Chapter Problem Solutions. a. b. c. γ + γ + BE + C + + γ + ( γ ( γ C γ + BE + BE γ BE and C γ ( γ + or C BE + C ma.5 kω.7 ( ma + 4. kω.5 kω C. (a ln C BE T S (i μ 6 A,.6 ln.588 μa C BE 4 (ii μ 6 A,.6 ln.5987 μa C BE 4 (iii ma (b (i C BE 4,.6 ln.6585 ma + β 9.65 μa BE BE T ln + S 5 6 9.65 (.6 ln 4.578 6 96.5 96.5 A.6 ln + 5.5977 (ii μ BE 4
.965.965 ma.6 ln + 5 (iii B B BE 4. + BE.5. K ( on.7 (.5 C C C C.49 ma + + β 6 4. μa.4 + BE 8. B B ( on 5.7 ( 5.58 ma ( 6.98 μa.6575.58 C C C C.4958 ma + + β 8.5 (a (b 5.7 ( 5 + BE on or 58.6 kω.5.7 ( 5 on 8.6 kω + BE.5 Adantage: equires smaller resistance. (c For part (a: 9. ( max.56 ma 58.6.95 9. min.476 ma ( 58.6(.5 Δ.56.476.5 ma ± 5% For part (b: 4. ( max.56 ma 8.6.95 4. min.476 ma ( 8.6(.5 Δ.5 ma ± 5%.6 a. + + or.4 ma β 5.7 7. kω.4
b. A 8 r 4 kω Δ Δ ( 9..5 ma ΔCE r 4 Δ.5 Δ.6%.7 n C C C + B + B C + + β β n + n C + + C + β β β n + n β + or. n + n + β.8 (.. ma + + 4 β 5.7 4..5 K.9 a. 5.7.9 ma 8.9. ma + 5 b. A 5 r 8 kω. Δ Δ (..597 ma.6 ma EC r 7 Δ..56 ma.45 ma 7 c.. a. 5.7 5 9. kω ma 5.7 For EC min.7 C C.5 kω b. c..
.5 ma.5 ma A B A +.5 + β 6.65 ma.5.7 6.86 K.65..7 7. K.5. and (a. ma,.5 ma (b.5 ma,.75 ma (c.67 ma,. ma.4 a.
E C and C + B C+ + β BE C BE E B + B + + β C C + + BE β ( + β ( + β BE + BE ( + β + β( + β ( + β β ( + β b..7.7 + ( 8( 8 + ( 8.76 +.864 (.7.7 ma.7 kω.5 a. ES i C and C + BS C + + β ES B + B + B +... + BN ( + N B ( + N C β ( + N C Then C + β( + β or i ( + N + β ( + β 6.5 +.5 ma b. ( 5( 5 5(.7 ( 5.6 7.6 kω.5
+ (.5.54 ma β ( β + + ( 5( 5 5(.7 ( 5 7.9 kω.54.7 + β ( β + For.8 ma (.8 +.84 ma 5( 7 8 (.7.69 kω.84.8
The analysis is exactly the same as in the text. We hae + β ( + β.9 ma, B.67 ma 75 C ma, B. ma 75 E B + B. +.67.4 ma E.4 B.56 ma + β 76 C + B.56 ma (.7 8.6 8.6 kω. (a β ro Assuming A A r 4 K.5 ( 4 MΩ (b Δ Δ 5 Δ Δ MΩ MΩ Δ.5 μa. + BE 5.7 9..464 ma T.6.464 ln ln E.5.464.7ln By trial and error 4.7 μa BE.7 E BE.7 (.47(.5.675 BE. (a
5.7 5 + BE 9 μ A 9 ma E T ln.6 ln By trial and error, 6.8 μ A ( r + g o o m E Now ro 4.4 MΩ 6.8.68 gm.65 ma /.6 (.6 rπ.68 8.4 kω So E rπ E 8 9.74 kω Then o 4.4 + (.6( 9.74 o 5.6 M Ω (b. Δ Δ.68.68 BE BE o E BE BE ( m A 8.76 MΩ 6.8 m T.6 ( 8(.6 π C r + g r g.68.65 ma/ rπ 6 kω.68 r 6 9.68 K E E E.76 +.65 9.68 4.54 MΩ Now Δ ( 5 Δ. μa 4.54.4 (a 5.7 5.5 8.6 K E T ln.6.5 E ln.5.5. K E
(b r [ + g ] (c c E m r E E π 75.6.5 rπ 9 K gm.9 ma/.5.6 A ro MΩ E. 9.64 K.5 + (.64(.9 ( 6.477 MΩ Δ 5 Δ.77 μa 6.477 Δ.77 %.54% 5.5 Let 5 k Ω, Then.7 4.66 ma 5 Now.6 4.66 E T ln E ln E kω...6 BE T ln S 5.7 (.6 ln S. A S At ma, BE (.6 ln. 5.78 5.78 7.4 kω T.6 E ln ln E.9 kω.5.5.7 a..7.465 ma Let BE T ln S Then.465 BE (.6 ln.68 5. Then 5.7 (.6 ln S. A S
.68.466 ma T.6.466 b. E ln ln E 4Ω...8.7 (.485 ma 4 BE T ln S 5.7 (.6 ln S. A S Now BE BE So.485 (.6 ln.68 5..68.68.48 ma 4 E T ln.48 (.6 ln By trial and error. 8.7 μa.68.87.5766 BE BE E BE.9 + + BE E BE E BE BE E E For matched transistors BE T ln S BE T ln S Then T ln E E utput resistance looking into the collector of Q is increased.. (a (b + 5.7 5 BE + E 7. +.74 ma.74 ma Using the same relation as for the widlar current source.
