Chapter 13 Problem Solutions Computer Simulation Computer Simulation ma/ V 80. r I (120)(0.026)

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1 Chapter 3 Pblem lutions 3. Computer Simulation 3. Computer Simulation 3.3 (a) ( Ri) g 0 C m T 0.0 r 80 o MΩ C 0 r 80 o MΩ C m/ Ri + ( + βn) R 7 (0)(0.0) 7 5. kω 0. BE ( on) 0. C m R 0 (0)(0.0) 0 kω Ri 0 + () MΩ 79(.) 55 R r ( β ) r β r R+ 7 R o βn( + βn) 7 ban b R + 7 Ri o βn( + βn) r o7 R v o Ri R kω 0. o c7 o7 n b7 o7 n o7 c C7 (0)()(00) 0 v v Overall gain v (55)( 83).59 0 (80)(0.0) (b) Ri r π an 0 kω 0.00 Ri 08 kω (c) fpd an CM (0)( + 83) 8,0 pf π R C eq M Req Ri MΩ fpd 7.7 Hz π ( )(8,0 0 )

2 Gain-Banwith Puct (7.7)(.59 0 ).3 MHz 3. a. Q 3 acts as the ptection evice. b. Same as part (a). 3.5 f we assume BE (on) 0.7, then in breakown voltage ( 5) (a) REF 0.50 R5 57. kω R5 REF C0R T ln C R ln R. kω ( 5) (b) REF REF 0.53 m C0 (.) (0.0)ln C0 By trial an err,. μ C0 3.7 (a) REF 0.50 m 3 REF ln (0.0) ln S ( 5) R5 R5 57. kω R ln R. kω BE0 0.0ln BE (b) Fm Pblem 3., REF 0.5m BE EB 0.0ln ( 5) REF REF m 0. μ fm Pblem 3. BE T BE EB C 3.8 a ( 5) REF REF 0 0. m REF C0R T ln C0 0. C0 (5) (0.0) ln C0

3 By trial an err;. μ C0 C0 C C 7.0μ m C7 REF C m C3 REF C3 (b) Using Example kω R 50 [3.5 + (0)(0.)] kω r E π βn T C an 0.5 (0.5)(0.) + 0. C m kω R 39 + (0)(5.) 5.5 MΩ r i π 73 kω m / r0 7.0 MΩ R 7.0[ + (0.73)( 73)] 8.9 MΩ act 50 r0 7.0 MΩ ( ) Gain of ifferential amp stage Using Example 3.5, an neglecting the input resistance to the output stage: 50 Ract 303 kω 0.5 C3B (00)(0)(50)(303) (303) v (5500)[ (0)(0.)] v 55 Gain of secon stage 3.9 C0 9 μ Fm Equation (3.) βp + βp + (0) + (0) + C0 βp + 3βP + (0) + 3(0) (9) 0.5 μ C 0.8 μ

4 μ + + β 0 P μ C9 C9 C9 B9 B9 β P μ B B ( + β P ) β P 0 C (0.8) C 9.35 μ + β P 3.0 (on) + () B5 BE C (0.0095)() C7 C7 C8 C9 REF E3 C μ 50 9 μ 0.7 m 0.7 m REF 38 μ + ( )[ C 7 C8 C9 REF E3 C] Power [ ] Power 8.8 mw + Current supplie by an C7 C8 C9 REF E3 C.3 m 3. (a) v (min) v cm cm (max) v. cm (b) v (min) 5 + (0.). v cm cm (max) v. cm 3. f v0 5, the base voltage of Q is pulle low, an Q 8 an Q 9 are effectively cut off. s a first appximation 0. C. m B 0. m 00 C5 C3 B m C5 BE5 T ln (0.0)ln

5 s a secon appximation C C.8 m B 0.09 m 00 an m C5 C5 3.3 a. Neglecting base currents: D BS D BB D T ln S (0.0) ln BB BB / CN CP S exp T CN exp (0.0) 0.5 m CP b. F v 5, v0 5 5 il.5 m s a first appximation i.5 m CN L BEN (0.0)ln Neglecting base currents,.089 BB EBP 0.58 CP 0.0 s a secon appximation, i m 5 0 exp CP 0.3 m CN L CP CN 3. BB.57 R + R.8 kω (0.) 0.08 BE BS C (0.9) BS T ln (0.0) ln S S (0.0) ln 0

