Getting J e (x), J h (x), E(x), and p'(x), knowing n'(x) Solving the diffusion equation for n'(x) (using p-type example)

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1 Electronic Devices and Circuits Lecture 4 - Non-uniform Injection (Flow) Problems - Outline Announcements Handouts - 1. Lecture Outline and Summary; 2. Thermoelectrics Review Thermoelectricity: temperature gradients as a driving force The 5 basic equations Non-uniform injection (flow) problems Five assumptions: 1. Uniform doping: r = e E/ x q(p' n'); n = n', p = p' 2. Low level injection: [np - n i2 ]r(t) n'/t min 3. Quasi-neutrality: n' p', n'/ x p'/ x 4. Negligible minority carrier drift: J e qd e n'/ x (assumes p-type) 5. Quasi-static: / t 0 The diffusion equation for minority carriers Getting J e (x), J h (x), E(x), and p'(x), knowing n'(x) Solving the diffusion equation for n'(x) (using p-type example) Homogeneous solutions; minority carrier diffusion length, L min Particular solution Boundary conditions: ohmic contact, reflecting boundary, internal boundary Total solution Clif Fonstad, 9/03 Lecture 4 - Slide 1

2 Thermoelectric effects - the Seebeck and Peltier effects (current fluxes caused by temperature gradients, and visa versa) Consider our electron current density expression with the drift term rewritten in terms of the potential and including the term that must be added if we have a temperature gradient*: J e = -q -m e n Ê ( ) Á - df Ë ˆ Ê + (-q)d dx e - dn ˆ Ê Á + (-q)s Ë dx e n - dt ˆ Á Ë dx This way of writing the current emphasizes the fact that the carriers respond to gradients in various quantities. Drift is the response to a gradient in the electrostatic potential, diffusion is the response to a gradient in the carrier concentration, and the Seebeck effect is the response to a gradient in temperature. The Seebeck effect is interesting because it can be used to generate electric power from a temperature difference. Conversely, a temperature difference can be generated by passing a current through a suitably designed semiconductor structure. This reverse effect is called the Peltier effect. * Don t panic. We will only consider isothermal situations in exams and problem sets. Clif Fonstad, 9/03 Lecture 4 - Slide 2

3 Thermoelectric effects - the Seebeck and Peltier effects (current fluxes caused by temperature gradients, and visa versa) Two examples: Right - The hot point probe, an apparatus for determining the carrier type of semiconductor samples. Below - A thermoelectric cooler array like that found in solidstate refrigerators. Clif Fonstad, 9/03 Ref.: Appendix B in the course text. Lecture 4 - Slide 3

4 Non-uniform material with non-uniform excitations A. General description To model devices we must understand semiconductors in which we have (1) spatial variation of the doping and (2) spatial and temporal variation of the generation function: N d (x), N a (x), g L (x,t) Because the doping and excitation are not uniform, we anticipate that the carrier population, currents, and electric field will also vary with space and time: n(x,t), p(x,t), J e (x,t), J h (x,t), E(x,t) To determine these five quantities, assuming we know the doping profile and generation function, we will need five equations relating these quantities. We have: 2 continuity equations (divergence of fluxes, generation, and recombination) 2 current equations (drift plus diffusion) 1 charge balance equation (Poisson's Eq.) Clif Fonstad, 9/03 Lecture 4 - Slide 4

5 B. The Five Equations Electron population n(x,t) - 1 J e (x,t) t q x Hole population p(x,t) + 1 J h (x,t) t q x Electron current density n(x,t) J e (x,t) = qm e n(x,t)e(x,t) + qd e x Hole current density p(x,t) J h (x,t) = qm h p(x,t)e(x,t) - qd h x Poisson's equation E(x,t) x = g L (x,t) - [ n(x,t) p(x,t) - n o (x) p o (x)]r(t) = g L (x,t) - [ n(x,t) p(x,t) - n o (x) p o (x)]r(t) = - q [ e p(x,t) - n(x,t) + N + d (x) - N ā (x)] Clif Fonstad, 9/03 Lecture 4 - Slide 5