( π r o + gm E r A 8.74 ro 5 K gm. ma/.74.6 (.6 rπ 8.9 K E rπ 8.9.68 K.74 5 + (.(.68 5. MΩ (c 5.7( 5.47 ma 7. A 8 ro 5 K.47. Assume all transistors are matched. a. BE BE + E BE T ln S BE T ln S T ln T ln E S S T ln ln E S S T ln E S b..7.6. BE at. ma BE (.6 ln.64 5..64 Since, then BE E E or E. 6.4 kω 5 BE.7 at ma S exp or S. A. (a 5.7( 5.8 ma.6 K.6.8 E ln E.44 K.5.5.6.8 E ln E 4.8 K.. (b BE.7 (.5(.44 BE.68.7. 4.8.64 BE BE
. (a BE BE + BE + Now BE + BE + E or E BE BE + We hae BE T ln and BE T ln S S (b Let and Then BE E ( E so + ( E + E + Then + ε (c Want.5 ma 5( 5 So E E kω.5 5(.7 ( 5 7. kω.5 Then 8.6 kω.4 a..7.7.55 ma. ma.55 ma 4.65 ma BE BE BE b. CE C ( (. + CE.8 EC C (.55( EC 5.5 4.65 5.5 EC C EC.5 a. st approximation
.4.5 ma 8. Now BE.7 (.6 ln BE EB.7 Then nd approximation (.7. ma 8 4.64 ma. ma 6.96 ma b. At the edge of saturation, CE BE.7.7 ( C C. kω 4.64.7 C C 4. kω..7 C C. kω 6.96.6.7.7 ( C C.86 ma C C4.86 ma.86 C5 (.5.6ln C5 By Trial and error. C5.6 ma C6 C7 C.8 + CE CE.86.8 CE 7. 5 EB6 + CE 5 + C5(.5 CE5 5 +.7 (.6(.5 CE5 4. 5 EC 7 + C 7(.8 EC 7 5 (.6(.8 4.89 EC 7.7.7.7 ( C C C C.86 ma C4 C5.86 ma C.86 CE T ln C(..6 ln C C By trial and error C.95 ma 5.86 C C6E T ln C6(.5.6 ln C6 C6 By trial and error C 6.6 ma
.8.7 ma 6. + BE ( Q.7 as assumed E E (( E E E E kω E E E E.5 kω E E E E.75 kω 4 ma ma 4 ma.8 DS ( sat GS TN GS.5 GS.5 μncox ( GS TN L 5 (.5.5.5 L L μncox ( GS TN L (.5.5.5 L L Now GS GS.5 6.5 So ( 6.5.5. L L.9.5 GS.8 GS 5.5.6 +.5 GS GS GS.6GS.6GS.6 GS.6 ± 6.76 +.4.6 ( 6(.5 GS..9.5.9 9. μa 9.5 5 o 9. μa DS ( sat GS TN.9.5 sat.69 DS.9 a. From Equation (.5,
5 5 5 5 GS GS ( 5 + (.5 5 5 + + 5 5.447.447 ( 5 + (.5 +.447 +.447 GS GS.74 Kn ( GS TN ( 8( 5(.74.5.69 ma b. μncox ( GS TN ( + λds L ( 8( 5(.74.5 + (. 45 4.4 ma c. 45 +. 4.448 ma.4 (a 8 5 (.5 GS L. GS.5 Design such that DS ( sat.5 GS.5 GS.75.75 So.5 5 K 8 5.75.5 L L L 5 4 W W L L L (b 667 K λ (.5(. Δ (c Δ.5 μa 666 Δ.5 % %.5%.4 8 GS GS.44 5μA at DS GS.44 λ K (a 5 ( ( (.(.5
(i (ii (iii Δ.44 Δ.8 μa 5.8 μa Δ 4.5.44 Δ. μa 6. μa Δ 6.44 Δ 7.8 μa 67.8 μa 4.5 5 75 μ A at DS.44. K λ (.(.75 Δ.44 Δ 4. μa. 79. μa Δ 4.5.44 Δ 5.5 μa. 9.5 μa Δ 6.44 Δ 6.7 μa. 4.7 μa (b (i (ii (iii.4 SD( sat.5 SG + TP SG.4 SG.65 k p W ( SG + TP L 4 5 (.65.4 L L ( WL / 75 μ A 6 ( WL / L k p W ( SG + TP L SG.65.5 4 Then 75 (.5.4.986 L L.4 (a.5 + + GS TN Kn.5 Kn K n K n Kn
