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EECS130 Integrated Circuit Devices
Professor Ali Javey8/30/2007
Semiconductor FundamentalsLecture 2
Read: Chapters 1 and 2
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Last Lecture: Energy Band Diagram
Conduction band Ec
Ev
Eg Band gap
Valence band
• Energy band diagram shows the bottom edge of conduction band, Ec , and top edge of valence band, Ev .
• Ec and Ev are separated by the band gap energy, Eg .
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Temperature Effect on Band Gap
Decreasing atomic separation
Ener
gy
p
s
isolated atoms lattice spacing
valence band
conduction band
How does the band gap change with temperature?
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Measuring the Band Gap Energy by Light Absorption
photons
photon energy: h v > Eg
Ec
Ev
Eg
electron
hole
Bandgap energies of selected semiconductors
• Eg can be determined from the minimum energy (hν) of photons that are absorbed by the semiconductor.
Material PbTe Ge Si GaAs GaP DiamondE g (eV) 0.31 0.67 1.12 1.42 2.25 6.0
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Semiconductors, Insulators, and Conductors
• Totally filled bands and totally empty bands do not allow
• Metal conduction band is half-filled.
E c
Ev
Eg=1.1 eV
E c
Eg= 9 eV empty
Si (Semiconductor) SiO2
(Insulator) Conductor
Ecfilled
Top ofconduction band
E v
current flow. (Just as there is no motion of liquid in a totally filled or totally empty bottle.).
• Semiconductors have lower Eg 's than insulators and can be doped.
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Donor and Acceptor Levels in the Band Model
Conduction Band Ec
EvValence Band
Donor Level
Acceptor Level
Ed
Ea
Donor ionization energy
Acceptor ionization energy
Ionization energy of selected donors and acceptors in siliconAcceptors
Dopant Sb P As B Al InIonization energy, E c –E d or E a –E v (meV) 39 44 54 45 57 160
Donors
Hydrogen: Eionm0 q4
13.6 eV==8ε0
2h2
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Dopants and Free Carriers
Dopant ionizationenergy ~50meV (very low).
Donorsn-type
Acceptorsp-type
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Effective Mass
In an electric field, , an electron or a hole accelerates.
Electron and hole effective massesSi Ge GaAs GaP
m n /m 0 0.26 0.12 0.068 0.82m p /m 0 0.39 0.30 0.50 0.60
electrons
holes
Remember :F=ma=-qE
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Density of StatesE
g c
g v
Ec
Ev
g(E)
Ec
Ev
ΔΕ
⎟⎠⎞
⎜⎝⎛
⋅⋅ΔΔ
≡ 3cmeV1
volumein states ofnumber )(
EEEgc
( )
2)( 32
**
hEEmm
Eg cnnc π
−≡
( )
2)( 32
**
hEEmm
Eg vppv π
−≡
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Thermal Equilibrium
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Thermal Equilibrium An Analogy for Thermal Equilibrium
• There is a certain probability for the electrons in theconduction band to occupy high-energy states under the agitation of thermal energy (vibrating atoms, etc.)
Dish
Vibrating Table
Sand particles
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At E=EF , f(E)=1/2
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Assume the two extremes:
High Energy (Large E): E-Ef >> kT, f(E) 0Low Energy (Small E): E-Ef << kT, f(E) 1
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Effect of T on f(E)
T=0K
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Question
• If f(E) is the probability of a state being occupied by an electron, what is the probability of a state being occupied by a hole?
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n = electron density : number of unbound electrons / cm3
p = hole density : number of holes / cm3
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Nc is called the effective density of states (of the conduction band) .
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Nv is called the effective density of states of the valence band.
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Intrinsic Semiconductor
• Extremely pure semiconductor sample containing an insignificant amount of impurity atoms.
n = p = ni
Material Ge Si GaAsEg (eV) 0.67 1.12 1.42ni (1/cm3) 2 x 1013 1 x 1010 2 x 106
Ef lies in the middle of the band gap
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intrinsic
n-type
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Remember: the closer Ef moves up to Ec , the larger n is; the closer Ef moves down to Ev , the larger p is.For Si, Nc = 2.8×1019cm-3 and Nv = 1.04×1019 cm-3.
