MIT6_012F09_lec05

MIT6_012F09_lec05 - 6.012 - Electronic Devices and Circuits...

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Unformatted text preview: 6.012 - Electronic Devices and Circuits Lecture 5 - p-n Junction Injection and Flow - Outline • Review Depletion approximation for an abrupt p-n junction Depletion charge storage and depletion capacitance (Rec. Fri.) qDP(vAB) = – AqNApxp = – A[2εq(φb-vAB){NApNDn/(NAp+NDn)}]1/2 Cdp(VAB) ≡ ∂ qDP/∂ vAB|VAB = A[εq{NApNDn/(NAp+NDn)}/2(φb-VAB)]1/2 • Biased p-n Diodes Depletion regions change Currents flow: two components – flow issues in quasi-neutral regions – boundary conditions on p' and n' at -xp and xn (Lecture 4) (Today) (Lecture 6) • Minority carrier flow in quasi-neutral regions The importance of minority carrier diffusion Boundary conditions Minority carrier profiles and currents in QNRs – Short base situations – Long base situations – Intermediate situations Clif Fonstad, 9/24/09 Lecture 5 - Slide 1 The Depletion Approximation: an informed first estimate of ρ(x) Assume full depletion for -xp < x < xn, where xp and xn are two unknowns yet to be determined. This leads to: $0 & &#qN Ap "( x ) = % & qN Dn &0 ' for x < #x p for # x p < x < 0 for 0 < x < x n for xn < x ρ(x) qNDn -xp x xn -qNAp Integrating the charge once gives the electric field $ 0 ! & qN &" Ap ( x + x p ) &# Si E ( x) = % & qN Dn ( x " x ) n & #Si & 0 ' Clif Fonstad, 9/24/09 ! for x < "xp for " xp < x < 0 -xp Ε(x) xn for 0 < x < xn for xn < x x E(0) = -qNApxp/εSi = -qNDnxn/εSi Lecture 5 - Slide 2 The Depletion Approximation, cont.: Integrating again gives the electrostatic potential: % "p ' qN Ap 2 ' "p + (x + x p ) ' 2$Si "( x) = & ' " # qN Dn ( x # x ) 2 n ' n 2$Si ' "n ( for x < #x p for - xp < x < 0 φ(x) φn -xp x for for xn 0 < x < xn φp φ(0) = φp + qNApxp 2/2εSi = φn − qNDnxn2/2εSi xn < x Insisting E(x) is continuous at x = 0 yields our first -x equation relating our unknowns, xn and xp: Ε(x) p N Ap x p = N Dn x n xn 1 Requiring that the potential be continuous at x = 0 gives us our second relationship between xn and xp: ! Clif Fonstad, 9/24/09 qN Ap 2 qN Dn 2 "p + x p = "n $ xn 2#Si 2#Si x E(0) = -qNApxp/εSi = -qNDnxn/εSi 2 Lecture 5 - Slide 3 Comparing the depletion approximation with a full solution: Example: An unbiased abrupt p-n junction with NAp= 1017 cm-3, NDn= 5 x 1016 cm-3 Charge Potential po(x), no(x) nie±qφ(x)/kT E-field Clif Fonstad, 9/24/09 Lecture 5 - Slide 4 Courtesy of Prof. Peter Hagelstein. Used with permission. Depletion approximation: Applied bias Forward bias, vAB > 0: φ vAB -w p (φb-vAB) x -xp 0 xn Reverse bias, vAB < 0: vAB wn φ -w p x (φb-vAB) -xp 0 xn No drop in wire No drop at contact wn No drop in wire No drop in QNR No drop in QNR No drop at contact In a well built diode, all the applied voltage appears as a change in the the voltage step crossing the SCL Note: With applied bias we are no longer in thermal equilibrium so it is no longer true that n(x) = ni eqφ(x)/kT and p(x) = ni e-qφ(x)/kT. Clif Fonstad, 9/24/09 Lecture 5 - Slide 5 The Depletion Approximation: Applied bias, cont. Adding vAB to our earlier sketches: assume reverse bias, vAB < 0 2"Si (# b $ v AB ) ( N Ap + N Dn ) w= q N Ap N Dn ρ(x) qNDn w -xp xp xn xn x -qNAp N Ap w N Dn w xp = , xn = (N Ap + N Dn ) (N Ap + N Dn ) Ε(x) ! -xp xn |Epk| x E pk ! 