.5 ( max (.5(.5 ( max.55 ma.5.5 ( min (.5(.95 ( min.475 ma.5 So.475.55 ma (b K + n TN TN Kn.5 min.5 +.5 (min.45 ma.5.5 ( max (.5 +.95 (max.55 ma.5 So.45.55 ma.4 x A ( x + g m gs r A ( x + gmgs r So ( o o, gs x gs A x x A + gm ro ro A + g r g r ( [ ] x m x A o x m x o
Then x x ro( x gmx + gm ro ro x x gm x ro + gmx x gmx ro ro ro g r + g + g r x x x m o x m x m o x ro x[ + gmro] x + gm + gmro ro Since gm >> ro x [ + gmro] x( gm( + gmro x + gmro Then o x gm( + gmro Usually, gmr o>>, so that o g.44 DS (sat GS.8 GS.8 6 (.8.8.67 L L L.4 L. W. (.67 L L GS 6.8. 6 4 (..8. L L.45 (a 6 6 GS + GS 5 ( GS.7 5 GS.7.58 6.7.75 6 (.75.7 66 μa (b 8.4 K λ (.5(.66 Δ.5.75 Δ. μa 8.4 6.5 μa (.7 ( (.7 GS GS GS GS GS 6 ( (.75.7 6.6 μ A at DS.75 m
(c Δ.75 Δ 7.6 μa 8.4 7.7 μa.46 5 5 + ( 5(.5 ( (.5 SG SG SG SG SG SG 5 ( SG.5 SG.5.6.68.8 SG SG 5 ( 5 (.8.5.9 ma.9 ma (sat +.8.5 (sat.78 SD SG TP SD.47 (sat..5.55 SD SG SG 5 (.55.5.78 L L ( W L ( W ( W L L.78 + 4.45 SG SG SG L 5.56 5 (.45.5.8 L L.48 8 8 SG SG+ SG SG ( 5(. ( 4(. SG SG 5 ( 4 6.8.7 SG Then SG.. SG SG 8 ( 5(.67..4 ma (sat +.7. sat.7 SD SG TP SD.49 (sat.8.4. SD SG SG ( W L ( W L ( W 8 (..4.77 L L L.77 L.54
Assume M and M 4 are matched.. SG + SG SG.4 8 (.4.4 L,4.5 L,4.5 (a k p ( SG + TP L k p ( SG + TP L But SG SG So 5( SG.4 5( SG.4 which yields SG.8 and SG.9 ( 5(.8.4 μ A ( W / L 5.6 ( W / L 5 Then.6 9 μ A (b DS ( sat.8.4.68.68. then. 6.7 kω.9.5 SD(sat.5 SG.4 SG.75 8 (.75.4.7 L L ( W ( W L 5 L ( W.4 ( W 8.7 L L SG.75.5 5 (.5.4.7 L L.5 a. Kn ( GS TN ( D4 GS GS For, μa
b. r + r ( + g r.5 r 4 m 4 4 r 5 kω λ (.(. g K.. ma/ m n GS TN 5 + 5 +. 5 5 MΩ 6 Δ Δ Δ.8 μa D4 5 r gs4 X g r + r S6 X m gs4 4 X ( + g r r + r X m X 4 X r + + g r r S6 X m 4 gs6 g + g + X S6 X X m gs6 S6 m r6 r6 r6 X X X gm + r + ( + gmr r4 r 6 r6 X X + gm + r + ( + gmr r4 r6 r6 X r6 + ( + gmr6 r + ( + gmr r4 X 4 6. ma. GS g K..4 ma/ m n GS TN r r r 5 kω GS λ (.(. 9 57575 kω.58 Ω { } 5 + +.4 5 5 + +.4 5 5
.54 k n k n L L k p W GS 4 + L GS TN GS TN 4 ( 5(.5 5( 5(.5 TP GS GS GS GS 4 ( 5(.5 (.5 ( 4 + + 6 SG GS GS From ( ( ( ( 4.5.5.5 +.5 GS GS GS GS From ( ( ( ( 5.5.5 5.5 +.5 GS GS 4 SG4 GS Then ( becomes 5.5 +.5 +.5 +.5 + 6 GS GS GS which yields.6 and.,.4 GS GS SG4 k Then 5.6.5 or.74 ma n GS TN L.6 GS GS sat.6.5 sat.86 DS GS TN DS.55 sat.5.5 DS GS TN GS GS k 5 μ A n GS TN L 5 (.5 4 L L k n GS GS ( GS TN L L L 5 5.5 + + 6 GS SG4 GS 6 5.5 GS GS 5 5 (.5.5.75 L L k W p SG4 + TP L 4 5 (.5.5.88 L L 4 4.56 a. As a first approximation GS GS 8 8 Then 4 DS The second approximation ( 8 8 GS +. 4 8 r ( GS GS.96 86.4
