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Unformatted text preview: Solutions to Chapter 4 Problems
Fundamentals of IIIV Devices, by W. Liu 1.1 When we consider the Early effect, then the commonbase yparameters of Eq. (470a)
should be replaced by, _} ,Vec yet"
b4!) —i ycc ycc ] .
_ gy( l +jm—w%~)+ij,‘p  aTagc 5( 121.32): )
t ‘ n ' 7—. ’
 O‘Toge( 1 '.I(1"77)“ag)—:— ) Mp (.Ime/2) ge :( 1 +1 31f)“ )‘l'chjc ya. and ycc are changed to include the Early effects, in accordance with (liqs. (4—64) and (4—
66). The common—emitter y—parameters should be modified as well. According to Eq. (4 83),
(Yb:
82(1' aTa)+ja‘Cje + Cjc)+ jgeggm‘l'jge 140‘ ' geE":(1 ' aTo) 'j 895 .LQ...ijjc
(Do we
. .(DT —  . . :7  2w . .
' aToge( 1 ‘Al'm)'(%:—'J'_2m)+ ge :( 1 +1 32a: )‘Jijc 8e *‘( 1 +1 3600 )+]wC}c To construct the hybrid1t model, we again ‘use the transformation from Fig. 411 to Fig. 4
12. We notice that the element — y12 in Fig. 412 is simply  y12e, or ' Yl2e =ge 5“" aT0)+j 86 E ﬁT‘j ijc ~
Comparing Fig. 4—12 to Fig.4—27, we find that, —y12‘e=;§:+jwcs+ jwcjc 
.— Therefore,
RE =____.L__.
ge 3(1 ' aTo)
X2.
C: = E '1 = E ' B
“ (Do 8e 2 D» 2. (a) According to Eq. (476), .
' [y] _[ Ybb ybc ]_[ 1.3 +j3.4 O.2j0.005
“ ycb ycc ’ 9.8 ~j 8.7 0.3 +j 2.5
Using the relationships of Eq. (473) and (475), the remaining 5 yparameters (all in the
unit of mS) characterizing a threeport are: ybe =  Ybb ~ch =  1.1 — j 3.395; yce =  ycb —
ya = ~101'j 62: M =  ybb m = 111+j53;yec =  ybc  ya: =  01j2.495;yee =
 yeb  yec = 11.2  j 2.8. The three—porty—parameters are:
yee yeb yec
[y] = J’be Ybb ch
y“ Yeb Yea
11.2 —j2.8 11.1+j5.3 —O.1—j2.495‘
[y] = [ 1.1 j 3.395 1.3 +j 3.4  0.2 j0.005 ] in the unit ofmS.
—10.1j6.2 9.8 —j 8.7 0.3 +j 2.5
The commonbase y—parameters are: ,
[y] _[ 11.2 j 2.8 0.1j2.495
1’ “ 10.1 j 6.2 0.3 +j 2.5
, (b) The common—collector y—parameters are:
LY] _ 11.2j2._8 11.1+j5.3 ]
"_ l.1j3.395 1.3 +j3.4
(c) According to Eq. (4148), in the unit of mS. 37 2 R6(y11)R€()’22) Rd M2921) ‘ .12’12'9’21 I
For the commonemitter configuration, k _.2x1.3><0.3 Re((O.2—j0.005)(9.8—j8.7))
W“ ‘ (— 0.2 j0.005)  ( 9.8 j8.7) _ 2 x13 x 0.3 .( — 0.2) x 9.8 — (— 0.005 x 8.7) 10.22 + 0.0052 x 19.82 + 8.72 , For the common—base configuration
k _ 2 x 11.2 X 0.3  Re( (0.1 j 2.495)  (—10.1 j 6.2))
W“ ‘ (—O.1j2.495)'(—10.lj6.2) _ 2x11.2><0.3 0.1><10.1~(—.2.495 ><6.2)_O77 . 1012+ 2.4952 X 110.12 + 6.22
((1) Applying the transformation listed on Fig. 4—17, we obtain h11 = 98.1 j257 9; I212 = 0.021 —j0.051; h21=—1.27 j3.37 Q; and h22 = 6.08 X 10'5 —j1;82' X 10'3 9'1. kRolleI = =1.06’ 3. (a) When N3 = 3 x 1019 cm'3, the minority carrier lifetime 1”,, = 1.16 x 10'10 5,