2q (" b # v AB ) N Ap N Dn = $Si (N Ap + N Dn ) φ(x) -xp (φb-vAB) Clif Fonstad, 9/24/09 ! x xn "# = # b $ v AB and kT N Dn N Ap #b = ln q n i2 Lecture 5 - Slide 6 The Depletion Approximation: comparison cont. Example: Same sample, reverse biased -2.4 V Charge E-field Potential p (x) n (x) Clif Fonstad, 9/24/09 nie±qφ(x)/kT Lecture 5 - Slide 7 Courtesy of Prof. Peter Hagelstein. Used with permission. The Depletion Approximation: comparison cont. Example: Same sample, forward biased 0.6 V Charge E-field Clif Fonstad, 9/24/09 Potential p (x) n (x) Courtesy of Prof. Peter Hagelstein. Used with permission. nie±qφ(x)/kT Lecture 5 - Slide 8 The value of the depletion approximation The plots look good, but equally important is that 1. It gives an excellent model for making hand calculations and gives us good values for quantities we care about: • Depletion region width • Peak electric field • Potential step 2. It gives us the proper dependences of these quantities on the doping levels (relative and absolute) and the bias voltage. Apply bias; what happens? Two things happen 1. The depletion width changes • (φb - vAB) replaces φb in the Depletion Approximation Eqs. 2. Currents flow • This is the main topic of today’s lecture Clif Fonstad, 9/24/09 Lecture 5 - Slide 9 Depletion capacitance: Comparing depletion charge stores with a parallel plate capacitor ρ(x) ρ(x) qNDn qA qB ( = -qA) -xp d/2 x x xn qA -d/2 -qNAp qB( = -qA) Depletion region charge store Parallel plate capacitor qA ,PP $qA ,PP C pp (VAB ) # $v AB A" = d ! qA ,DP (v AB ) = " AqN Ap x p (v AB ) " = A v AB d = " A 2q#Si [$ b " v AB ] v AB = VAB Cdp (VAB ) % Many similarities; important differences. Clif Fonstad, 9/24/09 =A &qA ,DP &v AB N Ap N Dn [N Ap + N Dn ] v AB = VAB N Ap N Dn q#Si 2[$ b " VAB ] [ N Ap + N Dn ] = A #Si w (VAB ) Lecture 5 - Slide 10 ! Bias applied, cont.: With vAB ≠ 0, it is not true that n(x) = ni eqφ(x)/kT and p(x) = ni e-qφ(x)/kT because we are no longer in TE. However, outside of the depletion region things are in quasi-equilibrium, and we can define local electrostatic potentials for which the equilibrium relationships hold for the majority carriers, assuming LLI. φQNRp Forward bias, vAB > 0: vAB vAB -w p (φb-vAB) -xp vAB -w p vAB 0 xn φQNRp Reverse bias, vAB < 0: vAB φQNRn wn φQNRn (φb-vAB) vAB -xp 0 xn x vAB x wn vAB In this region p(x) ≈ ni e-qφQNRp/kT Clif Fonstad, 9/24/09 In this region n(x) ≈ ni eqφQNRn/kT Lecture 5 - Slide 11 Current Flow qφ Unbiased junction Population in equilibrium with barrier x qφ Forward bias on junction Barrier lowered so carriers to left can cross over it. x qφ Reverse bias on junction Barrier raised so the few carriers on top spill back down it. x Clif Fonstad, 9/24/09 Lecture 5 - Slide 12 Current flow: finding the relationship between iD and vAB There are two pieces to the problem: • Minority carrier flow in the QNRs is what limits the current. • Carrier equilibrium across the SCR determines n'(-xp) and p'(xn), the boundary conditions of the QNR minority carrier flow problems. Ohmic contact A + Uniform p-type p iD Uniform n-type n - vAB -wp - x p 0 xn Quasineutral region I Minority carrier flow here determines the electron current - Today's Lecture - Clif Fonstad, 9/24/09 Ohmic contact Space charge region The values of n' at -xp and