Microelectronic Circuit Design Fifth Edition - Part I Solutions to Exercises
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- Ἰούδας Διδασκάλου
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1 Page Microelectronic Circuit Design Fifth Edition Part I Solutions to Exercises CHAPTER V LSB 5.V 0 bits 5.V 04bits 5.00 mv V MSB 5.V.560V V O mV or 5.V 3.95 V V O Page V LSB 5.0V 8 bits 5.00V 56bits 3.95 V.V 9.53 mv N 56bits 6.44bits 5.00V Page The dc component is V A 4V. The signal consists of the remaining portion of v A : v a 5 sin 000πt + 3 cos 000 πt) Volts. Page 3 5sin 000πt + 5 o 90 o ) v o 5cos 000πt + 5 o V o 5 65 o V i o A v Page 5 A v R 5 0kΩ R 0 kω R R [ ] 5sin 000πt 65 o 5 65o o o
2 Page 6 v s! 0.5sin 000πt) + sin 4000πt) +.5sin 6000πt) The three spectral component frequencies are f 000 Hz f 000 Hz f Hz a) The gain of the bandpass filter is zero at both f and f 3. At f, V o 0 V ) 0 V, and v o 0.0sin 4000πt) volts. b) The gain of the lowpass filter is zero at both f and f 3. At f, V o 5 0.5V ).5 V, and v o.50sin 000πt) volts. Page 7 R 39kΩ + 0.) or 35. kω R 4.9 kω R 3.6kΩ + 0.0) or 3.56 kω R 3.64 kω 39kΩ kΩ 0.0 Page 9 V I P P nom mw R + R 54kΩ P max.x5 ) 0.95x54kΩ 5.3 mw P min 0.9x5 3. mw.05x54kω Page 33 R 0kΩ o C o C ) 9.0 kω R 0kΩ o C o C ) 0.6 kω
3 Page 47 n i Page 47 n i n i.3x0 30 K 3 cm 6 ) 300K.08x0 3 K 3 cm 6 ) 50K 3 exp 3 exp.08x0 3 K 3 cm 6 ) 35K cm 3 ) m, L x0 39 * 0. cm 3 exp 3 CHAPTER 0.66eV 8.6x0 5 ev / K) 300K).7x03 cm 3.eV 8.6x0 5 ev / K) 50K) 4.34x0 39 cm 3 L 6.3x0 0 m.ev 8.6x0 5 ev / K) 35K) 4.0x00 cm 3 Page 48 v p µ p E 470 cm V s 0 V 4.70x0 3 cm cm s E V L x0 4 V cm 5.00x03 V cm v n µ n E 40 cm V s 000 V.4x0 6 cm cm s Page 48 a) From Fig..5: The drift velocity for Ge saturates at 6x0 6 cm / sec. At low fields the slope is constant. Choose E 00 V/cm µ n v n E 4.3x05 cm / s cm V / cm s b The velocity peaks at x0 7 cm / sec µ n v n E 8.5x05 cm / s 00V / cm 8500 cm s µ p v p E.x05 cm / s 00V / cm 00 cm s 3
4 Page 50 n i.08x exp ρ σ.60x0 9.3x0 n i.08x exp. 8.6x ) 5.40x04 / cm 6 n i.3x0 / cm 3 40 ρ σ.60x x0 39 Page 54 n i.08x exp x0 40 Ω cm. 8.6x0 5 50).88x0 77 / cm 6 n i 4.34x0 39 / cm x x ) 5.40x04 / cm 6 holes p N A N D 0 6 x0 5 8x0 5 n n i cm 3 p 5.40x04 8x0 5 Antimony Sb) is a Column V element, so it is a donor impurity. p n i n 00 x0 holes x03 n > p n type silicon cm 3 Page 55 a) N D x0 6 / cm 3 N A 0 / cm 3 N T x0 6 / cm 3.69x0 53 Ω cm 6.75x0 8 electrons cm 3 n N D x0 6 electrons cm µ n 5. +! x cm /V s µ p ! x cm /V s x0 6.3x0 6 N D 0 / cm 3 N A 3x0 7 / cm 3 N T 3x0 7 / cm µ n 5. +! 3x cm /V s µ p ! 3x cm /V s x0 6.3x0 6 b) N T N D + N A 4x0 6 / cm 3 + 6x0 6 / cm 3 x0 7 / cm µ n 5. +! x cm /V s µ p ! x cm /V s x0 6.3x0 6 4