according to Eq. (171), and the minority carrier mobility u" = 989 cmZ/Vs, according to
Eq. (175). D", found from the Einstein relationship, is/25.5 cmz/s. According to Eq. (4—
52), the base transport factor is, ‘ (800x 108 )2
2  25.5  1.16 x1010 coo, accOrding to Eq. (436), is we = 43153—4: 7.97 x 1011 rad‘ (800 x 108)2 an, = 1  = 0.989 (b) We assume the collector current ideality factor is 1. The emitter current, which is
approximately equal to the collector current, is equal to lg x A}; = 104  2 x 30 x 10‘8 = 6 , x 10'_3 A. The emitter transconductance, ge, according to Eq. (436), is, 6 X 10'3 1
g“ “ 0.0258 ' M33 5' The collector area AC = A5  2.5 = 1.5 x 10‘6 cm2. From Eq. (4—67), the base—collector
capacitance is, . 12
q,.1.5x1062:92<_19_.3.5xmMF . 0.5 x 10'4
At T: 300 K, the saturation velocity is calculated from Eq. (177) to be 8.3 x 106 cm/s. The parameter Tm is calculated from Eq. (459) to be,
__ 0.5 x I04 8.3 x 106 ,
c. Assuming that CjE = ”O, the commonemitter yparameters can be found_from Eq. (4
85b) as, ' rm = 6.02 x 1012 s 27t'1010 ~ 6.02 x10"12 ,y11= 0.233 (1  0.989) +1 2115'1010 (0 + 3.5 x 10”) + j 0.233 2 33’ V . 10 .
+j0.233._?“_12_._ =2.56+j49.1 mS 2n  7.97 ><10ll
m = j 2n10‘0( 3.5 x 1014 )= j 2.2 m3 . 10 ‘
m = 0.233 0.989 1—j( 13—)—_3“—L—i 2711010 . 6.02 x 1012
3 21t7.97><10” 2 j 2711010  3.5 x 1014 = 230 j 467 m5
yzz =j 27I~10‘°( 3.5 x10'14)=j 22 ms 4. (a) According to Eq. (1474), the majority carrier mobility in the base layer of 3 x 1019
cm 3 doping level 15 up— = 61. 3 cmz/V— s. The base layer resistivity is found from Eq. (3
53) to be: p3 = ———4—————1———————— = 3.4 ><10'3 .Qcm ._ 1.6 x 10‘19  61.3 3 3 x1019
. Rgiavsyis given by Eq. (3—52): 2 x 10‘4
Rm“) = 3.4 x 103 . ———,————————— = 28.3 9 30 x 104  800 x 10'8
The intrinsic base resistance, in the absence of current crowding, is given by Eq. (360): 113,:28—31—2369 12
The total base resistance is RBi + RBB/Z—  12. 36 £2. (b) We are still assuming that CE = 0. According to Eq. (4127), the emitter charging time
is J ‘
, 1 14 _. 13
T—e *ngC; 0233 .'35x10 1.5x10 s The minority carrier mobility IS un — 989 cm2/Vs, according to Eq. (1 —.75) DnB, found
from the Einstein relationship, is 25. 5 cm 2./s The base transit time is governed by Eq. (4 128), with v — 2 for a uniform base: X2 (800 x 10'3)2
——————————= ‘12
Tb: ——L2Dn8= 2 255 1.25X10 S The collector spacecharge transit time is found from Eq. (4 130):
15c=3§m=§93§ﬁ= 3 01 x 1012 s
Finally, the collector charging time is found from Eq. (4—131):
rc=(6+1) 35x1014=245x1013s
The sum of the transit/charging times are rec: 1. 5 x 10 13 + 1. 25 X 10'12 + 3 01 x 10 12 +
