SOUIONS: ECE 606 Homework Week 7 Mark udstrom Purdue Uiversity (revised 3/27/13) 1) Cosider a - type semicoductor for which the oly states i the badgap are door levels (i.e. ( E = E D ). Begi with the SRH formula: 2 p R = i ( + 1 )τ p + ( p + p 1 ) where ( 1 = i e E E i ) k B ( ad p 1 = i e E i E ) k B ad aswer the followig questios. 1a) Derive a expressio for the low- ijectio, miority hole lifetime. 2 ( )( p 0 + Δp) i R = 0 + Δ ( 0 + Δ + 1 )τ p + ( p 0 + Δp + p 1 ) ow level ijectio i a N- type semicoductor meas: 0 + Δ 0 ad p 0 + Δp Δp so: R 0 Δp ( 0 + 1 )τ p + ( Δp + p 1 ) Assume a moderately doped semicoductor, so the Fermi level is located well above Ei but well below EC ad ED. I this case: ( 1 = i e E D E i ) k B ( >> 0 = i e E F E i ) k B ( p 1 = i e E i E D ) k B ( << p 0 = i e E i E F ) k B ad p 0 << 0 so the recombiatio rate becomes: ECE- 606 1
Δp R = ( Δp τ 1 0 )τ p τ hole = ( 1 0 )τ p hole 1b) Explai mathematically why door levels are ot efficiet recombiatio ceters. For door level traps: ( τ hole = ( 1 0 )τ p = e E D E F ) k B τ p >> τ p he hole miority carrier lifetime is very log, so the recombiatio rate is low. 1c) Explai physically why door levels are ot efficiet recombiatio ceters. Door traps ca efficietly capture electros from the coductio bad, but they will quickly be emitted back to the coductio bad. he probability of emissio to the coductio bad is MUCH more efficiet tha the probably of capturig a hole from the valece bad (which is ecessary for recombiatio). 2) Whe computig SRH recombiatio rates, we usually focus o defects with eergy levels ear the middle of the badgap. Begiig with the SRH expressio, show that states ear the middle of the badgap have the largest effect o the SRH recombiatio rate. R = 2 p i ( 1 = i e ) ( + 1 )τ p + p + p 1 ( E E i ) k B ( p 1 = i e E i E ) k B Need to evaluate: R E = 0 R E 2 = ( p i ) 1 2 ( )τ p + ( p + p 1 ) 1 τ E p + p 1 τ E = 0 ( ) + 1 ECE- 606 2
1 τ E p + p 1 τ E = 0 1 = 1 E k B p 1 = p 1 E k B 1 τ p = p 1 1 = e 2 ( E E i ) k B = p 1 τ p E = E i + k B 2 l τ p If is ot too differet from τ p, the the most effective recombiatio ceters will be located ear the itrisic level. 3) Defect states come i two flavors. A door- like state is positive whe empty ad eutral whe filled. A acceptor- like state is eutral whe empty ad egative whe filled. hese states have differet cross sectios for electros ad holes. For example, whe empty, a door- like state has a large cross sectio for electro capture because there is a Coulombic attractio (a typical umber might be 3 10 13 cm 2 ). Whe a door like state is filled, it is eutral ad has a small cross sectio for holes (typically the radius of the defect itself, or about 10 15 cm - 2 ). 3a) Assume a N- type silico sample with a RG ceter cocetratio of 10 12 cm - 3. (Assume that the RG ceters are located ear the middle of the badgap, as is typically the case.) Assume room temperature, so that υ th 10 7 cm/s. Compute the miority hole lifetime assumig that the defects are door- like. Fermi level is well above the trap eergy, so the traps are filled, which meas eutral for door- like traps. his will give a small cross sectio, because there is o Coulombic attractio for holes. So (assumig that the thermal velocity is 10 7 cm/s): τ p = 1 1 = c p N 10 15 10 7 10 = 12 10 4 s = 100 µs ECE- 606 3
