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Unformatted text preview: Chapter 7 / Prob. 7.1 For the bzpolar junction transistor whose fabrication is described in the question, ﬁnd
where thepeaks and Mdths of the boron and phosphorus proﬁles prior to annealing are
emitter, and where the peaks and widths are after annealing, and what the collector
junction depths are aﬂer annealing. (a)
From Appendix IX 2
RP = 0.1631131 for 50 keV boron. ARP = 0.05m
width = ZARP = 6.11m) RP = 0.0411111 for 30 keV P.
ARI, = 0.018)»): width = ZARP = 0.03am (b)
Annealing causes diffusion. From Eq.(5  lb), 0'2 = +200
For boron, from Appendix IX we have
62 = [(0.05)2 + 200‘12 x10><103)],m2 = 0.0025 + 0.002 = 00045;me
=3 6 = 0.0671101.
The width of Gaussian distribution is
20‘ = 0434le
Peak locations are the same before and aﬁer annealing. 97 97 (G) Since the Base  Collector junction is far from thgemitter peak, we can
ignore the phosphorus concentration at the collector junction. 0: — RP )2 N(Boron) = ﬁe”: ( where y2 = W
p _ 5x10”
J21: [0.067 x10’4] 15 —14
WW 3 X .e‘y2 :1 2x10’scm"3 =9 y=3.l (x — RP? ~————2—, solve for x :
2(ARp + 21):) Since 312 = x = Name}, + 21):) +121, = 3.142 x0.0045 + 0.16 = 0.45pm
2) Collector junction = 0.45pm. (5
Peak concentration = ' =>
J21“) 5x10M B := W
own peak J'2F[(o.067) x10‘4] Phosphorus peak  wail—93:“
421: [(0.0364) x10'4] = 2.988 XlO‘gcm”3 = 1.096x1020cm”3 Atx=0,
_l 0.04‘
0208 2 Phosphorus concentration = l .096 x] 0'03“ ) = 5.9.9 x 10'9 CHI—3 Graphically, emitter junction = 0.10811m 98 98 2:212:12
Sketch ic vs. (—vcyfor the BJT of Fig. 74. and ﬁnd WVCE for [B = 0.1 mA. 18 (mA) Prob. 7.3
Plot 8p across the base Qfa p—n—p Wilh Wb/Lp = 0.5. Refer to Fig. 77. em 2155, 91/2 =0.606. M,=1.58. M2 =058 Atxn/Lp = . i=1.ss—0.58=1.0. A0): A! x /L = 0.5. £8 : !.58(0.606) — 0580.65) = 0.
n p APE Values can be ﬁlled into obtain a plot such as Fig‘ 7—7. with normalized axes. 99 99 trot). 7.4
Derive Igﬁom Qp/Tp. W
9,, = we ’6p(x,.>dx,.
= qALphem/‘P — 10— 6mm» 4)] [E = 9.3.
"p
~W /L 770/], "lb/L, qALp x[(APC "Apge ° ")(e ” *1) __ (Apge ~ApCXe'W‘xL' 4)] ll Tp eW,/L, __ eWJL, erg/Ly _ eWJL, L ._ e‘m/Lt' = 2% [mpg Maw“? + am»)— 21] P
AD
= Li’ <ApE +Apcxcumﬂcschﬂ)
LP LP LP Prob. 7.5 Extend Eq. (720a) to include the eﬁ‘ects ofnonunx'ty emitter injection efﬁciency (74).