5 Page 57 Use the mobilty expressions in Fig..8 with a spread sheet or MATLAB ρ σ σ x0 9 µ n n u n n 6.5x0 / cm ) 9.68x0 9 a) N D x0 6 / cm 3 N A 0 / cm 3 N T x0 6 / cm Ω cm 365 µ n 5. + x cm /V s µ p x cm /V s + * + * 9.68x0 6 ).3x0 6 ) n N D x06 p n i cm 3 n x0 cm ρ 3 qu n n.60x0 9 b) N T N D + N A x0 6 / cm 3 + 3x0 6 / cm 3 5x0 6 / cm x Ω cm 365 µ n x cm /V s µ p x cm /V s + * + * 9.68x0 6 ).3x0 6 ) p N A N D x06 n n i cm 3 p ρ 6 x0 cm 3 qu n n.60x0 9 Page 58 Boron B) is a ColumnIII element, so it is an acceptor impurity. 98 x0 6 p N A N D 4x0 8 holes cm n n i 3 p 00 electrons 5 ptype material 8 4x0 cm 3 N T 4x08 cm 3 and mobilities from expressions in Fig..8: µ n 53 cm V s µ p 54.8 cm V s ρ qµ p p.60x0 9 4x Indium In) is a ColumnIII element, so it is an acceptor impurity. p N A N D 7x0 9 holes cm n n i 3 p 00 electrons.43 ptype material 9 7x0 cm 3 N T 7x09 cm 3 µ n 67.5 cm V s µ p 46. cm V s ρ σ.93 mω cm.60x0 9 7x0 9 ) 46.) Ω cm 3.6 Ω cm 5
6 Page 59 V T kt q.38x mv V.60x0 9 T.38x mv.60x0 9 V T.38x mv V.60x0 9 T.38x mV.60x0 9 D n kt q µ n 5.8mV ) cm cm 36.6 V s s D p kt q µ p 5.8mV ) cm V s.cm s dn j n qd n dx.60x0 9 C 0 cm µm 30 A s cm 3 µm cm cm dp j p qd p dx.60x0 9 C 4 cm µm 64 A s cm 3 µm cm cm 6
7 CHAPTER 3 Page 77 φ j V T ln N N A D n i 0.05V ln x ) V w do Page 78 E max ε s ε s q + φ j.7 N A.60x0 9 x x p N D qn A dx qn x A p E max.6x0 9 C 0 7 / cm x0 4 ε s E max ε s qn D 0.3x0 5 cm) x0 4 F / cm x n 75 kv cm x0 6 cm µm dx qn D x n ε s For the values in Ex.3. : E max.05v kv x0 6 cm cm x p 0.063µm + x08 * µm x 0 0 n 0.063µm + 00 * ) x0 8 ) Page 8 n kt.38x T 300 q.60x K 5.6x0 4 µm Page 83 ) i D 40x0 5 A exp 0.55, ) µa i D 40x0 5 A exp 0.70, ma * 0.05 * 0.05 V D 0.05V ) ln + 6x0 3 A V 40x0 5 A ) i D 5x0 5 A exp 0.04, ) fa i D 5x0 5 A exp.0, fa * 0.05 *
8 Page 85! a) V BE V T ln + I! D 0.05V ln + 40x0 6 A V x0 5 A! V BE V T ln + I! D 0.05V ln + 400x0 6 A 0.65 V x0 5 A! b) V BE V T ln + I! D 0.059V ln + 40x0 6 A 0.64 V x0 5 A! V BE V T ln + I! D 0.059V ln + 400x0 6 A V x0 5 A ΔV BE 57.6 mv ΔV BE 59.6 mv Page 86! ln i D ln i D dv D dt k q ln! i D kt q ln ) Assume i D is constant, and E G E GO d dt v d T V T d dt d dt 3KBT! exp E! GO + KBT 3 exp E! GO + E GO kt kt kt d dt 3 T + E GO kt 3 T + qv GO 3 kt T + V GO V T T V D V D 75K) 0.680V.8x0 3 V K 350K) 0.680V.8x0 3 V K! Kn i KBT 3 exp E GO kt dv D dt v d T V T 300K 75K) 0.76 V 350K 300K) V d dt v V 3V d GO T T Page 87 w d 0.3µm + 0V 0.979V µm E V + φ j) max 0.979V ) w d 0.378x0 4 cm 58kV cm 0 fa + 0V 0.8V 36.7 fa 8
