MIT6_012F09_lec03

MIT6_012F09_lec03 - 6.012 - Electronic Devices and Circuits...

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Unformatted text preview: 6.012 - Electronic Devices and Circuits Lecture 3 - Solving The Five Equations - Outline • Announcements Handouts - 1. Lecture; 2. Photoconductivity; 3. Solving the 5 eqs. • Review See website for Items 2 and 3. 5 unknowns: n(x,t), p(x,t), Je(x,t), Jh(x,t), E(x,t) 5 equations: Gauss's law (1), Currents (2), Continuity (2) What isn't covered: Thermoelectric effects; Peltier cooling • Special cases we can solve (approximately) by hand Carrier concentrations in uniformly doped material Uniform electric field in uniform material (drift) Low level uniform optical injection (LLI, τmin) Photoconductivity Doping profile problems (depletion approximation) Non-uniform injection (QNR diffusion/flow) (Lect. 1) (Lect. 1) (Lect. 2) (Lect. 2) (Lects. 3,4) (Lect. 5) • Doping profile problems Electrostatic potential Poisson's equation Clif Fonstad, 9/17/09 Lecture 3 - Slide 1 Non-uniform doping/excitation: Summary What we have so far: Five things we care about (i.e. want to know): Hole and electron concentrations: Hole and electron currents: Electric field: p( x, t ) and n ( x, t ) J hx ( x, t ) and J ex ( x, t ) E x ( x, t ) And, amazingly, we already have five equations relating them: "p( x, t ) 1 "J h ( x, t ) + = G # R $ Gext ( x, t ) # [ n ( x, t ) p( x, t ) # n i2 ] r( t ) "t ! q "x "n( x, t ) 1 "J e ( x, t ) Electron continuity: # = G # R $ Gext ( x, t ) # [ n ( x, t ) p( x, t ) # n i2 ] r( t ) "t q "x "p( x, t ) J h ( x, t ) = qµ h p( x, t ) E ( x, t ) # qDh Hole current density: "x "n ( x , t ) Electron current density: J e ( x, t ) = qµ e n ( x, t ) E ( x, t ) + qDe "x " [&( x ) E x ( x, t )] Charge conservation: %( x, t ) = $ q[ p( x, t ) # n ( x, t ) + N d ( x ) # N a ( x )] "x Hole continuity: So...we're all set, right? No, and yes..... Clif Fonstad, 9/17/09 We'll see today that it isn't easy to get a general solution, but we can prevail. Lecture 3 - Slide 2 ! Thermoelectric effects* - the Seebeck and Peltier effects (current fluxes caused by temperature gradients, and visa versa) Hole current density, isothermal conditions: Drift Diffusion $ d [ q# ] ' $ dp ' J h = µ h p &" ) + qDh & " ) % dx ( % dx ( Hole potential energy gradient Concentration gradient Hole current density, non-isothermal conditions: ! Drift Diffusion Seebeck Effect $ d [q# ] ' $ dp ' $ dT ' J h = µ h p &" ) + q Dh & " ) + q Sh p&" ) % dx ( % dx ( % dx ( Temperature gradient Seebeck Effect: ! eltier Effect: P Clif Fonstad, 9/17/09 temperature gradient → current Generator current → temperature gradient Cooler/heater * A cultural item; we will only consider isothermal situations on 6.012 exams and problem sets. Lecture 3 - Slide 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 array like those in thermoelectric generators and solid-state refrigerators. Clif Fonstad, 9/17/09 Ref.: Appendix B in the course text. Lecture 3 - Slide 4 Thermoelectric Generators and Coolers - modern examples Electrical power for a trip to Pluto Cooling/heating for the necessities of life Image of thermoelectric wine cooler removed due to copyright restrictions. Source: NASA. "…electrical power for the New Horizons spacecraft and science instruments [is] provided by a single radioisotope thermoelectric generator, or RTG." Thermoelectric Wine Cooler 