Derive Eq. (7~25) for 'y. Eq.(7  203) is actually 151,. = ﬂnpeqy" “‘7 for V53 >> kT/q
Ln Thus, the total emitter current is: [En D W D V MT
1 =1 +1 = A —” cumJ+——"n e" E”
E Ep £11 ‘1 [ LP Pu LP L" p] Apply the charge control approach to the distribution in Fig. 781) loﬁnd the base
transpon factor B. 100 100 . 4 v1 W2 1
B: :Tﬂ—j= =il+ b
up 1, MP 20pr Using (1 + x)" = l — x for small x, we have:
I I W \2 W \2
3: 1+« J zx—l[—£
L 2 Lp 2 LP
Eq.(7—26)is 2
B = seek—Inf”; z 1 from Table 7.1. p 2 LP Erob. 7.6 A symmetrical Si p'n_p' BJT with area .10" cm‘: and base width of one micron has an
emitter with Na = 10" “ cm", 1,, = 0.] p5. and p“ = 700. In the base M, = [0’5 cm'3, IF = 101.15. and 1.1,, = 450. Find 15, and 13 with V55 =7 0.3 V and large reverse bias on the
collector junction. In the base: p" = nil/n" =(1.5><1o'°)2/10‘S 2225x105
13,, = 450(0.0259)=1L66. LP =(11.66><10‘5)V2 =1.oz;><10'2
Wb/LP = 10‘4/1.08x 10’2 = 9.26x10‘3
[ES = I“ = qA(DP/Lp) p” cmh(W,,/Lp)
=(1.6><10"9)(10“)(1Les/1.08 x10‘2X225x105) cmh 926x104
= 4.2x10'13A APE = Pnqu"M: AFC ‘7 0
AUE = Xe(0.3/0.0259) = 24x1010
18 = qA(Dp/Lp)ApE tanh(W,,/2LP) or 13 =§2¢L =qAWbAp5/2‘Cp =1.9x1o"”A
p Find 'y and Bfor a long emitter of the BJT. 101 101 In the emitter,
D" = 700 (0.0259)=18.13 L,, =(18.13xl0‘7)”2 = 1.35x10’3 \ E
= 2.49.". as 1
En Lf p IV . .
15!, = p I): ctnh—g eqhﬂd
P LP [ 18.13x1.08x10'’x10‘5
y = 1+ —1
e ,, tanh 9.26x10‘3 =0.99sss
11,66x1.35><10"’x10’ B = sech g1 = sec}: 9.26 x 10'3 = 0.99996
p on = B7 = (0.99885)(0.99996) = 0.9988 l—a 0.0012 Prob. 7.7 For (he diode connections shown. sketch 5p in the base. Which is the best
diode .9 (a) Is = Ic , 113 = 0. Since v is (b) V09 = 0: ms Ape = o. glance
large, the collector is strongly teat due 15' thenmow— ase
reverse biased, Ape = _—p,‘, diode dlstnbutxon. Since IE = 1c, Apg = aApc
= p" from Eq.(7—34). The area.
under 6p(a:,,) is zero. 102 102 (Sp 6p
APE
APE APO
“APE
x,‘ 3n
0 W; 0 W‘
(e) Since 10 = 0, Ape = aApE from (d) VI.m = van a V. in.” Am. = APE Eq.( 7—3411). Connection gives the best diode since the stored charge is least; (a) is not
a good diode since the current is small and symmetrical about V = 0. Prob. 7.8 For the transistor connection in Fig. P11 a, (a) show that V33 = (JCT/q) ln’Z;
(b) ﬁnd the expression for I when V 1» kT/q and sketch I as. V. While V > kT/q. APE = ‘APC = Pu
(a) m = pn(e'VEB"'T1)= p“ eaves/H __. 2p" , VEB = Elng ‘1 Pa: (b) I = [E = [C = [55(1+a) since 155 = Ics
(1,11)? W5 W1»
,. .1: 11— ~
LP p (rn Lp+csch LP The current is small and independent of V except when V drops below IcT/q. or I: 103 103 M2 (a) Find I ﬁvr the transistor connection of Fig. P77b; compare with the narrow base
diode. AD
Since Apc = 0, 1 = g—ﬁApE cmhﬂ
LP LP (b) How does 1 divide between the base lead and the collector lead? AD
15 = q ’7 ApE cschﬂ
LP LP
qADp Wb
1 = mun—
8 LP APE up where [C and 13 are the components in the collectors and base lead. respectively. Note that
these results are anaiogous to those ofProbs. 5.35 and 5.36. 104 104 Prob. 7.10 Suppose that V is negative in Fig. P7 7c. (a) Find I ﬁ‘om the EbenM011 equations; (b)
find the expression for‘ VCR; (c) sketch 5pm.) in the base. With V53<<kT/q and [c = 0: (a) From Eq.(736), 15 = (1— a2) [55 (—4) = 155 (a2 —1).
