
Carrier Diffusion
EE 4353/5353
Spring 2023
Dr. Ariful Haque
EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
What is the average net velocity in the direction of the electric field?
a) Velocity of electrons
b) Velocity of holes
c) Drift velocity
d) Collision velocity
A long specimen of p-type semiconductor material is:
a) Positively charged
b) Is electrically neutral
c) Has an electric field directed along its length
d) None of the above
EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
Carrier Diffusion
Particles diffuse from a higher-concentration location to a lower-
concentration location.
Current due to carriers moving from high concentration to a
low concentration is called diffusion current
EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
Diffusion Process
• Carriers in a s/c diffuse in a carrier gradient by random thermal motion and
scattering from lattice impurities
• A pulse of electrons injected at x-0 and t=0 will spread out in time
EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
Consider:
• Electron density that varies in the x-direction
• Electrons have thermal velocity vth and a mean
free path l in the x-direction
𝑙 = 𝑣𝑡ℎ𝜏𝑐 (𝜏𝑐= mean free time)
• At x=0, there is a cross-section plane:
• There is current flowing from x = -l
• There is current flowing from x = l
• Both currents come from different electron
densities
Carrier Diffusion
• Electrons at x = - l (one mean free path away on the left side) have an equal
chance of moving to the left or right with a mean free time 𝜏𝑐
• Half of them will move across the plane at x=0
• Avg. rate of electron flow/unit area F1 crossing plane at x=0 is
𝐹1 =
1
2
𝑛 −𝑙 . 𝑙
𝜏𝑐
=
1
2
𝑛 −𝑙 . 𝑣𝑡ℎ
EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
• Similarly, the average rate of electron flow/unit area of electrons at x=0 crossing
from the right is
• Net rate of carrier flow from left to right is
• We can approximate the densities at x = ± l by the 1st two terms of the Taylor
expansion:
• This carrier flow gives rise to a current (since charge is –q):
𝐹2 =
1
2
𝑛 𝑙 . 𝑣𝑡ℎ
𝐹 = 𝐹1 − 𝐹2 =
1
2
𝑣𝑡ℎ 𝑛 −𝑙 − 𝑛 𝑙
𝐹 =
1
2
𝑣𝑡ℎ 𝑛 0 − 𝑙
𝑑𝑛
𝑑𝑥
− 𝑛 0 + 𝑙
𝑑𝑛
𝑑𝑥
= 𝑣𝑡ℎ𝑙
𝑑𝑛
𝑑𝑥
≡ −𝐷𝑛
𝑑𝑛
𝑑𝑥
Dn is called the diffusion coefficient or diffusivity ≡ 𝑣𝑡ℎ𝑙
𝐽𝑛 = 𝑞𝐷𝑛
𝑑𝑛
𝑑𝑥
Diffusion current results from the random thermal motion of
carrier in a concentration gradient
EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
Problem
Assume that in an n-type semiconductor at 300K, the electron concentration
varies linearly from 1x1018 to 7x1017 cm-3 over a distance of 0.1 cm. Calculate
the diffusion current density if the electron diffusion coefficient is Dn =22.5
cm2/s
EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
Drift and Diffusion of Carriers
𝐽𝑛,𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑜𝑛 = 𝑞𝐷𝑛
𝑑𝑛
𝑑𝑥
𝐽𝑝,𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑜𝑛 = −𝑞𝐷𝑝
𝑑𝑝
𝑑𝑥
𝐽𝑝 (𝑥) = 𝑞𝑝(𝑥)𝜇𝑝ℰ(𝑥) − 𝑞𝐷𝑝
𝑑𝑝(𝑥)
𝑑𝑥
𝐽𝑛(𝑥) = 𝑞𝑛(𝑥)𝜇𝑛ℰ(𝑥) + 𝑞𝐷𝑛
𝑑𝑛(𝑥)
𝑑𝑥
𝐽𝑡𝑜𝑡𝑎𝑙(𝑥) = 𝐽𝑛(𝑥) + 𝐽𝑝(𝑥)
If an electric field is present in addition to a carrier gradient:
𝜙 = 𝑐𝑎𝑟𝑟𝑖𝑒𝑟 𝑓𝑙𝑢𝑥
EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
• Both carrier mobility µ and diffusion constants D are determined by the
frequency of the carrier collisions with phonons or dopant ions and lose their
momentum.
• They should be related to each other.
Einstein Relationship between D and µ
• Let’s re-write the expression for Dn using the equipartition theorem
• In one dimension,
1
2
𝑚𝑛𝑣𝑡ℎ
2 =
1
2
𝑘𝑇
𝐷𝑛 = 𝑣𝑡ℎ𝑙 = 𝑣𝑡ℎ 𝑣𝑡ℎ𝜏𝑐 = 𝑣𝑡ℎ
2
𝜇𝑛𝑚𝑛
𝑞
=
𝑘𝑇
𝑚𝑛
𝜇𝑛𝑚𝑛
𝑞
𝐷𝑛 =
𝑘𝑇
𝑞
𝜇𝑛
Einstein relation characterize carrier transport by diffusion
and drift in a semiconductor
EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
Problem
Minority carriers (holes) are injected into a homogeneous n-type semiconductor sample
at one point. An electric field of 50 V/cm is applied across the sample, and the field
moves minority carriers a distance of 1 cm in 100 µs. Find the drift velocity and the
diffusivity of the minority carriers. The temperature is 300K.
EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
Optical Absorption
• Electron-hole pair is created (a)
(Excess Carriers)
• Excited electron gives up energy
to lattice via scattering events (b)
• Electron recombines with a hole
in the VB (c)
−
𝑑𝐈 𝑥
𝑑𝑥
= α𝐈(𝑥)
Solution: 𝐈 𝑥 = 𝐈𝐨𝑒−𝛼𝑥
𝐈𝐭 = 𝐈𝐨𝑒−𝛼𝑙
Transmitted Intensity
𝐸 =
1.24
𝜆
EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
Semiconductors – Nonequilibrium conditions
➢ Excess electrons in the conduction band and excess holes in the valence band may
exist in addition to the thermal-equilibrium concentrations if an external excitation
is applied to the semiconductor.
➢ The creation of excess electrons and holes means that the semiconductor is no
longer in thermal equilibrium.
Carrier generation and recombination:
Generation ⇒ process whereby electrons and holes are created
Recombination ⇒ process whereby electrons and holes are coincided and lost.
Change in temperature or optical excitation can generate excess electrons and
holes creating a nonequilibrium condition.
EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
Generation and Recombination
➢ In thermal equilibrium, pn = ni
2.
➢ In a non-equilibrium situation excess carriers are introduced, in general
pn > ni
2.
➢ Excess carrier introduction is called carrier injection
➢ Excess carriers (generation) can be introduced by optical excitation or
forward-biasing a pn junction.
➢ Whenever the thermal-equilibrium condition is disturbed, processes exist
to restore the system to equilibrium by recombination.
Radiative and non-radiative recombination.
Direct and indirect recombination.
EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
Example
A 0.46 µm thick sample of GaAs is illuminated with monochromatic light pf hv=2eV.
The absorption coefficient is 5.4 x 104 cm-1. The power incident on the sample is
10mW.
(a) Find the total energy absorbed by the sample per second (J/s)
(b) Find the rate of excess thermal energy given up by the electrons to the lattice
before recombination (J/s)
(c) Find the number of photons per second given off from recombination events,
assuming perfect quantum efficiency.
EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
Example
𝐈 𝑥 = 𝐈𝐨𝑒−𝛼𝑙
= 10−2
exp(−5𝑥10−4
𝑥0.46𝑥10−4
)
A 0.46 µm thick sample of GaAs is illuminated with monochromatic light pf hv=2eV. The
absorption coefficient is 5.4 x 104 cm-1. The power incident on the sample is 10mW.
(a) Find the total energy absorbed by the sample per second (J/s)
(b) Find the rate of excess thermal energy given up by the electrons to the lattice before
recombination (J/s)
(c) Find the number of photons per second given off from recombination events, assuming
perfect quantum efficiency.
= 10−2
𝑒−2.3
= 10−3
𝑊
⇒ 𝑎𝑏𝑠𝑜𝑟𝑏𝑒𝑑 𝑝𝑜𝑤𝑒𝑟 = 10 − 1 = 9𝑚𝑊 = 9𝑥10−3 Τ
𝐽 𝑠
Fraction of each photon energy unit converted to heat
=
2 − 1.43
2
= 0.285
Amount of energy converted to heat per second
= 0.285𝑥9𝑥10−3
= 2.57𝑥10−3 Τ
𝐽 𝑠
Amount of radiation accounts for 6.43 mW of power
Each photon emitted has 1.6𝑥10−19
𝑥1.43 𝐽 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦
# of photons emitted per second is:
6.43𝑥10−3
1.6𝑥10−19𝑥1.43
= 2.81𝑥1016 Τ
𝑝ℎ𝑜𝑡𝑜𝑛𝑠 𝑠
EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
Generation and Recombination of Electron-Hole Pairs
Injection: process of introducing excess carriers in semiconductor
Generation:
⚫ Spontaneous Generation [Thermal Energy]
⚫ External Generation [Light]
⚫ Direct band-to-band generation and recombination
⚫ Recombination through allowed energy states within the bandgap,
referred to as traps or recombination centers
Recombination:
⚫ Spontaneous Recombination [Light or Heat]
⚫ Stimulated Recombination [light/laser]
⚫ Direct band-to-band generation and recombination
⚫ Recombination through allowed energy states within the bandgap,
referred to as traps or recombination centers
EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
Direct Generation and Recombination
Under thermal equilibrium the recombination rate must be balanced by the
generation rate
𝐺𝑡ℎ = 𝑅𝑡ℎ = 𝛽𝑛𝑛𝑜𝑝𝑛𝑜
n=type s/c
equilibrium quantity
Electron-hole
generation
Electron-hole
recombination
Under thermal equilibrium:
➢ The random generation-recombination of electrons-
holes occur continuously due to the thermal
excitation.