6 BE 0. R R.8 R 33.9 kω R 30.3 kω BE BB R+ R 0. (.57) 3.5 (a) ( Ri) Fm example μ /, 5. MΩ MΩ ssuming R 8 0, we fin Ri + ( + β n) R E 39 + (0) ( ).95 M Ω (35) ( ) 09 (b) Fm Equation (3.0), βn( + βn) R9 ( Ract Ri3 R07 ) v Ri{ R ( + βn) Rg } F Rg 0, Ri.95 MΩ Using the results of Example (0)(50) ( ) v v 79 (950){ } 3. Let C0 0 μ, then C C 0 μ. Using Example 3.5, Ri.07 MΩ (00)(0.0) 0 kω m/ r0.5 MΩ 0.0 Ract.5[ + (0.79)( 0)]. MΩ 50 r0.5 MΩ 0.0

7 CQ ( r0 Ract Ri ) T 0 (.5..07) Fm Pblem μ, C7 0.5 m, C m E7R8 + BE7 0.5 (0.5)(0.) + 0. C B R C 0.03 m (00)(0.0) K 0.5 RE R9 [ 7 + ( + β ) R8] 50 [3.5 + (0)(0.)] K (00)(0.0) 39 K 0.03 Ri + ( + β ) RE 39 + (0)(5.) 5.50 MΩ (00)(0.0) 73 K m/ MΩ Ract [ + ( R )] 7.0[ + (0.73)( 73)] 8.9 MΩ MΩ ( Ract Ri) 7.0 ( ) R act 303 K R o K Fm Eq. (3.0), assuming Ri 3 β( + β) R9 ( Ract R07 ) v Ri { R9 + [ 7 + ( + β ) R8] } 8 (00)(0)(50)( ) (5500)[ (0)(0.)] v Overall gain v ( 7)( 55) 3,75

8 3.8 Using results fm Ri 5.50 M Ω, Ract [ + (0.73)( 73)] 7.93 MΩ MΩ ( ) Ract 0 K R o 7 0 K v 5 v (00)(0)(50)(0 0).09 0 (5500)[ (0)(0.)] Overall gain ( 885)( 090) 9, 50 v 3.9 R e r + R + β π 0 P an R0 R + R e ssume series resistance of Q 8 an Q 9 is small. R0 r03 Re + R07 r03b where Re + β P an R07 r07[ + 7 ( R8 r π 7 )] Using results fm Example 3., kω 7. kω m/ r07 9. kω R07 9.[ + (0.8)(0. 9.3)] 83 kω 50 r03b 9. kω Re.5 kω 5 R0 r03 Re kω (50)(0.0) r π 0.5 kω R e 0.0 kω 5 R e. Ω R. + 7 R 9. Ω

9 Ri + + β βn 00, βp 0 (a) 9.5 μ C ( n ) 3 + β P (00)(0.0) 57 K (0)(0.0) 3 7. K (0)(7.) Ri 57+ R.095 MΩ i (b) 7.0 μ C (00)(0.0) 73 K (0)(0.0) 3 3. K (0)(3.) Ri 73+ R.80 MΩ i 3. We can write 0 ( f) f f + j + j fpd f 8, 0 f f + j + j 0.7 f Phase: f f φ tan tan 0.7 f F a Phase margin 70, φ 0 f f 0 tan tan 0.7 f ssuming f 0.7, we have tan f f f f t this frequency, ( f ), so

10 8,0 f + + (0.3) , 37 f f 70,37 f.8 MHz 0.7, secon pole at f f f 5 MHz a. Original g m an g m μpcox Kp Kp L (.5)(0) 5 μ / Q K p (0.5)(0) m/ f L is increase to 50, then K p Kp (50)(0) 500 μ / (0.5)(0.099) m/ b. Gain of first stage ( r0 r0) (0.995)( ) 50 oltage gain of secon stage remains the same, v 5 v v (50)(5) 5, a. K p (0)(0) 00 μ / 0. m / 0 SG ( 0) REF SET 00 k (.5) P SG 0 SG (0.)(00)( SG 3SG +.5) 0SG 9SG ± (9) (0)(70) SG (0) SG.7