Since an == 0:; = (x for a symmetrical device.
(b) From Eq. (7  32), awn/“4 = a(l) Thus VCB = q (c) From Eq. (734), Apc= aApg
Since Apg = to", Apr = M'Pn)» 5p Prog. Z.“ For the transistor connection of Fig. P7~7d, (a) ﬁnd 5p(xn) in the base: (1)) ﬁnd the
currentl. VER=VCB:V»APE=APC (a) In Eq. (713), C2 = JAG
mm, 1 e‘WL»
)W r 5p(x,,) = Ap£(ex'/L' + ew' L D
(b) I = qu—Ppnmxhgé—xeqm I) from Eq. (7 .19).
p P 105 105 \_/ Prob. 7.12
(a) Showﬁom Eq. (732) that [go and [co are the saturation currents afthe emitter and
collector junctions, respectivelv. with the opposite junction open circuited. For IC = 0, Eq. (73223) gives e414,” VulkT _1) 1cs(  1) =a2v1£s (8"
Substituting this into Eq. (7  32a) we obtain
[[5 = 155(1’Gﬂ~)(eqy”m 4) This is a diode equation with saturation current laxkam) = 1m
Similarly, for 15 = 0, Eq. (7323) gives [55 (eqyﬂlﬂ "‘ = allcs(qu“/" " Substituting this into Eq. (7 — 32b) we obtain .1C = [CS(1_alaNxqu(,/kr *1): 1co(qur,/k7' __ I) where the minus sign arises from the choica of 15 deﬁned in the reverse direction through
the collector junction. (1)) Find expressionsfor: Ape with the emitter junction forward biased and the collector
open; Ape with the collector junction forward biased and the emitter open. From Eq. (734): Apc= (X/Apg for [c '= 0
and Apg= uNApc for 15 = 0. (c) Sketch 6p(x,,) in the base for the two cases of part (b). 330 3‘?
408 @C
09% 09Vqu
0 Wb x" o Wb x"
106 106 W
(a) Show that the definitions oqu.{7~40) are correct: what does qN represent? av :19: QN/T’N \ =———————IPN
’ IE QMVTwH/TFN} Imﬂ’cplv
JEN=QN(1/T:;v+1/Tp2v)3183 APE
’1
thus 1E5 =q~(l/T,N+I/”CPN), whereqN=QN A?
E and similarly for the inverted mode. qN ~ V2qA Wbpn is the magnitude of the charge stored in the base when the emitter junction
is reverse biased and the collector junction is shorted. (b) Show that Eqs.(7—39) correspond to Eqs,(7'34). Qv 3" 15$ APE WENT my Q! = 1C5 AFC 4“”: I
1 ‘ pr: 141V +TpN pn Tll 4"th
Thus Eq.(7  39a) is \ ’E I .
IE=IES‘£§B§'”ICS£2§‘ p =Issﬂ‘allcs AFC
pn pn 1:2] +Tp1 pa pl!
Similarly, Eq.(7  39b) is
,i 1: N I.
1c = [ES Apt @“ICSEEQ‘S'CXNIES APE “103 Ape
pn Tw +tpN .pn pr: pa Prob. 7.14 (a) How can the. transit time across the base 13 be shorter than the hole life
time in the base 7,, $7 (Spun) is a steady state distribution and is replaced on average every 61’
7,, seconds. However, the distribution
is made up of indistinguishable APE holes in transit across the base,
each spending on average 7, seconds
in transit. 107 / (b) Explain why the turnon transient of a BJT is faster when the device is
driven tit/to oversaturation. Saturation (Qt, = Q.) is reached earlier in the exponential rise: Q5 = 18 7p (1 " CAM)
l is = 1 W' Tp n 1 10/513 use ﬁle > Ic(sat.) QB I31, “*"— Prob. 7 .15
Design an npn HBT with reasonable 'yand base resistance. Since this is an openended design problem, there is no unique solution. Students should
use the results of Eq.(781) with the band gap difference of 0.42 eV between GaAs and
Alo‘gGaojAS, to conclude that the base doping can be considerably higher than the emitter
doping while maintaining a good emitter injection efﬁciency for electrons. It is possible
to estimate the base spreading resistance with the higher doping concentration. This will
require using Fig.3—23 to estimate the GaAs electron mobility at that concentrations Note
that App. In only gives the value for light dopingi Clearly, much important information
will be lost in these estimates, because of the sparse information the students have to
work with. For example, real HBTs using AlGaAs/GaAs suitor from surface
recombination problems, and scaling to small dimensions is inhibited. Some students will
be interested enough to read current articles on HBTs and will therefore provide
comments to this effect. In fact, a good answer to this problem might be “I wouldn’t use
AlGaAs/GaAs at all. What I would do instead is . . ." 108 108 ‘0 (A) Prob. 7.16 Ampliﬁcation factor, 3. 0. 08 ~0. 0‘ ..