➢ The random generation-recombination of electrons-
holes occur continuously due to the thermal
excitation.
➢ Concentration of electrons and holes are
independent of time
EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
Direct Generation and Recombination
When excess carriers are introduced into a direct
bandgap s/c
➢ High probabilty that direct recombination can
occur, no additional momentum is needed
➢ Rate of direct recombination  # of electrons
available in the CB and the # of holes in the VB
𝑅 = 𝛽𝑛𝑝
𝛽 𝑖𝑠 𝑡ℎ𝑒 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛𝑎𝑙𝑖𝑡𝑦 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
Under thermal equilibrium the recombination rate must be balanced by the
generation rate
𝐺𝑡ℎ = 𝑅𝑡ℎ = 𝛽𝑛𝑛𝑜𝑝𝑛𝑜
n=type s/c
equilibrium quantity
Electron-hole
generation
Electron-hole
recombination
EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
➢ When we shine a light to create electron-hole pair at a rate GL, the carrier
concentrations are above their equilibrium values
➢ The recombination and generate rate becomes:
𝑅 = 𝛽𝑛𝑛𝑝𝑛 = 𝛽(𝑛𝑛𝑜 + ∆𝑛)(𝑝𝑛𝑜 + ∆𝑝)
𝐺 = 𝐺𝐿 + 𝐺𝑡ℎ
∆𝑛, ∆𝑝 are the excess carrier conc.
∆𝑛 = 𝑛𝑛 − 𝑛𝑛𝑜, ∆𝑝 = 𝑝𝑛 − 𝑝𝑛𝑜
∆𝑛 = ∆𝑝 to maintain charge neutrality
The net rate of change of the hole concentration:
In steady state
𝑑𝑝𝑛
𝑑𝑡
= 0 ⟹
𝑑𝑝𝑛
𝑑𝑡
= G − R = 𝐺𝐿 + 𝐺𝑡ℎ − R
𝐺𝐿 = 𝑅 − 𝐺𝑡ℎ ≡ 𝑈
U is the net recombination rate
𝑈 = 𝛽(𝑛𝑛𝑜 + 𝑝𝑛𝑜 + ∆𝑝)∆𝑝
Nonequilibrium excess carriers
Previous page
EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
Direct Recombination
➢ For low level injection, ∆𝑝, 𝑝𝑛𝑜 ≪ 𝑛𝑛𝑜
𝑈 = 𝛽(𝑛𝑛𝑜 + 𝑝𝑛𝑜 + ∆𝑝)∆𝑝
𝑈 ≅ 𝛽𝑛𝑛𝑜∆𝑝 =
𝑝𝑛 − 𝑝𝑛𝑜
1
𝛽𝑛𝑛𝑜
The net recombination rate is proportional to the excess minority carrier
concentration
(U=0 in thermal equilibrium)
1
𝛽𝑛𝑛𝑜
≡ 𝑙𝑖𝑓𝑒𝑡𝑖𝑚𝑒 𝜏𝑝 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑥𝑐𝑒𝑠𝑠 𝑚𝑖𝑛𝑜𝑟𝑖𝑡𝑦 𝑐𝑎𝑟𝑟𝑖𝑒𝑟𝑠
𝑈 =
𝑝𝑛 − 𝑝𝑛𝑜
𝜏𝑝
The net recombination rate is also determined by the excess minority lifetime
EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
Direct Recombination
Consider light illumination:
• Electron-hole pairs are generated with rate GL
𝐺𝐿 = 𝑈 =
𝑝𝑛 − 𝑝𝑛𝑜
𝜏𝑝
⇒ 𝑝𝑛 = 𝑝𝑛𝑜 + 𝜏𝑝 𝐺𝐿
∆𝑛 = ∆𝑝 = 𝜏𝑝 𝐺𝐿
At t=0 the light is now switched off, the
boundary condition is:
𝑝𝑛(𝑡 = 0) = 𝑝𝑛𝑜 + 𝜏𝑝 𝐺𝐿
𝑑𝑝𝑛
𝑑𝑡
= 𝐺𝑡ℎ − R = −U = −
𝑝𝑛 − 𝑝𝑛𝑜
𝜏𝑝
𝑑𝑝𝑛
𝑑𝑡
= G − R = 𝐺𝐿 + 𝐺𝑡ℎ − R
The following equation
becomes With a solution:
𝑝𝑛 𝑡 = 𝑝𝑛𝑜 + 𝜏𝑝 𝐺𝐿exp( Τ
−𝑡 𝜏𝑝 )
Note: 𝑝𝑛 𝑡 ⇢ ∞ = 𝑝𝑛𝑜
EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
Decay of excess electrons and holes
Assume GaAs doped with 1015 acceptors/cm-3
1014 EHP/cm-3 are created at t=0
We can then calculate the decay of these
carriers with time
Shown in the figure the decay far a carrier
recombination lifetime of 𝜏𝑛 = 𝜏𝑝 = 10ns
EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
Carrier Lifetime
We can use this idea to measure the carrier lifetime using photoconductivity.