6 C. Solving the five equations: easy to do in several important special cases 1. Uniform doping, thermal equilibrium: x = 0, t = 0, g L (x,t) = 0, J e = J h = 0 2. Drift in uniformly doped semiconductor Lecture 1 Lecture 2 x = 0, t = 0, g L (x,t) = 0, E x constant 3. Uniform optical injection in uniformly doped semiconductor, photoconductivity Lecture 2 x = 0, E x constant, n'<< p o 4. Flow problems: uniform doping, non-uniform injection N d Today x = N a x = 0, n'ª p', n'<< p n' o, J e ª qd e x, t ª 0 5. Junction problems: non-uniform doping in thermal equilibrium t = 0, g L (x,t) = 0, J e = J h = 0 Lecture 5 Clif Fonstad, 9/03 Lecture 4 - Slide 6

7 C. Solving the five equations, cont.: where the special cases we can solve appear in important semiconductor devices Junction diodes, LEDs: p-type n-type Bipolar transistors: Flow problem B Junctiion problem Junction problem Flow problem E n-type p n-type C MOS transistors: Flow problems S n+ Diodes G p-type Drift D n+ Depletion approximation Clif Fonstad, 9/03 Lecture 4 - Slide 7

8 D. Flow Problems: uniformly doped material, non-uniform carrier injection, and boundary conditions; no applied voltage Five unknowns, five equations, five flow problem assumptions: 1. Uniform doping (in p-type, for example) dn o dx = dp o dx 4. Negligible minority carrier drift Note: It is also always true that = 0 fi n x = n' x, p x = p' x n o - p o + N a - N d = 0 fi n - p + N a - N d = n'-p' 2. Quasineutrality n' n'ª p', x ª p' x 3. Low level injection 2 n'<< p o fi ( np - n i )r ª n' p o r = n' n t = n' t, J e (x,t) ª qd e n'(x,t) x p t = p' t Clif Fonstad, 9/03 Lecture 4 - Slide 8 t e

9 D. Flow Problems, cont. (where we are after imposing four conditions) Combining everything we have to date, with these four conditions our five equations become: n'(x,t) 1,2 : - 1 J e (x,t) = p'(x,t) + 1 J h (x,t) t q x t q x n'(x,t) 3 : J e (x,t) ª +qd e x p'(x,t) 4 : J h (x,t) = qm h p(x,t)e(x,t) + qd h x E(x,t) 5 : = q [ p'(x,t) - n'(x,t) ] x e = g L (x,t) - n'(x,t) t e Before continuing to our fifth condition, we note that already Equations 1 and 3 can be combined to give one equation in n'(x,t): n'(x,t) t - D e 2 n'(x,t) x 2 = g L (x,t) - n'(x,t) t e The time dependent diffusion equation Clif Fonstad, 9/03 Lecture 4 - Slide 9

10 D. Flow Problems, cont. (the fifth constraint) The time dependent diffusion equation, which is repeated below, is in general still very difficult to solve n'(x,t) 2 n'(x,t) - D e = g t x 2 L (x,t) - n'(x,t) t e Consequently we impose a final, fifth constraint. 5. Quasi-static excitation With this constraint the above equation becomes a second order linear differential equation: -D e d 2 n'(x) dx 2 g L (x,t) such that all = g L (x) - n'(x) t e which in turn becomes, after rearranging the terms : d 2 n'(x) - n'(x) = - 1 g dx 2 L (x) D e t e D e The steady state diffusion equation Clif Fonstad, 9/03 Lecture 4 - Slide 10 t ª 0

11 D. Flow Problems, cont. (the other unknowns) Solving the steady state diffusion equation gives n. Finding p, J e, J h, and E x knowing n : First find J e : J e (x) ª qd e dn'(t) dx Then find J h : J h (x) = J Tot - J e (x) Next find E x : E x (x) ª 1 qm h p o È J h (x) - D h Í Î D e J e (x) Then find p : p'(x) ª n'(x) + e q de x (x) dx Finally, go back and check that all of the five conditions are met by the solution. Clif Fonstad, 9/03 Lecture 4 - Slide 11

