6.012 - Electronic Devices and Circuits Lecture 5 - p-n Junction Injection and Flow - Outline Review Depletion approimation for an abrupt p-n junction Depletion charge storage and depletion capacitance (Rec. Fri.) q DP (v AB ) = AqN Ap p = A[2εq(φ b -v AB ){N Ap N Dn /(N Ap +N Dn )}] 1/2 C dp (V AB ) q DP / v AB VAB = A[εq{N Ap N Dn /(N Ap +N Dn )}/2(φ b -V AB )] 1/2 Biased p-n Diodes Depletion regions change (Lecture 4) Currents flow: two components flow issues in quasi-neutral regions (Today) boundary conditions on p' and n' at - p and n (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 Approimation: an informed first estimate of ρ() Assume full depletion for - p < < n, where p and n are two unknowns yet to be determined. This leads to: $ 0 for < # ρ() p qn Dn &#qn Ap for # p < < 0 "() = % & '& qn Dn 0 for for 0 < < n n < - p -qn Ap n Integrating the charge once gives the electric field $ 0 for < " p & " qn Ap - & ( + p ) for " p < < 0 p E() = % # Si & qn Dn ( " n ) for 0 < < & n # Si '& 0 for n < Ε() n E(0) = -qn Ap p /ε Si = -qn Dn n /ε Si Clif Fonstad, 9/24/09 Lecture 5 - Slide 2
The Depletion Approimation, cont.: Integrating again gives the electrostatic potential: % " p for < # p ' " p + qn Ap ( + p ) 2 ' for - p < < 0 "() = & 2$ Si ' " n # qn Dn ( # n ) 2 for 0 < < ' n 2$ Si (' " n for n < φ n φ() - p n Insisting E() is continuous at = 0 yields our first Ε() equation relating our unknowns, n and p : - p N Ap p = N Dn n Requiring that the potential be continuous at = 0 gives us our second relationship between n and p : " p + qn Ap 2 p = " n $ qn Dn 2 n 2# Si 2# Si 1 2 φ p φ(0) = φ p + qn Ap p 2 /2ε Si = φ n qn Dn n 2 /2ε Si n E(0) = -qn Ap p /ε Si = -qn Dn n /ε Si Clif Fonstad, 9/24/09 Lecture 5 - Slide 3
Comparing the depletion approimation with a full solution: Eample: An unbiased abrupt p-n junction with N Ap = 10 17 cm -3, N Dn = 5 10 16 cm -3 Charge Potential E-field p o (), n o () n i e ±qφ()/kt Clif Fonstad, 9/24/09 Lecture 5 - Slide 4 Courtesy of Prof. Peter Hagelstein. Used with permission.
Depletion approimation: Applied bias Forward bias, v AB > 0: φ v AB -w p (φ b -v AB ) - p 0 n w n Reverse bias, v AB < 0: φ v AB No drop in wire No drop at contact -w p (φ b -v AB ) - p 0 n w n 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() = n i e qφ()/kt and p() = n i e -qφ()/kt. Clif Fonstad, 9/24/09 Lecture 5 - Slide 5
The Depletion Approimation: Applied bias, cont. Adding v AB to our earlier sketches: assume reverse bias, v AB < 0 - p - p E pk p qn Dn w ρ() n -qn Ap Ε() φ() n n w = 2" Si (# b $ v AB ) q p = ( N Ap + N Dn ) N Ap N Dn N Dn w ( N Ap + N Dn ), n = E pk = 2q ( " # v b AB ) N Ap w ( N Ap + N Dn ) ( ) N N Ap Dn $ Si N Ap + N Dn - p (φ b -v AB ) n "# = # b $ v AB and # b = kt q ln N Dn N Ap n i 2 Clif Fonstad, 9/24/09 Lecture 5 - Slide 6
The Depletion Approimation: comparison cont. Eample: Same sample, reverse biased -2.4 V Charge Potential E-field p () n () n i e ±qφ()/kt Clif Fonstad, 9/24/09 Courtesy of Prof. Peter Hagelstein. Used with permission. Lecture 5 - Slide 7
The Depletion Approimation: comparison cont. Eample: Same sample, forward biased 0.6 V Charge Potential E-field p () n () n i e ±qφ()/kt Clif Fonstad, 9/24/09 Lecture 5 - Slide 8 Courtesy of Prof. Peter Hagelstein. Used with permission.
