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Chapter 5 Carrier transport phenomena W.K. Chen 1 Transport The net flow of electrons and holes in material is called transport Two basic transport mechanisms Electrophysics, NCTU Drift: movement of charged due to electric fields Diffusion: the flow of charges due to density gradient We implicitly assume the thermal equilibrium during the carrier transport is not substantially disturbed W.K. Chen Electrophysics, NCTU 2 Outline Carrier drift Carrier diffusion Graded impurity distribution The Hall effect W.K. Chen 3 Electrophysics, NCTU 5.1 Carrier drift Drift: the net movement of charge due to electric fields is called drift Drift current: The net drift of charge gives rise to a drift current l l N: total number of flow charge n: volume density of flow charge A: cross-sectional area υ: average drift velocity l: traveling length of carrier per Δt A N nAl nA(υΔt ) # = nAυ ( ) = = Δt Δt Δt sec Φ # ) Flux density φ = = nυ ( A sec ⋅ cm 2 Flux Φ = Drift current density J = qφ = qnυ d W.K. Chen A ( 2) cm Electrophysics, NCTU Ampere 4 mobility The average drift velocity for low electric fields is directly proportional to the electric field, similar to the terminal velocity case in “Fundamental Physics” r Fext Hole: F = m*p a = (+ e)E E: electric field r υ dp = μ p E r υdp: average drift velocity for holes μp: hole mobility, proportionality factor crystal The mobility describe how well a particle will move due to an electric field r Fext = m*a Drift current due to holes J p , drf = ( + e) pυ dp = ( + e) pμ p E W.K. Chen 5 Electrophysics, NCTU J = qφ = qnυ d ( A ) cm 2 F = (−e) E Flux φ p r υ dn = − μ n E r Flux φn J n ,drf = qnυ = (−e)nυ dp = (−e) p (− μ n E ) Drift current due to electrons Total drift current W.K. Chen F = (+ e) E E J p ,drf J n ,drf J n ,drf = enμ n E J = J n ,drf + J p ,drf = (enμ n + epμ p )E Electrophysics, NCTU 6 Example 5.1 drift current density GaAs sample at 300K, N a = 0, N d = 1016 cm −3 Assume complete ionization ⇒ Calulate the drift current density if applied electric field E = 10 V/cm Solution: complete ionization ⇒ n ≈ N d = 1016 cm -3 ni2 (1.8 ×1016 ) 2 = = 3.24 × 10 − 4 cm -3 p= 16 10 n ⇒ Calulate the drift current density if applied electric field E = 10 V/cm J dft = J n ,drf + J p ,drf = (enμ n + epμ p )E J dft = (1.6 × 10 −19 )(8500)(1016 )(10) = 136A/cm 2 W.K. Chen 5.1.2 Mobility effect F = m*p a = m*p 7 Electrophysics, NCTU F eE dυ = ( + e) E ⇒ υ = * t mp dt t Under thermal equilibrium Assume the net drift velocity is a small perturbation on the random thermal velocity, so the time between collision will not be altered appreciably ⎛ eτ cp ⎞ mean peak velocity υ d , peak = ⎜ * ⎟E τcp: mean time between collisions ⎜ m ⎟ ⎝ p ⎠ W.K. Chen Electrophysics, NCTU 8 The average drift velocity is one half the peak value 1 ⎛ eτ cp ⎞ average velocity υ d = ⎜ * ⎟E 2 ⎜⎝ m p ⎟⎠ Due to the statistic nature, the factor of ½ does not appear in a more accurate model ⎛ eτ cp ⎞ average velocity υ dp = ⎜ * ⎟E = μ p E ⎜ m ⎟ ⎝ p ⎠ μp = μn = υ dp E υ dn E = = eτ cp m*p , eτ cn mn* The less the collisions, the longer the mean collision time and the higher the mobility W.K. Chen 9 Electrophysics, NCTU Scattering (collision) mechanisms Two major scattering mechanisms Phonon (lattice) scattering A perfect periodic potential in a solid allows electrons to move unimpeded, or with no scattering, through the crystal The thermal vibrations of lattice atoms cause a disruption in the perfect periodic potential, resulting the interactions between the electrons or holes and the vibrating lattice atoms μ L ∝ T −3 / 2 lattice Ionized impurity scattering The impurites in semiconductor at higher temperatures. The coulomb interactions between the electrons or holes and the ionized impurities produce scattering or collisions. impurity W.K. Chen T −3 / 2 μI ∝ NI Electrophysics, NCTU total ionized impurity N I = ( N d+ + N a− ) 10 Mobilitis versus