2.45 x 1013: 4. 66 x 10'12 s. The cutoff frequency, inferred from Eq. (4126), is fT=——1—=—————1——'—= 3. 42x 101°Hz
. 2mg, 21: 4.66x1012 
c. The maximum oscillation frequency 1s given by Eq. (4— 159) as, 10
fm= 4 / ___3_£:<_19____ = 5.61 x 1010Hz
. 8n12.363.5x1014  . 5. (a) the emitter contact resistance is given by Eq. (4—112): 3‘1 REE: —————1—0—()———— — 1.67 9
_ 2 x‘ 10 4 30 x 10 4
(b) The base contact resistance is given by Eq. (4 120): 6 7 . 6 \
R33 =———'———Wco1h 11x1044/ﬂ)=—————M 1.13 = 5.32 12
30><10'4 106 30><10'4 (c) The collector resistance is given by Eq. (4—121): 4/?  6 1/9 . 6
Rccz—IL—l—Q—coth 1x104 20—)=—:9——10—1=1.49Q
30x10‘4 10'6 3O><10'4 6, We use the material parameters of § 1— 7 The election saturation velocity at room temperature (300 K) 18 found from Eq (1— 77) to be 8 3 x 106 cm/s. J], calculated from
Eq. (3 80) IS, 11:16x10‘9 8.3x1o6 3x1016=3.98x104 —A— cm2
At a doping level of 3 x 1016 cm’3, the majority electron mobility lS calculated from Eq. (1 73) to be 5735 cm2/V— s. The resistivity of the collector layer 15,
pC = 1 = 1 =363X10H29cm
qﬂnNc 16 x 1019  5735  3 x1016 JC, is 1 x 104 A/cmz. From Eq. (383), the depletion thickness is:
1.159 x10‘12~‘3.63 x1021x104 1 1x 104 ’1
1.6 x 1019  3 x 1016 3.98 x 104 +[(1.159x10'123.63x10'21x104)2(1_ 1x 104 )‘2
1.6 x 1019  3 x1016 3.98 x 104
21.159x10'12(4+1.3)(1_ 1x104 )“
1.6 x 1019 . 3 x 1016 3.98 x 104
1.159x 1012  3.63 x10'2~ 1 x104~ 1.3 x10'4(1_ 1x104 )“ r
2 1.6><1o19 3x1016 3.98x104 = 5. 87 x 10 5 cm.
The depletion thickness is 0.587 um. Xdep = + b. Tsc is found from Eq. (4130): 7 5
r“: w: 3. 52 x 1012 s 2 8 3 x 106
The basecollector junction capacitance is found from Eq. (4—67): 1.159 x 10 12
5.87 x 105 = 3.55 x 1014 F qt: (3 2x30x108) Tc is found from Eq. (4131) as,
c=(5+1 )3.55x1014=2.13x 1013s , 7. I (a) We follow the calculation of § 47, except noting that the emitter width is changed
from 2 to_1 11m: LiO 7 8 ‘ _. —8
_OOOXIO +0.07’HOOO 310)X10 =0.66Q Rm  = 1.3 ><10‘3 3
"m 1><10""30x10‘4 1x10'430x10‘4
7.4 0‘8 I
REE = . X 1 = 0.246 S2
,1><10'430><10‘4
W W 1 X 10'4
RBims) = £8 % = RSHB —— = 280' = 9.3 Q
. E E 30 x 10
_ _ Ramp.“ __ 9.3 _
1171— 12 I*———12——078$2
. 0.2 x 10'4
RBxlcpi) = RSI/B SLBE = 280 ——————— = 1.87 9. (unchanged)
E 30 X 10‘4
’V 280 '. . 0‘6 ’
R33 = 3 3 x 1 coth 1 x 10’4  q / ————————280 = 13.9 $2 (unchanged)
30 x 10‘4 3.3 x 10'6
0.2 x 104 + 0.2 x10'4 + 1x104
RSC(epi) = 85 ————————————‘———‘—‘— = 0.40 9 (unchanged). 30 x 10‘4
1.2 x 104 — 0.591 x 104 =7.4§2
1x10430x104 / RC(epi) = 3.63 X 102 RC5: _8 5 8 X 10 coth 20 X 10‘4  ————&'———— = 0.89 £2 (unchanged)
30 x104 1 8.4 x 107 Hence, .