p' at xn are established here. - Lecture 6 - wn B x Quasineutral region II Minority carrier flow here determines the hole current - Today's Lecture - Lecture 5 - Slide 13 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 5. Uniform doping, non-uniform LL injection (QNR diffusion) "N d "N a "n ' " = = 0, n ' # p', n' << p o , J e # qDe , #0 "x "x "t ! "x Clif Fonstad, 9/24/09 Lectures 3,4 TODAY Lecture 5 Lecture 5 - Slide 14 QNR Flow: Uniform doping, non-uniform LL injection What we have: Five things we care about (i.e. want to know): p( x, t ) and n ( x, t ) Hole and electron concentrations: Hole and electron currents: J hx ( x, t ) and J ex ( x, t ) Electric field: 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 Electron continuity: "n( x, t ) # 1 "J e ( x, t ) = G # R $ Gext ( x, t ) # [ n ( x, t ) p( x, t ) # n i2 ] r( t ) "t q "x "p( x, t ) Hole current density: J h ( x, t ) = qµ h p( x, t ) E ( x, t ) # qDh "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 if 5 conditions are met! Clif Fonstad, 9/24/09 Lecture 5 - Slide 15 ! QNR Flow, cont.: Uniform doping, non-uniform LL injection Five unknowns, five equations, five flow problem conditions: 1. Uniform doping dn o dpo # n # n ' # p # p' = =0 " = , = dx dx #x #x #x #x po " n o + N d " N a = 0 # $ = q( p " n + N d " N a ) = q( p' "n ') n' n ' << po " ( np # n ) r $ n ' po r = (in p-type, for example) %e 3. Quasineutrality holds n ' " p' , #n' " #p' # x #x #n' ( x, t ) ! 4. Minority carrier drift is negligible J e ( x, t ) " qDe #x (continuing to assume p-type) ! Note: It is also always true that "n "n ' "p "p' = , = "t "t ! "t "t 2. Low level injection ! ! Clif Fonstad, 9/24/09 2 i Lecture 5 - Slide 16 QNR Flow, cont.: Uniform doping, non-uniform LL injection With these first four conditions our five equations become: (assuming for purposes of discussion that we have a p-type sample) 1, 2 : 3: 4: 5: ! "p' ( x, t ) 1 "J h ( x, t ) "n ' ( x, t ) 1 "J e ( x, t ) n ' ( x, t ) + = # = gL ( x, t ) # "t q "x "t q "x $e "n ' ( x , t ) J e ( x, t ) % +qDe "x "p' ( x, t ) J h ( x, t ) = qµ h p( x, t ) E ( x, t ) + qDh "x "E ( x, t ) q = [ p' ( x, t ) # n ' ( x, t )] "x & In preparation for continuing to our fifth condition, we note that combining Equations 1 and 3 yields one equation in n'(x,t): 2 "n' ( x, t ) " n ' ( x, t ) n ' ( x, t ) # De = gL ( x, t ) # "t "x 2 $e The time dependent diffusion equation Clif Fonstad, 9/24/09 ! Lecture 5 - Slide 17 QNR Flow, cont.: Uniform doping, non-uniform LL injection The time dependent diffusion equation, which is repeated below, is in general still very difficult to solve "n' ( x, t ) " 2 n ' ( x, t ) n ' ( x, t ) # De = gL ( x, t ) # "t "x 2 $e but things get much easier if we impose a fifth constraint: 5. Quasi-static excitation gL ( x, t ) such that all " #0 "t ! With this constraint the above equation becomes a second order linear differential equation: d 2n' ( x ) n' ( x ) ! "De = gL ( x ) " 2 dx #e which in turn becomes, after rearranging the terms : ! d 2 n' ( x ) n' ( x ) 1 " =" gL ( x ) 2 dx De # e De The steady state diffusion equation Clif Fonstad, 9/24/09 Lecture 5 - Slide 18 ! QNR Flow, cont.: Solving the steady state diffusion equation The steady state diffusion equation in p-type material is: d 2 n' ( x ) n' ( x ) 1 " =" gL ( x ) 2 2 dx Le