Clif Fonstad, 9/29/09 Lecture 6 - Slide 1• AnnouncementsFirst Hour Exam - Oct. 7, 7:30-9:30 pm; thru 10/2/09, PS #4• ReviewMinority carrier flow in QNRs: 1. Lmin << w, 2. Lmin >> w• I-V relationship for an abrupt p-n junctionAssume: 1. Low level injection2. All applied voltage appears across junction:3. Majority carriers in quasi-equilibrium with barrier4. Negligible SCL generation and recombinationRelate minority populations at QNR edges, -xp and xn, to vABUse n'(-xp), p'(xn) to find hole and electron currents in QNRsConnect currents across SCL to get total junction current, iD• Features and limitations of the modelEngineering the minority carrier injection across a junctionDeviations at low and high current levelsDeviations at large reverse bias6.012 - Microelectronic Devices and CircuitsLecture 6 - p-n Junctions: I-V Relationship - OutlineClif Fonstad, 9/29/09 Lecture 6 - Slide 2QNR Flow: Uniform doping, non-uniform LL injectionWe use the 5 QNR flow conditions* to simplify our 5 equations...(assuming a p-type sample)! q"(x)p(x,t) # n(x,t) + Nd+(x) # Na#(x)[ ]=dEx(x,t)dx! Jh(x,t) = qµhp(x,t)E(x ,t)" qDh#p(x,t)#x! Je(x,t) = qµen(x,t)E(x,t) + qDe"n(x,t)"x! "p(x,t)"t+1q"Jh(x,t)"x="n(x,t)"t#1q"Je(x,t)"x$ gL(x,t) # n(x,t)p(x,t) # ni2[ ]r(t)! d2n'(x,t)dx2"n'(x,t)De#e= "1DegL(x,t)..and end up with one equation in n': the static diffusion equation!! "1qdJh(x,t)dx= #1qdJe(x,t)dx$ gL(x,t) #n'(x,t)%eQuasi-staticQuasi-staticLLILLIUniform dopingNegligible minority driftQuasi-neutrality! " qµhpo(x)E(x,t) # qDhdn'(x,t)dx! " Je(x,t) # qDedn'(x,t)dx! "q#p'(x,t) $ n'(x,t)[ ]* Five assumptions that define flow problems AND should be validated at the end.Clif Fonstad, 9/29/09 Lecture 6 - Slide 3QNR Flow, cont.: Solving the steady state diffusion equationThe steady state diffusion equation in p-type material is: and for n-type material it is:! d2n'(x)dx2"n'(x)Le2= "1DegL(x)! d2p'(x)dx2"p'(x)Lh2= "1DhgL(x)! d2p'(x)dx2"p'(x)Lh2= 0! d2n'(x)dx2"n'(x)Le2= 0In a basic p-n diode, we have gL = 0 which means we only needthe homogenous solutions, i.e. expressions that satisfy: n-side: p-side:! Le" De#eIn writing these expressions we have introduced Le and Lh,the minority carrier diffusion lengths for holes andelectrons, as:! Lh" Dh#hWe'll see that the minority carrier diffusion length tells us howfar the average minority carrier diffuses before it recombines.Clif Fonstad, 9/29/09 Lecture 6 - Slide 4QNR Flow, cont.: Solving the steady state diffusion equationWe seldom care about this general result. Instead, we findthat most diodes fall into one of two cases: Case I - Long-base diode: wn >> LhCase II - Short-base diode: Lh >> wnCase I: When wn >> Lh, which is the situation in an LED, forexample, the solution is! p'(x) " p'(xn)e# x#xn( )Lh for xn$ x $ wnThis profile decays from p'(xn) to 0 exponentially as e-x//Lh. The corresponding hole current for xn ≤ x ≤ wn in Case I is! Jh(x) " #qDhdp'(x)dx=qDhLhp'(xn)e# x#xn( )Lh for xn$ x $ wnThe current decays to zero also, indicating that all of the excessminority carriers have recombined before getting to the contact.Clif Fonstad, 9/29/09 Lecture 6 - Slide 5QNR Flow, cont.: Solving the steady state diffusion equationCase II: When Lh >> wn, which is the situation in integratedSi diodes, for example, the differential equation simplifiesto:We see immediately that p'(x) is linear:! d2p'(x)dx2=p'(x)Lh2" 0! p'(x) = A x + B! p'(x) " p'(xn) 1#x # xnwn# xn$ % & ' ( ) * + , - . / for xn0 x 0 wnThis profile is a straight line, decreasing from p'(xn) at xn to 0 at wn.Fitting the boundary conditions we find:In Case II the current is constant for xn ≤ x ≤ wn:! Jh(x) " #qDhdp'(x)dx=qDhwn# xnp'(xn) for xn$ x $ wnThe constant current indicates that no carriers recombinebefore reaching the contact.Clif Fonstad, 9/29/09 Lecture 6 - Slide 6wn0 xnp'n(x) [cm-3]x [cm]p'n(xn)Case II - Short base: wn << Ln (the situation in most Si diodes and transistors)x [cm]wn0 xnJh(x) [A/cm2]qDhp'n(xn) [wn-xn]QNR Flow, cont.: Uniform doping, non-uniform LL injection Case I - Long base: wn >> Ln (the situation in LEDs)x [cm]wnJh(x) [A/cm2]qDhp'n(xn) Lhe-x/Lh0 xn xn+Lhwn0 xn xn+Lhp'n(x) [cm-3]x [cm]p'n(xn)e-x/LhSketching and comparing the limiting cases: wn>>Lh, wn<<LhClif Fonstad, 9/29/09 Lecture 6 - Slide 7The four other unknowns⇒ In n-type the steady state diffusion equation gives p’.⇒ Knowing p', we can easily get n’, Je, Jh, and Ex:QNR Flow, cont.: Uniform doping, non-uniform LL injection! Ex(x) "1qµenoJe(x) +DeDhJh(x)# $ % & ' ( Next find Ex:! n'(x) " p'(x) #$qdEx(x)dxThen find n’:Finally, go back and check that all of the five conditions aremet by the solution.! Jh(x) " # qDhdp'(x)dxFirst find Jh:! Je(x) = JTot" Jh(x)Then find Je: Once we solve the diffusion equation and getthe minority carrier excess we know everything.Note: In Lec 5 we saw thisfor a p-type sample.Clif Fonstad, 9/29/09 Lecture 6 - Slide 8There 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), theboundary conditions of the QNR minority carrier flow problems.p nUniform p-type Uniform n-type-wpxwn -xp 0 xnOhmiccontactOhmiccontactA BiD+-vABQuasineutralregion IQuasineutralregion IISpace chargeregionMinority carrier flowhere determines theelectron currentMinority carrier flowhere determines thehole currentThe values of n' at-xp and p' at xn areestablished here.- Today's lecture topic -Current flow: finding the relationship between iD and vABClif Fonstad, 9/29/09 Lecture 6 - Slide 9The p-n Junction Diode: the game plan for getting iD(vAB)We have two QNR's and a flow problem in each:nxwn0 xnOhmiccontactB-Quasineutralregion IIxpOhmiccontactAiD+vABQuasineutralregion I-wp -xp 0-wp -xp 0 wn0 xnn'pp'nn'(-wp) = 0n'(-xp) = ?p'(wn) = 0p'(xn) = ?If we knew n'(-xp) and p'(xn), we could solve the flow problemsand we could get n'(x) for -wp<x<-xp, and p'(x) for xn<x<wn …xxClif Fonstad, 9/29/09 Lecture 6 - Slide 10….and knowing n'(x) for -wp<x<-xp, and p'(x) for xn<x<wn, we canfind Je(x) for -wp<x<-xp, and Jh(x) for xn<x<wn. -wp -xp 0 wn0 xnn'pp'nn'(-wp) = 0n'(-xp,vAB) = ?p'(wn) = 0p'(xn,vAB) = ? Having Je(x) for -wp<x<-xp, and Jh(x) for xn<x<wn, we can get iDbecause 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). We will do this
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