• Excess carriers generated by the light pulse cause a momentary increase in conductivity
• This cause a drop in voltage across the sample when a constant current is passed
through it.
• Using the decay in the signal is a measure of the lifetime of the excess minority carriers.
EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
Example
A GaAs sample with nno=1014 cm-3 is illuminated with light and 1013 electron-hole
pairs/cm3 are created every microsecond. If τn = τp = 2µs, find the change in the
minority carrier concentration.
EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
Example
A Si sample with nno=1014 cm-3 is illuminated with light and 1013 electron-hole
pairs/cm3 are created every microsecond. If τn = τp = 2µs, find the change in the
minority carrier concentration.
Before illumination:
𝑝𝑛𝑜 = Τ
𝑛𝑖
2
𝑛𝑛𝑜 = Τ
1𝑥1010 2
1014
= 1𝑥106
𝑐𝑚3
After illumination:
𝑝𝑛 = 𝑝𝑛𝑜 + 𝜏𝑝 𝐺𝐿 = 1𝑥106 + 2𝑥10−6
1013
1𝑥106
= 2𝑥1013𝑐𝑚−3
∆𝑝𝑛 = 𝜏𝑝 𝐺𝐿 = 2𝑥1013𝑐𝑚−3
EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
Quasi-Fermi Level
Under illumination excess carriers are introduced to the semiconductor
➢ Non equilibrium state:
𝑝𝑛 > 𝑛𝑖
2
➢ In the non-equilibrium state, the Fermi energy is strictly no longer
defined. Under non-equilibrium condition, we can assign individual
Fermi level for electrons and holes.
➢ These are called quasi-Fermi levels and the total electron or hole
concentration can be determined using these quasi-Fermi levels.
➢ We can write,
𝑛 = 𝑛𝑛𝑜 + ∆𝑛 = 𝑛𝑖𝑒 Τ
𝐸𝐹𝑛−𝐸𝑖 𝑘𝑇
𝑝 = 𝑝𝑛𝑜 + ∆𝑝 = 𝑛𝑖𝑒 Τ
𝐸𝑖−𝐸𝐹𝑝 𝑘𝑇
EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
Example
A GaAs sample with nno=1016 cm-3 is illuminated with light and 1013 electron-hole
pairs/cm3 are created every microsecond. If τn = τp = 2ns, find the quasi-Fermi level
at room temperature. (𝑛𝑖 = 2.25𝑥106𝑐𝑚−3)
EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
Indirect Recombination
➢ In real semiconductors, there are some crystal defects that create
discrete electronic energy states within the forbidden energy band.
Recombination through the defect (trap) states is called indirect
recombination
➢ The carrier lifetime due to the recombination through the defect energy
state is determined by the Shockley-Read-Hall theory of recombination.
Shockley-Read-Hall recombination:
➢ Shockley-Read-Hall theory of recombination assumes that a single trap
center exists at an energy Et within the bandgap.
EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
Indirect Recombination - trapping
• For indirect gap materials: small probability of direct electron-hole recombination
• Some band gap light but extremely weak
• Major recombination events occur via recombination levels with the gap: energy
given up as to the lattice as heat
• Any impurity or defect can result in recombination center
• Excess electron and holes created recombines at
Er in 2 steps: (a) hole capture and electron capture
• Since Er is filled, first process must be (a)
• Impurity or defect centers is usually referred to as
traps.
• Recombination lifetime depends on the time the
first carrier is held before the second one is
captured
• Effect of recombination and trapping can be
measured by a photoconductive decay experiment
𝜎 𝑡 = 𝑞 𝑛 𝑡 𝜇𝑛 + 𝑝(𝑡)𝜇𝑝
EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
Indirect Recombination
➢ Direct recombination process is unlikely because electrons at the bottom of CB have
non-zero momentum wrt holes at the top of the valence band.
