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Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-1
Chapter 2 Motion and Recombination
of Electrons and Holes
2.1 Thermal Motion
Average electron or hole kinetic energy 22
1
2
3thmvkT
kg101.926.0
K300JK1038.13331
123
eff
thm
kTv
cm/s103.2m/s103.2 75
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Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-2
2.1 Thermal Motion
Zig-zag motion is due to collisions or scatteringwith imperfections in the crystal. Netthermal velocity is zero. Mean time between collisions ism ~ 0.1ps
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Modern Semiconductor Devices for Integrated Circuits (C. Hu)Slide 2-3
Hot-point Probe can determine sample doing type
Thermoelectr ic Generator
(f rom heat to electricity )
and Cooler (fromelectr icity to refr igeration)
Hot-point Probe
distinguishes Nand P type
semiconductors.
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Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-4
2.2 Drif t
2.2.1 Electron and Hole Mobilities
Drift is the motion caused by an electric field.
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Modern Semiconductor Devices for Integrated Circuits (C. Hu)Slide 2-5
2.2.1 Electron and Hole Mobilities
p is the hole mobility andn is the electron mobility
mpp qvm
p
mp
m
qv
p
mp
pm
q
n
mn
n
m
q
Epv Env
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Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-6
Electron and ho le mob i l it ies o f selected
semiconductors
2.2.1 Electron and Hole Mobilities
Si Ge GaAs InAs
n (cm2/Vs) 1400 3900 8500 30000
p (cm
2
/Vs) 470 1900 400 500
.sV
cm
V/cm
cm/s 2
v =E; has the dimensions ofv/E
Based on the above table alone, which semiconductor and which carriers(electrons or holes) are attractive for applications in high-speed devices?
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Modern Semiconductor Devices for Integrated Circuits (C. Hu)Slide 2-8
There are two main causes of carrier scattering:1. Phonon Scattering2. Ionized-Impurity (Coulombic) Scattering
2/3
2/1
11
TTTvelocitythermalcarrierdensityphonon
phononphonon
Phonon scatter ingmobility decreases when temperature rises:
= q/m
vthT1/2
2.2.2 Mechanisms of Carrier Scattering
T
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Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-9
_+
- -Electron
Boron IonElectron
ArsenicIon
Impurity (Dopant)-I on Scatter ing or Coulombic Scatter ing
dada
thimpurity
NN
T
NN
v
2/33
There is less change in the direction of travel if the electron zips bythe ion at a higher speed.
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Modern Semiconductor Devices for Integrated Circuits (C. Hu)Slide 2-10
Total Mobil i ty
1E14 1E15 1E16 1E17 1E18 1E19 1E20
0
200
400
600
800
1000
1200
1400
1600
Holes
Electrons
Mobility(cm2
V-1s
-1)
Total Impurity Concenration (atoms cm-3)N
a+ N
d(cm-3)
impurityphonon
impurityphonon
111
111
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Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-11
Temperature Effect on Mobil i ty
101 5
Question:WhatNdwill makedn/dT= 0 at room
temperature?
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Modern Semiconductor Devices for Integrated Circuits (C. Hu)Slide 2-12
Velocity Saturation
When the kinetic energy of a carrier exceeds a critical value, itgenerates an optical phonon and loses the kinetic energy.
Therefore, the kinetic energy is capped at large E and the
velocity does not rise above a saturation velocity, vsat.Velocity saturationhas a deleterious effect on device speed asshown in Ch. 6.
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Modern Semiconductor Devices for Integrated Circuits (C. Hu)Slide 2-13
2.2.3 Drift Current and Conductivity
Jp
n
E
unitarea
Jp= qpv A/cm2or C/cm2sec
Ifp = 1015cm-3andv = 104 cm/s, thenJp= 1.610
-19C 1015cm-3104cm/s
= 22 A/cm1.6cmC/s6.1
EXAMPLE:
Hole current density
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Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-14
Jp,drift = qpv = qppE
Jn,drift =qnv = qnnE
Jdrift = Jn,drift + Jp,drift = E=(qnn+qpp)E
conductivity(1/ohm-cm) of a semiconductor is= qnn+ qpp
2.2.3 Drift Current and Conductivity
1/=is resistivity (ohm-cm)
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Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-15
N-type
P-type
Relationship between Resistivity and Dopant Density
= 1/
DOPANTDE
NSITYcm-3
RESISTIVITY (cm)
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Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-16
EXAMPLE: Temperature Dependence of Resistance
(a) What is the resistivity () of silicon dopedwith 1017cm-3 of arsenic?