28 bottles 12˚C - 18˚C Quiet, gas free, vibration free, environmentally friendly, LED display, interior light. Zhongshan Candor Electric Appl. Co. Launched 1/19/2006 http://www.alibaba.com/ Clif Fonstad, 9/17/09 http://pluto.jhuapl.edu/ Source: NASA/Johns Hopkins University Applied Physics Laboratory/Southwest Research Institute. Lecture 3 - Slide 5 Non-uniform doping/excitation: Back to work (laying the groundwork to model diodes and transistors) What we have: Five things we care about (i.e. want to know): Hole and electron concentrations: Hole and electron currents: Electric field: p( x, t ) and n ( x, t ) J hx ( x, t ) and J ex ( x, t ) E x ( x, t ) And, five equations relating them: "p( x, t ) 1 "J h ( x, t ) + = G # R $ Gext ( x, t ) # [ n ( x, t ) p( x, t ) # n i2 ] r( t ) "t !q "x "n( x, t ) 1 "J e ( x, t ) Electron continuity: # = G # R $ Gext ( x, t ) # [ n ( x, t ) p( x, t ) # n i2 ] r( t ) "t q "x "p( x, t ) J h ( x, t ) = qµ h p( x, t ) E ( x, t ) # qDh Hole current density: "x "n ( x , t ) Electron current density: J e ( x, t ) = qµ e n ( x, t ) E ( x, t ) + qDe "x " [&( x ) E x ( x, t )] Charge conservation: %( x, t ) = $ q[ p( x, t ) # n ( x, t ) + N d ( x ) # N a ( x )] "x Hole continuity: We can get approximate analytical solutions in 5 important cases! Clif Fonstad, 9/17/09 Lecture 3 - Slide 6 ! Solving the five equations: special cases we can handle 1. Uniform doping, thermal equilibrium (nopo product, no, po): " " = 0, = 0, gL ( x, t ) = 0, J e = J h = 0 "x "t Lecture 1 2. Uniform doping and E-field (drift conduction, Ohms law): " " = 0, = 0, gL ( x, t ) = 0, E x constant "x "t ! Lecture 1 3. Uniform doping and uniform low level optical injection (τmin): " = 0, gL ( t ), n ' << po "x ! Lecture 2 3'. Uniform doping, optical injection, and E-field (photoconductivity): ! " = 0, E x constant, gL ( t ) "x Lecture 2 4. Non-uniform doping in thermal equilibrium (junctions, interfaces) ! " = 0, gL ( x, t ) = 0, J e = J h = 0 "t Lectures 3,4 5. Uniform doping, non-uniform LL injection (QNR diffusion) "N d "N a "n ' " = = 0, n ' # p', n' << p o , J e # qDe , # 0 Lecture 5 " "x "x "t !x Clif Fonstad, 9/17/09 Lecture 3 - Slide 7 Non-uniform material with non-uniform excitations (laying the groundwork to model diodes and transistors) Where cases 2, 4, and 5 appear in important semiconductor devices Junction diodes, LEDs: p-type Flo 5 - Flow Case w problem Bipolar transistors: E B F ow r - lem Clasep5obFlow Junctiion- problem Case 4 Junction JCnctio4 problem uase n - Junctions p n-type n-type n-type C F ow r - lems Clasep5obFlow MOS transistors: G S n+ n+ p-type Diode Case 4 - Junctionss Clif Fonstad, 9/17/09 D Drift Case 2 - Drift Depletion approximation Case 4 - Interface (In subthrehsold: Case 5 - Flow) Lecture 3 - Slide 8 Case 4: Non-uniform doping in thermal equilibrium Doping Profiles and p-n Junctions in TE: Na (x), Nd (x) Any time the doping varies with position, we can no longer assume that there is charge neutrality everywhere and that ρ(x) = 0. The dopants are fixed, but the carriers are mobile and diffuse: no(x), Nd-Na Nd-Na Can't say: no ( x) = N D ( x ) ρ(x) < 0 no(x) ! ρ(x) > 0 E(x) x Electron diffusion Electron drift Hole drift Hole diffusion Clif Fonstad, 9/17/09 In T.E., the net fluxes must be zero Lecture 3 - Slide 9 Non-uniform doping in thermal equilibrium, cont. To treat non-uniformly doped materials we begin by looking at them in thermal equilibrium, as we've said. This is useful because in thermal equilibrium we must have: gL ( x, t ) = 0 n ( x, t ) = n o ( x ) p( x, t ) = po ( x ) J e ( x, t ) = 0 J h ( x, t ) = 0 Consequently, the 2 continuity equations in our 5 equations reduce to 0 = 0, e.g.: "n( x, t ) 1 "J e! , t ) (x # = gL ( x, t ) # [ n ( x, t ) $ p( x, t ) # n o ( x ) $ po ( x )] r(T ) "t q "x 0 Clif Fonstad, 9/17/09 ! 