Mark Lundstrom Spring SOLUTIONS: ECE 305 Homework: Week 5. Mark Lundstrom Purdue University

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1 Mark udstrom Sprig 2015 SOUTIONS: ECE 305 Homework: Week 5 Mark udstrom Purdue Uiversity The followig problems cocer the Miority Carrier Diffusio Equatio (MCDE) for electros: Δ t = D Δ + G For all the followig problems, assume p- type silico at room temperature, uiformly doped with N A = cm - 3, µ = 300 cm 2 /V sec, = 10 6 s. From these umbers, we fid: D = k B T q µ = 7.8 cm2 s = D = 27.9 µm Uless otherwise stated, these parameters apply to all of the problems below. 1) The sample is uiformly illumiated with light, resultig i a optical geeratio rate G = cm - 3 sec - 1. Fid the steady- state excess miority carrier cocetratio ad the QF s, F ad F p. Assume spatially uiform coditios, ad approach the problem as follows. 1a) Simplify the Miority Carrier Diffusio Equatio for this problem. Begi with: Δ = D Δ + G t Simplify for steady- state: 0 = D Δ + G Simplify for spatially uiform coditios: 0 = 0 Δ + G So the simplified MCDE equatio is: Δ + G = 0 1b) Specify the iitial ad boudary coditios, as appropriate for this problem. Sice there is o time depedece, there is o iitial coditio. Sice there is o spatial depedece, there are o boudary coditios. ECE Sprig 2015

2 Mark udstrom Sprig 2015 HW5 Solutios (cotiued): 1c) Solve the problem. I this case the solutio is straightforward: Δ = G = = cm -3 Now compute the QFs: Sice we are doped p- type ad i low level ijectio: ( p p 0 = N A = i e E i F p ) k B T F p = E i k B T l N A i = E l i = E 0.41 ev i ( Δ >> 0 = i e F E i ) k B T F = E i + k B T l Δ i = E l i = E ev i Note that sice we are usig the MCDE, we are assumig low- level ijectio. We should check the assumptio that Δ << p. From the solutio ad the give p- type dopig desity, we have: Δ = << p = We are, ideed, i low level ijectio. 1d) Provide a sketch of the solutio, ad explai it i words. The excess carrier desity is costat, idepedet of positio. So are the QF s, but they are split because we are ot i equilibrium. The hole QF is essetially where the equilibrium Fermi level was, because the hole cocetratio is virtually uchaged (low- level ijectio). But the electro QF is much closer to the coductio bad because there are orders of magitude more electros. ECE Sprig 2015

3 Mark udstrom Sprig 2015 HW5 Solutios (cotiued): 2) The sample has bee uiformly illumiated with light for a log time. The optical geeratio rate is G = cm - 3 sec - 1. At t = 0, the light is switched off. Fid the excess miority carrier cocetratio ad the QF s vs. time. Assume spatially uiform coditios, ad approach the problem as follows. 2a) Simplify the Miority Carrier Diffusio Equatio for this problem. Begi with: Δ t = D Δ + G Simplify for spatially uiform coditios with o geeratio: dδ dt = Δ 2b) Specify the iitial ad boudary coditios, as appropriate for this problem. Because there is o spatial depedece, there is o eed to specify boudary coditio. The iitial coditio is (from prob. 1): ( ) = cm -3 Δ t = 0 2c) Solve the problem. dδ = Δ dt The solutio is: Δ( t) = Ae t/ Now use the iitial coditio: Δ t = 0 Δ( t) = e t/ ( ) = = A ECE Sprig 2015