Solution:
(a) Using the N-type curve in the previous
figure, we find that= 0.084 -cm.
(b) What is the resistance (R) of a piece of this
silicon material 1m long and 0.1 m2in cross-
sectional area?
(b) R =L/A = 0.084 -cm 1 m / 0.1 m2
= 0.084 -cm 10-4cm/ 10-10cm2= 8.4 10-4
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Modern Semiconductor Devices for Integrated Circuits (C. Hu)Slide 2-17
EXAMPLE: Temperature Dependence of Resistance
By what factor will R increase or decrease from
T=300 K to T=400 K?
Solution:The temperature dependent factor in (and
therefore) is n. From the mobility vs. temperature
curve for 1017cm-3, we find that n decreases from 770
at 300K to 400 at 400K. As a result, R increases by
93.1400770
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Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-18
2.3 Diffusion Current
Particles diffuse from a higher-concentration locationto a lower-concentration location.
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Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-19
2.3 Di ffusion Current
dx
dn
qDJ ndiffusionn , dx
dp
qDJ pdiffusionp ,
D is called the diffusion constant. Signs explained:
n p
x x
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Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-20
Total Current Review of Four Cur rent Components
Jn= Jn,drift+ Jn,diffusion= qnnE+dx
dnqDn
Jp= Jp,drift+ Jp,diffusion= qppEdx
dpqDp
JTOTAL = Jn+ Jp
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Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-21
2.4 Relation Between the Energy
Diagram and V, E
E(x)=dx
dE
q
1
dx
dE
q
1
dx
dV vc
Ecand Evvary in the opposite
direction from the voltage. That
is, Ecand Evare higher where
the voltage is lower.
V(x)
0.7V
x0
x
Ef(x)
E
0.7V
-
+
Ec(x)
Ev(x)
N-+
0.7eV
N type Si
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Modern Semiconductor Devices for Integrated Circuits (C. Hu)Slide 2-22
2.5 Einstein Relationship between D and
dx
dEe
kT
N
dx
dn ckT/)EE(c fc
dx
dE
kT
n c
kT/)EE(
cfceNn
Consider a piece of non-uniformly doped semiconductor.
Ev(x)
Ec(x)
Ef
n-type semiconductor
Decreasing donor concentration
N-type semiconductor
EqkT
n
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Modern Semiconductor Devices for Integrated Circuits (C. Hu)Slide 2-23
2.5 Einstein Relationship between D and
EqkT
n
dx
dn
0dx
dnqDqnJ nnn E at equilibrium.
EkTqDqnqn nn 0
nn
q
kTD
These are known as the Einstein relationship.
pp
q
kTD Similarly,
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Modern Semiconductor Devices for Integrated Circuits (C. Hu)Slide 2-24
EXAMPLE: Di ffusion Constant
What is the hole diffusion constant in a piece of
silicon with p = 410 cm2 V-1s-1?
Solution:
Remember: kT/q = 26 mV at room temperature.
/scm11sVcm410)mV26( 2112
pp
q
kTD
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Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-25
2.6 Electron-Hole Recombination
The equilibrium carrier concentrations are denoted with
n0andp0.
The total electron and hole concentrations can be different
fromn0andp0.
The differences are called theexcess carr ier
concentrationsnandp.
'0 nnn
'0 ppp
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Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-26
Charge Neutrali ty
Charge neutrality is satisfied at equilibrium(n=p= 0).When a non-zeron is present, an equalp may
be assumed to be present to maintain charge
equality and vice-versa.If charge neutrality is not satisfied, the net chargewill attract or repel the (majority) carriers throughthe drift current until neutrality is restored.
'p'n
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Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-27
Recombination L ifetime
Assume light generatesn andp. If the light issuddenly turned off, n andp decay with timeuntil they become zero.
The process of decay is calledrecombination.The time constant of decay is therecombinationtime orcarr ier li fetime, .Recombination is natures way of restoring
equilibrium (n= p= 0).