➢ Direct transition that conserves both energy and momentum is not possible without
a lattice interaction
➢ Indirect transition is via localized energy states in the forbidden gap
Indirect Recombination
Indirect Recombination:
• Electron Capture
• Electron Emission
• Hole Capture
• Hole Emission
EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
Assume that an n-type semiconductor is uniformly illuminated, producing a uniform
excess generation rate G. Show that in steady state the change in the
semiconductor conductivity is given by:
Example
∆𝜎 = 𝑞(𝜇𝑛 + 𝜇𝑝)𝜏𝑝𝐺
EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
Thermal Equilibrium
Classifications:
• Si – Indirect Recombination
• GaAs – Direct Recombination
Direct vs Indirect
EE4353/5353 – Fundamentals of Advanced Semiconductor Technology
Assignment #2
A Si sample with 1016 cm-3 donors is optically excited such that 1019 cm-3 electron-hole
pairs are generated per second uniformly in the sample. The laser caused the sample
to heat up to 450K. Find the quasi-Fermi levels and the change in conductivity of the
sample upon shining the light. Electron and hole lifetimes are both 10µs. Dp=12 cm2/s,
Dn=36 cm2/s: ni= 1014cm-3 at 450K

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The Carrier Diffusion of a Semiconductor

  • 1. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology Lecture 5 Carrier Diffusion EE 4353/5353 Spring 2023 Dr. Ariful Haque
  • 2. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology What is the average net velocity in the direction of the electric field? a) Velocity of electrons b) Velocity of holes c) Drift velocity d) Collision velocity A long specimen of p-type semiconductor material is: a) Positively charged b) Is electrically neutral c) Has an electric field directed along its length d) None of the above
  • 3. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology Carrier Diffusion Particles diffuse from a higher-concentration location to a lower- concentration location. Current due to carriers moving from high concentration to a low concentration is called diffusion current
  • 4. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology Diffusion Process • Carriers in a s/c diffuse in a carrier gradient by random thermal motion and scattering from lattice impurities • A pulse of electrons injected at x-0 and t=0 will spread out in time
  • 5. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology Consider: • Electron density that varies in the x-direction • Electrons have thermal velocity vth and a mean free path l in the x-direction 𝑙 = 𝑣𝑡ℎ𝜏𝑐 (𝜏𝑐= mean free time) • At x=0, there is a cross-section plane: • There is current flowing from x = -l • There is current flowing from x = l • Both currents come from different electron densities Carrier Diffusion • Electrons at x = - l (one mean free path away on the left side) have an equal chance of moving to the left or right with a mean free time 𝜏𝑐 • Half of them will move across the plane at x=0 • Avg. rate of electron flow/unit area F1 crossing plane at x=0 is 𝐹1 = 1 2 𝑛 −𝑙 . 𝑙 𝜏𝑐 = 1 2 𝑛 −𝑙 . 𝑣𝑡ℎ
  • 6. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology • Similarly, the average rate of electron flow/unit area of electrons at x=0 crossing from the right is • Net rate of carrier flow from left to right is • We can approximate the densities at x = ± l by the 1st two terms of the Taylor expansion: • This carrier flow gives rise to a current (since charge is –q): 𝐹2 = 1 2 𝑛 𝑙 . 𝑣𝑡ℎ 𝐹 = 𝐹1 − 𝐹2 = 1 2 𝑣𝑡ℎ 𝑛 −𝑙 − 𝑛 𝑙 𝐹 = 1 2 𝑣𝑡ℎ 𝑛 0 − 𝑙 𝑑𝑛 𝑑𝑥 − 𝑛 0 + 𝑙 𝑑𝑛 𝑑𝑥 = 𝑣𝑡ℎ𝑙 𝑑𝑛 𝑑𝑥 ≡ −𝐷𝑛 𝑑𝑛 𝑑𝑥 Dn is called the diffusion coefficient or diffusivity ≡ 𝑣𝑡ℎ𝑙 𝐽𝑛 = 𝑞𝐷𝑛 𝑑𝑛 𝑑𝑥 Diffusion current results from the random thermal motion of carrier in a concentration gradient
  • 7. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology Problem Assume that in an n-type semiconductor at 300K, the electron concentration varies linearly from 1x1018 to 7x1017 cm-3 over a distance of 0.1 cm. Calculate the diffusion current density if the electron diffusion coefficient is Dn =22.5 cm2/s
  • 8. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology Drift and Diffusion of Carriers 𝐽𝑛,𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑜𝑛 = 𝑞𝐷𝑛 𝑑𝑛 𝑑𝑥 𝐽𝑝,𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑜𝑛 = −𝑞𝐷𝑝 𝑑𝑝 𝑑𝑥 𝐽𝑝 (𝑥) = 𝑞𝑝(𝑥)𝜇𝑝ℰ(𝑥) − 𝑞𝐷𝑝 𝑑𝑝(𝑥) 𝑑𝑥 𝐽𝑛(𝑥) = 𝑞𝑛(𝑥)𝜇𝑛ℰ(𝑥) + 𝑞𝐷𝑛 𝑑𝑛(𝑥) 𝑑𝑥 𝐽𝑡𝑜𝑡𝑎𝑙(𝑥) = 𝐽𝑛(𝑥) + 𝐽𝑝(𝑥) If an electric field is present in addition to a carrier gradient: 𝜙 = 𝑐𝑎𝑟𝑟𝑖𝑒𝑟 𝑓𝑙𝑢𝑥
  • 9. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology • Both carrier mobility µ and diffusion constants D are determined by the frequency of the carrier collisions with phonons or dopant ions and lose their momentum. • They should be related to each other. Einstein Relationship between D and µ • Let’s re-write the expression for Dn using the equipartition theorem • In one dimension, 1 2 𝑚𝑛𝑣𝑡ℎ 2 = 1 2 𝑘𝑇 𝐷𝑛 = 𝑣𝑡ℎ𝑙 = 𝑣𝑡ℎ 𝑣𝑡ℎ𝜏𝑐 = 𝑣𝑡ℎ 2 𝜇𝑛𝑚𝑛 𝑞 = 𝑘𝑇 𝑚𝑛 𝜇𝑛𝑚𝑛 𝑞 𝐷𝑛 = 𝑘𝑇 𝑞 𝜇𝑛 Einstein relation characterize carrier transport by diffusion and drift in a semiconductor
  • 10. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology Problem Minority carriers (holes) are injected into a homogeneous n-type semiconductor sample at one point. An electric field of 50 V/cm is applied across the sample, and the field moves minority carriers a distance of 1 cm in 100 µs. Find the drift velocity and the diffusivity of the minority carriers. The temperature is 300K.