9 Page 9 C jo x0 4 F / cm 9.7 nf 0.3x0 4 cm cm C j 0V.5pF C j 5V ) 4.64 pf 5V V Page nf cm 0 cm).5x0 cm).5 pf C D 0 5 A 0.05V 0 8 s 4.00 pf C D 8x0 4 A 0.05V 0 8 s 30 pf C D 5x0 A 0.05V 0 8 s 0.0 µf Page 97 Two points on the load line : V D 0, 5V 5kΩ ma; 0, V D 5V The intersection of the two curves occurs at the Q pt : 0.88 ma, 0.6 V Page V D V D V T ln + I D ln + I D Page 00 id fzero@id) 00000*id0.05*log+id/e3), 5e4) ans 9.458e004 Page 0 Using MATLAB: vd fzero@vd) 00000*e4)*exp40*vd))vd), 0.5) vd V id e4)*exp40*vd)) id e04 id fzero@id) 00000*idlog+id/e3)),.00) id 9.458e04 id fzero@id) 00000*idlog+id/e5)),.00) id 9.30e04 Page 08 9
10 From the answer, the diodes are on,on,off. + V B 0.6V 0V ) 0 V B.87 V 0kΩ 0VV I + I B D.5kΩ V B 0.6V 0V 0kΩ V ).3 ma 0kΩ V ).3 ma V D ).7 V 0kΩ > 0, > 0, V D3 < 0. These results are consistent with the assumptions. Page 0 R min 5kΩ.67 kω 0 5 kω V O 0V 5kΩ+kΩ 3.33 V V Z V O 5 V V Z is conducting) is off) Page V L 0V kω + V 5V L 0.kΩ + V L 5kΩ 0 V 6.5 V I 6.5V 5V L Z 0.kΩ +00Ω.5mA) 78. mw P Z 5V.5mA Line Regulation 0.kΩ mv kΩ + 5kΩ V π 60.5mA Load Regulation 0.kΩ 5kΩ 98.0 Ω Page 7 V dc V on V I dc 7.9V 0.5Ω 5.8 A V I dc r C T 5.8A 0.5F 60 s 0.57 V ΔT ω V r 0.57V * ), 0.9 ms θ c 0π 0.9ms) 80o 8.9V + π 9.7o V dc V on 0 3. V I dc 3.V Ω 6.57 A C I dc T 6.57A V r 0.F 60 s.0 F ΔT ω V r π 60 0.V * ), 0.36 ms θ c 0π 0.36ms 4.V + 80o π 6.8o Page 9 0
11 At 300K : V D kt q ln + I D.38x0 3 J / K 300K.60x0 9 C ln A V 0 5 A At 50 o C : V D kt q ln + I D.38x0 3 J / K 73K + 50K ln A.07 V.60x0 9 C 0 5 A Note : For V T 0.05V, V D kt q ln + I D 0.05V ln A 0.96 V 0 5 A Page 6 V rms V dc + V on C ωc π 60Hz ΔT ω V r V C I dc T V r 0.F) 6V ) 340 A 0.0 ) 5) π 60) 6 A 0.0 5V s 0.F,000 µf 60 T ms I P I dc ΔT A s ms 84 A
12 Page 5 K n µ n ε ox T ox 500 cm V s Scaling: K n K n K n W L 5x0 7 cm 5x0 7 cm CHAPTER x0 4 F / cm) 69.x0 6 C 5x0 7 cm V s 69.µA V 69.µA V 345 µa µa 0µm µa V µm V K µa 60µm n µa V 3µm V K n 50 µa 0µm 000 µa V 0.5µm V For 0 V and V, < V TN and the transistor is off. Therefore 0. V V TN V and V DS 0. V Triode region operation W K n L V V V DS GS TN V DS 5 µa 0µm V.3 µa V µm V TN 3.5.5V and V DS 0. V Triode region operation W K n L V TN V DS V DS 5 µa 0µm V 36.3 µa V µm K n 5 µa 00µm 50 µa V µm V Page 53 R on V TN + W K n L V TN ) K n R on V + 50 µa V 50 µa V ) 4.00 kω R on 000Ω) 3.00 V 50 µa V.00 kω 4 )
