6.720J/3.43J - Integrated Microelectronic Devices - Fall 2001 Lecture 8-1 Lecture 8 - Carrier Drift and Diffusion (cont.) September 21, 2001 Contents: 1. Non-uniformly doped semiconductor in thermal equilibrium Reading assignment: del Alamo, Ch. 4, 4.5
6.720J/3.43J - Integrated Microelectronic Devices - Fall 2001 Lecture 8-2 Key questions Is it possible to have an electric field in a semiconductor in thermal equilibrium? What would that imply for the electron and hole currents? Is there a relationship between mobility and diffusion coefficient? Given a certain non-uniform doping distribution, how does one compute the equilibrium carrier concentrations? Under what conditions does the equilibrium majority carrier concentration follow the doping level in a non-uniformly doped semiconductor?
6.720J/3.43J - Integrated Microelectronic Devices - Fall 2001 Lecture 8-3 1. Non-uniformly doped semiconductor in thermal equilibrium It is possible to have an electric field in a semiconductor in thermal equilibrium non-uniform doping distribution Gauss Law: electrical charge produces an electric field: de dx = ρ ɛ ρ volume charge density [C/cm 3 ] if ρ =0 E=0 if ρ =0 de dx =0 it is possible to have E 0withρ =0 In semiconductors: ρ = q(p n + N + D N A ) If N + D N D and N A N A, de dx = q ɛ (p n + N D N A )
6.720J/3.43J - Integrated Microelectronic Devices - Fall 2001 Lecture 8-4 de dx = q ɛ (p n + N D N A ) Uniformly-doped semiconductor in TE: far away from any surface charge neutrality: ρ o =0 de o dx =0 since no field applied from the outside E o =0
6.720J/3.43J - Integrated Microelectronic Devices - Fall 2001 Lecture 8-5 Non-uniformly doped semiconductor in TE (n-type): n o, N D N D (x) n o (x)? n o (x)? x de o dx = q ɛ (N D n o ) Three possibilities: n o (x) =N D (x) net diffusion current n o (x) uniform net drift current n o = f(x) but n o (x) N D (x) in a way that there is no net current
6.720J/3.43J - Integrated Microelectronic Devices - Fall 2001 Lecture 8-6 n o, N D partially uncompensated donor charge a) - N D (x) + n o (x) net electron charge ρ o x b) + x c) ε ο x φ o d) φ ref x E c q[φ(x)-φ ref ] e) E F E v
6.720J/3.43J - Integrated Microelectronic Devices - Fall 2001 Lecture 8-7 Goal: understanding physics of non-uniformly doped semiconductors in TE principle of detailed balance for currents in TE Einstein relation Boltzmann relations general solution quasi-neutral solution
6.720J/3.43J - Integrated Microelectronic Devices - Fall 2001 Lecture 8-8 In thermal equilibrium: J = J e + J h =0 Detailed balance further demands that: J e = J h =0 [study example of 4.5.2]
6.720J/3.43J - Integrated Microelectronic Devices - Fall 2001 Lecture 8-9 Einstein Relation µ relates to ease of carrier drift in an electric field. D relates to ease of carrier diffusion as a result of a concentration gradient. Is there a relationship between µ and D? Yes, Einstein relation: D e µ e = D h µ h = kt q Relationship between D and µ only depends on T. [study derivation and restrictions in 4.5.2]
6.720J/3.43J - Integrated Microelectronic Devices - Fall 2001 Lecture 8-10 Boltzmann Relations InTE,diffusion = drift there must be a conexion between electrostatics and carrier concentrations. InTE,J e =0: J e = qµ e n o E o + qd e dn o dx =0 Then, relationship between E o and equilibrium carrier concentration: E o = kt q Express in terms of φ: 1 n o dn o dx = kt q d(ln n o ) dx dφ dx = kt q d(ln n o ) dx Integrate: φ(x) φ(ref) = kt q ln n o(x) n o (ref) Relationship between the ratio of equilibrium carrier concentration and difference of electrostatic potential at two different points. At 300 K: factor of 10 in n o kt q ln 10 = 60 mv of φ 60 mv/decade rule
6.720J/3.43J - Integrated Microelectronic Devices - Fall 2001 Lecture 8-11 If φ(ref) =0wheren o = n i : Similarly: n o = n i exp qφ kt p o = n i exp qφ kt Note: n o p o = n 2 i
6.720J/3.43J - Integrated Microelectronic Devices - Fall 2001 Lecture 8-12 Equilibrium carrier concentration: general solution in the presence of doping gradient (n-type) de o dx = q ɛ (N D n o ) Relationship between E o and n o in thermal equilibrium: E o = kt q d(ln n o ) dx Then: ɛkt d 2 (ln n o ) q 2 dx 2 = n o N D One equation, one unknown. Given N D (x), can solve for n o (x) (but in general, require numerical solution).
6.720J/3.43J - Integrated Microelectronic Devices - Fall 2001 Lecture 8-13 Quasi-neutral situation ɛkt d 2 (ln n o ) q 2 dx 2 = n o N D If N D (x) changes slowly with x n o (x) will also change slowly, then: n o (x) N D (x) The majority carrier concentration closely tracks the doping level. When does this simple result apply? ɛkt d 2 (ln n o ) q 2 dx 2 N D or ɛkt d 2 (ln n o ) 1 q 2 N D dx 2 Alternatively: n o N D N D 1
6.720J/3.43J - Integrated Microelectronic Devices - Fall 2001 Lecture 8-14 Define Debye length: L D = ɛkt q 2 N D Rewrite condition: L 2 D de o dx kt q Note: L D de o dx is change in electric field over a Debye length L 2 D de o dx is change in electrostatic potential over a Debye length For a non-uniformly doped profile to be quasi-neutral in TE, the change in the electrostatic potential over a Debye length must be smaller than the thermal voltage.
6.720J/3.43J - Integrated Microelectronic Devices - Fall 2001 Lecture 8-15 L D = ɛkt q 2 N D L D characteristic length for electrostatic problems in semiconductors. Similar arguments for p-type material. 1E-04 Si at 300K 1E-05 LD (cm) 1E-06 1E-07 1E-08 1E+15 1E+16 1E+17 1E+18 1E+19 1E+20 1E+21 N (cm -3 ) Since L D 1/ N: N L D QN condition easier to fulfill.
6.720J/3.43J - Integrated Microelectronic Devices - Fall 2001 Lecture 8-16 Key conclusions Detailed balance in TE demands that J e = J h = 0 everywhere. Einstein relation: D µ = kt q Boltzmann relations: in TE, difference of electrostatic potential between two points is proportional to ratio of carrier concentration at same points: φ(x) φ(ref) = kt q ln n o(x) n o (ref) = kt q ln p o(x) p o (ref) In quasi-neutral non-uniformly doped semiconductor in TE: n o (x) N D (x) Debye length: key characteristic length for electrostatics in quasineutral semiconductor. Order of magnitude of key parameters for Si at 300K: Debye length: L D 10 8 10 5 cm (depends on doping)
6.720J/3.43J - Integrated Microelectronic Devices - Fall 2001 Lecture 8-17 Self-study Derive Einstein relation. Study example supporting detailed balance for J e and J h. Work out exercises 4.5-4.7.