O _/
Relative dopiﬂg: “rt/Pp an“! '
_ agaa: V Nomiizcd base width‘ Wbll.p .30 L20 1.6:: 21m)
VBCW) 109 109 M
(a) What is 9;, in Fig. 74 aghe d—c bias? 9,): 1,, 1,, = (10" A) (10'5 s)=10'9C
also 1C1,=(10'2)(10'7)= 104’ c (1)) Why is B dzﬁzrem in the normal and inverted mode of a dxﬁhsed 8.17? The base transport factor is affected by the
gradient in the base. This ﬁeld assists trans
in the inverted mode. 1 builtin ﬁeld resulting ﬁom the doping
port the normal mode, but opposes transpon 10 110 rob. 7.19 For the given p—np transistor. calculate the neutral base width W; The emitter doping is so high that Epp ~ Em Therefore,
The built  in potential at the base  emitter junction is approximately given by : E , . 16
’0 _ = ——&+—k’:tn—Iii = 0.55+o.0259 In #175 = 0.897V
"‘ 2 q n,» 1.5 X 10
The built  in potential at the collector  base junction is given by :
1
to .zﬂhﬂhmkj
"‘ q ’1: "i
z 16 ‘ f :6
= 0.0259 lnl ~—‘°—m— +1n‘ 49—15 : 0.695V
g_ \V1.5x10 I [1.5x10 Next calculate the width of the base — emitter and base  collector space chargeregions: 26
W = I S V —V
£3 qub(o,,,_ £5) Since N 5 >> N3 and the base  emitter junction is forward biased, W 2(ll.8)(8.85><10"”)
W = W0397~V
E” \I 1.6x10“9x10‘° ( E”) For V53 = 0.2 v. WEB = 3.02x10~5 cm
For VLF,3 = 0.6 v, WEB = 1.97 x10‘5 cm The width of the collector — base space charge region is given by : I25. A +N 4) W = ———=L
CB 1’ q NCNB T
where (Dr = V0 + VCB is the voltage drop at the base  collector junction.
Note : One  sided step junction cannot be assumed since for this problem N 3 = NC 11'] 111 Given: VCB = O, (1)7 = V0” = 0‘69SV Hence: WCB = 4.26x10'5cm
Calculate width of neutral base region :
Given : W = metallurgical base width = 1.5m
N ~
W = W — W ~—L——W
[1 EB NC + NB CB
For VEB : WI, =15 — 0.302 — 94:32 = 0.985pm For V55 = 0.6V Wb=1.s—0.197—9%2—‘i=1.09pm. Emitter 23321.29 For the BJT in Prob. (7.19), calculate the base lransportfacz’or and the emitter injection qﬁ'icz’ency for V5,; = 02 and 0.6 V. At ﬁrst, we determine the electrons and hole diﬁlsion lengths : Given 1,, =11, =10 =10”7s
and Dn=Dp=10cm2/s Ln =‘/D,,t,, =\/10x1(r7 =10‘3 cm Lp=Ln=L=loum 112 112 Calculate the base transport factor, B. 2
For pZyl’«<< l, B=——l——== high] p coshFE} ‘9 LP
For V 33 = 0.2V,
2 B==l~l =0.995 2 10
For V55 = 0.6V : 2
B = — 1 50—9— = 0.994
2 10 Calculate the emitter efﬁciency 7 : I
7 = ————E" 15,, ‘4' [En where 15,, is the current for holes injected from the emitter to the base; I 5,, is the current for electrons injected from the base to the emitter.