11 0.7 REF REF 89. μ 00 M 5, M, M 8 matche transists so that Q D7 REF 89. μ b. Small-signal voltage gain of input stage: ( ) Kp Q r0. MΩ λ 89. PD (0.0) r0. MΩ λn D 89. (0.0) (00)(89.) (..) Small-signal voltage gain of secon stage: v 7 ( r07 r08) Kn7 (0)(0) 00 μ / 7 Kn7D7 (0.)(0.089) m/ r08 5 kω λp D7 (0.0)(0.089) r07 kω λn D7 (0.0)(0.089) v (0.378)( 5) v overall voltage gain v v ()() v 9, Small-signal voltage gain of input stage: Kp Q ( ) Kp (0)(0) 00 μ / r0 000 kω Q 0. λ (0.0) P r0 000 kω Q 0. λ (0.005) n (0.)(0.) ( ) 33 Small-signal voltage gain of secon stage: v 7 ( r07 r08 ) Kn7 (0)(0) 00 μ /

12 g K (0.)(0.) 0.5 m/ r m7 n7 D kω λ (0.0)(0.) P D7 r kω λn D7 (0.005)(0.) (0.5)( ) 89 v v overall voltage gain is v v (33)(89) v 5,37 3. fpd π R C eq i where Req r0 r0 an Ci C( + v ) We can fin that v 5 an r 0 r MΩ R MΩ eq an C i ( + 5) 30 pf f PD π (.5 0 )(30 0 ) f.0 Hz PD 3.7 fpd π R C eq i where Req r0 r0 Fm Pblem 3., r 0. M Ω, r 0. MΩ an v 8 π (..) 0 Ci 8 C. 0 C ( + ) C () i C 88 pf 3.8 R r r We can fin that r07 r08.5 MΩ R R 0. MΩ v

13 3.9 a. ( g )( r r ) 0 m gs gs 0 0 r0 r0 0 ( r0 r0) v + g ( r r ) ( )( ) m g + an X X b. X m gs gs X r0 r0 R r r (a) [ ] Q Q 80 (0) μ 80 (b) D (5) ( GS 0.7 ) 5 GS D7 (50) ( SG7 0.7 ) 5 SG Set SG8P GS 8N L L 80 ( ) 30 8P 8P 80 L L 80 ( ) N 8N

14 80 ( ) GS 00 (0) GS 0.7 GS.0 Let M transists in series. Than 5.0 GS ( ) 3.7 L L L B 3.3 (a) 80 ( ) Q 50 μ (5) GS8 0.7 GS GS SG D D7 (5)( ) 3.7 μ (b) (5) m/ 800K ( 0.0)( 0.5) r K ( 0.05)( 0.5) ( ) ( 0.577)( ) 75 Secon stage:

15 g g ( r r ) m5 o5 o9 m5 r r (80)(50).5 m/.7 K (0.05)(0.5) 00 K (0.0)(0.5) (.5)(.7 00) 0 ssume the gain of the output stage, then v (75)( 0) 35, 350 v 3.33 (a) Ro Ro8 D ( ) KnDQ (0.5)(0.05) μ / 8 (0.5)(0.05) μ / MΩ λ DQ (0.05)(5).33MΩ λ ( 0.05)( 50) Ro8 8( 8 0) ()(.7)(.7) 597 MΩ Ro ( )( ) ()(.7)(.7.33) Ro 53 MΩ ()(53 597) 89, (b) Ro Ro Ro Ro 398 MΩ (c) fpd f 80 PD Hz π RC π o L ( )( ) GBW (89, )(80) GBW 7. MHz 3.3 (a) 8 0 MΩ λp D (0.0)(5).7 MΩ λn D (0.05)(5).33 MΩ λn D (0.05)(50) 35 (5).8 8 L L 80 (5) 3. L L Ro Ro Ro8 [ ( )( )] [ 8( 8 0)]