τ p = 100 µs 3b) Repeat 3a) assumig that the defects are acceptor- like. (Assume that the RG ceters are located ear the middle of the badgap, as is typically the case.) Whe acceptor like traps are filled, they are egatively charged, which creates a strog attractio for holes. he cross sectio will be large. τ p = 1 1 = c p N 3 10 13 10 7 10 = 12 3.33 10 7 s = 0.33 µs τ p = 0.33 µs 4) he Fermi fuctio gives the probability that a state i the coductio or valece bad is occupied. Oe might thik that the probability that a state i the forbidde gap (i.e. a trap or recombiatio ceter) would also be give by the Fermi fuctio, but this is ot quite right. Begi with eq. (5.9a) i ASF, simplify it for equilibrium, ad obtai the correct result. From eq. (5.9a): r N = c p e I equilibrium: r N 0 = c 0 p 0 0 e 0 0 = 0 From eq. (5.11a): e 0 = c 0 1 so r N 0 = c 0 p 0 0 c 0 1 0 = 0 p 0 0 = 1 0 ECE- 606 4
( N 0 ) 0 = 1 0 ( ) N 0 = 0 1 + 0 0 N = f = 0 1 + 0 = 1 1+ 1 0 1 = e ( E E i ) k B i ( 0 i e E F E i ) = e ( E E i ) k B k B f ( E ) = 1 ( ) k B 1+ e E E F E = E ± k B l g (plus sig for acceptor- like defects ad mius sig for door- like defects) For the plus sig (acceptor- like states): f ( E ) = 1 ( 1+ g e E E F ) k B ad for the mius sig (door- like states): f ( E ) = 1 1+ 1 ( e E E F ) k B g hese are almost Fermi fuctios except for the trap degeeracy factor (like the degeeracy factors we saw for doors ad acceptors). - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Problems 5) 13): he followig problems cocer the Miority Carrier Diffusio equatio for electros as follows: Δ t = D Δ + G For all the followig problems, assume silico at room temperature, uiformly doped with N A = 10 17 cm - 3, µ = 300 cm 2 /V sec, = 10 6 s. From these umbers, we fid: ECE- 606 5
D = k B q µ = 7.8 cm2 s = D = 27.9 µm Uless otherwise stated, these parameters apply to all of the problems below. 5) he sample is uiformly illumiated with light, resultig i a optical geeratio rate G = 10 20 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. 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 Simplify for spatially uiform coditios: 0 = 0 Δ So the simplified MDE equatio is: Δ + G = 0 + G 5b) 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. 5c) Solve the problem. I this case the solutio is trivial: Δ = G = 10 20 10 6 = 10 14 cm -3 ECE- 606 6
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 F p = E i k B l N A i ( Δ >> 0 = i e F E i ) k B = E 1017 0.026l i 10 10 = E 0.41 ev i F = E i + k B l Δ = E 1014 + 0.026l i 10 10 = E + 0.24 ev i i 5d) Provide a sketch of the solutio, ad explai it i words. he excess carrier desity is just costat, idepedet of positio. So are the QF s, but they split because we are ot i equilibrium. he hole QF is essetially where the equilibrium Fermi level was, because the hole cocetratio is virtually uchaged. But the electro QF is much closer to the coductio bad because there are orders of magitude more electros. 6) he sample has bee uiformly illumiated with light for a log time. he optical geeratio rate is G = 10 20 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. 6a) Simplify the Miority Carrier Diffusio Equatio for this problem. Begi with: Δ t = D Δ + G ECE- 606 7
Simplify for spatially uiform coditios with o geeratio: dδ dt = Δ 6b) 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. he iitial coditio is (from prob. 5): Δ( t = 0) = 10 14 cm -3 6c) Solve the problem. dδ dt = Δ he solutio is: Δ( t) = Ae t/ Now use the iitial cotitio: Δ( 0) = 10 14 = A Δ( t) = 10 14 e t/ 6d) Provide a sketch of the solutio, ad explai it i words. For the electro QF: F ( t) = E i + k B l Δ t ( ) + 0 i ECE- 606 8
