Schroder Solution

8/2/2019 Schroder Solution

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.Chapter 11.1 The function y=xn is plotted in Fig. PI. I. Detennine n.

n= dlog(y) = -0.5 - (-2) =-0.5dlog(x) 1-4

n=-O.5)()2 103 )()4

x1.2 Detennine Yo and Xl in the equationy =Yo expx / XI) - I) plotted in Fig. Pl.2.

yAt x = 0, y = yo = 10-4

dx .XI = - - -dln(y) dx = 0.92 =0.1In(IO)dlog(y) 2.3 x4 x1.3 Plot the !og(y) - x data ofFig. P 1.3(a) on the x-y figure in Fig. P L3(b). Write numeric values on the yaxis ofFig.P 1.3(b). Take vertical axis values from (a) and raise to "10" power, e.g., 1.S ==> 1015 = 31.6.

2 1001.5 80

S 60bI ) Y..8 40

0.5 200 o

1.4 Using the equations leading to Eq. (1.8) we find

v= pI (.!.___ __ _+.!.)= pI 2 - . f i ==> p = ~ ! : . . 2" s . fis ..{is s 2" s 2 - . f i I

I.S Using the equations from the notes we find

v=pI (.!._ _.2.. +.2..) =pI ==> \p =2.Ss VI2" s Ss 3s 3s 2" Ss I1.6 Yes, since the voltage is measured on the n-portion of the sample. The

3

1I/L---V (b)0.25 0.5

X

1 2 s I

0.75

3s I 3s 4


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region for the current probe does not affect the measurement.1.7 The resistance is given by R=ptlA + 2pc/A. Hence p=(dR/dt)A and 2Pc=RinterceptA. A=7.85xlO's cm2 The resistance,plotted as a function ofwafer thickness 1, is:

. . ".From this plot we find

2000 I1500

g: 1000!X:500

--y=12 .7+15280x R=l

0.025 0.05 0.075 OIt (cm)

p = 1.2 a 'em, Pc = 5xlO-4 a 'em2

~ ! ! ! . . . = I d In I = lin 10 d log I =2.31 d log I 10.3 . . . . , . . . . . ~ . . - - ~ " ' T ' " ~ . . . . , ~ r - r - r ~ ...........dV dV dV dV

dlogJ = _ 8_ = 2.3xlO-7 x _ 8_ =4xIO-6 SdV 0.46 g 0.46jg =4x1O-6 S I

1.9 The resistance is generally given by R=pL/A; in this case,since A varies with x (vertical dimension), we writedR=(p/A(x))dxwhere A(x)=7t(r+x)2 . So we have

R p I dx P(l I)R - - ~ R - - ----J - ; r J ( r+x ) 2 -;r r r+ fFor r t, we can approximate this equation asR"" p(,!,_!) and m = ~ " " p

; r r I d(l/r) ; r

10-4 "" .""" '"' ' ' '10.5 . " .

: ~ ~ f . .':..:..." ':.10.8 . . " . ' . . , . .. .IO9r ; . .: ~ : : ~ [ ';"'" . .. ,,'o 0.1 0.2 OJ 0.4 0.5V(V)

p

The approximation r


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LR - P J dxt 0 Wo(l - x/2L) P [-2Lln(l-x/2L)]5 == 2pLln(2) 1R=1.39x10sOltWo tWo

1.11 The resistance is generally given by R=pLlA; in this case, since A varies with r(radial dimension), we write dR=(p/A(rdr where A(r)=21tft. So we have

R r, ( )dr P r2JdR==- J - ~ R = - I nO 2m r 2m r lr,

IR = 219.9 ohmsI1.12 First determine I!RJ,. Next differentiate the IIR", - x curve to obtain:p(x) =- 1 . Then determine n(x): n(x) = 1d(lI RSh)/ dx qPnP(x)

1.110-' I lO16 'l - , - - - -- . - - - , - . . - - - , . ,110-'RHr '

. 61o-s'" I o - ~ 210"s

8 10 15

:::;' 6 IOU t- t41 0 lS r2 1 0 "V '

S l(}-s 7.5 1 O - ~ I J()-'" () ' - - - ~ - - ' - ~ - - l . __ - -_- 'o 1()Il 2.S 10-' 510" 7.510" I 10-' Ole" c.. __ ' - - ~ - L . . _ - - 010' 2.510" 5 10" 7.510-' I 10-'x(cm) x(cm) X (cm)