4 Mark udstrom Sprig 2015 HW5 Solutios (cotiued): 2d) Provide a sketch of the solutio, ad explai it i words. F ( ) + 0 For the electro QF: F ( t) = E i + k B T l Δ t i Iitially, Δ( t) >> 0 ad Δ( t) = Δ( 0)e t/, so ( t) iitially drops liearly with time towards E F. 3) The sample is uiformly illumiated with light, resultig i a optical geeratio rate G = cm - 3 sec - 1. The miority carrier lifetime is 1 μsec, except for a thi layer (10 m wide ear x = 0 where the lifetime is 0.1 sec. Fid the steady state excess miority carrier cocetratio ad QF s vs. positio. You may assume that the sample exteds to x = +. HINT: treat the thi layer at the surface as a boudary coditio do ot try to resolve Δ( x) iside this thi layer. Approach the problem as follows. 3a) Simplify the Miority Carrier Diffusio Equatio for this problem. Begi with: Δ t = D Δ + G Simplify for steady- state coditios: 0 = D Δ + G The simplified MCDE equatio is: D Δ + G = 0 Δ + G = 0 2 D = D ECE Sprig 2015

5 Mark udstrom Sprig 2015 HW5 Solutios (cotiued): Δ + G = 0 where 2 D = legth. D is the miority carrier diffusio 3b) Specify the iitial ad boudary coditios as appropriate for this problem. Sice this is a steady- state problem, there is o iitial coditio. As x, we have a uiform semicoductor with a uiform geeratio rate. I a uiform semicoductor uder illumiatio, Δ = G (recall prob. 1)), so ( ) = G Δ x I the thi layer at the surface, the total umber of e- h pairs recombiig per cm 2 per secod is the recombiatio rate per cm 3 per sec, which is R = Δ( 0) τ S cm -3 s -1, times the thickess of the thi layer at the surface, Δx i cm. If we multiply these two quatities, we get the total umber of miority carriers recombiig per cm 2 per s i the surface layer, which we will call R S. Δx R S = R( x) dx Δ 0 Δx cm -2 -s -1. τ S 0 ( ) Rearragig this equatio, we ca write R S = Δx Δ( 0) = S τ F Δ( 0) cm -2 -s -1 S where: S F Δx cm/s τ S is a quatity with the uits of velocity. I practice, we usually do t kow the thickess of the low- lifetime layer at the surface or the lifetime i this layer, so istead, we just specify the frot surface recombiatio velocity. Typically, 0 < S F 10 7 cm/s. For this specific problem, S F = Δx τ S = 10 6 cm s = 104 cm/s The surface recombiatio velocity is simply a way to specify the stregth of the recombiatio rate (i cm s - 1 ) at the surface: R S = Δx τ S Δ 0 ( ) = S F Δ( 0) cm -2 -s -1 ECE Sprig 2015

6 Mark udstrom Sprig 2015 HW5 Solutios (cotiued): I steady- state, carriers must diffuse to the surface at the same rate that they are recombiig there so that the excess miority carrier cocetratio at the surface stays costat with time. The diffusio curret of electros at the surface is dδ J = qd dx A/cm2. The flux of electros i the +x directio is J q = D dδ dx cm -2 -s -1 The flux of electros i the x directio (to the surface where they are recombiig) is: J q = +D dδ cm -2 -s -1 dx I steady- state, this flux of electros flows to the surface at exactly the same rate that they recombie at the surface, so the boudary coditio is dδ +D = R dx S = S F Δ( 0) x=0 Note: Specifyig surface recombiatio by just givig the surface recombiatio velocity ot the lifetime ad thickess of the thi layer at the surface, is commo practice i semicoductor work. 3c) Solve the problem. Δ + G = 0 2 D To solve the homogeeous problem first, we set G = 0. Δ = 0 The solutio is Δ ( x 2 )= Ae x/ + Be+ x/ Now solve for a particular solutio by lettig x + where everythig is uiform: Δ + G = 0 The solutio is: Δ = 2 G 2 = G D D Add the two solutios: Δ( x)= Ae x/ + Be+ x/ + G ECE Sprig 2015