Ec
Ev
EfEc
EvEf
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Example: The Fermi Level and Carrier Concentrations
kTEEc
fceNn /)( −−=
( ) ( ) eV 614.010/108.2ln026.0ln 1719 =×==− nNkTEE cfc
E cEf
E v
0.146 eV
Where is Ef for n =1017 cm-3? Solution:
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The np Product and the Intrinsic Carrier Concentration
• In an intrinsic (undoped) semiconductor, n = p = ni .
kTEvci
geNNn 2/−=
2innp =
kTEEc
fceNn /)( −−= kTEEv
vfeNp /)( −−=andMultiply
kTEvc
kTEEvc
gvc eNNeNNnp //)( −−− ==
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Question: What is the hole concentration in an N-type semiconductor with 1015 cm-3 of donors?
Solution: n = 1015 cm-3.
After increasing T by 60°C, n remains the same at 1015 cm-3 while p increases by about a factor of 2300 because .
Question: What is n if p = 1017cm-3 in a P-type silicon wafer?
Solution:
EXAMPLE: Carrier Concentrations
3-5315
-3202
cm10cm10cm10
=≈= −nnp i
kTEi
gen /2 −∝
3-3317
-3202
cm10cm10cm10
=≈= −pnn i
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EXAMPLE: Complete ionization of the dopant atomsNd = 1017 cm-3 and Ec -Ed =45 meV. What fraction of the donors are not ionized?
Solution: First assume that all the donors are ionized.
Ec
Ef
Ev
146 meVEd
45meV
Probability of non-ionization ≈ 02.01
11
1meV26/)meV)45146((/)( =
+=
+ −− ee kTEE fd
Therefore, it is reasonable to assume complete ionization, i.e., n = Nd .
meV146cm10 317 −=⇒== −cfd EENn
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Doped Si and Charge
• What is the net charge of your Si when it is electron and hole doped?
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Bond Model of Electrons and Holes (Intrinsic Si)• Silicon crystal in
a two-dimensionalrepresentation.
Si Si Si
Si Si Si
Si Si Si
Si Si Si
Si Si Si
Si Si Si
Si Si Si
Si Si Si
Si Si Si
• When an electron breaks loose and becomes a conduction electron, a hole is also created.
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Dopants in Silicon
Si Si Si
Si Si
Si Si Si
Si Si Si
Si Si
Si Si Si
As B
• As (Arsenic), a Group V element, introduces conduction electrons and creates N-type silicon,
• B (Boron), a Group III element, introduces holes and creates P-type silicon, and is called an acceptor.
• Donors and acceptors are known as dopants.
and is called a donor.
N-type Si P-type Si
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General Effects of Doping on n and p
Charge neutrality: da NpNn −−+ +_= 0
da NpNn −−+ = 0
Assuming total ionization of acceptors and donors:
aN_
: number of ionized acceptors /cm3
dN +: number of ionized donors /cm3
aN : number of ionized acceptors /cm3
dN +: number of ionized donors /cm3
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I. (i.e., N-type)
If ,
II. (i.e., P-type)
If ,
ad NNn −=
nnp i2=
iad nNN >>−
ad NN >> dNn = di Nnp 2=and
ida nNN >>− da NNp −=
pnn i2=
da NN >> aNp = ai Nnn 2=and
General Effects of Doping on n and p
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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 ?
(a)
(b) Na = 2×1016 + 6×1016 = 8×1016 cm-3 > Nd !
+ + + + + + + + + + + +. . . . . .
. . . . . . . . . . . - - - - - - - -
. . . . . .
. . . . . .
316 cm104 −×=−= ad NNn3316202 cm105.2104/10/ −×=×== nnp i
3161616 cm102106108 −×=×−×=−= da NNp3316202 cm105102/10/ −×=×== pnn i
Nd = 6×1016 cm-3
Na = 2×1016 cm-3
n = 4×1016 cm-3
Nd = 6×1016 cm-3
Na = 8×1016 cm-3
p = 2×1016 cm-3
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Carrier Concentrations at Extremely High and Low Temperatures
intrinsic regime
n = Nd
freeze-out regime
lnn
1/Thigh temp.
roomtemperature
cryogenictemperature
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Infrared Detector Based on Freeze-out
To image the black-body radiation emitted by tumors requires a photodetector that responds to hν’s around 0.1 eV. In doped Si operating in the freeze-out mode, conduction electrons are created when the infrared photons provide the energy to ionized the donor atoms.
photon Ec
Ev
electron
Ed
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Chapter Summary
Energy band diagram. Acceptor. Donor. mn , mp . Fermi function. Ef .
kTEEc
fceNn /)( −−=kTEE
vvfeNp /)( −−=
ad NNn −=
da NNp −=
2innp =