V = = A = ln V

Transcription

1 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 (.6 ln A.6 ln (ii μ BE 4

2 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 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 min.476 ma ( 58.6(.5 Δ ma ± 5% For part (b: 4. ( max.56 ma min.476 ma ( 8.6(.5 Δ.5 ma ± 5%.6 a. + + or.4 ma β kω.4

3 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 β K.9 a 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 kω ma 5.7 For EC min.7 C C.5 kω b. c..

4 .5 ma.5 ma A B A β 6.65 ma K K.5. and (a. ma,.5 ma (b.5 ma,.75 ma (c.67 ma,. ma.4 a.

5 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 ( 8( 8 + ( (.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 ma b. ( 5( 5 5(.7 ( kω.5

6 + (.5.54 ma β ( β + + ( 5( 5 5(.7 ( kω β ( β + For.8 ma ( ma 5( 7 8 (.7.69 kω.84.8

7 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 ma E.4 B.56 ma + β 76 C + B.56 ma ( 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 ma T ln ln E ln By trial and error 4.7 μa BE.7 E BE.7 (.47( BE. (a

8 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 gm.65 ma /.6 (.6 rπ kω So E rπ E kω Then o (.6( 9.74 o 5.6 M Ω (b. Δ Δ BE BE o E BE BE ( m A 8.76 MΩ 6.8 m T.6 ( 8(.6 π C r + g r g ma/ rπ 6 kω.68 r K E E E MΩ Now Δ ( 5 Δ. μa (a K E T ln.6.5 E ln.5.5. K E

9 (b r [ + g ] (c c E m r E E π rπ 9 K gm.9 ma/.5.6 A ro MΩ E K.5 + (.64(.9 ( MΩ Δ 5 Δ.77 μa Δ.77 %.54% 5.5 Let 5 k Ω, Then ma 5 Now E T ln E ln E kω...6 BE T ln S 5.7 (.6 ln S. A S At ma, BE (.6 ln kω T.6 E ln ln E.9 kω a ma Let BE T ln S Then.465 BE (.6 ln Then 5.7 (.6 ln S. A S

10 ma T b. E ln ln E 4Ω (.485 ma 4 BE T ln S 5.7 (.6 ln S. A S Now BE BE So.485 (.6 ln ma 4 E T ln.48 (.6 ln By trial and error. 8.7 μa BE BE E BE 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 BE + E ma.74 ma Using the same relation as for the widlar current source.

11 ( π r o + gm E r A 8.74 ro 5 K gm. ma/.74.6 (.6 rπ 8.9 K E rπ K (.( 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 BE at. ma BE (.6 ln 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 E ln E 4.8 K.. (b BE.7 (.5(.44 BE BE BE

12 . (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 ma. ma.55 ma 4.65 ma BE BE BE b. CE C ( (. + CE.8 EC C (.55( EC EC C EC.5 a. st approximation

13 .4.5 ma 8. Now BE.7 (.6 ln BE EB.7 Then nd approximation (.7. ma ma. ma 6.96 ma b. At the edge of saturation, CE BE.7.7 ( C C. kω C C 4. kω..7 C C. kω ( 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 CE (.6(.5 CE EC 7 + C 7(.8 EC 7 5 (.6( EC ( 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

14 .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 ( L L μncox ( GS TN L ( L L Now GS GS So ( L L.9.5 GS.8 GS GS GS GS.6GS.6GS.6 GS.6 ± ( 6(.5 GS μa o 9. μa DS ( sat GS TN.9.5 sat.69 DS.9 a. From Equation (.5,

15 GS GS ( 5 + ( ( 5 + ( GS GS.74 Kn ( GS TN ( 8( 5( ma b. μncox ( GS TN ( + λds L ( 8( 5( ( ma c ma.4 (a 8 5 (.5 GS L. GS.5 Design such that DS ( sat.5 GS.5 GS So.5 5 K 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

16 (i (ii (iii Δ.44 Δ.8 μa 5.8 μa Δ Δ. μa 6. μa Δ 6.44 Δ 7.8 μa 67.8 μa μ A at DS.44. K λ (.(.75 Δ.44 Δ 4. μa. 79. μa Δ 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 Then 75 ( L L.4 (a GS TN Kn.5 Kn K n K n Kn

17 .5 ( max (.5(.5 ( max.55 ma.5.5 ( min (.5(.95 ( min.475 ma.5 So ma (b K + n TN TN Kn.5 min (min.45 ma.5.5 ( max ( (max.55 ma.5 So 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

18 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 ( L L L.4 L. W. (.67 L L GS (..8. L L.45 (a 6 6 GS + GS 5 ( GS.7 5 GS ( μa (b 8.4 K λ (.5(.66 Δ.5.75 Δ. μa μa (.7 ( (.7 GS GS GS GS GS 6 ( ( μ A at DS.75 m

