some calculations at room temperature kt = 0.0259ev at room temperature n i for si = 1.5 x 10 10 /cm...
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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
Electrons and holes in intrinsic [no impurities] semiconductor materials
Electrons and holes in extrinsic [“doped”] semiconductor materials
Some Terminology and Definitions
Electron and Hole Concentrations at Equilibrium
Calculating Concentrations
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
Fermi-Dirac plus Energy Band
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 – EV = 1.1eVWhat is the value of EC – 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 – EV = 1.1eVWhat is the value of Ei – EF if Na = 2 x 1015 /cm3 at T= 300K
The band gap of Si at room temp is 1.1eV or EC – EV = 1.1eVWhat is the value of EF – Ei if Nd = 2 x 1015 /cm3 at T= 300K
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?
Si with 1015/cm3 donor impurity
Conductivity
Excess Carriers
Photoluminescence
Diffusion of Carriers
Drift and Diffusion
Diffusion Processesn(x)
n1 n2
x0
x0 - l x0 + l
𝜑𝑛 (𝑥0 )= 𝑙2𝑡
(𝑛1−𝑛2)
Since the mean free path is a small differential,we can write:
(𝑛1−𝑛2 )=𝑛 (𝑥 )−𝑛 (𝑥+∆𝑥 )
∆ 𝑥𝑙
Where x is at the center of segment 1 and ∆ 𝑥=𝑙In the limit of small∆ 𝑥
𝜑𝑛 (𝑥 )= 𝑙2
2𝑡lim∆ 𝑥→0
𝑛 (𝑥 )−𝑛 (𝑥+∆ 𝑥 )∆ 𝑥
= 𝑙2
2 𝑡𝑑𝑛(𝑥 )𝑑𝑥
or
Diffusion Current Equations
Combine Drift and Diffusion
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
Energy Bands when there is an Electric Field
𝑉 (𝑥 )=𝐸 (𝑥 )−𝑞
¿𝑑𝑉 (𝑥 )𝑑𝑥
E(x) ¿𝑑𝑉 (𝑥 )𝑑𝑥
=− 𝑑𝑑𝑥 [ 𝐸𝑖
−𝑞 ]= 1𝑞 𝑑 𝐸𝑖
𝑑𝑥E(x)
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:
𝐽𝑝 (𝑥 )=0=𝑞𝜇𝑝𝑝 (𝑥 )𝐸 (𝑥 )−𝑞𝐷𝑝
𝑑𝑝 (𝑥 )𝑑𝑥
¿𝐷𝑝
𝜇𝑝
1𝑝(𝑥 )
𝑑𝑝(𝑥)𝑑𝑥
Using for p(x) 𝑝0=𝑛𝑖𝑒(𝐸𝑖−𝐸𝐹 ) /𝑘𝑇
¿𝐷𝑝
𝜇𝑝
1𝑘𝑇 (𝑑𝐸 𝑖
𝑑𝑥−𝑑𝐸𝐹
𝑑𝑥 ) The equilibrium Fermi Level does not vary with x.
E(x)
E(x)
0qE(x)
Finally:𝐷𝑝
𝜇𝑝
=𝑘𝑇𝑞
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
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:
𝑁 𝑑=𝑁0𝑒−𝑎𝑥 Find an expression for E(x) and evaluate when a=1(μm)-1
Sketch band Diagram
Diffusion & Recombination
x x + Δx
Jp(x) Jp (x + Δx)
Rate of Hole buildup =
Increase in hole concIn differential volumePer unit time
- RecombinationRate
𝜕𝑝𝜕𝑡 𝑥→𝑥+∆ 𝑥
= 1𝑞𝐽𝑝 (𝑥 )− 𝐽𝑝 (𝑥+∆ 𝑥 )
∆𝑥−𝛿𝑝𝜏𝑝
𝜕𝛿𝑝𝜕𝑡
=− 1𝑞𝜕 𝐽𝑝𝜕𝑥
−𝛿𝑝𝜏𝑝
𝜕𝛿𝑛𝜕𝑡
=− 1𝑞𝜕 𝐽𝑛𝜕 𝑥
−𝛿𝑛𝜏𝑛
If current is exclusively Diffusion
𝐽𝑛 (𝑑𝑖𝑓𝑓 )=𝑞𝐷𝑛𝜕 𝛿𝑛𝜕 𝑥
𝜕𝛿𝑛𝜕𝑡
=𝐷𝑛𝜕2 𝛿𝑛𝜕 𝑥2
−𝛿𝑛𝜏𝑛
And the same for holes
And Finally, the steady-stateDetermining Diffusion Length
𝜕𝛿𝑛𝜕𝑡
=𝐷𝑛𝜕2 𝛿𝑛𝜕 𝑥2
−𝛿𝑛𝜏𝑛
=0 𝜕2𝛿𝑛𝜕𝑥2
= 𝛿𝑛𝐷𝑛𝜏𝑛
= 𝛿𝑛𝐿❑2 𝐿𝑛=√𝐷𝑛𝜏𝑛
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