0 0 0 Lecture 3 - Slide 10 Non-uniform doping in thermal equilibrium, cont. The third and fourth equations, the current equations, give: dn o ( x ) 0 = qµ e n o ( x ) E ( x ) + qDe dx dpo ( x ) 0 = qµ h po ( x ) E ( x ) " qDh dx ! " # d# De 1 dn o ( x ) = dx µ e n o ( x ) dx d$ Dh 1 dpo ( x ) =" dx µ h po ( x ) dx And Poisson’s equation becomes: dE ( x ) d 2# ( x ) q =" = [ po ( x ) " n o ( x ) + N d ( x ) " N a ( x )] dx dx 2 $ ! ! In the end, we have three equations in our three remaining unknowns, no(x), po(x), and φ(x), so all is right with the world. Clif Fonstad, 9/17/09 Lecture 3 - Slide 11 Non-uniform doping in thermal equilibrium, cont. Looking initially at the first of our new set of equations, we note that both sides can be easily integrated with respect to position : x d" D x 1 dn o ( x ) dx = e # x dx # xo dx µ e o n o ( x ) dx "( x) $ " ( xo ) = De D n ( x) [ln n o ( x ) $ ln n o ( x o )] = e ln o µe µe no ( x o ) Next, raising both sides to the e power yields: n o ( x ) = n o ( x o )e ! µe [ " ( x )#" ( x o )] De We chose intrinsic material as our zero reference for the electrostatic potential: ! and arrive at : Clif Fonstad, 9/17/09 ! " ( x ) = 0 where n o ( x ) = n i n o ( x ) = n ie µe " (x ) De Lecture 3 - Slide 12 Non-uniform doping in thermal equilibrium, cont. From the corresponding equation for holes we also find : po ( x ) = n ie " Next use the Einstein relation: µh # (x ) Dh µh µe q = = Dh De kT Incredibly Multilingually rhyming Note: this relationship rhymes as written, as well as when inverted, and also ! either way in Spanish. It is a very fundamental, and important, relationship! Note : @ R.T. q kT " 40 V #1 and kT q " 25 mV ! Using the Einstein relation we have: n o ( x ) = n ie q" ( x ) kT ! and po ( x ) = n ie# q" ( x ) kT Finally, putting these in Poisson’s equation, a single equation for φ(x) in a doped semiconductor in TE materializes: ! [ d 2" ( x ) q = # n i (e# q" ( x ) / kT # e q" ( x ) / kT ) + N d ( x ) # N a ( x ) dx 2 $ Clif Fonstad, 9/17/09 Lecture 3 - Slide 13 Non-uniform doping in thermal equilibrium, cont. (an aside) What do these equations say? n o ( x ) = n ie q" ( x ) kT and po ( x ) = n ie# q" ( x ) kT To see, consider what they tell us about the ratio of the hole concentration at x2, where the electrostatic potential is φ2, and that at x1, φ1: " q [ # ( x 2 )"# ( x1 )] / kT ! po ( x 2 ) = po ( x1 )e The thermal energy is kT, and the change in potential energy of a hole moved from x1 to x2 is q(φ2 - φ1), so have: ! po ( x 2 ) = po ( x1 )e " #PE x1 $x2 / kT If the potential energy is higher at x2, than at x1, then the population is lower at x2 by a factor e-ΔPE/kT. That is, the population is lower at the top of a potential hill. ! If the potential energy is lower, then the population is higher. That is, the population is, conversely, higher at the bottom of a potential hill. Clif Fonstad, 9/17/09 Lecture 3 - Slide 14 Non-uniform doping in thermal equilibrium, cont. (continuing the aside) The factor e-ΔPE/kT is called a Boltzman factor. It is a factor relating the population densities of particles in many situations, such as gas molecules in an ideal gas under the influence of gravity (i.e, the air above the surface of the earth) and conduction electrons and holes in a semiconductor.