1.13 The sheet resistances are:Pi 10 5 P2 I 4R,hl=-=--_S =1.67x10 O.lsquare; Rsh2=-= - -_-=3.33xIO n/squareII 6xlO 12 3xlO

The resistances are:

RI I = R s h l ( ~ - = 4.2x104 ( l ' RI2 = R s h l ~ = 8.35xl05 ( l ' R2 = R s h 2 ~ = 1.67xiOS (lw . w ' WRII is in series with RI211R2 in parallel; hence

4.18xlOS +1.39x10 s =5.56xIOs n

IR'hl = 1.67xlOs ohms/square; R,h2= 3.33xl04 ohms/square; RAB = 5.56x10s o h m ~ 1.14 Rsh= -,---J(lI p)dx

PoIJ( l -kx/ I )- I dx Iln(l- k)o o

1.15 For an n-type layer on a p-type substrate: (i) Rsh = -,- -- , 5 X 10 155.5x10 IS. JqnPndx Jndxo 05


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(ii) calculate and plot: (J versus x (linear-linear plot), p versus x (linear-linear plot), R.t, versus x (log-linear plot) and l/Rmversus x (log-linear plot). UseRsh =-, - -- ,

Jqllnndx Jortxand let x vary from 0 to t.

2....,-.. 1.5Su,c'- '

'ii'..COl=...,-..g""r : r . ~

0.5oo 10 2 10. 5 4 10.5 6 10.5 8 10 .5 I 10.4

X (cm)

106IDs

10 4

103o 10 2 10.5 4 10.5 6 10.5 8 10 .5 1 10-4x (cm)

5

E 4't 3C'-'2a..oo 10 2 10 .5 4 10 .5 6 10's 8 10.5 I 10"

X (cm)

-10-4 r - - e : : : : : : : : : : : = : : : : : : : : : : : : - ~ = 10.510-6- I 0.7 ' - - ~ . L - . ~ J . . . . . . ~ - ' - ~ . . i - ~ - ' o10 2 10.5 4 10.5 6 10-5 8 10.5 I 10""

x (cm)

4.5321F(R12 34 + R23 41)LI6 p= ' .2where F is obtained from the relationship Rr -I = ~ a r COSh(eXP[ln(2)/ FJ)Rr + 1 In(2) 2where Rr=R12,341R23,4I. For this problem, Rr=74/6=12.33 leading to F=O.664 and

p = 6 a'cm, R.h = pIt = 120 a/square

4.532tF( R12 34 + R 23 41 )p= .' .21.17Rr -I = ~ a r c o s h ( e X p [ l n ( 2 ) / F])Rr +1 In(2) 2

where Rr=R12,341R23,41. For this problem, Rr=59111 =5_634 leading to F=O.8055 and

p= 4.47 a'cm, R,h = pIt = 127.8 O/square4.532tF(R I234 + R 23 41 )p= ..1.18

Rr - \ F h(exPfin(2)i F1J'--=--arCOSRr+1 In(2) 26


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where Rr=R12,341R23,41.For this problem, R r=90/9=10 leading to F=0.7 and

p= 7.84 O'em, Rsi. = pIt = 157 O/square

1.19 4.532V34 = 4.532x58 263- t= L 1.84 x 10-2 =7xlO - 5112 1 ' Rsh 263w= RshLll3 = 263xO.05xl0-4 =7.5xlO-4

V23 1.75

Rsh = 263 OIsquare, t= 0.7 pm, W= 7.5 Jlm

1.20

Junction depth Xj is where No = NA, i.e.,

The sheet resistance isRsh == 1 => I sh =62.6 n JsquareI

qr (x)Pndx1.21Junction depth Xj is where No = NA , i.e.,

The sheet resistance isI =>IR'h = 25 OJ square I

qr (x)Pndx1.22 4.532x18 =81.5; t= L 4xlO-3 =4.9xlO- s

I Rsh 81.5

(b)

Rsb = 81.5 Qlsquare. t = 0.49 11m, W = 51 flm, W' = 18.4 flm7


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1.23 R = 4.532V34 =4.532 xS8 = 263- I=L= 1.84 x 10-2 7 x 10 -5sh 112 1 ' Rsh 263

