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集成电路课程回顾
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1 . Lecture #31 ANNOUNCEMENTS • TA’s will hold a review session on Thursday May 15: – 2-5 PM, 277 Cory • Final Exam will take place on Friday May 23: – 12:30-3:30 PM, Sibley Auditorium (Bechtel Bldg.) – Closed book; 7 pgs of notes + calculator allowed OUTLINE – Review of Fundamental Concepts Spring 2003 EE130 Lecture 31, Slide 1 − E g / kT Intrinsic Carrier Concentration ni = N c N v e conduction ni ≅ 1010 cm-3 at room temperature Spring 2003 EE130 Lecture 31, Slide 2 1
2 . Charge-Carrier Concentrations ND: ionized donor concentration (cm-3) NA: ionized acceptor concentration (cm-3) Note: Carrier concentrations depend on net dopant concentration (ND - NA) ! Spring 2003 EE130 Lecture 31, Slide 3 Energy Band Diagram increasing electron energy increasing hole energy electron kinetic energy Ec Ev hole kinetic energy distance in semiconductor Spring 2003 EE130 Lecture 31, Slide 4 2
3 . Fermi Function 1 f (E) = 1 + e ( E − EF ) / kT f ( E ) ≅ e −( E − E F ) / kT if E-E F > 3kT Spring 2003 EE130 Lecture 31, Slide 5 Relationship between EF and n,p n = ni e ( E F − Ei ) / kT = N c e ( Ec − E F ) / kT p = ni e ( Ei − E F ) / kT = N v e ( E F − Ev ) / kT Spring 2003 EE130 Lecture 31, Slide 6 3
4 . Free Carriers in Semiconductors • Three primary types of carrier action occur inside a semiconductor: – drift D kT = µ q – diffusion – recombination-generation Spring 2003 EE130 Lecture 31, Slide 7 Net Generation Rate • The net generation rate is given by 2 ∂p ∂n ni − np = = ∂t ∂t τ p (n + n1 ) + τ n ( p + p1 ) where n1 ≡ ni e ( ET − Ei ) / kT and p1 ≡ ni e ( Ei − ET ) / kT ET = trap - state energy level Spring 2003 EE130 Lecture 31, Slide 8 4
5 . Drift and Resistivity • Electrons and holes moving under the influence of an electric field can be modelled as quasi-classical particles with average drift velocity |vd| = µ • The conductivity of a semiconductor is dependent on the carrier concentrations and mobilities σ = qnµn + qpµp 1 1 • Resistivity ρ= = σ qnµ n + qpµ p Spring 2003 EE130 Lecture 31, Slide 9 Mobility Dependence on Doping 1600 1400 1 1 1 = + 1200 τ τ phonon τ impurity E lectrons 1 1 1 = + Mobility (cm V s ) -1 1000 µ µ phonon µ impurity -1 800 2 600 400 H o les 200 0 1E 14 1E 15 1E 16 1E 17 1E 18 1E 19 1E 20 -3 Total Doping T otal Concentration Im p urity C on ce nra tio nN(a A + ND to m s (cm cm -3) ) Spring 2003 EE130 Lecture 31, Slide 10 5
6 . Total Current J = JN + JP dn JN = JN,drift + JN,diff = qnµn + qDN dx dp JP = JP,drift + JP,diff = qpµp – qDP dx Spring 2003 EE130 Lecture 31, Slide 11 Electrostatic Variables 1 V= ( Ereference − Ec ) q ε = − dV = 1 dE dx q dx c ρ dε = ε dx Spring 2003 EE130 Lecture 31, Slide 12 6
7 . ∂n 1 ∂J n ( x ) ∆n = − + GL Continuity ∂t q ∂x τn Equations: ∂p 1 ∂J p ( x ) ∆p =− − + GL ∂t q ∂x τp ∂∆n p ∂ 2 ∆n p ∆n p Minority = DN − + GL Carrier ∂t ∂x 2 τn Diffusion ∂∆pn 1 ∂ 2 ∆pn ∆pn Equations: = DP − + GL ∂t q ∂x 2 τp Spring 2003 EE130 Lecture 31, Slide 13 Work Function Ε0: vacuum energy level ΦM: metal work function ΦS: semiconductor work function Spring 2003 EE130 Lecture 31, Slide 14 7
8 . Schottky Diode VA > 0 ΦBn = ΦM – χ VA < 0 2ε s (Vbi − V A ) W= qN D I = AT 2 J S (e qV A / kT − 1) εs C = A where J S = 120e − qΦ B / kT A/cm 2 W Fermi level splits into two levels (EFM and EFS) separated by qVA Spring 2003 EE130 Lecture 31, Slide 15 pn Junction Electrostatics Spring 2003 EE130 Lecture 31, Slide 16 8
9 . pn Junction Electrostatics, VA ≠ 0 • Built-in potential Vbi (non-degenerate doping): kT N A kT N D kT N A N D Vbi = ln + ln = ln q ni q ni q ni 2 • Depletion width W : 2ε s 1 1 W = x p + xn = (Vbi − V A ) + q N A ND ND NA xp = W xn = W N A + ND N A + ND Spring 2003 EE130 Lecture 31, Slide 17 Avalanche Breakdown Mechanism ε sε crit High E-field: 2 VBR ≈ if VBR >> Vbi 2qN ε crit increases slightly with N: For 1014 cm-3 < N < 1018 cm-3, Small E-field: 105 V/cm < ε crit < 106 V/cm Spring 2003 EE130 Lecture 31, Slide 18 9