  • 11. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology Optical Absorption • Electron-hole pair is created (a) (Excess Carriers) • Excited electron gives up energy to lattice via scattering events (b) • Electron recombines with a hole in the VB (c) − 𝑑𝐈 𝑥 𝑑𝑥 = α𝐈(𝑥) Solution: 𝐈 𝑥 = 𝐈𝐨𝑒−𝛼𝑥 𝐈𝐭 = 𝐈𝐨𝑒−𝛼𝑙 Transmitted Intensity 𝐸 = 1.24 𝜆
  • 12. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology Semiconductors – Nonequilibrium conditions ➢ Excess electrons in the conduction band and excess holes in the valence band may exist in addition to the thermal-equilibrium concentrations if an external excitation is applied to the semiconductor. ➢ The creation of excess electrons and holes means that the semiconductor is no longer in thermal equilibrium. Carrier generation and recombination: Generation ⇒ process whereby electrons and holes are created Recombination ⇒ process whereby electrons and holes are coincided and lost. Change in temperature or optical excitation can generate excess electrons and holes creating a nonequilibrium condition.
  • 13. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology Generation and Recombination ➢ In thermal equilibrium, pn = ni 2. ➢ In a non-equilibrium situation excess carriers are introduced, in general pn > ni 2. ➢ Excess carrier introduction is called carrier injection ➢ Excess carriers (generation) can be introduced by optical excitation or forward-biasing a pn junction. ➢ Whenever the thermal-equilibrium condition is disturbed, processes exist to restore the system to equilibrium by recombination. Radiative and non-radiative recombination. Direct and indirect recombination.
  • 14. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology Example A 0.46 µm thick sample of GaAs is illuminated with monochromatic light pf hv=2eV. The absorption coefficient is 5.4 x 104 cm-1. The power incident on the sample is 10mW. (a) Find the total energy absorbed by the sample per second (J/s) (b) Find the rate of excess thermal energy given up by the electrons to the lattice before recombination (J/s) (c) Find the number of photons per second given off from recombination events, assuming perfect quantum efficiency.
  • 15. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology Example 𝐈 𝑥 = 𝐈𝐨𝑒−𝛼𝑙 = 10−2 exp(−5𝑥10−4 𝑥0.46𝑥10−4 ) A 0.46 µm thick sample of GaAs is illuminated with monochromatic light pf hv=2eV. The absorption coefficient is 5.4 x 104 cm-1. The power incident on the sample is 10mW. (a) Find the total energy absorbed by the sample per second (J/s) (b) Find the rate of excess thermal energy given up by the electrons to the lattice before recombination (J/s) (c) Find the number of photons per second given off from recombination events, assuming perfect quantum efficiency. = 10−2 𝑒−2.3 = 10−3 𝑊 ⇒ 𝑎𝑏𝑠𝑜𝑟𝑏𝑒𝑑 𝑝𝑜𝑤𝑒𝑟 = 10 − 1 = 9𝑚𝑊 = 9𝑥10−3 Τ 𝐽 𝑠 Fraction of each photon energy unit converted to heat = 2 − 1.43 2 = 0.285 Amount of energy converted to heat per second = 0.285𝑥9𝑥10−3 = 2.57𝑥10−3 Τ 𝐽 𝑠 Amount of radiation accounts for 6.43 mW of power Each photon emitted has 1.6𝑥10−19 𝑥1.43 𝐽 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 # of photons emitted per second is: 6.43𝑥10−3 1.6𝑥10−19𝑥1.43 = 2.81𝑥1016 Τ 𝑝ℎ𝑜𝑡𝑜𝑛𝑠 𝑠
  • 16. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology Generation and Recombination of Electron-Hole Pairs Injection: process of introducing excess carriers in semiconductor Generation: ⚫ Spontaneous Generation [Thermal Energy] ⚫ External Generation [Light] ⚫ Direct band-to-band generation and recombination ⚫ Recombination through allowed energy states within the bandgap, referred to as traps or recombination centers Recombination: ⚫ Spontaneous Recombination [Light or Heat] ⚫ Stimulated Recombination [light/laser] ⚫ Direct band-to-band generation and recombination ⚫ Recombination through allowed energy states within the bandgap, referred to as traps or recombination centers