263x 0.05 x 10-4 = 7.5x 10 - 41.75

R,h = 263 O/square, t =0_7 f.lm, W =7.5 f.1IDWe have W=R.I,LIR. Let us consider three cases(a) R.J, is uniform, hence W=R.I,LIRo.(b) Fig. (a) of he problem:

R _ R'R" _ (2RshLl W)( 2R shl LI W) 2RshLl W1 - R'+R"- 2RshL/W+2RshIL/W I+Rsh/Rshl

(c) Fig. (b) ofthe problem:

1.24

1.25.(a)

R, =R"'+R""= RshL RshlL = RShL(.!.+ Rshl )=\.5Ro =>W2 = RshL- 2W 2W W 2 2Rsh 1.5Ro 1.5WI/W= 0.75; W2/W = 0.67

R - - - ----sh - 10-'qJ.L p J10 19 e-krdxo

6.25 X 10-3 x 10 51- e- IO 625

n 1 6'"'5 "10-3 D T T 6 " ~ 1 (\-2 1(\-3V _l\sh'12 _ L. " - 018' V _ "sh'"'" 16 _ " . J x . J ~ ' V v =31.334 - 4.532 - 4.532 - . , 45 - W - 10-3

R,h = 625 Q/square, V34 = 0.18 V, V45 =31.3 V

4.532V34 4.532xl IRsh = = =1001(2 0.5 6t=L= 5xlO- 3 =5xlO-5

Rsh 100

Rs h = 100 n /square, 1= 0.5 f.lm, W = 0.2 f.lm(b) Now we have AX) = 0.5x+0.005 ohm-cm, with X in cm.

1 IS' 4 RshL IOOxlo-2 -5Rm e =- (O.5x +O,OOS)dx = 7.5x10 ; Wef f = -- = 4 = 1.33xlOWt 0 RUne 7.5x10Iweer= 0.133 fJ.ml8


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1.26 The thinnest layer is 0.1 I 1 m = 1 0 ~ 5 cm, the probe steps 2 JIm and we want 20 points or 19 intervals. ~ z = L l x . s i n { e ) . ~ 1 5/20=5.26xlO,7. ~ x = 2 x l O " " cm:Hence

1.27 T h ~ space-charge region in the pnjunction gives high at the junction location. For the n+n junction there is no suchspace-charge region.1.28 First find an expression for p(x) from the relationshipp(x)-n(x)-NA(X)+ND(X)=O.

N{X(I + ~ I + (2n; IN A)2 )Solving for p(x) gives: p(x) = - ~ ' - - - 2where N=NA{x)-No. Then n{x)=n j2/p(x).Find the sheet resistance from Rsh = - 1- - - - - - - - -

q J[p(x)p p (x ) + n(x)Pn (x)]tb-o(a) t is the wafer thickness, which is 400 11m or 4xlO'2 cm.

x. Depth

(b) t is the p-Iayer thickness, i.e., the depth at which NA (t)=No, t=Lln(NAolNo). For this case t=I1.5 11m.The results are: I R,h(a) = 326 Q/square; R.h(b) = 520.5 Q/square1.29 The p-Iayer is uniformly doped, hence R.J, = pIt = 1.512xlO"" = 7.5x103 Q/square.After the charge is depOsited, a depleted space-charge region is created. 10 12 cm,2 charges on the surface induce 10 12 cm'2charges in the semiconductor, i.e., 10 12 = NAW, where W is the induced space-charge region width. This gives W = 10"'; cm,or the new thickness is 111m.Hence Rsh = p/t= 1.5/1xI0-4 = 1.5xlO4 Q/square.