7 Mark udstrom Sprig 2015 HW5 Solutios (cotiued): To satisfy the first boudary coditio as x, B = 0. Now cosider the boudary coditio at x = 0 : dδ +D = D A = S dx x=0 F Δ( 0) = S F ( A + G ) A = S G τ G F = D + S F 1+ D ( ) = G 1 Δ x Check some limits. ( ) S F e x/ ( ) S F 1+ D i) SF = 0 cm/s, which implies that there is o recombiatio at the surface. The we fid: Δ( x) = G, which make sese, sice we have spatial uiformity. ( ) = 0, ( ) = G. The trasitio from 0 to ii) S F. Strog recombiatio at the surface should force Δ x = 0 but i the bulk we should still have Δ x a fiite value i the bulk should take a diffusio legth or two. From the solutio: e Δ( x) x/ = G 1 1+ D For S F, we fid ( ) G 1 e x/ Δ x ( ) S F which behaves as expected. ECE Sprig 2015

8 Mark udstrom Sprig 2015 HW5 Solutios (cotiued): 3d) Provide a sketch of the solutio, ad explai it i words. The cocetratio is G i the bulk, but less at the surface, because of surface recombiatio. The trasitio from the surface to the bulk takes place over a distace that is a few diffusio legths log. The hole QF is costat ad almost exactly where the equilibrium Fermi level was, because we are i low level ijectio (the hole cocetratio is very, very ear its equilibrium value). But the electro QF is much closer to the coductio bad edge. It moves away from EC ear the surface, because surface recombiatio reduces Δ x liear, because Δ x ( ) ear the surface. The variatio with positio is ( ) varies expoetially with positio. Note: Typically, i semicoductor work, these kids of problems are stated directly i terms of surface recombiatio velocities ad ot i terms of very thi layers with low lifetime. The way problem 3) would usually be stated is as follows: ECE Sprig 2015

9 Mark udstrom Sprig 2015 HW5 Solutios (cotiued): 3 ) The sample is uiformly illumiated with light, resultig i a optical geeratio rate G = cm - 3 sec - 1. The miority carrier lifetime is 1 μsec. At x = 0 the surface recombiatio velocity is S F = 10 4 cm/s. Fid the steady state excess miority carrier cocetratio ad QF s vs. positio. You may assume that the sample exteds to x = +. Approach the problem as follows. 3a) Simplify the Miority Carrier Diffusio Equatio for this problem. 3b) Specify the iitial ad boudary coditios, as appropriate for this problem. 3c) Solve the problem. 3d) Provide a sketch of the solutio, ad explai it i words. 4) The sample is i the dark, but the excess carrier cocetratio at x = 0 is held costat at Δ 0 ( ) = cm- 3. Fid the steady state excess miority carrier cocetratio ad QF s vs. positio. You may assume that the sample exteds to x = +. Make reasoable approximatios, ad approach the problem as follows. 4a) Simplify the Miority Carrier Diffusio Equatio for this problem. Begi with: Δ t = D Δ + G Simplify for steady- state: 0 = D Δ + G No geeratio: G = 0 the simplified MCDE equatio is: D Δ = 0 Δ = 0 D Δ = 0 where = D τ 2 Δ = 0 = D τ 2 is the miority carrier diffusio legth. 4b) Specify the iitial ad boudary coditios, as appropriate for this problem. Sice this is a steady- state problem, there is o iitial coditio. As x, we expect all of the miority carriers to have recombied, so: ( ) = 0 Δ x + ECE Sprig 2015

10 Mark udstrom Sprig 2015 HW5 Solutios (cotiued): At the surface, the excess electro cocetratio is held costat, so Δ( x = 0) = cm -3 4c) Solve the problem. Δ = 0 The solutio is Δ ( x 2 )= Ae x/ + Be+ x/ To satisfy the first boudary coditio i 4b): B = 0. Now cosider the secod: Δ( 0) = cm -3 Δ( x) = Δ( 0)e x/ = ( 1012 )e x/ 4d) Provide a sketch of the solutio, ad explai it i words. ECE Sprig 2015