Then Kn( GS TN ( + λngs 8(.96 + (.(.96 r 76.94 μa b. From a PSpice analysis, 77.9 μa for D and 77.4 μa for D. The change is Δ.5 μa or.65%..57 a. For a first approximation, 8 8 GS 4 GS 4 As a second approximation 8 8( GS 4 + (. r GS 4.98 GS K + λ n GS TN GS To a ery good approximation 8 μa b. From a PSpice analysis, 8. μa for and the output resistance is 76.9 M Ω. Then For D + 4 Δ D 76.9 8.5 μa.58.5 μa (a DS sat GS TN or GS DS sat + TN. +.8. k n D ( GS TN L 5 48 (. 6 L L GS 5 TN GS TN.8 +.. (b GS 5 GS 5 (c min sat. min.4 D DS D.59 (a k n Kn 5( 5 5 μ A/ L ( W / L K ( W / L n D 5 8.944.688 (.5(.5 5 6. kω (b D
SD GS + SD + sat + sat SD SG TP 5 5.5.7 D SG SG Then sat..5.77 D GS GS Also 5 5 5.5.947 + Then.7+.947.66 (c min 5 5 (.947.5.5 L L 5 5 75 (.7.5 7.5 L L 6 6.6 GS ( 5.667 C μn ox GS TN L W W W W (.667.5 L L L L μ C 4 5 n ox GS TN L r L W L. (.5.5 L L. L. L L. L And.5.75.6 4 SGP GSN (( GSN 8(( SGP. ( ( Also 4. Then 4. 8. GSN SGP 6.5 which yields SGP ( GSN. +. 4.4
4 ( ( Then.4 GSN. 4 GSN.49 GSN.. which yields.69 GSN and.4 SGP 4.4.69 Now 89.4 μ A 89.4 7.9 μ A 5.5 89.4 μ A.8 89.4 7.5 μ A 4 89.4 58 μ A.6 a. gm( M Kn g ( M g ( M r m.5..447 ma/ m r 5 kω (.(. n n λn r r 67 kω (.(. p p λp b. A g r r.447 5 67 A 44.8 m n p c. L rn rp L 5 67 or kω.6 We hae.69 and.4.69.4.88 So 9.4 μ A Then. 9.4.88 μ A 4 GSN 4( 9.4 77.6.5 9.4 4. μ A.8 9.4 5.5 μ A μ A SGP.6 ( W L ( W L ( W L 9 D ( D μ A 5 SG4 4 D ( 67 μ A W 9 L 4 66.7 SG4.6.46 ( SD4 SD4 (sat.46.6 sat.86.64 4 5 SG.6 L.75 SG 5 (
SD(sat.5 SG.6 SG.95.75.95 6 K.5 4 5 (.95.6.4 L L ( W L 49 5 (.4 L ( W D 5 L. 5 (.4 L DS 5(sat.5 GS 5.4 GS 5.75 (.75.4 5 4.5 L 5 L 5 ( W D4 D 5 L 4 4.8 5 4.5 L.65 For GS, id DSS ( + λds a. D 5, DS 5 id + (.5( 5 id.5 ma b. D, DS id + (.5 id ma c. D 5, DS 5 id + (.5( 5 id.5 ma.66 GS DSS P GS 4 P GS.9 P 4 So GS (.9( 4.7 S Then and S GS (.7 GS.586 kω Finish solution: See solution.66 Completion of solution Need DS DS ( sat GS P.7 ( 4 DS.8 So sat +.8 +.7 4 D DS S D 4