The value of the depletion approimation The plots look good, but equally important is that 1. It gives an ecellent 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 (φ b - v AB ) replaces φ b in the Depletion Approimation 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() q A ρ() qn Dn -d/2 d/2 q B ( = -q A ) Parallel plate capacitor q A,PP = A " d v AB C pp (V AB ) # $q A,PP $v AB v AB =V AB = A" d - p q A q B ( = -q A ) -qn Ap n Depletion region charge store q A,DP (v AB ) = "AqN Ap p C dp (V AB ) % &q A,DP &v AB ( ) v AB = "A 2q# Si [ $ b " v AB ] v AB =V AB N Ap N Dn [ N Ap + N Dn ] Many similarities; important differences. = A q# Si [ ] 2 $ b "V AB N Ap N Dn [ N Ap + N Dn ] = A # Si w(v AB ) Clif Fonstad, 9/24/09 Lecture 5 - Slide 10
Bias applied, cont.: With v AB 0, it is not true that n() = n i e qφ()/kt and p() = n i e -qφ()/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. Forward bias, v AB > 0: φ QNRp φ QNRn v AB v AB -w p (φ b -v AB ) - p v AB 0 n w n v AB Reverse bias, v AB < 0: v AB (φ b -v AB ) φ QNRp φ QNRn v AB -w p - p 0 n v AB w n v AB In this region p() n i e -qφqnrp/kt In this region n() n i e qφqnrn/kt Clif Fonstad, 9/24/09 Lecture 5 - Slide 11
Current Flow Unbiased junction Population in equilibrium with barrier Forward bias on junction Barrier lowered so carriers to left can cross over it. qφ qφ qφ Reverse bias on junction Barrier raised so the few carriers on top spill back down it. Clif Fonstad, 9/24/09 Lecture 5 - Slide 12
Current flow: finding the relationship between i D and v AB 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'(- p ) and p'( n ), the boundary conditions of the QNR minority carrier flow problems. Ohmic Uniform p-type Uniform n-type contact Ohmic contact i D p A B + - v AB -w p - p 0 n w n n Minority carrier flow here determines the electron current - Today's Lecture - Quasineutral Space charge Quasineutral region I region region II The values of n' at - p and p' at n are established here. Minority carrier flow here determines the hole current - Today's Lecture - Clif Fonstad, 9/24/09 - Lecture 6 - Lecture 5 - Slide 13
Solving the five equations: special cases we can handle 1. Uniform doping, thermal equilibrium (n o p o product, n o, p o ): " " = 0, " "t = 0, g L (,t) = 0, J e = J h = 0 2. Uniform doping and E-field (drift conduction, Ohms law): " " = 0, " "t = 0, g L (,t) = 0, E constant 3. Uniform doping and uniform low level optical injection ( τ min ): " " = 0, g L (t), n'<< p o Lecture 1 Lecture 1 Lecture 2 3'. Uniform doping, optical injection, and E-field (photoconductivity): " " = 0, E constant, g L (t) Lecture 2 4. Non-uniform doping in thermal equilibrium (junctions, interfaces) J = J = 0 e h " "t = 0, g (,t) = 0, L TODAY 5. Uniform doping, non-uniform LL injection (QNR diffusion) "N d " = "N a " = 0, n'# p', n'<< p "n' o, J e # qd e ", " "t # 0 Lectures 3,4 Lecture 5 Clif Fonstad, 9/24/09 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): Hole and electron concentrations: Hole and electron currents: Electric field: And, five equations relating them: Hole continuity: Electron continuity: Hole current density: Electron current density: Charge conservation: "p(,t) "t "n(,t) "t + 1 q # 1 q "J h (,t) " "J e (,t) " p(,t) J h (,t) E (,t) and n(,t) and J e (,t) J h (,t) = qµ h p(,t)e(,t) # qd h "p(,t) " J