16 Define X an X L L Ro 3.X(.7)(.33 ).8X( )( ), 539XX 3.8X 7.X 3.8X + 7.X Ro, 539XX (.8 X) 3.8 X + 7. X 0,000 X 0.7X L. L We then fin X.0 L L p an.85 L n 3.35 Let + 5, 5 P T (0) 3 T 0.3 m REF 0. m 00 μ 8 0 MΩ (0.0)(50).33 MΩ (0.05)(50) 0.7 MΩ (0.05)(00) 35 (50) 59.X 8 L where X L ssume all with-to-length ratios are the same. 80 (50) 89.X L Ro Ro Ro8 ( )( r ) o 8( 8 0) 89.X(.33)( 0.7 ) 59. X()() ( 7.X)( 59.X) [ 7.X] [ 59.X] 7.X + 59.X Ro.X Ro ( 59.X)(.X) 5,000 W that X f all transists L 3.3

17 (a) B 8 ( ) MΩ λ (0.05)(90) DQ k n D (500)(30) 5 μ / L (3)(5)( ) 7 (b) R r r R 37 kω o o o o f fpd 85.8 khz (37 0 )(5 0 ) (c) PD 3 πrc o π 3 GBW (7)( ) GBW 3.3 MHz 3.37 (a) r o r o8 0.5 MΩ (0.0)(.5)(0) 0.7 MΩ (0.05)(.5)(0) Bg r r m o o8 ( ) m ( ) 00 (.5) g g 50 μ / m (0) 9 L L ssume all (W/L) ratios are the same except f M5an M..5 L L 5 (b) ssume the bias voltages are + 5, 5. ssume 9 L L B 80 (9)( 0.5) Q GS 80 GS REF 80 ( GSC 0.5) L C F four transists

18 L L GSC 80 ( ) 0.0 C (c) f3 B Ro MΩ π RC f 3 B 3 3 (00)(85 0 ) 7 o 85 khz π (8 0 )(3 0 ) GBW MHz C 3.38 (a) Fm previous results, we can write Ro 0 0 ( 0 ) Ro ( 8 ) Bg R R ( ) m o0 o MΩ λ ( ) (0.0)(.5)(0) PB Q/ MΩ λ B (0.05)(.5)(0) n ( Q/ ) ssume all transists have the same with-to-length ratios except f M 5 an M. Let X L k p 35 0 ( DQ 0 ) X (.5)(0) L X k n 80 ( DQ ) X (.5)(0) L.5 X 80 X (0) 80 X Ro 0 (83.7 X)(0.5)(0.5) 0.9 X MΩ Ro (.5 X)(0.7)(0.7) 5.3 X MΩ We want 0,000 (.5)(80 X)[0.9X 5.3 X] (0.9 X)(5.3 X) 00X 308X 0.9X + 5.3X X.5 L (.5)(.5). L L 5 (b) ssume bias voltages are + 5, 5

19 ssume.5 L L B Q (.5)( GS 0.5) GS.05 Nee 5 transists in series 0.05 GSC REF 80 ( ).0 L L C (c) f3 B where Ro Ro 0 Ro π RC o Ro MΩ Ro MΩ Ro MΩ f3 B.3 khz π (39 0 )(3 0 ) 3 GBW (0,000)(.3x0 ) GBW 7. MHz 3.39 ( M) ( M) ( Q) 0 ( M ) (5)(00) 7 μ / ( M) 500 kω λ DQ (0.0)(0.) 0 ( Q) 00 kω 0. CQ 7(0.5.) C

20 ( M) ( M) ( Q) 80 ( M ) (5)(00) 3 μ / ( M) 7 kω λ DQ (0.05)(0.) 80 ( Q) 800 kω CQ 0. (3) (a) ( ) ( ) m o o8 REF 00 μ Kn Kp 0.5 m/ λn λp 0.05 g R R where Ro8 8( 8 0) Ro ( ) ( ) 8 KPD8 (0.5)(0.) 0.7 m/ 8 7 kω λp D8 (0.05)(0.) 0 7 kω λp D8 C m/ T kω C kω λn D (0.05)(0.) 7 kω λp D (0.05)(0.) g K (0.5)(0.) 0.7 m/ m P D Ro8 (0.7)(7)(7) 98.9 MΩ Ro (3.8)(800)(333 7) 83. MΩ 7( ) 8,85 3. ssume biase at + 0, 0. P 3 REF (0) 0 REF 7 μ ( Ro Ro8) 5, 000 k n 80 μ/, k p 35 μ/ λ 0.05, λ 0.0 n ssume. L L p p n