.67 a. exp EB S T EB T EB S 5 or ln.6 ln.5568 b. 5.5568 4.44 kω c. From Equations (.79 and (.8 and letting.5.5 +.5 exp 8 +.5568 T + 8.5.8 exp ( T.696 9 Then.6ln (. So.589 ( / T d. A ( / AN + ( / AP.6 8.46 A +.8 +.5 8 A 846.68.5 BE T BE S a. ln.6 ln.58 b. 5.58 8.96 kω.5 c. Modify Eqs..79 and.8 to apply to pnp and npn, and set the two equation equal to each other. EB EC BE CE C S exp + C S exp + T AP T AN EB.5 BE.5 5 exp + exp + T 8 T EB BE 5.565 exp.8 exp T T EB exp T EB BE.9798 exp BE T exp T EB BE + T ln (.9798.58 + (.6 ln (.9798.586 5.586 4.46 EB CE EC
d. A ( / T ( / + ( / AN.6 8.46 A +.8 +.5 8 A 846 AP.69 a. M and M matched. For, we hae SD SG SG DS.5 For M and M : μpcox ( SG + TP ( + λpsd L (.5 + (.(.5 4. L L L L For M : μ C + λ L n ox GS TN n DS ( + (.(.5 4.76 L L b. r n p r 5 kω λ (.(. g K μ C L g m n n ox o m. 4.76..95 ma/ ( A g r r.95 5 5 A 48.8 m n p.7 a. Kp ( SG + TP ( b. From Eq..89 + + λp( SG Kn( TN λ + λ λ + λ ( SG SG (. +. (. +. ( +. 5 5 9.4.96.98 n p n p
c. A gm( rn rp rn rp 5 kω λ (.(. gm K n... ma / A. 5 5 5 A.7 5.6 5.6. ma From Eq..96 C. T A.6.. C C + + + + 4 L 9 AN L AP 8.465 8.465.574 + +.444.46 + (a L, A 7 (b L 5 K, A 56 (c K, A 64 L L.7 5.6.57 ma 5 Then C.54 ma From Eq..96.54.6 9.669 A.54.54 + +.95 +.45 + 8 L L 9.669 A.575 + (a L A 846 (b 5 K, A 47 L L.7 (a To a good approximation, output resistance is the same as the widlar current source. r + gm( rπ E A g r (b m L.74 utput resistance of Wilson source L
β r Then m m ( A g r r AP 8 4 kω. AN r 6 kω. g. 7.69 ma/.6 T ( 8( 4 A 7.69 6 7.69[ 6 6, ] A 4448.75 (a μa D D For M; ro 5 K λ g g m P D m P D (.(..4 K 5..748 ma/ (.5(. For M; r K λ g mo n Do.8 (. gmo.8 ma/ (b A g ( r r (.8( 5 (c mo o oo A 4. ( Want A 57.5.8 4.8 4.8L 4.8 L 7.75 L 4 K 4.8 +.76 Assume M, M matched μa D Do r o r oo 5 K λ p D (.(. K λ n D A g r r mo o oo (.5(. ( g g 5.7 ma/ g mo mo.8 (..7 L 5. L mo L L
k n k p Now L L 8 4 ( 5. L.77.6 L L Since sg, the circuit becomes
g r x sg x m sg + and sg x o r Then x so that o x + gmro + ro ro x r o o ro + g mro + x ro or ( r + r + g r o o o m o A g r o m o o i Now g r m o r.5..6 ma / 5 kω λ n DQ (.(. g K. 8..8 ma/ m p DQ ro ro 5 kω λ Then p DQ (.(. o 5 + 5 +.8 5 MΩ A.6 5 A 5.78 From Eq..5 gm A + r r r r g g m m o o o4 o o k n L D.5.8.5657 ma/ r 65 K A λd (.(.8 (.5657. + (.56 ( 65 ( 65 A 6, 5.79
( ( g + + g + π π m i m π rπ ro ro ( π + + m π ro g ( ( gm i π + + gm + + rπ ro ro ro π ( + + π + gm ro ro gm >> r o + β ( gm i π + rπ ro ( + + π gm ro ( π + gm ro Then + β ( gm i + + gm ro rπ ro + β + + r β r o o + β β β gm i + β From Equation (. β r So
A g r β m o gm i + β.6 o ( 9.65(.5 8 K.5 A 66,65 r 9.65 ma/