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4 - Solutions to Chapter 4 Problems Fundamentals of III-V...

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Unformatted text preview: Solutions to Chapter 4 Problems Fundamentals of III-V Devices, by W. Liu 1.1 When we consider the Early effect, then the common-base y-parameters of Eq. (4-70a) 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 hybrid-1t model, we again ‘use the transformation from Fig. 4-11 to Fig. 4- 12. We notice that the element — y12 in Fig. 4-12 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. (4-76), . ' [y] _[ Ybb ybc ]_[ 1.3 +j3.4 -O.2-j0.005 “ ycb ycc ’ 9.8 ~j 8.7 0.3 +j 2.5 Using the relationships of Eq. (4-73) and (4-75), the remaining 5 y-parameters (all in the unit of mS) characterizing a three-port are: ybe = - Ybb ~ch = - 1.1 — j 3.395; yce = - ycb — ya = ~10-1'-j 62: M = - ybb -m = -11-1+j5-3;yec = - ybc - ya: = - 0-1-j2.495;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. —10.1-j6.2 9.8 —j 8.7 0.3 +j 2.5 The common-base y—parameters are: , [y] _[ 11.2 -j 2.8 -0.1-j2.495 1’ “ -10.1 -j 6.2 0.3 +j 2.5 , (b) The common—collector y—parameters are: LY] _ 11.2-j2._8 -11.1+j5.3 ] "_ -l.1-j3.395 1.3 +j3.4 (c) According to Eq. (4-148), in the unit of mS. 37 2 R6(y11)R€()’22)- Rd M2921) ‘ .12’12'9’21 I For the common-emitter 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.1-j2.495)'(—10.l-j6.2)| _ 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 -j-257 9; I212 = 0.021 —j-0.051; h21=—1.27 -j-3.37 Q; and h22 = 6.08 X 10'5 —j-1;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. (1-71), and the minority carrier mobility u" = 989 cmZ/V-s, according to Eq. (1-75). D", found from the Einstein relationship, is/25.5 cmz/s. According to Eq. (4— 52), the base transport factor is, ‘ (800x 10-8 )2 2 - 25.5 - 1.16 x1010 coo, accOrding to Eq. (4-36), is we = 43153—4: 7.97 x 1011 rad‘ (800 x 10-8)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. (4-36), 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,.1.5x10-62:92<_19_.3.5xm-MF . 0.5 x 10'4 At T: 300 K, the saturation velocity is calculated from Eq. (1-77) to be 8.3 x 106 cm/s. The parameter Tm is calculated from Eq. (4-59) to be, __ 0.5 x I04 8.3 x 106 , c. Assuming that CjE = ”O, the common-emitter y-parameters can be found_from Eq. (4- 85b) as, ' rm = 6.02 x 10-12 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 2n-10‘0( 3.5 x 10-14 )= -j 2.2 m3 . 10 ‘ m = 0.233 -0.989 1—j( 1-3—)—_3“—L—-i 271-1010 . 6.02 x 10-12 3 21t-7.97><10” 2 -j 2711010 - 3.5 x 10-14 = 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 .Q-cm ._ 1.6 x 10‘19 - 61.3 3 3 x1019 . Rgiavsyis given by Eq. (3—52): 2 x 10‘4 Rm“) = 3.4 x 10-3 . ———,—-———————— = 28.3 9 30 x 10-4 - 800 x 10'8 The intrinsic base resistance, in the absence of current crowding, is given by Eq. (3-60): 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. (4-127), the emitter charging time is J ‘ , 1 -14 _. -13 T—e *ng-C; 0233 -.'35x10 1.5x10 s The minority carrier mobility IS un- — 989 cm2/V-s, 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 space-charge 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 10-13: 4. 66 x 10'12 s. The cutoff frequency, inferred from Eq. (4-126), is fT=——-1-—=-—————-1——'—= 3. 42x 101°Hz . 2mg, 21: 4.66x10-12 - c. The maximum oscillation frequency 1s given by Eq. (4— 159) as, 10 fm= 4 / ___3_-£:<_19____ = 5.61 x 1010Hz . 8n-12.36-3.5x10-14 - . 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 11x10-4-4/fl)=—————M- 1.13 = 5.32 12 30><10'4 10-6 30><10'4 (c) The collector resistance is given by Eq. (4—121): 4/? - -6 1/9 . -6 Rccz—IL—l—Q—coth 1x10-4- 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 10-19 - 5735 - 3 x1016 JC, is 1 x 104 A/cmz. From Eq. (3-83), the depletion thickness is: 1.159 x10‘12~‘3.63 x10-2-1x104 1 1x 104 ’1 1.6 x 10-19 - 3 x 1016 3.98 x 104 +[(1.159x10'12-3.63x10'-2-1x104)2(1_ 1x 104 )‘2 1.6 x 10-19 - 3 x1016 3.98 x 104 2-1.159x10'12(4+1.3)(1_ 1x104 )“ 1.6 x 10-19 . 3 x 1016 3.98 x 104 1.159x 10-12 - 3.63 x10'2~ 1 x104~ 1.3 x10'4(1_ 1x104 )“ r 2- 1.6><1o--19 3x1016 3.98x104 = 5. 87 x 10 5 cm. The depletion thickness is 0.587 um. Xdep = + b. Tsc is found from Eq. (4-130): 7 -5 r“: w: 3. 