UNIVERSITY OF CALIFORNIA, BERKELEY DEPARTMENT OF ELECTRICAL ENGINEERING AND COMPUTER SCIENCES. Midterm I

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1 UNIVERSITY OF CALIFORNIA, BERKELEY EPARTMENT OF ELECTRICAL ENGINEERING AN COMPUTER SCIENCES EECS 130 Professor Chemg Hu Fall 009 Mdterm I Name: Closed book. Oe sheet of otes s allowed. There are 8 pages of ths exam cludg ths page. Problem 1 30 Problem 40 Problem 3 30 Total 100 1

2 Physcal Costats Electroc charge q 1.60 x C Permttvty of vacuum ε x F cm -1 Relatve permttvty of slco ε s/ ε Relatve permttvty of SO ε ox/ ε Boltzma s costat k x 10-5 ev K -1 or 1.38 x 10-3 J K -1 Thermal voltage at T = 300K /q 0.06V Effectve desty of states (S) N c.8 x cm -3 Effectve desty of states (S) N v 1.04 x cm -3 Slco Bad Gap E g 1.1eV Itrsc carrer cocetrato of S at 300K 1.5 x cm -3 GaAs Bad Gap E g 1.4eV

3 Problem 1: Compare Slco wth GaAs a) Wthout dopg, whch materal has a larger trsc coductvty at room temperature Coductvty σ = q..(µ + µ p ) For GaAs, E g s lower => s orders of magtude lower. ( Pts) (moblty for GaAs s oly ~ 5 tmes hgher) So GaAs has lower coductvty. (1 Pts) b) Based o bad gap, estmate order of magtude for GaAs. E ( GaAs) E E = => = ( Pts) g,, 5..exp( ) g S g GaAs Nc Nv exp( ) 10 ( S) => ( GaAs) 10 cm ( Pts) 15 3 Assumed N c ad N v are the same for both the materals cm s accepted too. c) GaAs has about the same lattce costat as S. What s the rato of mass destes (g/cm 3 ) for GaAs to that of S. (Atomc Weghts of {(S 8.09), (Ga 69.7), (As 74.9)} I a ut lattce cell there are 8 atoms of S. Ad 4 of Ga ad 4 of As case of GaAs. Both of them are gve to have same lattce cotats =>Volume s same ρgaas 4 M Ga + 4 M As Rato of destes, = = 0.5 =. 57 ρ 8 M 8.09 S S (3 Pts) d) Assume that effectve mass (m * ) s 0.07m 0 for GaAs ad 0.6m 0 for S, re-estmate the value for GaAs. 3 N c would chage as N ( m*) c c v 3 g ( GaAs) m * g, S GaAs g, GaAs ( ) * S E E E = N. N.exp( ) => = exp( ) S m ( Pts) => ( GaAs) 10 cm ( Pts) cm s also accepted. e) Assume detcal mea free tme, usg the effectve mass d), gve the moblty of electros Slco s 100cm /(V s), what s the electro moblty GaAs? 3

4 qτ m * S µ = => µ ( GaAs) = µ ( S) (3 Pts) m * m * GaAs µ = 4457 /( ) (1 Pts) cm V s f) What s the rato of electro dffuso costats for GaAs ad S? Este Relato, = => s drectly proportoal to µ. µ q qτ As µ =, * m µ m = = (3 Pts),GaAs,GaAs *, S *,S µ,s m, GaAs,GaAs,S = (1 Pts) g) What s the rato of the average hole ketc eergy of GaAs ad S at room temperature? Average Hole Ketc Eergy = 3/ (oes t deped o materal) (3 Pts) Rato =1 (1 Pts) h) Gve the E-k dagram below I geeral do you expect GaAs to have loger or shorter recombato lfetme., assumg both the materals have very few trap states the bad gap. Expla your cocluso cocsely. GaAs wll have a shorter recombato lfetme. ( Pts) GaAs has a drect badgap ad S has a drect badgap. Recombato S has to happe through trap states, whle GaAs drect bad to bad recombato s possble. So gve we have few trap states both the materals, the recombato lfetme s perhaps shorter GaAs due to drect bad to bad jump. ( Pts) 4

5 Problem Hgh Power Rectfer P + -I-N + Part A Fabrcato Process Gve below s a meu cosstg of varous fabrcato processes you are famlar wth. Usg oly these gve processes, complete the sequece of steps to fabrcate the structure gve below. (Ht: All gve processes may or may ot be requred, ad the same process maybe used multple tmes.) Processes Avalable: Thermal Oxdato Lthography Eptaxy Io Implatato Asotropc Etchg Isotropc Etchg Chemcal Vapor eposto Nao-Imprt Sputterg Chemcal Mechacal Polshg Thermal Aealg Process Flow: 1. Eptaxy (Grow Itrsc Slco above N+ Slco Substrate). Thermal Oxdato (Form SO th flm) 5