13 Page 54 V DS should be 0.5 V V TN.5.5 V and V DS 0.8 V Triode region operation W K n L V V V DS GS TN V DS 000 µa V 344 µa V g m K n V DS 000 µa V 0.5V 50 µs Page 56 V TN 5 4 V and V DS 0 V Saturation region operation K n V TN ) ma V W K n K n L W L ma 40µA 5 Assuming saturation region operation, ma V K n V TN ) g m K n V TN ) ma V 5 ) V 8.00 ma W 5L 8.75 µm.5 ) V.3 ma.5 )V.50 ms.5 I Checking : g m D.5mA V TN V.50 ms Page 57 V TN 5 4 V and V DS 0 V Saturation region operation K n V TN ) + λv DS ) + λv DS ) K n V TN ma 5 ) V V ma 5 ) V V [ ] 9.60 ma [ ] 8.00 ma V TN 4 3 V and V DS 5 V Saturation region operation K n V TN ) + λv DS ) 5 K n V TN ) + λv DS ) 5 µa 4 ) V V V [ ] 8 µa [ ] 0 µa µa 5 ) V
14 Page Assuming pinchoff region operation, K n 0 V TN V TN + K n V + 00 µa) 50µA/V 4.00 V Assuming pinchoff region operation, K n V TN ) 50 Page 60 V TN V TO + γ V SB + 0.6V 0.6V µa V +V 00 µa µa V [ V )] 5 µa ) V.5 V V TN ).84 V V TN a) is less than the threshold voltage, so the transistor is cut off and 0. b) V TN V and V DS 0.5 V Triode region operation K n V TN V DS * V DS ma 0.5 * 0.5V 375 µa ) V ) c) V TN V and V DS V Saturation region operation K n V TN ) + λv DS ) 0.5 ma V [ ] 50 µa ) V Page 6 a is greater than the threshold voltage, so the transistor is cut off and 0. b) V TN + V and V DS 0.5 V Triode region operation K n V TN V DS V DS 0.4 ma )V 50 µa V c) V TN + V and V DS V Saturation region operation K n V TN ) + λv DS ) 0.4 ma V + V )* +, 08 µa 4
15 Page 67 C GC 00 µf 5x0 6 m) 0.5x0 6 m) ff m Triode region : C GD C GS C GC + C W 0.5 ff pf GSO m Saturation region : C GS 3 C + C W ff pf GC GSO m C GD C GSO W 300 pf 5x0 6 m).50 ff m 5x0 6 m 5x0 6 m.75 ff.83 ff Page 68 KP K n 50U LAMBDA λ VTO V TN PHI φ F 0.6 W W.5U L L 0.5U Page 70 Circuits/cm α 0.8µm nm 0.8µm 0.0µm 8.8) 66.9 PowerDelay Product α ; Rounding 550 times improvement i * ε D µ ox W /α n T ox /α L /α P * A * αp W /α) L /α) α 3 P A v GS V TN v DS v DS α i D P * V DD i * D V DD α i D i * D α i D P * α P P * A α 3 P * A α P Page 7 a) f T 500cm /V s π 0.5x0 4 cm b) f T π V 500cm /V s V 40nm) 7 GHz 500cm /V s π * 40x0 9 m ) 00 cm m 4.97 THz + V, 5
