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Lecture 4, Sept. 6, 2007 EECS130 Integrated Circuit Devices Professor Ali Javey 9/06/2007 Semiconductor Fundamentals Lecture 4 Reading: Chapter 3 Announcements • 1st HW due Tuesday. • New lecture room: 3106 Etcheverry, starting next Thursday (9/13) • TA OHs will be held in Cory 353 What to memorize? • Learn the concepts • Understand the equations • Only memorize the most important and fundamental equations Drift Electron and Hole Mobilities • Drift is the motion caused by an electric field. Effective Mass In an electric field, , an electron or a hole accelerates. electrons Remember : F=ma=-qE holes Electron and hole effective masses Si Ge GaAs GaP m n /m 0 0.26 0.12 0.068 0.82 m p /m 0 0.39 0.30 0.50 0.60 Remember : F=ma=mV/t = -qE Electron and Hole Mobilities mpv = q τ mp q τ v = mp mp = μ v p v = −μn qτ mp qτmn μ p = μn = mp mn • μp is the hole mobility and μn is the electron mobility Electron and Hole Mobilities v =μ ; μ has the dimensions of v/ ⎡ cm/s cm2 ⎤ ⎢ = ⎥ . ⎣V/cm ⋅sV ⎦ Electron and hole mobilities of selected semiconductors Si Ge GaAs InAs 2 μ n (cm /V·s) 1400 3900 8500 30000 2 μ p (cm /V·s) 470 1900 400 500 Based on the above table alone, which semiconductor and which carriers (electrons or holes) are attractive for applications in high-speed devices? Drift Velocity, Mean Free Time, Mean Free Path 2 EXAMPLE: Given μp = 470 cm /V·s, what is the hole drift velocity at 3 = 10 V/cm? What is τmp and what is the distance traveled between collisions (called the mean free path)? Hint: When in doubt, use the MKS system of units. Drift Velocity, Mean Free Time, Mean Free Path 2 EXAMPLE: Given μp = 470 cm /V·s, what is the hole drift velocity at 3 = 10 V/cm? What is τmp and what is the distance traveled between collisions (called the mean free path)? Hint: When in doubt, use the MKS system of units. 2 3 5 Solution: ν = μp = 470 cm /V·s × 10 V/cm = 4.7× 10 cm/s 2 -31 -19 τmp = μp mp /q =470 cm /V ·s × 0.39 × 9.1×10 kg/1.6 ×10 C = 0.047 m2/V ·s × 2.2×10-12 kg/C = 1×10-13s = 0.1 ps -13 7 mean free path = τmhνth ~ 1× 10 s × 2.2×10 cm/s = 2.2×10-6 cm = 220 Å = 22 nm This is smaller than the typical dimensions of devices, but getting close. Mechanisms of Carrier Scattering There are two main causes of carrier scattering: 1. Phonon Scattering 2. Impurity (Dopant) Ion Scattering Phonon scattering mobility decreases when temperature rises: 1 1 μ∝ τ ∝ ∝ ∝ T3− / 2 phonon phonon phonon density × carrier thermalvelocity× 1 / T 2 T μ = qτ/m ∝ T v ∝ T1/2 th purity (Dopant)-Ion Scattering or Coulombic ScatteringIm Boron Ion _ Electron - - Electron + Arsenic Ion There is less change in the direction of travel if the electron zips by the ion at a higher speed. 3 3 / 2 vth T μ impurity ∝ ∝ NNa+ d NNa+ d Total Mobility 1600 1 1 1 1400 = + τ τ τ 1200 onsrtecEl phonon impurity ) 1 1 1 -1 1000 = + s -1 μ μphonon impurity μ V 800 2 600 400 selHo Mobility (cm 200 0 1E14 1E15 1E16 1E17 1E18 1E19 1E20 -3 -3 Tota ImlpurtyiNa + Concenraotin Nd (cm ) (atoms cm ) Temperature Effect on Mobility 10 15 Question: What Nd will make dμn /dT = 0 at room temperature? Velocity Saturation When the kinetic energy of a carrier exceeds a critical value, it generates an optical phonon and loses the kinetic energy. Therefore, the kinetic energy is capped and the velocity does not rise above a saturation velocity, vsat , no matter how large is. Velocity saturation has a deleterious effect on device speed as we will see in the later chapters. Drift Current and Conductivity Jp unit area ν 2 2 Current density Jp = qpv A/cm or C/cm ·sec EXAMPLE: If p = 1015cm-3 and v = 104 cm/s, then -19 15 -3 4 Jp = 1.6×10 C × 10 cm × 10 cm/s 2 2 1 . 6 C/s= ⋅ cm = 1.6 A/cm Drift Current and Conductivity Jp unit area ν + - Remember: • Holes travel in the direction of the Electric field • Electrons travel in the direction opposite to that of the E-field Drift Current and Conductivity Jp,drift = qpv = qpμp Jn,drift = –qnv = qnμn Jdrift = Jn,drift + Jp,drift = (qnμn +qpμp) σ ∴ conductivity of a semiconductor is σ = qnμn + qpμp ∴ resistivity of a semiconductor is ρ = 1/σ Diffusion Current Particles diffuse from a higher-concentration location to a lower-concentration location. Diffusion Current dn dp J = qD J = −qD n, diffusion n dx p, diffusion p dx D is called the diffusion constant. Signs explained: n p x x Total Current – Review of Four Current Components JTOTAL = Jn + Jp dn J = J + J = qnμ + qD n n,drift n,diff n n dx dp J = J + J = qpμ – qD p p,drift p,diff p p dx JTOTAL = Jn,drift + Jn,diff + Jp,drift + Jp,diff Ereference V Ec and Ev vary in the opposite direction from the voltage. That x E is, Ec and Ev are higher where c the voltage is lower. E Ev E -E = -qV c reference x Variation in Ec with position is called band bending Einstein Relationship between D and μ Consider a piece of non-uniformly doped semiconductor. ( E−c E − f ) / kT Nype-t smecionducotr y n-t smpeceionducort n= Nc e dn N(c E−c E − f ) /dE kT c Derecsngai donor conctnietrano = − e dx kT dx E (x) n dE c = − c kT dx Ef n = − q kT Ev (x) Einstein Relationship between D and μ dn n = − q dx kT dn J = qnμ + qD = 0 at equilibrium. n n n dx qD 0 = qnμ − qn n n kT kT kT D = μ Similarly, D = μ n q n p q p These are known as the Einstein relationship. EXAMPLE: Diffusion Constant What is the hole diffusion constant in a piece of 2 -1 -1 silicon with μp = 410 cm V s ? Solution: ⎛ kT ⎞ 2− 1 − 1 2 ( 26D p = mV⎜ ⎟μ p )= 410 ⋅ cm= V s 1 1 cm /s ⎝ q ⎠ = 26 mV at room temperature.Remember: kT/q .
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