6 3. Lthography (Coat photoresst, patter cotact area ad develop photoresst) 4. Asotropc Etchg (Etch SO whch s above P+ S rego) 5. Io Implatato (Form P+ dopg rego) 6. Thermal Aealg (Repar damage ad dffuso to acheve rght jucto depth) 7. Chemcal Vapor eposto or Sputter (Form Metal Cotact) 8. Lthography (Patter trech area of metal) 9. Asotropc Etchg (Form the trech of metal) ( Pts each step). You may splt the formato of metal treches two separate steps. Part B (doable wthout solvg Part A) The structure Part A s a hgh voltage ad hgh curret rectfer wth a 500V breakdow voltage. Assume that the crtcal electrc feld s x10 5 V/cm. a) raw the electrc feld profle through the P + -I-N + jucto uder reverse bas codto. E-Feld has to be costat trsc rego (3 pots) refer HW Assgmet ad depleto regos P + ad N + regos are eglgble ( pots). (A threshold for the steepess of edges was set ad gradg was based o that => steep edges were accepted. ) b) What s the mmum thckess of trsc rego to acheve the breakdow voltage? From the plot above all the E-Feld s the trsc rego => All the voltage drops ths rego. 6

7 Thckess, t trsc = Breakdow Voltage / Crtcal Feld ( pots) =500 V / (x10 5 V/cm) =.5 x10-3 cm ( pots) c) What s the advatage of dopg substrate N + rather tha lghtly doped N-type substrate? Resstace ( pots) More dopg => lower resstace of the Quas-Neutral rego. Remember the lower part of the substrate s the secod cotact to ths dode. Ths resstace teds to lower your curret for a gve voltage. ( pots for reaso) Capactace creases (as W dep decreases), so ths actually teds to make thgs slower. So ths s ot accepted as a aswer. d) What s the possble advatage of startg wth N + substrate rather tha P + substrate the fabrcato process Part A? Lower Resstace aga ( pots) moblty of holes < moblty of electros. So the P+ substrate would have a lower coductvty. ( pots for correct reaso) Questo was uclear. Everyoe was awarded full marks e) Calculate the capactace of the dode F/cm. Capactace, C = ε 0.ε S / W dep ( pots) But W dep = t trsc ( pots) C = x F cm -1 x11.8/.5 x10-3 cm=4.174x10-10 F cm - (1 pot) 7

8 Problem 3 Forward-based dode Cosder a oe-sded P + N dode as show the fgure (uform dopg) a. raw the electro eergy bad dagram throughout the dode at V = 0V. Iclude the Ferm Level(s) the fgure. Fgure (crude) ( pots) Labels mportat. E C, E V ad E F. ( pots) Also relatve posto of E F to E C ad E V o ether sde mportat (1 pot) b. How would the bad dagram look for a forward bas V F appled across the dode. Iclude the quas-ferm level(s) through the depleto rego too. Fgure (crude) ( pots) Labels mportat. E C, E V, E F ad E Fp. ( pots) qv F also eeds to be dcated. (1 pot) 8

9 c. Sketch the excess morty carrer cotet versus x for two forward bas codtos, oe at some value V F ad aother at V F - 60 mv. Show the crtcal rato(s). 9

10 Fgure (crude) (1 pot) Need to show p <<p ( pots) Labels ad Rato p ad p /10 ( pots) d. Label the stored charge at V F ad V F -60 mv the above questo. Calculate the rato of the stored charges at the two bases. Label ( pots) Rato (3 pots) Need to show steps Else (1.5 pots) For log base dode Q = I.τ p Q( VF ) = Q( V 60mV ) F τ. I p τ. I p 0. e qvf 0. e q( VF 60mV ) = e q.60mv 10, e. raw the morty, majorty ad total curret destes versus x. Label the compoets clearly. Label J total, J maj ad J m atleast for lowly doped sde; ( pots) 10

11 J tot ad J maj for hghly doped sde. ( pots) For dcatg J m o hghly doped sde as small / ot preset ad J m at edge of depleto for lghtly doped sde to be as close to J total. (1 pot) f. For a forward bas of 0.7 V, fd the rato of stored charge (up to the order of magtude) at T = 300 K ad T = 500 K. Not partcular about the aswer but depedeces. Temperature epedece p ( pots), (T 3 ad exp(-e g /))(1 pot + 1 pot) ad exp(qv/) (1 pot) But f assumptos were made ad metoed for p ad (T 3 ) parts t s accepted. Q = I τ I I 0.( e = N c Q(300K) = Q(500K) 300 = 500 qv. N 3 ; I v 0 0 q 1.1 exp k 0 E. e g = qa N I (300K)( e I (500K)( e k.300k k.500k 1 ( 300 Assumg µ p s roughly costat p ; τ = p p 1 ) 500 = µ p. q (300K). (500K) ( e ( e k.300k k.500k ( e ( e k.300k k.500k =