temperarure Inserts show the temperature dependence for “almost” intrinsic silicon The inserts show that the parameter n is not equal to 3/2, but is 2.2, as the first-order scattering theory predicted. However , the mobilites do increase as the temperature increases W.K. Chen 11 Electrophysics, NCTU Mobilties versus impurity concentrations at 300K Ge Si GaAs W.K. Chen Electrophysics, NCTU 12 Net mobility The probability of a scattering even in the differential time dt is the sum of individual events dt τ Qμp = υ dp E = eτ cp m * p = dt τI , μn = υ dn E + dt τL = eτ cn ⇒ μ ∝τ mn* The net mobility due to the ionized and lattice scattering processes 1 μ = 1 μI + 1 μL High effective mass of carrier results in low mobility The mobility will increase with the increasing collision time W.K. Chen 13 Electrophysics, NCTU 5.1.3 Conductivity J = e(nμ n + epμ p )E = σE V = IR, R = J= ρL A = L σA I V / R EL 1 = = ⋅ = σE A A A ⎛ L ⎞ ⎜ ⎟ ⎝ σA ⎠ J = σE σ = (enμ n + epμ p ) ρ= 1 σ = 1 e(nμ n + epμ p ) The conductivity and resistivity of an extrinsic semiconductor are a function primarily of the majority carrier parameters, such carrier concentrations and mobilities W.K. Chen Electrophysics, NCTU 14 Resistivity versus impurity concentration at 300K Si W.K. Chen Ge, GaAs and GaP Electrophysics, NCTU 15 σ i = (eμ n + eμ p )ni In the midtemperature range (extrinsic range) We have complete ionization, the electron concentration remains essentially constant, However, the mobility decreases with increasing temperature At higher temperatures The intrinsic carrier concentration begins to dominate the electron W.K. Chen 16 Electrophysics, NCTU concentration as well as the conductivity Example 5.2 mobility Compensated n - type Si at 300K, σ = 16 (Ω - cm) −1 , Na = 1016 cm −3 ⇒ Determine the donor concentration and electron mobility Solution: compenstaed semiconductor and complete ionization at 300K ⇒ n ≈ (Nd − Na ) σ ≈ eμ n n = eμ n ( N d − N a ) 16 = (1.6 × 10 −19 ) μ n ( N d − 1017 ) Use the left figure with trial and error For example if we choose N I = N d+ − N a− = 3 × 1017 cm -3 (i.e., N d = 2 × 1017 cm -3 ) Then μ n = 510 cm 2 /V - s ⇒ σ = 8.16 (Ω - cm)-1 W.K. Chen Electrophysics, NCTU 17 If we choose N I = N d+ − N a− = 6 ×1017 cm -3 (i.e., N d = 5 × 1017 cm -3 ) Then μ n = 325 cm 2 /V - s ⇒ σ = 20.8 (Ω - cm)-1 Further tial and error yields N I = N d+ + N a− = 4.5 ×1017 cm -3 N d = 3.5 ×1017 cm -3 ) μ n = 400 cm 2 /V - s ⇒ σ = 16 (Ω - cm)-1 (agree with given value) W.K. Chen Electrophysics, NCTU 18 5.1.4 Velocity saturation 1 3 mυth2 = kT 2 2 at T = 300 K 1 3 mυth2 = (0.0259) = 0.03885eV 2 2 2(0.03885eV) 2(0.03885)(1.6 ×10 −19 ) υth = = ≈ 105 m/s −31 m (9.11×10 ) υth ≈ 107 cm/s W.K. Chen Electrophysics, NCTU 19 Figure 5.8 For Si At low electric fields, there is linear variation of velocity with electric field At high electric fields, the velocity saturated at approximately 107cm/s W.K. Chen Electrophysics, NCTU 20 GaAs : mn* = 0.067mo Si : mn* = 1.08mo For GaAs Due to low effective mass, the low-electric field electron velocity in GaAs is much larger than in Si. At high electric fields, negative differential mobility occurs due to the scattering of electrons into upper valley. Because of larger effective mass in the upper valley (0.55 mo vs. 0.067 mo), the intervalley transfer mechanism results in decreasing average drift velocity of electrons with electric field. W.K. Chen 21 Electrophysics, NCTU 5.2 Carrier diffusion (1) xo xo − l n(x) xo (2) xo xo + l x Fn = −υthl W.K. Chen l = υthτ cn Electrophysics, NCTU dn dx 22 During electron travel between collisions In a mean free time, One half of electrons at segment (1) will move to the right and cross the xo plane into segment (2) One half of electrons at segment (2) will move to the left and cross the xo plane into (segment (1)) n1 Net rate of electron flow to the right (electron flux) xo − l l n2 (1) 1 1 1 n1υth − n2υth = (n1 − n2 )υth 2 2 2 dn n1 − n2 ≈ − l dx 1 dn ⇒ Fn ( x) = υth ⋅ (− l ) 2 dx Fn ( xo ) = (2) xo xo + l l = υthτ cn 1 dn net electron flux Fn = − υthl 2 dx W.K. Chen Electrophysics, NCTU 23 Electron diffusion current 1 dn J n = (−e) Fn = + eυthl 2 dx J n ,dif = eDn dn dx electron diffusion coefficient 1 Dn = υthll = υthτ cn 2 Hole diffusion current J p ,dif = −eD p W.K. Chen dp dx hole diffusion coefficent l = υthτ cp 1 D p = υthl 2 Electrophysics, NCTU 24 Example 5.4 n - type GaAs at 300K, The electron concentration varies linearly from 1× 1018 to 7 × 1018 cm −3 over a distance of 0.10 cm ⇒ Calculate the diffusion current if diffusion coefficient Dn = 225 cm 2 /s Solution: Δn dn ≈ eDn Δx dx ⎛ 1×1018 − 7 ×1017 ⎞ −19 ⎟⎟ = 108 A/cm2 = (1.6 ×10 )(225)⎜⎜ 0.10 ⎝ ⎠ J n ,dif = eDn W.K. Chen Electrophysics, NCTU 25 5.2.2 Total current density J = enμ n E + epμ p E + J n ,dif + eDn dp dn − eD p dx dx J = enμ n E + epμ p E + J n ,dif + eDn∇n − eD p ∇p W.K. Chen Electrophysics, NCTU 26 5.3 Graded impurity distribution In nonuniform doped semiconductor, there will be a diffusion of majority electrons from the region of high concentration to the region of low concentration. The flow of electrons leave behind positively charged donor ions. The separation of positive and negative charges induces a electric field The electric field is defined as The electric potential is related to the electric potential energy by the charge ( -e) φ= Ec ( − e) 1 ⇒ φ = ( EF − EFi ) e dφ 1 dEFi = Ex = − dx e dx W.K. Chen Electrophysics, NCTU 27 ⎛ N d ( x) ⎞ ⎡ E f − E fi ⎤ ⎜⎜ ⎟⎟ ≈ ⇒ − = no = ni exp ⎢ N ( x ) E E kT ln d f fi ⎥ n kT ⎦ ⎣ i ⎝ ⎠ − dE fi dx = kT dN d ( x) N d ( x) dx The induced electric field due to the nonuniform doping ⎛ kT ⎞ 1 dN d ( x) E x = −⎜ ⎟ ⎝ e ⎠ N d ( x) dx W.K. Chen Electrophysics, NCTU 28 Example 5.5 N d ( x) = 1016 − 1019 x (cm -3 ) (0 ≤ x ≤ 1μm) Solution: (0.0259)(1019 ) ⎛ kT ⎞ 1 dN d ( x) E x = −⎜ ⎟ =− (1016 − 1019 x) ⎝ e ⎠ N d ( x) dx At x = 0, we find E x ( x = 0) = 25.9 V/cm W.K. Chen 29 Electrophysics, NCTU 5.3.2 The Einstein relation In thermal equilibrium, the individual electron and hole current must be zero dn dx dN d ( x) J n = 0 = enμ n E + eDn dx J n = 0 = enμ n E + eDn ⎧ ⎛ kT ⎞ 1 dN d ( x) ⎫ dN d ( x) 0 = enμ n ⎨− ⎜ ⎟ ⎬ + eDn dx ⎩ ⎝ e ⎠ N d ( x) dx ⎭ Q n( x ) ≈ N d ( x ) Dn μn W.K. Chen = kT e Dn μn = Dp μp Electrophysics, NCTU = kT e 30 The hole current must also be zero Einstein relation μ D Dn μn = Dp μp = kT e 0.0256 at T = 300 K ≈ 40 at T = 300K W.K. Chen 31 Electrophysics, NCTU 5.4 The Hall effect The Hall effect is used to distinguish whether a semiconductor is n-type or ptype To measure the majority carrier concentration and Majority carrier mobility FB = qυ × B F = q[E + υ × B] = 0 qE y = qE H = qυ x Bz The induced electric field in the y-direction is called the Hall field The induced electric field produce a voltage in the y-direction is called the Hall Voltage VH = + E H W W.K. Chen VH = υ xWBz Electrophysics, NCTU υ x : drift velocity 32 Semiconductor type & concentration For p-type semiconductor, the drift velocity is related to material parameters υ dx = Jx Ix = ep (ep)(Wd ) ⇒ VH = I x Bz epd p= I x Bz edVH ⇒ VH = I x Bz ned n= I x Bz edVH W.K. Chen Electrophysics, NCTU 33 mobility Once the majority carrier concentration is determined, we can calculate the low-field majority carrier mobility Q J x = epμ p E x Ix V = epμ p ⋅ x Wd L μp = W.K. Chen IxL epVxWd μn = IxL enVxWd Electrophysics, NCTU 34 Example 5.7 carrier concnetration & mobility L = 0.1 cm, W = 10 −2 cm, d = 10 −3 cm I x = 1.0 mA, Vx = 12.5 V, Bz = 500 gauss = 5 ×10 − 2 tesla, and VH = −6.25 mV Solution: I B n= x z edVH − (10 −3 )(5 ×10 −2 ) n= = 5 ×10 21 m 3 = 5 × 1015 cm 3 −19 −5 −3 (1.6 ×10 )(10 )(−6.25 ×10 ) (10 −3 )(10 −3 ) IxL = 0.10 m 2 /V - s μn = μn = −19 −4 −5 21 (1.6 ×10 )(5 ×10 )(12.5)(10 )(10 ) enVxWd μ n = 1000 cm 2 /V - s W.K. Chen Electrophysics, NCTU 35 Figure 5.14 Figure for Problem 5.22 W.K. Chen Electrophysics, NCTU 36