electronics ii vlsi design fall 2013
DESCRIPTION
ELECTRONICS II VLSI DESIGN Fall 2013. The Hydrogen Atom. Allowable States for the Electron of the Hydrogen Atom. The Periodic Table. From Single Atoms to Solids. Energy bands and energy gaps Silicon. Band Structures at ~0K. Atomic Bonds. - PowerPoint PPT PresentationTRANSCRIPT
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ELECTRONICS II VLSI DESIGN
Fall 2013
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The Hydrogen Atom
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Allowable States for the Electron of the Hydrogen Atom
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The Periodic Table
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From Single Atoms to Solids
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Energy bands and energy gaps
Silicon
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Band Structures at ~0K
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Atomic Bonds
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Electrons and holes in intrinsic [no impurities] semiconductor materials
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Electrons and holes in extrinsic [βdopedβ] semiconductor materials
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Some Terminology and Definitions
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Electron and Hole Concentrations at Equilibrium
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Calculating Concentrations
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Some CalculationsAt room temperature kT = 0.0259eVAt room temperature ni for Si = 1.5 x 1010/cm3
Solve this equation for E = EF
Let T = 300K and EF = 0.5eV plot f(E) for 0 < E < 1
Let find f(E<EF) and f(E>EF)
EC
EV
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Fermi-Dirac plus Energy Band
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More Calculations
If Na = 2 x 1015 /cm3 find po and no
The band gap of Si at room temp is 1.1eV or EC β EV = 1.1eVWhat is the value of EC β EF for intrinsic Si at T= 300K
At room temperature kT = 0.0259eVAt room temperature ni for Si = 1.5 x 1010/cm3
The band gap of Si at room temp is 1.1eV or EC β EV = 1.1eVWhat is the value of Ei β EF if Na = 2 x 1015 /cm3 at T= 300K
The band gap of Si at room temp is 1.1eV or EC β EV = 1.1eVWhat is the value of EF β Ei if Nd = 2 x 1015 /cm3 at T= 300K
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Intrinsic Carrier ConcentrationsSEMICONDUCTOR ni
Ge 2.5 x 1013/cm3
Si 1.5 x 1010/cm3
GaAs 2 x 106/cm3
Which element has the largest Eg?
What is the value of pi for each of these elements?
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Si with 1015/cm3 donor impurity
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Conductivity
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Excess Carriers
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Photoluminescence
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Diffusion of Carriers
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Drift and Diffusion
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Diffusion Processesn(x)
n1 n2
x0
x0 - l x0 + l
ππ (π₯0 )= π2π‘ (π1βπ2)
Since the mean free path is a small differential,we can write:
(π1βπ2 )=π (π₯ )βπ (π₯+βπ₯ )
β π₯ π
Where x is at the center of segment 1 and β π₯=πIn the limit of smallβ π₯ππ (π₯ )= π2
2π‘ limβ π₯β0
π (π₯ )βπ (π₯+β π₯ )β π₯ =
π22 π‘
ππ(π₯ )ππ₯
or
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Diffusion Current Equations
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Combine Drift and Diffusion
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Drift and Diffusion Currents
E(x)
n(x)
p(x)
Electron drift
Hole drift
Electron & HoleDrift current
Electron diffusion
Hole diffusion
Electron Diff current
Hole Diff current
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Energy Bands when there is an Electric Field
π (π₯ )=πΈ (π₯ )βπ
ΒΏππ (π₯ )ππ₯E(x) ΒΏ
ππ (π₯ )ππ₯ =β π
ππ₯ [ πΈπ
βπ ]= 1π π πΈπ
ππ₯E(x)
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The Einstein Relation
At equilibrium no net current flows so any concentration gradient would be accompanied by an electric field generated internally. Set the hole current equal to 0:
π½π (π₯ )=0=ππππ (π₯ )πΈ (π₯ )βππ·πππ (π₯ )ππ₯
ΒΏπ·π
ππ
1π(π₯ )
ππ (π₯)ππ₯ Using for p(x) π0=πππ
(πΈπβπΈπΉ ) /ππ
ΒΏπ·π
ππ
1ππ (ππΈ π
ππ₯ βππΈπΉ
ππ₯ ) The equilibrium Fermi Level does not vary with x.
E(x)
E(x)
0qE(x)
Finally:π·π
ππ=ππ
π
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D and mu
Dn
(cm2/s)Dp mun
(cm2/V-s)mup
Ge 100 50 3900 1900
Si 35 12.5 1350 480
GaAs 220 10 8500 400
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Message from Previous AnalysisAn important result of the balance between drift and diffusion at equilibrium is that built-in fields accompany gradients in Ei. Such gradients in the bands at equilibrium (EF constant) can arise when the band gap varies due to changes in alloy composition. More commonly built-in fields result from doping gradients. For example a donor distribution Nd(x) causes a gradient in no(x) which must be balanced by a built-in electric field E(x).
Example: An intrinsic sample is doped with donors from one side such that:
π π=π0πβππ₯ Find an expression for E(x) and evaluate when a=1(ΞΌm)-1
Sketch band Diagram
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Diffusion & Recombination
x x + Ξx
Jp(x) Jp (x + Ξx)
Rate of Hole buildup =
Increase in hole concIn differential volumePer unit time
- RecombinationRate
ππππ‘ π₯βπ₯+β π₯
= 1ππ½π (π₯ )β π½π (π₯+β π₯ )
βπ₯ β πΏπππ
ππΏπππ‘ =β 1π
π π½πππ₯ β
πΏπππ
ππΏπππ‘ =β 1π
π π½ππ π₯ β
πΏπππ
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If current is exclusively Diffusion
π½π (ππππ )=ππ·ππ πΏππ π₯
ππΏπππ‘ =π·π
π2 πΏπππ₯2
β πΏπππAnd the same for holes
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And Finally, the steady-stateDetermining Diffusion Length
ππΏπππ‘ =π·π
π2 πΏππ π₯2
β πΏπππ=0 π2πΏπ
ππ₯2=
πΏππ·πππ
=πΏππΏβ2 πΏπ=βπ·πππ