De and for n-type material it is: d 2 p' ( x ) p' ( x ) 1 " =" gL ( x ) 2 2 dx Lh Dh ! In writing these expressions we have introduced Le and Lh, the minority carrier diffusion lengths for holes and electrons, as: Lh " Dh # h Le " De # e ! We'll see that the minority carrier diffusion length tells us how far the average minority carrier diffuses before it recombines. In a basic p! diode, we have gL = 0 which means we only need -n ! the homogenous solutions, i.e. expressions that satisfy: n-side: d 2 p' ( x ) dx 2 Clif Fonstad, 9/24/09 p' ( x ) " =0 2 Lh p-side: d 2 n' ( x ) n' ( x ) " =0 2 2 dx Le Lecture 5 - Slide 19 QNR Flow, cont.: Solving the steady state diffusion equation For convenience, we focus on the n-side to start with and find p'(x) for xn ≤ x ≤ wn. p'(x) satisfies d 2 p' ( x ) p' ( x ) = dx 2 L2 h subject to the boundary conditions: p' ( w n ) = 0 and p' ( x n ) = something we' ll find next time ! The general solution to this static diffusion equation is: p' ( x ) = Ae" x L h + Be + x L h ! where where A and B are constants that satisfy the boundary conditions. Solving for them and putting them into this equation yields the final general result: ! p' ( x n )e( wn " x n ) Lh p' ( x n )e"( wn " x n ) Lh p' ( x ) = ( wn " x n ) Lh e "( x " x n ) L h " ( w n " x n ) L h e +( x " x n ) L h e " e"( w n " x n ) L h e " e"( w n " x n ) L h for x n # x # w n Clif Fonstad, 9/24/09 Lecture 5 - Slide 20 QNR Flow, cont.: Solving the steady state diffusion equation We seldom care about this general result. Instead, we find that most diodes fall into one of two cases: Case I - Long-base diode: wn >> Lh Case II - Short-base diode: Lh >> wn Case I: When wn >> Lh, which is the situation in an LED, for example, the solution is p' ( x ) " p' ( x n ) e#( x # x n ) L h for x n $ x $ wn This profile decays from p'(xn) to 0 exponentially as e-x//Lh. The corresponding hole current for xn ≤ x ≤ wn in Case I is ! dp' ( x ) qDh J h ( x ) " #qDh = p' ( x n )e#( x # x n ) Lh dx Lh for x n $ x $ w n The current decays to zero also, indicating that all of the excess minority carriers have recombined before getting to the contact. ! Clif Fonstad, 9/24/09 Lecture 5 - Slide 21 QNR Flow, cont.: Solving the steady state diffusion equation Case II: When Lh >> wn, which is the situation in integrated Si diodes, for example, the differential equation simplifies to: d 2 p' ( x ) p' ( x ) dx 2 = 2 h L "0 We see immediately that p'(x) is linear: p' ( x ) = A x + B Fitting the boundary conditions we find: ! * $ x # x 'n p' ( x ) " p' ( x n ),1 # & )/! for x n 0 x 0 w n + % w n # x n (. This profile is a straight line, decreasing from p'(xn) at xn to 0 at wn. ! In Case II the current is constant for xn ≤ x ≤ wn: dp' ( x ) qDh J h ( x ) " #qDh = p' ( x n ) dx wn # x n for x n $ x $ wn The constant current indicates that no carriers recombine before reaching the contact. Clif Fonstad, 9/24/09 ! Lecture 5 - Slide 22 QNR Flow, cont.: Uniform doping, non-uniform LL injection Sketching and comparing the limiting cases: wn>>Lh, wn<<Lh Case I - Long base: wn >> Ln (the situation in LEDs) p'n(x) [cm-3] p'n(xn) Jh(x) [A/cm2] qDhp'n(xn) Lh e-x/Lh 0 xn xn+Lh wn x [cm-3] e-x/Lh 0 xn xn+Lh wn x [cm-3] Case II - Short base: wn << Ln (the situation in most Si diodes and