R5 = RE(epi) + REE = 0.66 + 0.246 = 0.91 9. _ ». Ramp.) R33 1.87 13.9
Tb—Rgll 2 +12 —0.78+ 2 + 2 —8.679.. 510c+§§§gza=74+9§2+9§2=805§2 Now, let us work on the capacitances, again bearing in mind that the emitter width is reduced from 2 to 1 pm. , /.2 eaves +.¢ca) ( Jo) 4/2 _ .5
Xdep _ q NC .  J]  5.91 X 10 cm (unchanged) 12
CF: 1x10‘430 x10'4 1'159 X10 = 1.12 ><10'13 F
‘ 0.031><104 1.159 x1012 . 0.591 x 10'4
Finally, we are ready to calculate the transit time components: I 0.0258 1.12x 10'13 2,0 1 14
Te: k (Cje+Cjc)=—————(——————‘:—X_'9_')'=l.14ps Cjc = 2( 0.5 + 0.2 +1)$<10430 x 104 = 2.0 x10'14 F qJCWELE 1 XV'1041 X 10430 X 10_4
2 800 x 108)2
= ___2£B____ = _(_____________ =
 Tb V'kT/qwns 2  0.0258 . 988.62 1‘25 PS (“Chang“) Xdep _0.591x10‘4 : 3.69 s (unchanged)
2”sat 28x106 p 41 r, =‘( R5 + RC )C,, = ( 0.91 + 8.05 ) 2.0 ><10'14 = 0.18 ps
T“ = 1,, + 1,, + I“ + TC 2 1.14 +1.25 + 3.69 + 0.18 = 6.26 ps
fr: 7 1. =————1————,= 25.4 GHz
J‘Tec 2  3.14  6.26x 1012
The percentage change infTis (25.4  26)/26 = — 2.3 %. b. The maximum oscillation frequency is,
1/2 ’ . . 25.5 9 ~
f”m = ./ f7 =( X 10 = 76.5 GHz‘
SWIaCjc 8  3.14  8.67  2.0 ><10"4 . The percentage change inﬁnax is (76.5  65)/65 t 17.7 %.' 8. According to the question, NB was 3 x 1019 cm'3 and RSHB was 280 Q/sq. However, with the doping change, N3 = 1 x 1020 cm‘3 and RSHB = 8.4 Q/sq. (In reality, when the
base doping is this high, not all of the dopants are ionized, causing the effective electrical
density to be less than 1 x 1020 cm'3. Consequently, measured RSHB hardly goes below
100 Q/sq.) (a) We now follow the calculation of § 47.
RE(epi) = 0.33 S2 (unchanged)
R55 2 0.123 S2 (unchanged) 2 ><10‘4
R810”): RSHB—iWL: 84 ' ——————= 5 6 £2
E 30 ><10‘4
Ram”) 5.6 _
r , 12 12 — 0 47 Q
Sm; 0 2 x 10 4
RB( )=Rsys——~—=84 =056Q
”I" LE 30 x10‘4
1/ . 6 .
R33=’Wcoth 1X10'4 ' ———4— =11.9 Q
30 x 104 , . 3.3 x 10—6 quepi) = 0.40 52 (unchanged)
Raepi) = 3.67 £2 (unchanged)
RCC = 0.89 £2 (unchanged) Hence,
RE = RE(epi) + REE = 0.33 + 0.123 = 0.45 9 RB_._x2<epi> + 5281 = 0.47 + 225—6 + LZQ = 6.7 52 RcﬁRCei +KQQ+RSC(epi)=3.67+9__8_9_+§L49=4.3252
(P) 2 2 2 2 Now, let us work on the capacitances, again bearing in mind that the base doping has
increased and the base sheet resistance has decreased. Xdep = 5.91 x 10'5 cm (unchanged)
C}. = 2.24 x 1013 F (unchanged) 6,, = 2.59 x 1014 F (unchanged)
Finally, we are ready to calculate the transit time components: =_kZ'___ .  . = d
I; chWELE (C1,,+ C10) 1.08 ps (unchange ) rb= R3i+ 4'2. X32 _ (800x 108 )2 = ’—‘_‘—  ———————— = 1.25 h d
Tb V'kT/wnn 20.0258988.62 1’5 (”“C ange )
X .5, 04
r“ = def) 2 1.9.1.5.; = 3.69 p5 (Unchanged) 2%: 28x106 TC =( R5 + R( )~Cjc = 0.124 ps (unchanged) ,
Ta = r. + 75,, + r“ + rc. = 1.08 + 1.25 + 3.69 + 0.124 = 6.14 ps (unchanged) f7: 1 = I = 26.0 GHz (unchanged)
2“ch 23.146.14><10"3
The percentage change infT is 0 %. The base doping change does not affecth because the
depletion of the base—emitter and basecollector junctions occurs in the emitter and the
collector layers. As long as the base doping is high, the baseemitter and basecollector
capacitances are determined by the emitter and the collector doping levels. b. The maximum oscillation frequency is, f = FT. = '26x109
max Sﬂrijc _ 8  3.14 . 6.7  2.59 X 10‘14
The percentage change in fmax is (77 — 65)/65 = 18%. ' 1/2
= 77 GHz 9. We brieﬂy describe each of the change.