52 x 10-12 s 2- 8 3 x 106 The base-collector junction capacitance is found from Eq. (4—67): 1.159 x 10 12 5.87 x 10-5 = 3.55 x 10-14 F qt: (3 2x30x108) Tc is found from Eq. (4-131) as, c=(5+1 )-3.55x10-14=2.13x 10-13s , 7. I (a) We follow the calculation of § 4-7, except noting that the emitter width is changed from 2 to_1 11m: Li-O 7 -8 ‘ _. —8 _OOOXIO +0.07’HOOO 310)X10 =0.66Q Rm - = 1.3 ><10‘3 -3 "m 1><10""30x10‘4 1x10'4-30x10‘4 7.4 0‘8 I REE = . X 1 = 0.246 S2 ,1><10'4-30><10‘4 W W 1 X 10'4 RBim-s) = £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 $2 (unchanged) 30 x 10‘4 3.3 x 10'6 0.2 x 10-4 + 0.2 x10'4 + 1x104 RSC(epi) = 8-5 ——-———————-———-‘———-‘—‘— = 0.40 9 (unchanged). 30 x 10‘4 1.2 x 10-4 — 0.591 x 10-4 =7.4§2 1x10-4-30x10-4 / RC(epi) = 3.63 X 10-2 RC5: _8 5 8 X 10 coth 20 X 10‘4 - ————&'————- = 0.89 £2 (unchanged) 30 x104 1 8.4 x 10-7 Hence, . R5 = RE(epi) + REE = 0.66 + 0.246 = 0.91 9. _ ». Ramp.) R33- 1.87 13.9- Tb—Rgl-l- 2 +12 —0.78+ 2 + 2 —8.679.. 510c+§§§gza=74+9§2+9§2=805§2 Now, let us work on the capacitances, again bearing in mind that the emitter width is reduced from 2 to 1 pm. , /.2 eaves +.¢ca) ( Jo) 4/2 _ .5 Xdep _- q NC . - J] - 5.91 X 10 cm (unchanged) -12 CF: 1x10‘4-30 x10'4 1'159 X10 = 1.12 ><10'13 F ‘ 0.031><10-4 1.159 x10-12 . 0.591 x 10'4 Finally, we are ready to calculate the transit time components: I 0.0258 1.12x 10'13 2,0 1 -14 Te: k (Cje+Cjc)=—————-(—————-—-‘:—X_'9_')'=l.14ps Cjc = 2-( 0.5 + 0.2 +1)$<104-30 x 10-4 = 2.0 x10'14 F qJCWELE 1 XV'1041 X 10-430 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 2-8x106 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 10-12 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 § 4-7. 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 10-4 , . 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 10-13 F (unchanged) 6,, = 2.59 x 10-14 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 10-8 )2 = ’—‘_‘— - —————-—-——- = 1.25 h d Tb V'kT/wnn 2-0.0258-988.62 1’5 (”“C ange ) X .5, 04 r“ = def) 2 1.9.1.5.; = 3.69 p5 (Unchanged) 2%: 2-8x106 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 2-3.14-6.14><10"3 The percentage change infT is 0 %. The base doping change does not affecth because the depletion of the base—emitter and base-collector junctions occurs in the emitter and the collector layers. As long as the base doping is high, the base-emitter and base-collector 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 non-uniform. 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 base-emitter 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 space-charge 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 § 4-7, 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 § 4-7 was 9.45 9. With an increment of 9.4 - 1.87 = 7.5 Q from widening the base-emitter 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 § 4-7 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 § 4-7 was dominated by the space-charge transit time. However, now that the base transit time has increased dramatically, the base transit time is now a significant component. From § 4-7, 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. (4-120), 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 § 4-7 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. (4-159) 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 § 4-7.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 space-charge 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 base-emitter 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 base-collector 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 built-in potential of the base-collector junction. According to Eq. (3-71), 16 ¢CB = M244 0.0258 - In A: 1.34 V. 2 ‘ 1.79 x 106 The depletion thickness is obtained from Eq. (3-78). We assume that 12 » JC. With J 1 = q-Nc'vmt, we find: _, /255(VCB+¢CB)( lg) 4/2 Xdep , (INC 1‘11 _ 2-1.159x-10-12(4+1.34) 1 104 1.6x10-19-7x10l6 ' 1.6x10-19-7x1016-8x106 = 3.53 x10‘5 cm . So, Xdep = 0.353 um. According to Eq. (4-99), 1.2 0-4 - 0.353 10*4 Ram) = 3.63 ><10'2 ——)—‘—1——-—-————X—— = 5.12 9. 2 x 10-4 - 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 base-emitter junction remains unchanged. Cje = 2.24 x 10‘13 F (unchanged) ‘ -4 1.159 x10-12 0.353. x 10-4 Finally, we are ready to calculate the transit time components: 0.0258(2.24x 10-13 +4.33 x10‘14)‘_ 1 15 ps 1 x 1042 x 10-4-30 x 10-4 ' Cjc=2-( 1+0.2+1)x10’4-30>< 10 =4.33 x 10-14F kT _. 1' =—-————- C- + C- .. e V qJCWELE ( 13 JC) 2 _ . 1b = a—XL—...
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