* The potential energy difference for holes is qΔφ, while that for electrons is -qΔφ. Thus when we look at the electron and hole populations at a point where the electrostatic potential is φ, relative to those where the potential is zero (and both populations are ni) we have: n o ( x ) = n ie q" ( x ) kT and po ( x ) = n ie# q" ( x ) kT We will return to this picture of populations on either side of a potential hill when we examine at the minority carrier populations on either side of a biased p-n junction. ! * Until the doping levels are very high, in which case the Boltzman factor must be replaced by a Fermi factor.** Clif Fonstad, 9/17/09 ** Don’t worry about it. Lecture 3 - Slide 15 Doing the numbers: I. D to µ conversions, and visa versa To convert between D and µ it is convenient to say in which case q/kT ≈ 40 V-1: Example 1: µe = 1600 cm2/V-s, µh = 600 kT/q ≈ 25 mV, 17˚C/62˚F cm2/V-s De = µe (q kT ) = 1600 / 40 = 40 cm 2 / s Dh = µh (q kT ) = 600 / 40 = 15 cm 2 / s II. Relating φ to n and p, and visa versa To calculate φ knowing n or p it is better to say that ! because then (kT/q)ln10 ≈ 60 mV: kT/q ≈ 26 mV, 28˚C/83˚F Example 1: n-type, ND = Nd - Na = 1016 cm-3 kT 1016 kT kT "n = ln 10 = ln10 6 = ln10 # log10 6 $ 0.06 ln10 6 = 0.36 eV q 10 q q ! Example 2: p-type, NA = Na - Nd = 1017 cm-3 kT 1017 kT " p = # ln 10 = # ln10 $ log10 7 % #0.06 $ 7 = #0.42 eV q 10 q ! Example 3: 60 mV rule: For every order of magnitude the doping is above (below) ni, the potential increases (decreases) by 60 meV. Clif Fonstad, 9/17/09 Lecture 3 - Slide 16 More numbers no[cm-3] po[cm-3] φ [V] 1019 − 1 03 − 0.42 − 1016 − 1 04 − 0.36 − 1015 − 1 05 − 0.30 − 1 06 − 0.24 − 1013 − 1 07 − 0.18 − 1012 − 1 08 − 0.12 − 1011 − 1 09 − 0.06 − 1010 − 1010 − 0.00 − 109 − 1011 − -0.06 − 108 − 1012 − -0.12 − 107 − 1013 − -0.18 − 106 − 1014 − -0.24 − 105 − 1015 − -0.30 − 104 − 1016 − -0.36 − 103 − 1017 − -0.42 − 102 − 1018 − -0.48 − 101 − Clif Fonstad, 9/17/09 0.48 − 1014 − p-type 1 02 − 1017 − Intrinsic 0.54 − 1018 − n-type 1 01 − 1019 − -0.54 − Typical range Typical range Lecture 3 - Slide 17 Non-uniform doping in thermal equilibrium,cont: We have reduced our problem to solving one equation for one unknown, in this case φ(x): [ d 2" ( x ) q = # n i (e# q" ( x ) / kT # e q" ( x ) / kT ) + N d ( x ) # N a ( x ) dx 2 $ Once we find φ(x) we can find no and po from: n o ( x ) = n ie q" ( x ) kT ! and po ( x ) = n ie# q" ( x ) kT Solving Poisson’s equation for φ(x) is in general non-trivial, and for precise answers a "Poisson Solver" program must ! be employed. However, in two special cases we can find very useful, insightful approximate analytical solutions: Case I: Abrupt changes from p- to n-type (i.e., junctions) also: surfaces (Si to air or other insulator) interfaces (Si to metal, Si to insulator, or Si to insulator to metal) Case II: Slowly varying doping profiles. Clif Fonstad, 9/17/09 Lecture 3 - Slide 18 Non-uniform doping in thermal equilibrium, cont.: Case I: Abrupt p-n junctions Consider the profile below: Nd-Na NDn x - NAp p-type po = N Ap , n o = n i2 N Ap "= # kT ln( N Ap / n i ) $ " p q Clif Fonstad, 9/17/09 n-type ? no ( x) = ? po ( x ) = ? "( x) = ? ! n o = N Dn , po = n i2 N Dn kT "= ln( N Dn / n i ) # " n q Lecture 3 - Slide 19 Abrupt p-n junctions, cont: First look why there is a dipole layer in the vicinity of the junction, and a "built-in" electric field. no , po NDn NDn NAp NAp ni2/NAp ni2/NDn x Hole diffusion Electron diffusion ρ(x) Drift balances diffusion in the steady state. qNDn +Q 0 0 x -Q -qNAp Hole drift Clif Fonstad, 9/17/09 Electron drift Lecture 3 - Slide 20 Abrupt p-n junctions, cont: If the charge density is no longer zero there must be an electric field: εEx(x) = ∫ρ(x)dx Ex 0 0 x Epk and an electrostatic potential step: φ(x) = -∫Ex(x)dx φ(x) φn x φp Clif Fonstad, 9/17/09 Ok, but how do we find φ(x)? Lecture 3 - Slide 21 Abrupt p-n Junctions: the general strategy We have to solve an non-linear, second order differential equation for φ: [ ! d 2" ( x ) $( x ) q =# = # n i (e# q" ( x ) / kT # e q" ( x ) / kT ) + N d ( x ) # N a ( x ) dx 2 % % $( x ) Or, alternatively " ( x ) = # && dx + Ax 2 + Bx % In the case of an abrupt p-n junction we have a pretty good idea of what ρ(x) must look like, and we know the details will be lost anyway after integrating twice, so we can try the following iteration strategy: Guess a starting ρ(x). Integrated once to get E(x), and again to get φ(x). Use φ(x) to find po(x), no(x), and, ultimately, a new ρ(x). Compare the new ρ(x) to the starting ρ(x). - If it is not close enough, use the new ρ(x) to iterate again. - If it is close enough, quit. Clif Fonstad, 9/17/09 Lecture 3 - Slide 22 To figure out a good first guess for ρ(x), look at how quickly no and po must change by looking first at how φ changes: φ(x) φn 60 mV φn 700 to 900 mV x 60 mV φp A 60 mV change in φ decreases no and po 10x and ρ increases to 90% of its final value. φp …and what it means for ρ(x): ρ(x) qNDn 90% +Q 0 90% Clif Fonstad, 9/17/09 The change in ρ must be much more abrupt! 0 x -Q -qNAp The observation that ρ changes a lot, when φ changes a little, is the key to the depletion approximation. Lecture 3 - Slide 23 6.012 - Electronic Devices and Circuits Lecture 3 - Solving The Five Equations - Summary • Non-uniform excitation in non-uniform samples The 5 unkowns: n(x,t), p(x,t), Je(x,t), Jh(x,t), E(x,t) The 5 equations: coupled, non-linear differential equations • Special cases we can solve (approximately) Carrier concentrations: Drift: Jdrift = Je,drift + Jh,drift = q (µe no + µh po) E = σo E Low level optical injection: dn'/dt – n'/tmin ≈ gL(t) Doping profile problems: junctions and interfaces Non-uniform injection: QNR flow problems (Lect. 1) (Lect. 2) (Lect. 2) • Using the hand solutions to model devices pn Diodes: two flow problems and a depl. approx. BJTs: three flow problems and two depl. approx.’s MOSFETs: three depl. approx.’s and one drift • Non-uniform doping in T.E. Relating no, po, and electrostatic potential, φ Poisson's equation: two situations important in devices Clif Fonstad, 9/17/09 Lecture 3 - Slide 24 MIT OpenCourseWare http://ocw.mit.edu 6.012 Microelectronic Devices and Circuits Fall 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. ...
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This note was uploaded on 11/07/2011 for the course COMPUTERSC 6.012 taught by Professor Charlesg.sodini during the Fall '09 term at MIT.

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