RSh(a) = 7.5xl03 a/square, R.h(b) = 1.5x104 O/square

1.30 For the hot probe positive with respect to thecold probe: n-typeThe current in an n-type semiconductor isI n = nPndEFn / tb--qnPnPndT I dxFor In=O (measured with high-impedancevoltmeter)

Ec- , - , - . - . - . - . . EFEi

Equilibrium

9

------ - - . - ~ Non-equilibrium


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Chapter 11R = 4.28xl 0-6 EI.7511.1 e 1.59 X 10-6 E175 for Al with p=2.7 g/crn3 For E=IO keY, R,,=8.9xI0s crn or 0.89 IJ.rn .p

11.2 Nehp = ~ ( I - aEhS ) = 10 4 (1 - 0.1 x !04)= 2500; Re(Si) = 1.84xlO-6 EI.75 = 10-4 em;Eehp E 3.64 10 .

IhNehp 10-9 x 2500 17 3 17 -5 12-3=>G=---= 19 =7x10 ehplem s; n=G'n =7x10 xlO =7x10 emqVol 1.6 x 10- x 2.2 X 10-5

11.3

Solving (3) for s i n ( ~ ) gives. '" /vI IvI f} '" ; . 2 '"SIO'f'=--Stn ;COS ' f '=V I - slO ' f '=M 2V2

Substituting for c o s ( ~ ) into (2) gives(

u .. ,2 .1 m I"]) . L f}- -- SIOM 2V2Mlva =Mlvl cosf} + M 2v2 1-( Mlv l )2 sin2 f} = MIVj c o s f } + ~ ( M 2 V 2 ) 2 _(M lvl )2 sin 2 f)M 2v2I 2 2' 2 2 M ~ ( v ~ - 2 v a v l c o s f } + v ~ ) or MI(va-v]cosf})=,,(M2V2) - (Mlv l ) sin-f};solvingforv2 gives v2 =_.!.......C....::.-_-=--;;-__ '- -M;

where R=M]1M 2 Solving for VI gives2Rva cosf} ~ 4 R 2 V ~ cos 2 f} + 4(1 - R)(I + R ) v ~ Rcosf} .J I - R2 sin 2 f}

VI =------'---2-(-1+ R-)- ------ =va 1+ RFo r MI


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11.4 The He ion energy is 2 MeV and the angle e s 164.Yield

EFor Xl on Y: [So]=45.5 eV/A, EI=O.332 MeV, E2=O.91 MeV, Ey1.l38 MeV,For X2 on Y: [So]=133.6 eV/A, El=0.338 MeV, E2=1.578 MeV, E3=1.847 MeV

[ ~ I - ( R S i n e ) 2 + Rcoser [JI-0.076R 2 -0 .961RrK = " '----------=-(l+R)2 (l+R)2JI-0.076R 2 - 0.961R = (1+ R)jK;

Through trial and error or using quadratic equation:I) KI=O.569; Rl=0.143=mIM1; MI=4/0 .143=28 Si2) K2=0.924; R2=0.02=mIM2; M2=4/0.02=200 Au

From Eq.( I0.25) Llli=E3-E2=[SoJd or d=Llli/[So], also Llli=(1+K)Eloss assuming AEin=Llliout=Eloss; henceEloss=[SoJt/( I+K)

I ) Lllil=2.28x105 eV d=2.28x105/45.5=5000 A2) 6 9 x 1 0 5 eV d=2.69xl051133.6=2000 A

Substrate:From Eq.(l0.24): El = EOK3 - ( ~ E i n K 3 + L l l i o u t ) = (EO-Eloss)K3-Eloss K3=(EI +Eloss)/(EO-Eloss)

1) K3=(0.332xl 06+0.145xl06)/(2xl06_0.145xl 06)=0.2572) K3=(0.338xl 06+0. 14x106)1(2x 106_0 .14xl 06)=0.257

K3=0.25 7 corresponds to R3=0.333, hence M3=4/0.333=12 which is carbon.

Layer 1: Silicon, 5000 A; Layer 2: Gold, 2000 A; Substrate: Carhon11.5 The He ion energy is 2 Me V and the angle e is 170.For X on Y: [So] = 193 eV/A, EI = 0.657 MeV, E2 = 1.256 MeV, E3 = 1.835 MeV.