Calculate 15p and 15,, as a function of V53. [5,, = Diffusion current injected across B  E junction by the emitter (holes
for the p  n  p transistor). 113 113 Forthegivenpmp:
D n2 V
I =A "'ex l—E—‘i holeurrent
5p qNBWb Pi kT)( C ) 10‘s><l.6xlo“9><10><(l.5><lo”’)2 M
10'“ ><o.9zzs><lo'*1 Similarly, at V58 = 0.6V, 15,, = 3.8 x l WA 15?: \
ex 0‘2 1:8.251xlo“2A
0.0259 , IE. = Diffusion current injected across 8  E junction by the base (electrons for the p  n — p transistor) IE}:qu exp(q—V%’) (electron current) Here we have WE rather than L" in the denominator because WE << L". For V58 = 0.2V, 5 49 102
210 xl.6><109 x10x§L5xlO ) “J 0.2 =2_709xm,5A
10‘ ><3><lo~ @0259 Similarly, at V5,; = 0.6V, 15,, = 1.38x10‘8A 15 I! Now that the various current components are known. we can calculate
the emitter injection eﬂiciency,
v = “~15” [in + 11.5,l For VEB = 0.2V, 8.251x10‘” 8.251 x10‘42 +2.7o9x1cr'5
For VEB = 3.8x10"5
3.8x10’5 +l.38><10'8 = 0.9997. 7: = 0.9996. M For the BJT in Prob. (7.19), calculate a. lg, 13, [c and the base Gummel number. 114 114 To calculate the common base current gaina : on = By. For V53 = 0.2V, (1 = (0.995X0.9997) = 0.9947.
Similarly, for V5,, = 0.6V, 0: = (O.994)(0.9996) = 0.9936. To calculate l3 : B =g—« l — a
09947 For VEB = B = W = 187.7. For VEB =,0.6V, B = = 155.3. To calculate current 15, 18, and [C for V5,, = 0.2 and 0.6V, the emitter
current is given by : 15 = 15;; + [En where 151, and 15,, are the hole and electron currents. respectively,
injected across the base  emitter junction. At VEB = 0.2V, IE = 8.251x10‘12 +2.709x112)"5 = 8.254x10“2A = 3.2pr At V53 = 0.6V, IE = 3.8x10‘5 +1.38X10“ = 3.8x10“"A = 38M The collector and base current can be determined by
3 Dion,
[C =(XIE 01' 1C =BIEP 1V8”? quE,/k7' and [3:(1’a)15=[£‘lc For V53 = 0.2V, 0: = 0.9947 and [E = 8.254pA,
IC = 0.9947 x8254 = 8.21pA [B = 8.254—— 8.21 = 0.044;». For Vm = 0.6V, 0: = 0.9936 and 15 = 38%, [C = 0.9936 x 38 = 37.8% 18 = 38 — 37.8 a 0.2;». Base Gummcl number = NBWb = 1016 x 1.09x 10'4 = 1.09 x iolzcm‘2 (for V53 = 0.2V)
For V55 = 0.6V, Base Gummel number = NBWb =1016 x0.985 ><10’4 = 9.85 x loucm'2 115 115 \ , Prob. 7.22 For the given Si pn—p BJT, calculate the {3 of the transistor in terms of B and y, and using
the charge control model. Comment on the results. In emitter, Lf =,/D,,z,, = In” 571:” =x/(150)(0.0259)(1X10"6)=1.97X10'3cm=19.7w!)