11 Mark udstrom Sprig 2015 HW5 Solutios (cotiued): 5) The sample is i the dark, ad the excess carrier cocetratio at x = 0 is held costat at Δ 0 ( ) = cm - 3. Fid the steady state excess miority carrier cocetratio ad QF s vs. positio. Assume that the semicoductor is oly 5 μm log. You may also assume that there is a ideal ohmic cotact at x = = 5μm, which eforces equilibrium coditios at all times. Make reasoable approximatios, ad approach the problem as follows. 5a) Simplify the Miority Carrier Diffusio Equatio for this problem. Begi with: Δ t = D Δ + G Simplify for steady- state: 0 = D Δ + G Geeratio is zero for this problem: G = 0 ; The simplified MCDE equatio is: D Δ = 0 Δ = 0 2 = D Sice the sample is much thier tha a diffusio legth, we ca igore recombiatio, so = 0. 5b) Specify the iitial ad boudary coditios, as appropriate for this problem. Sice this is a steady- state problem, there is o iitial coditio. As x, we expect all of the miority carriers to have recombied, so: ( ) = 0 Δ x = At the surface: Δ( 0) = cm -2 5c) Solve the problem. = 0 The geeral solutio is Δ( x)= Ax + B To satisfy the first boudary coditio i 5b): Δ( )= A + B = 0. A = B Δ( x)= B x + B = B 1 x ( ) ECE Sprig 2015

12 Mark udstrom Sprig 2015 HW5 Solutios (cotiued): Now cosider the secod boudary coditio: Δ( 0)= B = cm -3 Δ x ( ) 1 x ( ) = Δ( 0) ( 1 x ) = ( ) 5d) Provide a sketch of the solutio, ad explai it i words. 6) The sample is i the dark, ad the excess carrier cocetratio at x = 0 is held costat at Δ 0 ( ) = cm - 3. Fid the steady state excess miority carrier cocetratio ad QF s vs. positio. Assume that the semicoductor is 30 μm log. You may also assume that there is a ideal ohmic cotact at x = = 30 μm, which eforces equilibrium coditios at all times. Make reasoable approximatios, ad approach the problem as follows. 6a) Simplify the Miority Carrier Diffusio Equatio for this problem. Begi with: Δ = D Δ + G t Simplify for steady- state ad o geeratio: D Δ = 0 Δ = 0 D Δ = 0 = D τ 2 Δ = 0 where = D τ 2 is the miority carrier diffusio legth. ECE Sprig 2015

13 Mark udstrom Sprig 2015 HW5 Solutios (cotiued): 6b) Specify the iitial ad boudary coditios, as appropriate for this problem. Steady state, so o iitial coditios are ecessary. The boudary coditios are: Δ 0 ( ) = cm - 3 Δ( 30µm) = 0 6c) Solve the problem. Δ( x)= Ae x/ + Be+ x/ Because the regio is about oe diffusio legth log, we eed to retai both solutios. Δ( 0)= A + B Δ( = 30 µm)= Ae / + Be+ / = 0 Solve for A ad B to fid: A = Δ( 0)e+ / e / / ( e+ ) ( )e / B = Δ 0 e / e+ / ( ) So the solutio is: Δ( 0) Δ( x)= e / e+ / Δ( x)= Δ 0 ( ) e x ( ) sih ( x ) / ( ) sih / ( )/ + e + ( x )/ ECE Sprig 2015

14 Mark udstrom Sprig 2015 HW5 Solutios (cotiued): 6d) Provide a sketch of the solutio, ad explai it i words. The short base result is liear, but i this case, the slope i a little steeper iitially ad a little shallower at the ed. Sice the diffusio curret is proportioal to the slope, this meas that iflow greater tha outflow. This occurs because some of the electros that flow i, recombie i the structure, so the same umber caot flow out. ECE Sprig 2015