e (,t) = qµ e n(,t)e(,t) + qd e "n(,t) " %(,t) = " [ &()E (,t) ] " 2 = G # R $ G et (,t) # [ n(,t) p(,t) # n i ]r(t) 2 = G # R $ G et (,t) # [ n(,t) p(,t) # n i ]r(t) $ q[ p(,t) # n(,t) + N d () # N a ()] We can get approimate 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 d = dp o d = 0 " #n # = #n' #, #p # = #p' # p o " n o + N d " N a = 0 # $ = q( p " n + N d " N a ) = q( p'"n' ) 2. Low level injection 2 n'<< p o " ( np # n i )r $ n' p o r = n' (in p-type, for eample) % e 3. Quasineutrality holds #n' n'" p', # " #p' # 4. Minority carrier drift is negligible (continuing to assume p-type) J e (,t) " qd e #n'(,t) # Note: It is also always true that "n "t = "n' "t, "p "t = "p' "t Clif Fonstad, 9/24/09 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 : "p'(,t) "t + 1 q "J h (,t) " 3 : J e (,t) % +qd e "n'(,t) " = "n'(,t) "t In preparation for continuing to our fifth condition, we note that combining Equations 1 and 3 yields one equation in n'(,t): "n'(,t) " 2 n'(,t) # D e = g "t " 2 L (,t) # n'(,t) $ e The time dependent diffusion equation # 1 q "p'(,t) 4 : J h (,t) = qµ h p(,t)e(,t) + qd h " "E(,t) 5 : = q [ p'(,t) # n'(,t) ] " & "J e (,t) " = g L (,t) # n'(,t) $ e 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'(,t) " 2 n'(,t) # D e = g "t " 2 L (,t) # n'(,t) $ e but things get much easier if we impose a fifth constraint: 5. Quasi-static ecitation g L (,t) such that all " "t # 0 With this constraint the above equation becomes a second order linear differential equation: d 2 n'() "D e = g d 2 L () " n'() # e which in turn becomes, after rearranging the terms : d 2 n'() d 2 " n'() D e # e = " 1 D e g L () 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'() " n'() = " 1 g d 2 2 L () L e D e and for n-type material it is: d 2 p'() " p'() = " 1 g d 2 2 L () L h D h In writing these epressions we have introduced L e and L h, the minority carrier diffusion lengths for holes and electrons, as: L e " D e # e L h " D h # h 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-n diode, we have g L = 0 which means we only need the homogenous solutions, i.e. epressions that satisfy: n-side: d 2 p'() d 2 " p'() L h 2 = 0 p-side: d 2 n'() d 2 " n'() L e 2 = 0 Clif Fonstad, 9/24/09 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'() for n w n. p'() satisfies d 2 p'() = p'() d 2 2 L h subject to the boundary conditions: p'(w n ) = 0 and p'( n ) = something we'll find net time The general solution to this static diffusion equation is: p'() = Ae " L h + Be+ L h 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'( p'() = n )e w n " n ) L h ( ) L h " e " ( w n " n ) L h e " ( " n) L h p'( " n )e " ( w n " n ) L h ( ) L h " e " w n " n e w n " n for n # # w n e w n " n ( ) L h e + ( " n) L h 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: w n >> L h Case II - Short-base diode: L h >> w n Case I: When w n >> L h, which is the situation in an LED, for eample, the solution is p'() " p'( n )e # ( # n ) L h for n $ $ w n This profile decays from p'( n ) to 0 eponentially as e -//Lh. The corresponding hole current for n w n in Case I is J h () " #qd h dp'() d = qd h L h p'( n )e # ( # n ) L h for n $ $ w n The current decays to zero also, indicating that all of the ecess 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 L h >> w n, which is the situation in