(a) Changing the emitter length reduces the resistance values, but also increases the capacitance values by the same factor. Therefore, there is zero percentage change infT and
fmax. (In practice, elongating finger too long causes the current density to be nonuniform.
Making the finger too short causes the transistor behavior to depend on the edge effects.
Therefore, there is an optimum range of emitter length. However, in general, the
dependence of the transistor characteristics on the emitter length is small.) (b) Changing W3 from 1 to 3 mm does not change RE, RB and RC significantly.
However, Cjc increases dramatically. The emitter width is 2 mm, and the baseemitter spacing is 0.2 pm. The collector area was [ 2 + 2(1 + 0.2) ] x LE, whereas now the collector area increases to [ 2 + 2(3 + 0.2) ] x LE. Therefore, the collector area becomes
larger by a factor of 1.91. This means the collector capacitance becomes larger than before '
by a factor of 1.91. ‘ This larger capacitance value will degrade both f1 and fmax. However, the degradation
in f7 is on the 5 % range since the dOminant transit/charging time is the spacecharge transit
time. Its magnitude, 3.69 ps, is much higher than other time constants, causing fT to be
relatively unmodiﬁed. The capacitance's impact on fmax, in contrast, is significant. Since
fmax has a square root dependence on the inverse of the capacitance, fmax is likely to
become 1/sqrt(1.91) times the original fmax. That is, fmax becomes a factor of 0.72
smaller. This is equivalent to 28 % reduction in fmax. (c) Increasing the base—emitter spacing will increase the extrinsic base resistance.
However, none of the transit/charging time constant depends on the base resistance.
Therefore, its effect on fT is nil. However, base resistance is a significant factor in the
_ determination of fmax. From the calculation of RBx(epi) in § 47, we see that the resistance increases from 1.87 9 by a factor of 5 (which is the ratio of 1 um and 0.2 pm). Therefore,
RBx(epi) becomes 9.4 Q. The total base resistance r], calculated in § 47 was 9.45 9. With
an increment of 9.4  1.87 = 7.5 Q from widening the baseemitter spacing, the total base
reslstance in the new structure is 9.45 + 7.5 = 17 $2. This represents a factor of 17/945 =
1.8 increase from the previous r1, value. Because fmax is inversely proportional to rb, fmax
in. the transistor with 2 urn base—emitter spacing is decreased by a factor of 0.74. This represents 26 % decrease of fmax. Lia (d) When the base thickness is changed from 800 to 1500 A, two main factors determining
the high frequency performance are modified. The first is the base transit time, and the
second is the base resistance. Let us first discuss about the base transit time, which, given
by Eq. (4—128), is proportional to the square of X3. At X3 7: 800 A, the base transit time
Was calculated in § 47 to be 1.25 ps. With the new base thickness of 1500 A is used, the base transit time will become 1.25 ' (1500/800)2 = 4.39 ps. This represents an increase of
3.14 ps. The overall transit time calculated in § 47 was dominated by the spacecharge
transit time. However, now that the base transit time has increased dramatically, the base
transit time is now a significant component. From § 47, owe se that the total emitter—
collector transit time was found to be 6.14 ps for X3 = 800 A. With the increase of 3.14
ps when we thickened the base, the total emitter—collector transit time becomes 9.28 ps for
X3 = 1500 A. This is a factor of 9.28/6.14 = 1.5 increase. Since fT is inversely
proportional to the overall transit time,fT is expected to decrease by a factor of 1.5, or 50 % reduction. We first focus on X 3'3 effect on the base resistance. When X3 decreases, the base
sheet resistance decreases by the same factor that the base thickness decreases. That is, by
a factor of 1500/ 800 = 1.88. Of the three base resistance components, R3i and R 31(epi)/2
are directly proportional to the base sheet resistance. However, R33 is only indirectly
related. From a crude estimate from Eq. (4120), we see that R33 is proportional to the
square root of the base sheet resistance. Because the base sheet resistance decreases by a
factor of 1.88, we expect R33 to decrease by a factor of 1.37. The calculation of § 47
demonstrates that R33 dominates the base resistance. We/can therefore estimates r1, to be
decreased by the same factor that R33 decreases. Now, fmax itself is inversely proportional
to the square root of rb, However, besides the effect from rb, Eq. (4159) shows that fmax
is also proportional to the square root of fT. Since the improvement of fmax from the r],
reduction is 1.37 yet the degradation of fmax due to fr reduction is 1.5, we find fmax to decrease somewhat. The arriount of the decrease is not too Significant because the 1.37
improvement factor nearly cancels out the 1.5 degradation factor. (e) The depletion thickness in the collector is 0.591 um, as .calculated in § 47.Changing
subcollector thickness from 1.2 to 1 pm does not affect the collector capacitance, which
would be an important factor in the determination of f7 and fmax. Only a relatively
unimportant parameter is affected by the change in the collector thickness, an action which
reduces the amount of the undepleted collector thickness. It is the undepleted collector
resistance, which is now reduced as the subcollector thickness decreases. However, the
reduction of f7 due to the decrease of this resistance is small, mainly because fr is primarily
determined by the transit time in the spacecharge region. The effect on fmax is also
negligible since the collector resistance does not play a role in the determination of fmax. We summarizes the above discussion.