16 Page 7 V V L 05 cm L V cm cm L 0 5 V 0 5 cm cm 0 V V L 0 5 V x0 7 cm cm 0. V V MIN min*+ V TN ),V DS,V SAT, min[ 3.5, 4,.5].5 V K n V TN V MIN V MIN + λv DS V MIN min*+ V TN ),V DS,V SAT, min 3.5, 4, 0 K n V TN V MIN V MIN + λv DS Page 74 a [ ] 3.5 V From the graph for 0.5 V, 0 8 A b) From the graph for V TN 0.5 V, 3x0 5 A c) I 0 9 transistors 3x0 5 A/transistors) 3 µa Page 76 L Λ 0.50 µm W 0Λ.5 µm Active area 0Λ Λ) 0Λ 0 0.5µm).88 µm Gate area Λ 0Λ) 0Λ 0 0.5µm) 0.33 µm.5) * ), µa 3.5) * ), 66.5 µa Transistor area 0Λ + Λ + Λ) Λ + Λ + Λ) 4Λ 4 0.5µm) 3.50 µm N 04 µm) 3.50µm 8.6 x 06 transistors 6
17 Page 83 Upper group + V TN K n V TN K n ± +V TN K n Assume saturation region operation. ) + K n V EQ V TN K n 30µA V EQ 0 V TN ± K n K n V TN K n V TN V TN +V TN V TN + V EQ V TN + + K n V EQ V TN K n K n K n 36.8 µa + V EQ K n ) 39kΩ) + 30µA) 39kΩ) 4 ) 5.8 V Saturation region is correct. QPoint: 36.8 µa, 5.8 V ) V DS 0 4kΩ 36.8µA Assume saturation region operation. 5µA 6.7 µa 39kΩ) + 5µA) 39kΩ) 4.5) 6.96 V Saturation region is correct. QPoint: 6.7 µa, 6.96 V ) V DS 0 4kΩ 6.7µA Assume saturation region operation. 5µA 5.4 µa 6kΩ) + 5µA) 6kΩ) 4 ) 6.5 V Saturation region is correct. QPoint: 5.4 µa, 6.5 V ) V DS 0 37kΩ 5.4µA 7
18 Page 83 Second group R V EQ MΩ V DD 0V 4 V Assume saturation region operation. R + R MΩ+.5MΩ ) + K n V EQ V TN K n 5µA 99.5 µa.8kω) + 5µA).8kΩ) 4 ) 5.94 V Saturation region is correct. QPoint: 99.5 µa, 5.94 V ) V DS kΩ 99.5µA R V EQ.5MΩ V DD 0V 6 V Assume saturation region operation. R + R.5MΩ+MΩ 5µA 99. µa kω) + 5µA) kω) 6 ) 6.03 V Saturation region is correct. QPoint: 99. µa, 6.03 V ) V DS 0 40kΩ 99.µA I Bias V EQ V DD R + R R + R V DD I Bias 0V µa 5 MΩ R V DD R R + R ) V EQ 5MΩ 4V R + R V DD 0V MΩ R 5MΩ R 3 MΩ R EQ R Page 87 Should refer to Fig R. MΩ V SB V TN V SB Spreadsheet iteration yields 8. µa. 5µA V TN ) 8
19 Page 90 Assume saturation region operation x0 6 6x0 4 +) V GS V, +.76V 5.4 µa V DS 0 37kΩ 5.4µA) 5x Saturation region is correct. Q Point : 5.4 µa, 6.5 V [ ] 6.5 V Page 94 a V DS 3 V, V. The transistor is pinched off. SS V ) GS ma V, µa Pinchoff requires V DS +.5 V * 3.5V b) V DS 6 V, 3.5) +.5 V. The transistor is pinched off. SS V GS ma V ), µa Pinchoff requires V DS +.5 V * 3.5V c) V DS 0.5 V, 3.5) +.5 V. The transistor is in the triode region. SS V DS V DS ma 3.5 Pinchoff requires V DS +.5 V a) V DS 0.5 V, 4 SS V DS V DS 0.mA b) V DS 6 V, 4 SS V GS µa + V. The transistor is in the triode region µa 4) +3 V. The transistor is pinched off. ) 0.mA V, +. 3 µa * 4V 9
20 Page 96 a V DS 3 V, 3 4 V. The transistor is pinched off. SS V GS.5mA 3V ), µa Pinchoff requires V DS V * 4V b) V DS 6 V, 4) 3 V. The transistor is pinched off. SS V GS.5mA V ), +..4 ma Pinchoff requires V DS 3 V * 4V c) V DS 0.5 V, 4) V. The transistor is in the triode region. SS V DS V DS.5mA Pinchoff requires V DS V 73 µa Page 97 BETA SS BETA SS.5mA 0.65 ma VTO V LAMBDA λ 0.05 V 5mA.5 ma VTO V LAMBDA λ 0.0 V 0