Iitially, Δ( t) >> 0 ad Δ( t) = Δ( 0)e t /, so F(t) iitially drops liearly with time towards EF. 7) he sample is uiformly illumiated with light, resultig i a optical geeratio rate G = 10 20 cm - 3 sec- 1. he 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 = +. HIN: 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. 7a) 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 he simplified MDE equatio is: D Δ + G = 0 Δ + G = 0 2 D = D Δ + G = 0 where 2 D = legth. D is the miority carrier diffusio 7b) 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, so Δ( x ) = G ECE- 606 9
At the surface, the total umber of e- h pairs recombiig per cm 2 per secod is ( ) R S = Δ 0 τ S Δx = Δx Δ( 0) = S τ F Δ( 0) cm - 2 - s - 1 where S F = Δx S τ S surface recombiatio velocity. cm/s is the frot S F = Δx τ S = 10 6 10 10 = 104 cm/s I steady- state, carriers must diffuse to the surface at the same rate that there are recombiig there so that the excess miority carrier cocetratio at the surface stays costat with time. dδ +D = R dx S = S F Δ( 0) x=0 Note: We usually specify surface recombiatio by just givig the surface recombiatio velocity ot the lifetime ad thickess of the thi layer at the surface. 7c) Solve the problem. Δ + G = 0 2 D Solve the homogeeous problem first G = 0. Δ = 0 solutio is Δ ( x 2 )= Ae x/ + Be+ x/ Now solve for a particular solutio by lettig x where everythig is uiform: Δ 2 + G = 0 he solutio is: Δ = 2 G = G D D ECE- 606 10
Add the two solutios: Δ( x)= Ae x/ + Be+ x/ + G o satisfy the first boudary coditio i 7b): B = 0. Now cosider the secod: dδ +D = D A = S dx x=0 F Δ( 0) = S F ( A + G ) A = D S G τ G F = + S F 1+ D ( ) S F ( ) = G 1 Δ x e x/ ( ) S F 1+ D Check some limits. i) SF = 0 cm/s, which implies that there is o recombiatio at the surface. he we fid: Δ x ( ) = G, which make sese, sice we have spatial uiformity. ( ) = 0, ( ) = G. he 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 o the bulk should take a diffusio legth or two. From the solutio: ( ) = G 1 Δ x For S F, we fid e x/ ( ) S F 1+ D ( ) G 1 e x/ Δ x which behaves as expected. ECE- 606 11
7d) Provide a sketch of the solutio, ad explai it i words. Cocetratio is G i the bulk, but less at the surface, because of surface recombiatio. he trasitio from the surface to the bulk takes place over a distace that is roughly the diffusio legth. he 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. he variatio with positio is ( ) varies expoetially with positio. ECE- 606 12
8) he sample is uiformly illumiated with light, resultig i a optical geeratio rate G = 10 24 cm - 3 sec - 1, but all of the photos are absorbed i a thi layer (10 m wide ear x = 0). Fid the steady state excess miority carrier cocetratio ad QF s vs. positio. You may assume that the sample exteds to x = +. HIN: 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. 8a) Simplify the Miority Carrier Diffusio Equatio for this problem. Begi with: Δ t = D Δ + G Simplify for steady- state: 0 = D Δ + G et s treat the geeratio i a thi surface layer as a boudary coditio, G = 0 he simplified MDE equatio is: D Δ = 0 Δ = 0 D Δ = 0 = D τ 2 Δ = 0 where = D τ 2 is the miority carrier diffusio legth. 8b) 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: Δ( x ) = 0 At the surface, the total umber of e- h pairs geeratio per cm 2 per secod is G S = G Δx = 10 24 10 6 = 10 18 cm -2 s -1. I steady- state, these must diffuse away at the same rate that they are geerated, so ECE- 606 13