21 ( ) Ro Ro ( )( ) MΩ λp D8 (0.0)(83.3) 0 λp D8 0.0 MΩ g k 35 X 3.3X p m8 D8 (.) (83.3) L 8 where X L n MΩ C MΩ λn D (0.05)(7) 0.0 MΩ λp D (0.0)(83.3) C μ / T 0.0 k p 35 D (.) X (83.3) L 3.3X Ro (30)(0.90) MΩ Ro8 (3.3 X)(0.0)(0.0) 0.8 X MΩ 5, 000 (3.3 X) X 30,0X (3.3 X ) X which yiels X.8 X. L n an + (.)(.).3 L P 3.3 F v cm (max), assume CB ( Q 5) 0. S D9 D0 0.8 m Using parameters given in Example 3. D9 0.8 SG TP +..7 KP 0.0 v (max) v (max). cm cm

22 F v cm (min), assume SD ( M9 ) SD ( sat) SG + TP D 0 D 0 (0.5) D 0 (0.5) D 0.8 vcm (min).8 + SD (sat) SG , common-moe voltage range 5.8 v cm. Or, assuming the input is limite to ± 5, then 5 v cm. 3. F 300 μ, SG BE + (0.3)(8) KP( SG + TP) 0.3 K (3.) K 0.7 m/ P P 3.5 F CB 0 f both Q an Q 7, then S SG + ( S) S.+ SG 0. + R SG + TP an KP lso KP( SG + TP) so + + SG SG SG SG + SG SG SG + 0. (0.5)(8)(.) 0. (.8.9) ± (.) ()(.5) SG.33 () S an.75 S 3. C5 C 300 μ Using the parameters fm Examples 3. an 3.3, we have βnt (00)(0.0) Ri kω 0.3 C3 KnQ5 ( Ri) (0.)(0.3) (7.3) 0.38

23 g 0.3 C3 m3 T m/ 50 r03 7 kω C3 0.3 v 3 r03 (.5)(7) v 97 Overall gain: v (0.38)(97) 9, ssuming the resistances looking into Q an into the output stage are very large, we have β R03 v 3 + ( + β ) RE3 where R03 r ( RE3 r π 3 ) 50 C3 300 μ,r03 7 kω m / 0.0 (00)(0.0) r π kω 0.3 R03 (7) + (.5)( 7.3).98 MΩ (00)(980) v (0)() Ci C( + v ) [ + 8] C,780 pf fpd π ReqCi R R r r eq i i 0 00 Neglecting R, 3 r kω λ D0 (0.0)(0.5) Neglecting R 5, 50 r0 333 kω 0.5 Ri 3 + ( + β ) RE (0)() 8 kω f PD π (, 780) 0 f PD 77. Hz Unity-Gain Banwith Gain of first stage: 3

24 K ( R r r ) n Qs o o0 (0.)(0.3) ( ) (0.)( ) 5. Overall gain: v (5.)(8) 0,7 unity-gain banwith (77.)(0,7) 7.95 MHz 3.8 Since GS 0 in, 0.8 m 3.9 DSS J REF DSS a. R r + ( + β) [ r + ( + β) R ] i π5 π (00)(0.0) 3 kω 0. C 00 μ C5 μ β 00 (00)(0.0) r π kω 0.00 R i (0) 3 + (0)(0.3) R i 5.7 MΩ [ ] b. v ( r0 r0 Ri) D DSS (0.)(0.) P m / r0 500 kω λ D (0.0)(0.) 5.0 r0 500 kω C 0. v (0.093)[ ]. v 3.50 a. Nee SD ( QE ) SD ( sat) P F minimum bias ± 3 Set 3 an 3 P ZK ZK D REF R so that R3 R3 kω 0. Set bias in QE REF + Z m Therefe, E

25 DSS 0. m b. Neglecting base currents 0. 0 REF 0.5 m R so that R.8 kω 3.5 a. We have D DSS (0.5)() P 0.35 m/ r0 00 kω λ D (0.0)(0.5) 00 r0 00 kω D m/ 0.0 (00)(0.0) r π 0. kω 0.5 R0 r 0 + ( r R ) π 00 + ( 9.3)( ) 035 kω ( ) r0 R0 RL F RL 0.35( ) 33.7 With these parameter values, gain can never reach 500. b. Similarly f this part, gain can never reach 700.

V = = A = ln V

V = = A = ln V 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