21 Page 99 VTO 5 V BETA SS 5mA 0. ma LAMBDA λ 0.0 V 5 V G V S I G R G 0 K n V TN ) K n V TN 0.4mA 5) kω 5 V R GS S V TN 5 + V GS 5 SS V GS Assuming pinchoff,.9 ma and V DS 9.9mA kω+kω 3.09 V, V DS >, and pinchoff is correct. Assuming pinchoff, SS V GS.5 V SS V GS V DS.5mA kω+ kω.5 5 5mA.5 5 for SS K n V TN and V TN V GS and the rest is identical. 3.7 V. 5x0 3 ) x0 3 ) +.5 ma 5 V GS V. +.5 V, V DS >, and pinchoff is correct. QPoint:.5 ma, 7.00 V ) Page 00 a V G I G R G 0nA 680kΩ 6.80 mv. V G V S I G R G SS V GS x0 3 ) 0 3 ) + 5 V GS The value of V G is insignificant with respect to the constant term of 5. So the answers are the same to 3 significant digits. b) V G I G R G µa 680kΩ V G V S I G R G SS V GS.6 V SS V GS V DS.54mA kω+kω V and now cannot be neglected x0 3 ) 0 3 ) + 5mA ma 7.38 V QPoint:.54 ma, 7.38 V ) 5 V GS
22 CHAPTER 5 Page a) β F α F α F β F β F b) α F β F β F α 00 F α 3 F Page 3 ) i C 0 5 A exp exp 9.30, A) exp 9.30, ma * * 0.05 ) i E 0 5 A exp exp 9.30, A) exp 0.700, ma * * 0.05 i B 0 5 A 00 ) exp 0.700, A) exp 9.30, µa * * 0.05 Page 5 ) i C 0 6 A exp exp 0.700, A) exp 0.700, µa * * 0.05 ) i E 0 6 A exp exp 0.700, A) exp 0.750, µa * * 0.05 i B 0 6 A 75 ) exp 0.750, A) exp 0.700, µa * * 0.05 ) i T 0 5 A exp 0.750, + exp. 0.7 ma * ) i T 0 6 A exp exp 4.5, ma *
23 Page 7 V BE V T ln I C V ln 0 4 A 0 6 A V V BE V T ln I C V ln 0 3 A 0 6 A V Page 8 npn :V BE > 0, V BC < 0 Forward Active Region pnp :V EB > 0, V CB > 0 Saturation Region Page 3 β F α F α F β α R R 0.5 α R V BE 0, V BC << 0 : I C + * 0 6 A + * fa ) 0.333) I E 0.00 fa β R I B 0 6 A fa β R V BE << 0, V BC << 0 : I C β R 3x0 6 A fa I E β F 0 6 A aa I B 0 6 A β F β R A fa / 3 Page 33 a The currents do not depend upon V CC as long as the collector base junction is reverse biased by more than 0. V. Later when Early votlage V A is discussed, one should revisit this problem.) b) I E 00 µa I B I E β F + 00µA.96 µa I C β F I B 50I B 98.0 µa 5 V BE V T ln I C V ln 98.0µA 0 6 A V 3