D dδ dx x=0 = G S 8c) Solve the problem. Δ = 0 solutios is Δ ( x 2 )= Ae x/ + Be+ x/ o satisfy the first boudary coditio i 8b): B = 0. Now cosider the secod: D dδ dx x=0 + D A = G S A = G S D ( ) = 10 18 7.8 27.9 10 4 = 3.6 1014 cm -3 Δ( x) = G S D ( ) e x/ = ( 3.6 1014 )e x/ 8d) Provide a sketch of the solutio, ad explai it i words. Electros are geerated at the surface ad diffuse ito the bulk, so the cocetratio is high at the surface ad approaches zero several diffusio legths ito the bulk. ECE- 606 14
Deep i the bulk, there are o excess carriers so F = F p = E F. he electro QF must get closer to the coductio bad ear the surface, because the excess electro cocetratio is larger there. he variatio is liear with positio because Δ x ( ) varies expoetially with positio. 9) he sample is uiformly illumiated with light, resultig i a optical geeratio rate G = 10 24 cm - 3 sec - 1, but all of the photos are absorbed i a thi layer (10 m wide ear x = 0). 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. HIN: treat the thi layer at the surface as a boudary coditio do ot try to resolve Δ x ( ) iside this thi layer. 9a) Simplify the Miority Carrier Diffusio Equatio for this problem. Begi with: Δ t = D Δ + G Simplify for steady- state: 0 = D Δ + G et s treat the geeratio i a thi surface layer as a boudary coditio, so G = 0 ; the simplified MDE equatio is: ECE- 606 15
D Δ = 0 Δ = 0 D Δ = 0 = D τ 2 Sice the sample is much thier tha a diffusio legth, we ca igore recombiatio, so = 0. 9b) Specify the iitial ad boudary coditios, as appropriate for this problem. Sice this is a steady- state problem, there is o iitial coditio. At x =, we expect all of the miority carriers to have recombied, so: Δ( x = ) = 0 At the surface, the total umber of e- h pairs geeratio per cm 2 per secod is G S = G Δx = 10 24 10 6 = 10 18 cm -2 s -1. I steady- state, these must diffuse away at the same rate that they are geerated, so D dδ dx x=0 = G S 9c) Solve the problem. = 0 solutios is Δ( x)= Ax + B o satisfy the first boudary coditio i 9b): Δ( )= A + B = 0. B = A Now cosider the secod: dδ D D dx x=0 A = G A = S G S ( ) = 10 18 7.8 5 10 = 6.4 4 1013 cm -3 D ECE- 606 16
Δ( x) = G S ( ) x D ( ) x ( ) = 6.4 10 13 ( ) 9d) Provide a sketch of the solutio, ad explai it i words. Cocetratio icreases towards surface, because geeratio occurs the. Is zero at x = because of the boudary coditio there. Variatio is liear with positio because there is o recombiatio. Electro QF(x) follows from ( x) Δ( x) = i e ( F ( x) E i )/k B. 10) he sample is i the dark, but the excess carrier cocetratio at x = 0 is held costat at Δ 0 ( ) = 10 12 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. 10a) 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 ; ECE- 606 17
the simplified MDE equatio is: D Δ = 0 Δ = 0 D Δ = 0 = D τ 2 Δ = 0 where = D τ 2 is the miority carrier diffusio legth. 10b) 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: Δ( x ) = 0 At the surface, the excess electro cocetratio is held costat, so Δ( x = 0) = 10 12 cm -3 10c) Solve the problem. Δ = 0 solutios is Δ ( x 2 )= Ae x/ + Be+ x/ o satisfy the first boudary coditio i 10b): B = 0. Now cosider the secod: Δ( 0) = 10 12 cm -3 Δ( x) = Δ( 0)e x/ = ( 1012 )e x/ 10d) Provide a sketch of the solutio, ad explai it i words. ECE- 606 18
ooks just like the solutios for prob. 8). Oly differece is that istead of creatig Δ 0 ( ) by geeratio at the surface, we just specify Δ( 0) directly. 11) he sample is i the dark, ad the excess carrier cocetratio at x = 0 is held costat at Δ 0 ( ) = 10 12 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. 11a) 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 MDE equatio is: D Δ = 0 Δ = 0 = D τ 2 Sice the sample is much thier tha a diffusio legth, we ca igore recombiatio, so = 0. 11b) 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: ECE- 606 19