10 . “Law of the Junction” The voltage VA applied to a pn junction falls mostly across the depletion region (assuming that low-level injection conditions prevail in the quasi-neutral regions). We can draw 2 quasi-Fermi levels in the depletion region: p = ni e ( Ei − FP ) / kT n = ni e ( FN − Ei ) / kT pn = ni2e( Ei − FP ) / kT e( FN − Ei ) / kT = ni2e( FN − FP ) / kT pn = ni2e qVA / kT Spring 2003 EE130 Lecture 31, Slide 19 Excess Carrier Concentrations at –xp, xn p-side n-side p p (− x p ) = N A nn ( xn ) = N D ni2 e qVA / kT ni2 e qVA / kT n p (− x p ) = p n ( xn ) = NA ND = n p 0 e qVA / kT = pn 0 e qVA / kT ∆n p (− x p ) = e ( ni2 qVA / kT −1 ) ∆pn ( xn ) = ni2 qVA / kT ND e −1 ( ) NA Spring 2003 EE130 Lecture 31, Slide 20 10
11 . pn Diode I-V Characteristic d∆n p ( x' ' ) Dn p-side: J n = − qDn =q n p 0 (e qVA kT − 1)e − x '' Ln dx ' ' Ln d∆pn ( x' ) Dp −x' Lp n-side: J p = − qD p =q pn 0 (e qVA kT − 1)e dx' Lp J = Jn x=− x p + Jp = Jn x ′′=0 + Jp x = xn x ′= 0 D D p qVA J = qni2 n + ( e kT − 1) L N n A L N p D Spring 2003 EE130 Lecture 31, Slide 21 pn Junction Capacitance 2 types of capacitance associated with a pn junction: 1. CJ depletion capacitance dQdep εs CJ ≡ =A dVA W 2. CD diffusion capacitance (due to variation of stored minority charge in the quasi-neutral regions) For a one-sided p+n junction (QP >> QN ): dQ dI τ p I DC CD = = τp = τ pG = dV A dV A kT / q Spring 2003 EE130 Lecture 31, Slide 22 11
12 . Deviations from the Ideal I-V Behavior Resulting from • recombination/generation in the depletion region • series resistance • high-level injection Spring 2003 EE130 Lecture 31, Slide 23 Transient Response of pn Diode • Because of CD, the voltage across the pn junction depletion region cannot be changed instantaneously. (The delay in switching between the ON and OFF states is due to the time required to change the amount of excess minority carriers stored in the quasi-neutral regions.) Turn-off transient: Spring 2003 EE130 Lecture 31, Slide 24 12
13 . Minority-Carrier Injection & Collection • Under forward bias, minority carriers are injected into the quasi-neutral regions of the diode. Æ Current flowing across junction is comprised of hole and electron components • Under reverse bias, minority carriers are collected into the quasi-neutral regions of the diode. (Minority carriers within a diffusion length of the depletion region will diffuse into the depletion region and then be swept across the junction by the electric field) → Current flowing depends on the rate at which minority carriers are supplied Spring 2003 EE130 Lecture 31, Slide 25 MOS Band Diagrams (n-type Si) Decrease VG (toward more negative values) -> move the gate energy-bands up, relative to the Si decrease VG decrease VG • Accumulation • Depletion • Inversion – VG > VFB – VG < VFB – VG < VT – Electrons – Electrons – Surface accumulate at repelled becomes surface from surface p-type Spring 2003 EE130 Lecture 31, Slide 26 VG = VFB + Vox + ψ s 13
14 . Biasing Conditions for p-type Si increase VG increase VG VG = VFB VG < VFB VT > VG > VFB 2ε Siψ s Spring 2003 EE130 Lecture 31, Slide 27 Wd = qN A MOS Charge & Capacitance (p-type Si) Qacc = −Cox (VG − VFB ) accumulation depletion inversion dQs C= VG dVG VFB VT Ideal C-V curve Qinv = −Cox (VG − VT ) C slope = -Cox Cox QF VFB = φ MS − Cox VG 2qN Aε Si (2ψ B ) VFB VT VT = VFB + 2ψ B + accumulation depletion inversion Cox Spring 2003 EE130 Lecture 31, Slide 28 14
15 . VT Adjustment by Back Biasing • In some IC products, VT is dynamically adjusted by applying a back bias: – When a MOS capacitor is biased into inversion, a pn junction exists between the surface and the bulk. – If the inversion layer contacts a heavily doped region of the same type, it is possible to apply a bias to this pn junction N+ poly-Si • VG biased so surface is inverted + + + + + + + + • Inversion layer contacted by N+ region SiO2 - - - - - - - - - • Bias VC applied to channel N+ Æ Reverse bias VB-VC applied btwn p-type Si channel & body Spring 2003 EE130 Lecture 31, Slide 29 Effect of VCB on ψs and VT • Application of reverse bias -> non-equilibrium – 2 Fermi levels (one for n-region, one for p-region) • Separation = qVBC Îψs increased by VC • Reverse bias widens Wd, increases Qdep ÎQinv decreases with increasing VCB, for a given VGB 2qN Aε Si (2ψ B + VCB ) VT = VFB + VC + 2ψ B + Cox Spring 2003 EE130 Lecture 31, Slide 30 15