Case ’ Change in fT Change in fmax (a) 0% 0%
,(b) —5%orso 28%
(c) 0% " 26%
'(d) —50% ~ 5%orso .
(e) ~O% ~0% Therefore, for f7, the increase of base thickness in (d) produces the most profound effect .
whereas all the other four actions have little effect. As far as fmax is concerned, changing
the base width and the baseemitter spacing has the same order of negative effects. This is
followed by the changing of the base thickness. Actions in (a) and (e) does not affect fmax. 10. When the collector doping level is changed, the most important consequence is the change of the depletion thickness. This leads to a change in the space—charge transit time,
which affects fT, as well as a change of the basecollector capacitance, which affects fmax. 4‘4 (a) We now follow the calculation of § 4—7.
R1507») = 0.33 S2 (unchanged)
R55 2 0.123 £2 (unchanged)
R310”, = 18.67 S2 (unchange)
 rb; = __l_81__76_7_ = 1.56 S2 (unchanged) Rana,” 2.1.87 9 (unchanged)
R3,; = 13.9 $2 (unchanged)
Rsmpn = 0.40 S2 (unchanged) Ra = 0.89 $2 (unchanged)
We are now ready to determine the resistance associated with the undepleted portion of the
collector. For this, we need to establish the depletion thickness when the collector doping is 7 x 1016 cm‘3. To do this, we firstfind the builtin potential of the basecollector
junction. According to Eq. (371), 16
¢CB = M244 0.0258  In A: 1.34 V.
2 ‘ 1.79 x 106
The depletion thickness is obtained from Eq. (378). We assume that 12 » JC. With J 1 =
qNc'vmt, we find: _, /255(VCB+¢CB)( lg) 4/2
Xdep , (INC 1‘11 _ 21.159x1012(4+1.34) 1 104
1.6x10197x10l6 ' 1.6x10197x10168x106 = 3.53 x10‘5 cm .
So, Xdep = 0.353 um. According to Eq. (499), 1.2 04  0.353 10*4
Ram) = 3.63 ><10'2 ——)—‘—1———————X—— = 5.12 9. 2 x 104  30 ><10'4  1/2 Hence, . _ '
' R5 = Rmpi) + R5,»; = 0.33 + 0.123 = 0.453 S2 (unchanged) n, = R3, + 53129—“1 + 55—51 = 1.56 + L381 + 1—3242 = 9.45 52 (unchanged) R¢=Raepﬂ+§§£+RS—C2‘eﬂil=5.12+0—§2+Q§—0=5.77sz Now, let us work on the capacitances, again bearing in mind that the depletion thickness in
the base—collector junction has changed. The depletion thickness in the baseemitter junction remains unchanged.
Cje = 2.24 x 10‘13 F (unchanged)
‘ 4 1.159 x1012 0.353. x 104
Finally, we are ready to calculate the transit time components: 0.0258(2.24x 1013 +4.33 x10‘14)‘_ 1 15 ps
1 x 1042 x 10430 x 104 ' Cjc=2( 1+0.2+1)x10’430>< 10 =4.33 x 1014F kT _.
1' =————— C + C ..
e V qJCWELE ( 13 JC) 2 _ .
1b = a—XL—...
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