Δ( x = ) = 0 At the surface: Δ( 0) = 10 12 cm -2 11c) Solve the problem. = 0 the solutio is Δ( x)= Ax + B o satisfy the first boudary coditio i 9b): Δ( )= A + B = 0. B = A Now cosider the secod boudary coditio: Δ( 0)= B = 10 12 cm -3 Δ( x) = Δ( 0) ( x) = ( 10 ) 12 ( x) 11d) Provide a sketch of the solutio, ad explai it i words. Just like problem 9) 12) he sample is i the dark, ad the excess carrier cocetratio at x = 0 is held costat at Δ 0 ( ) = 10 12 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. 12a) Simplify the Miority Carrier Diffusio Equatio for this problem. ECE- 606 20
Begi with: Δ t = D Δ + G 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. 12b) Specify the iitial ad boudary coditios, as appropriate for this problem. Steady state, so o iitial coditios are ecessary. he boudary coditios are: Δ( 0) = 10 12 cm - 3 Δ( 30µm) = 0 12c) 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: ECE- 606 21
( )e+ / e / / e+ A = Δ 0 ( ) ( )e / B = Δ 0 e / e+ / ( ) So the solutio is: Δ( x)= Δ( x)= Δ 0 ( ) Δ 0 e / e+ / ( ) e x ( ) sih ( x ) / ( ) sih / ( )/ + e + ( x )/ 12d) Provide a sketch of the solutio, ad explai it i words. he 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. his occurs because some of the electros that flow i, recombie i the structure, so the same umber caot flow out. 13) Cosider a sample that exteds from 5 x 200 μm. he sample is illumiated with light, resultig i a optical geeratio rate G = 10 24 cm - 3 sec - 1, but all of the photos are absorbed i a very thi layer (10 m wide cetered about x = 0). Assume ECE- 606 22
that half of the carriers geerated i this regio diffuse to the left ad half to the right. You may also assume that there are ideal ohmic cotacts, which eforce equilibrium coditios at all times located at x = 5 μm ad at x = 200 μm. Fid the steady state excess miority carrier cocetratio ad QF s vs. positio. Make reasoable approximatios, ad approach the problem as follows. 13a) Simplify the Miority Carrier Diffusio Equatio for this problem. Δ t = D Δ + G Steady- state, o geeratio (treat geeratio ear x = 0 as a boudary coditio). For x > 0: Δ = 0 2 For x < 0, we could also solve this equatio, but the regio is thi for x < 0, so we ca igore recombiatio ad solve: = 0 13b) Specify the iitial ad boudary coditios, as appropriate for this problem. Steady- state, so o iitial coditio. At the boudaries: Δ( x = 5 µm)= 0 Δ( x = 200 µm)= 0 At x = 0 - ad x = 0 + we ca assume that oe- half of the carriers geerated i the thi regio diffuse to the left ad oe- half to the right. ECE- 606 23
dδ D = G Δx = G S dx x=0 2 2 dδ D = G S dx x=0 + 2 13c) Solve the problem. For x < 0 -, the solutios is: Δ( x)= Ax + B At x = - 5 micrometers: Δ( 5 10 4 )= 0 = A 5 10 4 At x = 0 - : dδ( x) G D = D dx A = S A = G S 2 2D x=0 Δ( x)= G S ( x + 5 µm) 2D ( ) + B B= 5 10 4 ( )A For x > 0+: the solutio is: Δ( x)= Ae x/ + Be+ x/ Δ( x)= Ae x/ Because for x > 0, we ca treat the regio as beig ifiitely log. dδ D = D A = G S dx x=0 + 2 G S A = 2 D ( ) Δ( x)= G S ( ) e x/ 2 D Note that Δ( 0 )= G S G ( 5 µm) = S 2D 2( D 5 µm) ad ote that Δ( 0 + )= G S ( ) = G S 2( D 27.9 µm) 2 D ECE- 606 24
so ( ) ( ) = 27.9 Δ 0 + Δ 0 5 = 5.6 13d) Provide a sketch of the solutio, ad explai it i words. Carriers are geerated at x = 0 ad diffuse away to both sides. hey diffuse slower for x > 0, so the carrier desity for x > 0 is higher tha for x < 0. ECE- 606 25