E) [ ~ I - ( R S i n ( e 2 +RCOS(O)Y [JI-0.076R 2 -0 .96IRrKl =-= 1 =0.9 17Eo (I + R t (I + R)2

J I - 0.076Rl - 0.961R = (I + R)jK;Solving this equation: KI = 0 .917; R = 0.022 = mIMI; MI = 4/0.022 = 182 => W or Ta

tlEFrom Eq.(11.25) tlE = E3 - E2 =[So]d => d = [,'>'0] ' tlEAE = S.8xI05 eV d = 5.8x\05/193 = 3000 A

62


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"strafe:i f m Eq.( 11 .24): Let Ein = E o ~ t ' = E'

E = (E - E')K - E'=::" K = EI + E' . E'= [Sold == 193x3000 == 3.0Sx105 eVI 0 2 2 Eo-E" l+KI 1+0.917

K2 == 41 + 0.S7Eo - E": = 0 57 corresponds to R2 = 0.143, hence M2 = 4/0.143 = 28 which is silicon.

!L8yer: Tungsten,0.3 lim; Substrate: Silicont11.6 The He ion energy is 2 MeV and the angle e is 170.fo r X on Y: [So] == 108 eV/A, EI = 0.845 MeV, E2 = 1.23 MeV, E3 = 1.S6 MeV,

E [ ~ I - ( R S i n ( B 2 + RCOS(B)r [ ~ 1 - 0 . 0 7 6 R 2 -O.961RrK =_3 = =0.78 Eo (I+R)2 (l+R)2

0.076R2 - 0.961R = (I +R).jK;"Solving this equation: KI = 0.78; R = 0.063 = mIMI; MI = 4/0.063 = 63 .S =::" CuFrom Eq.(11.2S) M = E3 - E2 = [Sa]t =::" t = M , M = (1 + K. )EJoss assuming Min = MoUl = EJass[So]

Lill = 3.3xIOS eV t = 3.3xl05/I08 = 3050 ASubstrate:From Eq.(11.24): EI = EoK2 - (Mm K2 + M oU l ) = (Eo - EJoss )K2 - EJass =::" K 2 == (E. + EJoss )/(Eo - EJoss)

E = (Solt == 108x30S0 = 1.8Sxl05 eVJ().fS 1+ K 1+0.78K2 == E. + EJO.fS = 0.57

Eo - EJassK2= 0.57 corresponds to R2 = 0.143, hence M2 = 4/0.143 = 28 which is silicon.

ILayer: Copper, 0.3051'm; Substrate: Siliconl

11.7 The He ion energy is 2 MeV and the angle e is 164.For X on Y: (Sol = 109.2 eV/A, EI = 0.954 MeV, E2 = 1.51 MeV, E3 = 1.73 MeV,

. [ ~ I - ( R S i n ( B 2 + RCOS(B)f [.JI-0.076R 2 -0 .961RfK I = . =0 .87(I+R)2 (I+R)2

0.076R2 - 0.961R = (I + R).jK;"Solving this equation: K\ = 0.&7; R = 0.037 = miMI ; M, = 4/0.037 = 10& =::" AgFrom Eq.( 11.25) M = E3 - E2 =[Solt =::" t = IJ.E , M = (I + K. )EJoss assuming Min = MaUl = EJO.fS[S()]

AE = 2.2xlOS eV t = 2.2xlOSIl09.2 = 2000 ASubstrate:From Eq.( 11.24) : E, == EoK2 - (MinK 2 + IJ.EoUl ) = (EO - Elo,-,)K2 - E/oss =::" K2 = (EI + E/o,-,) /(0 - Elan )

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E - [So]t _ 109.2x2000 1.17xl05 eVloss - 1+ K - 1+0.87

-: = 0.57 corresponds to R2 = 0.143, hence M2 = 4/0.143 = 28 which is silicon.

II

IV!

ILayer: Silver, 0.2 !J.ffi; Substrate: Silicolil

(i ) (ii) (iii) (iv) (v) Electron 0 Ion X Ray

l Auger; (ii) Electron Microprobe; (iii) X-Ray Photoelectron Spectroscopy; (iv) X-Ray Fluorescence;is . Rutherford Backscattering.

Eo = 2 MeV, K2 = 06, K3 = 0.9, ~ i n = M:ou = 0.2 MeV.2 = K2 E C = 0.6x 2 = 1.2 MeV; E} = K}Eo =0.9x 2= \.8 MeV

Eo - Mm)K ) -MQUI =1.8xO.9 - 0.2=1.42 MeV; E; = (Eo -M;n)K2 -Maul =1.8xO.6 - 0.2=O.88 MeV

o IE(MeV)

2

64

o 1E(MeV)

2

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