12 D. Flow Problems, cont. (General solutions) The homogeneous solution of the steady state diffusion equation, is d 2 n'(x) dx 2 - n'(x) D e t e = - 1 D e g L (x) ' n HS (x) = Ae x / L e + Be-x / L e where L D e et e L e is the called the minority carrier diffusion length and it is the characteristic length over which returns to its equilibrium value. It is thus the spatial equivalent of the minority carrier lifetime, t e. A and B are unknowns at this point. The particular solution of this 2nd order differential equation looks functionally like g L (x). Beyond that it is difficult to say much in general without specifying g L (x). The unknowns A and B are found by matching the sum of the homogeneous and particular solutions to the boundary conditions. Clif Fonstad, 9/03 Lecture 4 - Slide 12

13 D. Flow Problems, cont. (Boundary conditions - sample ends) At sample ends we have three possible situations we deal with in 6.012, each of which imposes a different condition on either the excess minority carrier population or its gradient: An ohmic contact: here there is infinite recombination and the excess population is identically zero n'(x) = 0 at an Ohmic contact A reflecting surface: here there is no additional recombination and thus the flux into the surface is zero dn' = 0 at a reflecting surface dx An injecting contact: here we can either have the excess minority carrier population specified Option I : n'(x) specified at injecting contact or the minority carrier flux specified Option II : dn' specified at injecting contact dx Clif Fonstad, 9/03 Lecture 4 - Slide 13

14 D. Flow Problems, cont. (Boundary conditions - internal) At any internal point we must always have continuity of the carrier population: At x = X : n'(x + ) = n'(x - ) and, if the generation function, g L, is finite at that point, the gradient of the population is also continuous: dn' g L (X) finite fi = dn' dx x= X + dx x= X - If the generation function is infinite at that point, as is the case if it is modeled as an impulse, then we have: g L (X) = Md(x - X) fi dn' dx x= X + = dn' dx x= X - - M D e Clif Fonstad, 9/03 Lecture 4 - Slide 14

15 D. Flow Problems, cont. (Special solutions relevant to devices) 1. Infinite lifetime or long base situations: These terms refer to situations in which the minority carrier diffusion length is much greater than the sample. In this case we will always have L min >> x, and the n /L min term can be neglected in the steady state diffusion equation, leaving: d 2 n'(x) ª - 1 g dx 2 L (x) D e Integrating twice we arrive immediately at: n'(x) ª - 1 D e ÚÚ g L (x)dxdx + Ax + B 2. Impulse excitation : When g L can be modeled as a single impulse*, then n (x) is only the homogeneous solution: g L (x) = Gd(X) fi n ' (x) = Ae (x-x )/ L e + Be-(x-X )/ L e * Note: Multiple impulses are be handled using superposition. Clif Fonstad, 9/03 Lecture 4 - Slide 15

16 Electronic Devices and Circuits Lecture 4 - Non-uniform Injection (Flow) Problems - Summary Review Thermoelectric soda cooler Non-uniform injection (flow) problems Five assumptions The diffusion equation for minority carriers 2 n'/ x 2 n'/l 2 e = g L (x)/d e (assumes p-type) Getting J e (x), J h (x), E(x), and p'(x), knowing n'(x) J e (x) q D e dn'/dx J h (x) = J tot - J e (x) E(x) [J h (x) + (D h /D e )J e (x)]/q µ h p o p'(x) r(x)/q + n'(x), and r(x) = e de/dx Solving the diffusion equation for n'(x) Homogeneous solutions: exp ±x/l min where L min (D min t min ) 1/2 Particular solution: "looks" like drive Boundary conditions: ohmic contact (n' = p' = 0); reflecting boundary (dn'/dx = dp'/dx = 0); internal boundary (n', p' always continuous; dn'/dx, dp'/dx continuous unless impulse generation) Total solution: HS + PS fit to BCs Clif Fonstad, 9/03 Lecture 4 - Slide 16