Presentation is loading. Please wait.

Presentation is loading. Please wait.

ELECTRONICS II VLSI DESIGN Fall 2013

Similar presentations


Presentation on theme: "ELECTRONICS II VLSI DESIGN Fall 2013"— Presentation transcript:

1 ELECTRONICS II VLSI DESIGN Fall 2013

2 The Hydrogen Atom

3 Allowable States for the Electron of the Hydrogen Atom

4 The Periodic Table

5 From Single Atoms to Solids

6 Energy bands and energy gaps Silicon

7 Band Structures at ~0K

8 Atomic Bonds

9 Electrons and holes in intrinsic [no impurities] semiconductor materials

10 Electrons and holes in extrinsic [“doped”] semiconductor materials

11

12 Some Terminology and Definitions

13 Electron and Hole Concentrations at Equilibrium

14 Calculating Concentrations

15 Some Calculations At room temperature kT = 0.0259eV
At room temperature ni for Si = 1.5 x 1010/cm3 Solve this equation for E = EF 𝑓 𝐸 = 𝑒 (𝐸− 𝐸 𝐹 )/𝑘𝑇 Let 𝑇→0𝐾 find f(E<EF) and f(E>EF) Let T = 300K and EF = 0.5eV plot f(E) for 0 < E < 1 EC EV

16 Fermi-Dirac plus Energy Band

17 More Calculations At room temperature kT = 0.0259eV
At room temperature ni for Si = 1.5 x 1010/cm3 If Na = 2 x 1015 /cm3 find po and no The band gap of Si at room temp is 1.1eV or EC – EV = 1.1eV What is the value of EC – EF for intrinsic Si at T= 300K The band gap of Si at room temp is 1.1eV or EC – EV = 1.1eV What 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 = 1.1eV What is the value of EF – Ei if Nd = 2 x 1015 /cm3 at T= 300K

18 Intrinsic Carrier Concentrations
SEMICONDUCTOR 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?

19 Si with 1015/cm3 donor impurity

20 Conductivity

21 Excess Carriers

22 Photoluminescence

23 Diffusion of Carriers

24 Drift and Diffusion

25 Diffusion Processes 𝜑 𝑛 𝑥 0 = 𝑙 2 𝑡 ( 𝑛 1 − 𝑛 2 ) n(x) n1 n2
𝜑 𝑛 𝑥 0 = 𝑙 2 𝑡 ( 𝑛 1 − 𝑛 2 ) n(x) n1 n2 Since the mean free path is a small differential, we can write: 𝑛 1 − 𝑛 2 = 𝑛 𝑥 −𝑛(𝑥+∆𝑥) ∆𝑥 𝑙 x0 Where x is at the center of segment 1 and ∆𝑥= 𝑙 x0 - l x0 + l In the limit of small ∆𝑥 𝜑 𝑛 𝑥 = 𝑙 𝑡 lim ∆𝑥→0 𝑛 𝑥 −𝑛 𝑥+∆𝑥 ∆𝑥 = 𝑙 𝑡 𝑑𝑛(𝑥) 𝑑𝑥 𝑙 𝑡 ≡ 𝐷 𝑛 or 𝐷 𝑝

26 Diffusion Current Equations

27 Combine Drift and Diffusion

28 Drift and Diffusion Currents
Electron drift Hole drift Electron & Hole Drift current E(x) n(x) Electron diffusion Hole diffusion Electron Diff current Hole Diff current p(x)

29 Energy Bands when there is an Electric Field
𝑉 𝑥 = 𝐸(𝑥) −𝑞 = 𝑑𝑉(𝑥) 𝑑𝑥 = 𝑑𝑉(𝑥) 𝑑𝑥 =− 𝑑 𝑑𝑥 𝐸 𝑖 −𝑞 = 1 𝑞 𝑑 𝐸 𝑖 𝑑𝑥 E(x) E(x)

30 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: 𝐽 𝑝 𝑥 =0=𝑞 𝜇 𝑝 𝑝 𝑥 𝐸 𝑥 −𝑞 𝐷 𝑝 𝑑𝑝(𝑥) 𝑑𝑥 = 𝐷 𝑝 𝜇 𝑝 1 𝑝(𝑥) 𝑑𝑝(𝑥) 𝑑𝑥 E(x) Using for p(x) 𝑝 0 = 𝑛 𝑖 𝑒 𝐸 𝑖 − 𝐸 𝐹 /𝑘𝑇 qE(x) = 𝐷 𝑝 𝜇 𝑝 1 𝑘𝑇 𝑑 𝐸 𝑖 𝑑𝑥 − 𝑑 𝐸 𝐹 𝑑𝑥 E(x) The equilibrium Fermi Level does not vary with x. 𝐷 𝑝 𝜇 𝑝 = 𝑘𝑇 𝑞 Finally:

31 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

32 Message from Previous Analysis
An 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 = 𝑁 0 𝑒 −𝑎𝑥 Find an expression for E(x) and evaluate when a=1(μm)-1 Sketch band Diagram

33 Diffusion & Recombination
Jp(x) Jp (x + Δx) x x + Δx Increase in hole conc In differential volume Per unit time Rate of Hole buildup Recombination Rate = - 𝜕𝑝 𝜕𝑡 𝑥→𝑥+∆𝑥 = 1 𝑞 𝐽 𝑝 𝑥 − 𝐽 𝑝 𝑥+∆𝑥 ∆𝑥 − 𝛿𝑝 𝜏 𝑝 𝜕𝛿𝑝 𝜕𝑡 =− 1 𝑞 𝜕 𝐽 𝑝 𝜕𝑥 − 𝛿𝑝 𝜏 𝑝 𝜕𝛿𝑛 𝜕𝑡 =− 1 𝑞 𝜕 𝐽 𝑛 𝜕𝑥 − 𝛿𝑛 𝜏 𝑛

34 If current is exclusively Diffusion
𝐽 𝑛 𝑑𝑖𝑓𝑓 =𝑞 𝐷 𝑛 𝜕𝛿𝑛 𝜕𝑥 𝜕𝛿𝑛 𝜕𝑡 = 𝐷 𝑛 𝜕 2 𝛿𝑛 𝜕 𝑥 2 − 𝛿𝑛 𝜏 𝑛 And the same for holes

35 And Finally, the steady-state Determining Diffusion Length
𝜕𝛿𝑛 𝜕𝑡 = 𝐷 𝑛 𝜕 2 𝛿𝑛 𝜕 𝑥 2 − 𝛿𝑛 𝜏 𝑛 =0 𝜕 2 𝛿𝑛 𝜕 𝑥 2 = 𝛿𝑛 𝐷 𝑛 𝜏 𝑛 = 𝛿𝑛 𝐿 2 𝐿 𝑛 = 𝐷 𝑛 𝜏 𝑛


Download ppt "ELECTRONICS II VLSI DESIGN Fall 2013"

Similar presentations


Ads by Google