  • 17. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology Direct Generation and Recombination Under thermal equilibrium the recombination rate must be balanced by the generation rate 𝐺𝑡ℎ = 𝑅𝑡ℎ = 𝛽𝑛𝑛𝑜𝑝𝑛𝑜 n=type s/c equilibrium quantity Electron-hole generation Electron-hole recombination Under thermal equilibrium: ➢ The random generation-recombination of electrons- holes occur continuously due to the thermal excitation. ➢ The random generation-recombination of electrons- holes occur continuously due to the thermal excitation. ➢ Concentration of electrons and holes are independent of time
  • 18. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology Direct Generation and Recombination When excess carriers are introduced into a direct bandgap s/c ➢ High probabilty that direct recombination can occur, no additional momentum is needed ➢ Rate of direct recombination  # of electrons available in the CB and the # of holes in the VB 𝑅 = 𝛽𝑛𝑝 𝛽 𝑖𝑠 𝑡ℎ𝑒 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛𝑎𝑙𝑖𝑡𝑦 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 Under thermal equilibrium the recombination rate must be balanced by the generation rate 𝐺𝑡ℎ = 𝑅𝑡ℎ = 𝛽𝑛𝑛𝑜𝑝𝑛𝑜 n=type s/c equilibrium quantity Electron-hole generation Electron-hole recombination
  • 19. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology ➢ When we shine a light to create electron-hole pair at a rate GL, the carrier concentrations are above their equilibrium values ➢ The recombination and generate rate becomes: 𝑅 = 𝛽𝑛𝑛𝑝𝑛 = 𝛽(𝑛𝑛𝑜 + ∆𝑛)(𝑝𝑛𝑜 + ∆𝑝) 𝐺 = 𝐺𝐿 + 𝐺𝑡ℎ ∆𝑛, ∆𝑝 are the excess carrier conc. ∆𝑛 = 𝑛𝑛 − 𝑛𝑛𝑜, ∆𝑝 = 𝑝𝑛 − 𝑝𝑛𝑜 ∆𝑛 = ∆𝑝 to maintain charge neutrality The net rate of change of the hole concentration: In steady state 𝑑𝑝𝑛 𝑑𝑡 = 0 ⟹ 𝑑𝑝𝑛 𝑑𝑡 = G − R = 𝐺𝐿 + 𝐺𝑡ℎ − R 𝐺𝐿 = 𝑅 − 𝐺𝑡ℎ ≡ 𝑈 U is the net recombination rate 𝑈 = 𝛽(𝑛𝑛𝑜 + 𝑝𝑛𝑜 + ∆𝑝)∆𝑝 Nonequilibrium excess carriers Previous page
  • 20. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology Direct Recombination ➢ For low level injection, ∆𝑝, 𝑝𝑛𝑜 ≪ 𝑛𝑛𝑜 𝑈 = 𝛽(𝑛𝑛𝑜 + 𝑝𝑛𝑜 + ∆𝑝)∆𝑝 𝑈 ≅ 𝛽𝑛𝑛𝑜∆𝑝 = 𝑝𝑛 − 𝑝𝑛𝑜 1 𝛽𝑛𝑛𝑜 The net recombination rate is proportional to the excess minority carrier concentration (U=0 in thermal equilibrium) 1 𝛽𝑛𝑛𝑜 ≡ 𝑙𝑖𝑓𝑒𝑡𝑖𝑚𝑒 𝜏𝑝 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑥𝑐𝑒𝑠𝑠 𝑚𝑖𝑛𝑜𝑟𝑖𝑡𝑦 𝑐𝑎𝑟𝑟𝑖𝑒𝑟𝑠 𝑈 = 𝑝𝑛 − 𝑝𝑛𝑜 𝜏𝑝 The net recombination rate is also determined by the excess minority lifetime
  • 21. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology Direct Recombination Consider light illumination: • Electron-hole pairs are generated with rate GL 𝐺𝐿 = 𝑈 = 𝑝𝑛 − 𝑝𝑛𝑜 𝜏𝑝 ⇒ 𝑝𝑛 = 𝑝𝑛𝑜 + 𝜏𝑝 𝐺𝐿 ∆𝑛 = ∆𝑝 = 𝜏𝑝 𝐺𝐿 At t=0 the light is now switched off, the boundary condition is: 𝑝𝑛(𝑡 = 0) = 𝑝𝑛𝑜 + 𝜏𝑝 𝐺𝐿 𝑑𝑝𝑛 𝑑𝑡 = 𝐺𝑡ℎ − R = −U = − 𝑝𝑛 − 𝑝𝑛𝑜 𝜏𝑝 𝑑𝑝𝑛 𝑑𝑡 = G − R = 𝐺𝐿 + 𝐺𝑡ℎ − R The following equation becomes With a solution: 𝑝𝑛 𝑡 = 𝑝𝑛𝑜 + 𝜏𝑝 𝐺𝐿exp( Τ −𝑡 𝜏𝑝 ) Note: 𝑝𝑛 𝑡 ⇢ ∞ = 𝑝𝑛𝑜
  • 22. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology Decay of excess electrons and holes Assume GaAs doped with 1015 acceptors/cm-3 1014 EHP/cm-3 are created at t=0 We can then calculate the decay of these carriers with time Shown in the figure the decay far a carrier recombination lifetime of 𝜏𝑛 = 𝜏𝑝 = 10ns
  • 23. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology Carrier Lifetime We can use this idea to measure the carrier lifetime using photoconductivity. • Excess carriers generated by the light pulse cause a momentary increase in conductivity • This cause a drop in voltage across the sample when a constant current is passed through it. • Using the decay in the signal is a measure of the lifetime of the excess minority carriers.