integrated Si diodes, for eample, the differential equation simplifies to: d 2 p'() = p'() " 0 d 2 2 L h We see immediately that p'() is linear: Fitting the boundary conditions we find: p'() = A + B * $ p'() " p'( n ) 1# # '- n, & )/ for n 0 0 w n + % w n # n (. This profile is a straight line, decreasing from p'( n ) at n to 0 at w n. In Case II the current is constant for n w n : dp'() J h () " #qd h = qd h p'( n ) for n $ $ w n d w n # n 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: w n >>L h, w n <<L h Case I - Long base: w n >> L n (the situation in LEDs) p' () [cm -3 ] J h () [A/cm 2 ] n p' n ( n ) e -/Lh qd h p' n ( n ) L h e -/Lh 0 n n +L h w n [cm -3 ] 0 n n +L h w n [cm -3 ] Case II - Short base: w << L n n (the situation in most Si diodes and transistors) p' () [cm -3 ] J h () [A/cm 2 ] n p' n ( n ) qd h p' n ( n ) [w n - n ] 0 w [cm -3 ] 0 w [cm -3 ] n n n n Clif Fonstad, 9/24/09 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, J e, J h, and E : First find J e : Then find J h : J e () " qd e dn'(t) d J h () = J Tot " J e () Net find E : E () " 1 $ J h () # D h & qµ h p o % D e ' J e ()) ( Then find p : p'() " n'() + # q de () d Finally, go back and check that all of the five conditions are met by the solution. Once we solve the diffusion equation to get the minority ecess, n', we know everything. Clif Fonstad, 9/24/09 Lecture 5 - Slide 24
Current flow: finding the relationship between i D and v AB 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'(- p ) and p'( n ), the boundary conditions of the QNR minority carrier flow problems. Ohmic Uniform p-type Uniform n-type contact Ohmic contact i D p A B + - v AB -w p - p 0 n w n n Minority carrier flow here determines the electron current Quasineutral Space charge Quasineutral region I region region II The values of n' at - p and p' at n are established here. Minority carrier flow here determines the hole current Clif Fonstad, 9/24/09 - Lecture 6 net Tuesday - Lecture 5 - Slide 25
The p-n Junction Diode: the game plan for getting i D (v AB ) We have two QNR's and a flow problem in each: Ohmic contact Quasineutral region I Quasineutral region II Ohmic contact A i D + vab p n - B -w p - p 0 0 n w n n'(-w p ) = 0 n'(- p ) =? n' p p' n p'( n ) =? -w p - p 0 0 n w n p'(w n ) = 0 If we knew n'(- p ) and p'( n ), we could solve the flow problems and we could get n'() for -w p <<- p, and p'() for n <<w n Clif Fonstad, 9/24/09 Lecture 5 - Slide 26
.and knowing n'() for -w p <<- p, and p'() for n <<w n, we can find J e () for -w p <<- p, and J h () for n <<w n. n'(-w p ) = 0 n'(- p,v AB ) =? n' p p' n p'( n,v AB ) =? p'(w n ) = 0 -w p - p 0 0 n w n Je (-w p <<- p )=qd e (dn'/d) J e J h J h ( n <<w n )=-qd h (dp'/d) -w p - p 0 0 n w n Having J e () for -w p <<- p, and J h () for n <<w n, we can get i D because we will argue that i D (v AB ) = A[J e (- p,v AB )+J h ( n,v AB )] but first we need to know n'(- p,v AB ) and p'( n,v AB ). 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 (Lecture 4) Currents flow: two components flow issues in quasi-neutral regions boundary conditions on p' and n' at - p and n 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 - p and n to v AB Ecess minority carriers at - p and n Clif Fonstad, 9/24/09 Lecture 5 - Slide 28
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