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Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-28
ranges from 1ns to 1ms in Si anddepends on
the density of metal impurities (contaminants)such as Au and Pt.Thesedeep trapscapture electrons and holes tofacilitate recombination and are calledrecombination centers.
Recombination L ifetime
Ec
Ev
DirectRecombinationis unfavorable insilicon
Recombinationcenters
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Direct and Indirect Band Gap
Direct band gapExample: GaAs
Direct recombination is efficientas k conservation is satisfied.
Indirect band gap
Example: Si
Direct recombination is rare as kconservation is not satisfied
Trap
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Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-30
n
dt
nd
dt
pdpn
dt
nd
Rate of recombination (s-1cm-3)
pn
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Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-31
EXAMPLE: Photoconductors
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/scm3. The recombination lifetime is
10s. What are (a) p0, (b) n0, (c) p,(d) n,(e) p ,
(f) n, and (g) the np product?
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Modern Semiconductor Devices for Integrated Circuits (C. Hu)Slide 2-32
EXAMPLE: Photoconductors
Solution:
(a) What is p0?
p0= Na = 1015 cm-3
(b) What is n0 ?n0= ni
2/p0= 105 cm-3
(c) What is p?
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
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Modern Semiconductor Devices for Integrated Circuits (C. Hu)Slide 2-33
EXAMPLE: Photoconductors
(d) What is n?n= p= 1015 cm-3
(e) What is p?
p= p0+ p= 1015cm-3 + 1015cm-3= 21015cm-3
(f) What is n?
n = n0+ n= 105cm-3 + 1015cm-3~ 1015cm-3 since n0> ni2 = 1020 cm-6.
The np product can be very different from ni2.
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Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-34
2.7 Thermal Generation
Ifn is negative, there are fewerelectrons than the equilibrium value.
As a result, there is a net rate ofthermal generation at the rate of |n|/.
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Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-35
2.8 Quasi-equi l ibrium and Quasi-Fermi Levels
Whenevern = p 0, np ni2. We would like to preserve
and use the simple relations:
But these equations lead tonp = ni2. The solution is to introduce
twoquasi-Fermi levelsEfnandEfp such that
kTEEc
fceNn /)(
kTEE
vvfeNp
/)(
kTEE
cfnceNn
/)(
kTEE
vvfpeNp
/)(
Even when electrons and holes are not at equilibrium, within
each groupthe carriers can be at equilibrium. Electrons areclosely linked to other electrons but only loosely to holes.
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Modern Semiconductor Devices for Integrated Circuits (C. Hu)Slide 2-37
Now assume n= p= 1015cm-3.(b) Find Efnand Efp.
n = 1.011017cm-3=
EcEfn= kT ln(Nc/1.011017cm-3)
= 26 meV ln(2.81019cm-3/1.011017cm-3)= 0.15 eV
Efnis nearly identical toEfbecause nn0 .
kTEE
cfnceN
/)(
EXAMPLE: Quasi-Fermi Levels and Low-Level I njection
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Modern Semiconductor Devices for Integrated Circuits (C. Hu) Slide 2-38
EXAMPLE: Quasi-Fermi Levels
p = 1015 cm-3=
EfpEv= kT ln(Nv/1015cm-3)
= 26 meV ln(1.041019cm-3/1015cm-3)= 0.24 eV
kTEEv
vfpeN /)(
Ec
Ev
Efp
Ef Efn
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Modern Semiconductor Devices for Integrated Circuits (C. Hu)Slide 2-39
2.9 Chapter Summary
dx
dnqDJ ndiffusionn ,
dx
dpqDJ pdiffusionp ,
nnq
kTD
ppq
kTD
Eppv
Ennv
Epdriftp qpJ ,
Endriftn qnJ ,
-
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Modern Semiconductor Devices for Integrated Circuits (C. Hu)Slid 2 40
2.9 Chapter Summary
is the recombination lifetime.n andp are the excess carr ier concentrations.
n = n0+ n
p = p0+ p
Charge neutrality requiresn= p.
rate of recombination = n/= p/
Efn
andEfp
are the quasi-Fermi levels of electrons andholes.
kTEE
vvfpeNp
/)(
kTEE
cfnceNn
/)(
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