  • 24. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology Example A GaAs sample with nno=1014 cm-3 is illuminated with light and 1013 electron-hole pairs/cm3 are created every microsecond. If τn = τp = 2µs, find the change in the minority carrier concentration.
  • 25. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology Example A Si sample with nno=1014 cm-3 is illuminated with light and 1013 electron-hole pairs/cm3 are created every microsecond. If τn = τp = 2µs, find the change in the minority carrier concentration. Before illumination: 𝑝𝑛𝑜 = Τ 𝑛𝑖 2 𝑛𝑛𝑜 = Τ 1𝑥1010 2 1014 = 1𝑥106 𝑐𝑚3 After illumination: 𝑝𝑛 = 𝑝𝑛𝑜 + 𝜏𝑝 𝐺𝐿 = 1𝑥106 + 2𝑥10−6 1013 1𝑥106 = 2𝑥1013𝑐𝑚−3 ∆𝑝𝑛 = 𝜏𝑝 𝐺𝐿 = 2𝑥1013𝑐𝑚−3
  • 26. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology Quasi-Fermi Level Under illumination excess carriers are introduced to the semiconductor ➢ Non equilibrium state: 𝑝𝑛 > 𝑛𝑖 2 ➢ In the non-equilibrium state, the Fermi energy is strictly no longer defined. Under non-equilibrium condition, we can assign individual Fermi level for electrons and holes. ➢ These are called quasi-Fermi levels and the total electron or hole concentration can be determined using these quasi-Fermi levels. ➢ We can write, 𝑛 = 𝑛𝑛𝑜 + ∆𝑛 = 𝑛𝑖𝑒 Τ 𝐸𝐹𝑛−𝐸𝑖 𝑘𝑇 𝑝 = 𝑝𝑛𝑜 + ∆𝑝 = 𝑛𝑖𝑒 Τ 𝐸𝑖−𝐸𝐹𝑝 𝑘𝑇
  • 27. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology Example A GaAs sample with nno=1016 cm-3 is illuminated with light and 1013 electron-hole pairs/cm3 are created every microsecond. If τn = τp = 2ns, find the quasi-Fermi level at room temperature. (𝑛𝑖 = 2.25𝑥106𝑐𝑚−3)
  • 28. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology Indirect Recombination ➢ In real semiconductors, there are some crystal defects that create discrete electronic energy states within the forbidden energy band. Recombination through the defect (trap) states is called indirect recombination ➢ The carrier lifetime due to the recombination through the defect energy state is determined by the Shockley-Read-Hall theory of recombination. Shockley-Read-Hall recombination: ➢ Shockley-Read-Hall theory of recombination assumes that a single trap center exists at an energy Et within the bandgap.
  • 29. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology Indirect Recombination - trapping • For indirect gap materials: small probability of direct electron-hole recombination • Some band gap light but extremely weak • Major recombination events occur via recombination levels with the gap: energy given up as to the lattice as heat • Any impurity or defect can result in recombination center • Excess electron and holes created recombines at Er in 2 steps: (a) hole capture and electron capture • Since Er is filled, first process must be (a) • Impurity or defect centers is usually referred to as traps. • Recombination lifetime depends on the time the first carrier is held before the second one is captured • Effect of recombination and trapping can be measured by a photoconductive decay experiment 𝜎 𝑡 = 𝑞 𝑛 𝑡 𝜇𝑛 + 𝑝(𝑡)𝜇𝑝
  • 30. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology Indirect Recombination ➢ Direct recombination process is unlikely because electrons at the bottom of CB have non-zero momentum wrt holes at the top of the valence band. ➢ Direct transition that conserves both energy and momentum is not possible without a lattice interaction ➢ Indirect transition is via localized energy states in the forbidden gap Indirect Recombination Indirect Recombination: • Electron Capture • Electron Emission • Hole Capture • Hole Emission
  • 31. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology Assume that an n-type semiconductor is uniformly illuminated, producing a uniform excess generation rate G. Show that in steady state the change in the semiconductor conductivity is given by: Example ∆𝜎 = 𝑞(𝜇𝑛 + 𝜇𝑝)𝜏𝑝𝐺
  • 32. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology Thermal Equilibrium Classifications: • Si – Indirect Recombination • GaAs – Direct Recombination Direct vs Indirect
  • 33. EE4353/5353 – Fundamentals of Advanced Semiconductor Technology Assignment #2 A Si sample with 1016 cm-3 donors is optically excited such that 1019 cm-3 electron-hole pairs are generated per second uniformly in the sample. The laser caused the sample to heat up to 450K. Find the quasi-Fermi levels and the change in conductivity of the sample upon shining the light. Electron and hole lifetimes are both 10µs. Dp=12 cm2/s, Dn=36 cm2/s: ni= 1014cm-3 at 450K