q In base,
I
Lf, = .[Dprp = «or git,” =‘!(4OO)(0.0259)(25x10'6)=1.61x10’2cm =161p.m Assuming the emitter width is much greater than LHE, 1—1 V LLfobe ' +1 36
y=31+ 8 TE F =1+Wl =o._999992
L ppm, L,” L (400)(3x10 )(l9.7)_]
2 2
B=1——WL,= 1— (0'2) , =0.9999992
2L; zuexr
B: By 2 (0.9999992)(0.999992) =le6st
1  By 1 * (0.9999992x0999992)
Charge control approach
2 4 2 —4 2
FT”; =———~—~(0'2X12T) =J2$x1§m§9 =1.93x10““s
P 205—» 4 X ‘ ~ ~’
q
o “'6
B =:&=_:5_>.<L.=1.3x106 r, 1.93x10‘” By the two methods, the values of 8 calculated are different because of the different
approximations made in the two approaches. For the second (charge control) approach we erroneously assume 7 = l which, therefore, gives a higher value of B. In the ﬁrst
approach, we have a better approximation for'y (y < l). 116 116 Prob. 7.23
Far the BJT in Prob.{7.22}, calculate the charge stored in the base when Va; 3 0V and
V53 = 0. 7V. Findfr, if the base transit time is the dominant delav componentfor this INT. Q}; ” é'qAWiapaé’qraikr) \_
2.:(1.6x10"9)ﬁ0“4)€).2x10‘4lL—Z1‘5X10m lemma”)  to” a
l
=1.968x1o“"c
t Jezwzmst
’ zap 2(400x0.0259) fr = “L = ..t..._.~....lw = 8.25 x 1 09112 2m, 2x3.14x1.93x30““ Prob. 7.24 Given 3 apt: transistors which are identical except that transistor #2 has a base region
mice as long as transistOr #1 and transistor #3 has a base region doped twice as heavily
as transistor #1. All other dopings and lengths are identical for transistors #1. #2 and
#3. Give the number (s) of the transistors or tramistars which have the largest value of
each parameter, and give clear mathematical reasons for each: (a) emitter injection
efﬁciency (b) Base transport factor (c) punchthrough voltage (d) collector junction
capacitance with Va; reverse biased at 101’ (a) common emitter current gain. 1 2 3 +1Wk k—zw—q ﬁfth n W
N1 2N, N1 117 l
(a) 7=~————£"
IEniIEP ’7‘“ >7” because base doping is higher in #3 2 15,, less is in 3.
7,“ >7” because base width is larger in #2 => 15,, less is in 2. (b) 15:2; x=W3 B,“ > 832 because base width is larger in #2 :9 cause more recombination.
B“ > 13,8 because base doping is higher in #3 => cause more recombination. 2
7 _ QNBWI:
(‘3) pp! " 283 22> #2 > #1 because Wb > 2W6 (four times more)
#3 > #1 because Na —> 2N“ (two times more). Therefore, #2 highest. ('3) Ci“(Ndopmg}l/2 =9 #3highest. (e) 13 = 39; = 1 87m , Since (a)\and (b) Show #1 to be higher, #1 has highest [3.
w T, = 100 ps in the base of an n—pn, and electrons drift at v, through the
1 pm collector depletion region. The emitter junction charges in 30 ps and
the collector has Cc = 0.1 pF and rc 5— 109. Find fr. The total delay time for the parameters given is 10" rd 2 100ps + + 30135 +10(U.1)ps = l41ps
fT : —1 = 27er 118 118 Prob.7.26
An n—p‘n Si BJT has A"? 10‘s and NE a 10” cm”. AI. whai VB}; 21%:
Am; : NE ? Comment on 7. Am; = ilpquBE/ﬂ Set this equal to NE  , ,r __ g: B __ 10“ _.
This occurs for ‘31.; — g in N:‘ /np w 0.0259111 215x“), ~ 0.695V Wiﬁh z 100, high injection is not reached unii1 the emitter junction is N
biased to nearly 0.7 V. Since: the contact, po‘wntial V9 z: 0.0259111 0.81V, this is a very high bias. Thus 7 :01!on due to high injection is not.
likele in the normai operating range. Prob. 7.27
Find ApEU) if the emitter has an applied unitage 22mg} V5.3 —:« 2285(3)
117mm V53 >> IcT/q and veg, «<1 kT/q. Amt) = pnie‘W’M—l) 2 p” WW using v(t) = 1:53 :. V513 + v“, we have
5115(1) a In; expiwngasM/kfl’l 2m{expWVm/k’l'ﬂ{WWW/W} = Apsidc) expwvnil k1“)
If ad. << JCT/q , qvcb 3 APEU} = APE(dC)i1“*' H 119 ...
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This homework help was uploaded on 03/19/2008 for the course EE 339 taught by Professor Banjeree during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Banjeree

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