transistors) p'n(x) [cm-3] p'n(xn) Jh(x) [A/cm2] qDhp'n(xn) [wn-xn] 0 xn Clif Fonstad, 9/24/09 wn x [cm-3] 0 xn wn x [cm-3] Lecture 5 - Slide 23 QNR Flow, cont.: Uniform doping, non-uniform LL injection The four other unknowns ⇒ Solving the steady state diffusion equation gives n’. ⇒ Knowing n'..... we can easily get p’, Je, Jh, and Ex: First find Je: Then find Jh: ! Next find Ex: ! J e ( x ) " qDe dn ' ( t ) dx J h ( x ) = JTot " J e ( x ) ' 1$ Dh E x ( x) " J e ( x )) &J h ( x ) # qµ h po % De ( # dE x ( x ) q dx ! Finally, go back and check that all of the five conditions are met by the solution. Then find p’: Clif Fonstad, 9/24/09 p' ( x ) " n ' ( x ) + Once we solve the diffusion equation to get ! the minority excess, n', we know everything. Lecture 5 - Slide 24 Current flow: finding the relationship between iD and vAB There are two pieces to the problem: • Minority carrier flow in the QNRs is what limits the current. • Carrier equilibrium across the SCR determines n'(-xp) and p'(xn), the boundary conditions of the QNR minority carrier flow problems. Ohmic contact A + Uniform p-type p iD Uniform n-type n - vAB -wp - x p 0 xn Quasineutral region I Minority carrier flow here determines the electron current Clif Fonstad, 9/24/09 Ohmic contact Space charge region The values of n' at -xp and p' at xn are established here. - Lecture 6 next Tuesday - wn B x Quasineutral region II Minority carrier flow here determines the hole current Lecture 5 - Slide 25 The p-n Junction Diode: the game plan for getting iD(vAB) We have two QNR's and a flow problem in each: Quasineutral region I Ohmic contact A + iD Quasineutral region II p Ohmic contact B n - vAB x -w p -xp 0 n'(-xp) = ? x 0 xn n' p wn p' n p'(xn) = ? p' ( w n ) = 0 n'(-wp) = 0 x -w p -xp 0 x 0 xn wn If we knew n'(-xp) and p'(xn), we could solve the flow problems and we could get n'(x) for -wp<x<-xp, and p'(x) for xn<x<wn … Clif Fonstad, 9/24/09 Lecture 5 - Slide 26 ….and knowing n'(x) for -wp<x<-xp, and p'(x) for xn<x<wn, we can find Je(x) for -wp<x<-xp, and Jh(x) for xn<x<wn. n'(-xp,vAB) = ? n' p p' n p'(xn,vAB) = ? p' ( w n ) = 0 n'(-wp) = 0 -xp 0 x -w p Je(-wp<x<-xp)=qDe(dn'/dx) -wp Je 0 xn x wn Jh Jh(xn<x<wn)=-qDh(dp'/dx) -xp 0 x 0 xn x wn Having Je(x) for -wp<x<-xp, and Jh(x) for xn<x<wn, we can get iD because we will argue that iD(vAB) = A[Je(-xp,vAB)+Jh(xn,vAB)]… …but first we need to know n'(-xp,vAB) and p'(xn,vAB). Clif Fonstad, 9/24/09 We will do this in Lecture 6. Lecture 5 - Slide 27 6.012 - Electronic Devices and Circuits Lecture 5 - p-n Junction Injection and Flow - Summary • Biased p-n Diodes Depletion regions change Currents flow: two components (Lecture 4) – flow issues in quasi-neutral regions – boundary conditions on p' and n' at -xp and xn • Minority carrier flow in quasi-neutral regions The importance of minority carrier diffusion – minority carrier drift is negligible Boundary conditions Minority carrier profiles and currents in QNRs – Short base situations – Long base situations • Carrier populations across the depletion region (Lecture 6) Potential barriers and carrier populations Relating minority populations at -xp and xn to vAB Excess minority carriers at -xp and xn Clif Fonstad, 9/24/09 Lecture 5 - Slide 28 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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