19 (c Δ.75 Δ 7.6 μa μa ( 5(.5 ( (.5 SG SG SG SG SG SG 5 ( SG.5 SG SG SG 5 ( 5 ( ma.9 ma (sat (sat.78 SD SG TP SD.47 (sat SD SG SG 5 ( L L ( W L ( W ( W L L SG SG SG L ( L L SG SG+ SG SG ( 5(. ( 4(. SG SG 5 ( SG Then SG.. SG SG 8 ( 5( ma (sat +.7. sat.7 SD SG TP SD.49 (sat.8.4. SD SG SG ( W L ( W L ( W 8 ( L L L.77 L.54

20 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 then. 6.7 kω.9.5 SD(sat.5 SG.4 SG.75 8 ( L L ( W ( W L 5 L ( W.4 ( W 8.7 L L SG ( L L.5 a. Kn ( GS TN ( D4 GS GS For, μa

21 b. r + r ( + g r.5 r 4 m 4 4 r 5 kω λ (.(. g K.. ma/ m n GS TN 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 (.( kω.58 Ω { }

22 .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 ( SG GS GS From ( ( ( ( GS GS GS GS From ( ( ( ( GS GS 4 SG4 GS Then ( becomes GS GS GS which yields.6 and.,.4 GS GS SG4 k Then 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 GS SG4 GS GS GS 5 5 ( L L k W p SG4 + TP L 4 5 ( L L a. As a first approximation GS GS 8 8 Then 4 DS The second approximation ( 8 8 GS r ( GS GS

23 Then Kn( GS TN ( + λngs 8(.96 + (.(.96 r 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 μa.58.5 μa (a DS sat GS TN or GS DS sat + TN k n D ( GS TN L 5 48 (. 6 L L GS 5 TN GS TN (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( kω (b D

24 SD GS + SD + sat + sat SD SG TP D SG SG Then sat D GS GS Also Then (c min 5 5 ( L L ( L L GS ( C μn ox GS TN L W W W W ( 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 SGP GSN (( GSN 8(( SGP. ( ( Also 4. Then GSN SGP 6.5 which yields SGP ( GSN

25 4 ( ( Then.4 GSN. 4 GSN.49 GSN.. which yields.69 GSN and.4 SGP Now 89.4 μ A μ A μ A μ A μ A.6 a. gm( M Kn g ( M g ( M r m ma/ m r 5 kω (.(. n n λn r r 67 kω (.(. p p λp b. A g r r A 44.8 m n p c. L rn rp L 5 67 or kω.6 We hae.69 and So 9.4 μ A Then μ A 4 GSN 4( μ A 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 SG ( SD4 SD4 (sat.46.6 sat SG.6 L.75 SG 5 (

26 SD(sat.5 SG.6 SG K ( 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 ( L 5 L 5 ( W D4 D 5 L 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 D DS S D 4

27 .67 a. exp EB S T EB T EB S 5 or ln.6 ln.5568 b kω c. From Equations (.79 and (.8 and letting exp T exp ( T Then.6ln (. So.589 ( / T d. A ( / AN + ( / AP A A BE T BE S a. ln.6 ln.58 b 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 exp.8 exp T T EB exp T EB BE.9798 exp BE T exp T EB BE + T ln ( (.6 ln ( EB CE EC

28 d. A ( / T ( / + ( / AN A 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 ( + (.( L L b. r n p r 5 kω λ (.(. g K μ C L g m n n ox o m ma/ ( A g r r A 48.8 m n p.7 a. Kp ( SG + TP ( b. From Eq λp( SG Kn( TN λ + λ λ + λ ( SG SG (. +. (. +. ( n p n p

29 c. A gm( rn rp rn rp 5 kω λ (.(. gm K n... ma / A A ma From Eq..96 C. T A.6.. C C L 9 AN L AP (a L, A 7 (b L 5 K, A 56 (c K, A 64 L L ma 5 Then C.54 ma From Eq A L L A (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

30 β r Then m m ( A g r r AP 8 4 kω. AN r 6 kω. g ma/.6 T ( 8( 4 A [ 6 6, ] A (a μa D D For M; ro 5 K λ g g m P D m P D (.(..4 K 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 L 4.8 L 7.75 L 4 K 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

31 k n k p Now L L 8 4 ( 5. L.77.6 L L Since sg, the circuit becomes

32 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 ma/ m p DQ ro ro 5 kω λ Then p DQ (.(. o 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 ma/ r 65 K A λd (.(.8 ( (.56 ( 65 ( 65 A 6, 5.79

33 ( ( 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

34 A g r β m o gm i + β.6 o ( 9.65(.5 8 K.5 A 66,65 r 9.65 ma/