24 Page 34 a The currents do not depend upon V CC as long as the collector base junction is reverse biased by more than 0. V. Later when Early voltage V A is discussed, one should revisit this problem.) b) Forward active region : I B 00 µa I E β F +)I B 5.0 ma I C β F I B 5.00 ma V BE V T ln I C V ln 5.00mA 0 6 A V Checking : V BC Forward active region with V CB 0 requires V CC V BE or V CC V Page V 9V a) I E I B 5.6kΩ.48 ma I E β F + I E 5 9. µa I C β F I B 50I B.45 ma V CE V C V E I C b) I E β F + 0.7) 3.47 V QPoint:.45 ma, 3.47 V ) I C 5 β F 50 00µA 0µA R 0.7V 9V 0µA The nearest 5 value is 8 kω. Page V 9V I E 5.6kΩ 8.3V 8.4 kω 0µA.48 ma I B I E β F + I E µa I β I 50I C F B B.45 ma I E β + F I C 5 β F 50 I.0I V V ln I C C C BE T +* 0.05ln x0 5 I C +) ) + V BE x0 6 )exp V. BE * 0 9 V BE V using a calculator solver, 0.05) / or spreadsheet. I C 5x0 6 exp * 99 µa V CE I C ) D BJT α F x0 4 A fa 5.44 V 4
25 Page V 9V I C 5.6kΩ.48 ma I B I C β R + I C 0.74 ma I β I E R B I B 0.74 ma Page 4 ma + * +! 40µA V CESAT 0.05V )ln* 99.7 mv * 0.5 ma ) * 50 40µA), + * 0.mA + 0.5)mA V BESAT 0.05V )ln*! 0 5 A mv * )* 50, 0.mA ma + * V BCSAT 0.05V )ln* 50! 0 5 A 0.67 mv *! )* , Note: V CESAT V BESAT V BCSAT 67.7mV Page 45.5 cm / s a) D n kt q µ n 0.05V 500cm /V s b) qad nn i N AB W.6x0 9 C 50µm Page 48 V T.38x0 3 J / K.60x0 9 C f β f T β F 373K 300MHz 5.40 MHz 0 4 cm / µm).5cm / s) 0 0 / cm 6 ) 0 8 / cm 3 ) µm) 3.mV C D I C V T τ F 0 8 A aa 0A 0.03V 4x0 9 s).4 µf 5
26 Page 49 I C 0 5 Aexp ma β F I B.74mA 9.3 µa I C 0 5 Aexp ma β F 75 I B.45mA 9.3 µa Page 5 g m 40 V 0 4 A C D 4.00mS 5ps Page ms g m ms V 0 3 A 0.00 pf C D 40.0mS 5ps).00 pf V T kt q.38x mv IS.60x0 9 BF 80 VAF 70 V I C 350µA exp V BE exp fa Page 58 8kΩ V EQ 8kΩ + 36kΩ V 4.00 V R EQ 8kΩ 36kΩ kω V I B )6 kω.687 µa I C 75I B 0 µa I E 76I B 04 µa V CE 000I C 6000I E 4.9 V Q po int : 0 µa, 4.9 V V T 80kΩ V EQ 80kΩ + 360kΩ V 4.00 V R EQ 80kΩ 360kΩ 0 kω V I B )6 kω.470 µa I 75I 85 µa I 76I C B E B 88 µa V CE 000I C 6000I E 4.9 V Q po int : 85 µa, 4.93 V 6
27 Page 59 I I C 5 50I B 0I B 5 8kΩ V EQ 8kΩ + 36kΩ V 4.00 V R EQ 8kΩ 36kΩ kω V I B )6 kω 0.4 µa I C 500I B 05.6 µa I E 76I B 06.0 µa V CE 000I C 6000I E 4.8 V Q point : 06 µa, 4.8 V Page 60 The voltages all remain the same, and the currents are reduced by a factor of 0. Hence all the resistors are just scaled up by a factor of 0. 0 kω. MΩ 8 kω 80 kω 6.8 kω 68 kω Page V I B ) kω 95.4 µa I C 50I B 4.77 ma I E 5I B 4.87 ma V CE I C + I B 4.3 V Q point : 4.77 ma, 4.3 V ) Page 66 VBE V) IC A) VBE V) E E E E
Microelectronic Circuit Design Third Edition - Part I Solutions to Exercises
Microelectronic Circuit Design Third Edition - Part I Solutions to Exercises Page 11 CHAPTER 1 V LSB 5.1V 10 bits 5.1V 104bits 5.00 mv V 5.1V MSB.560V 1100010001 9 + 8 + 4 + 0 785 10 V O 786 5.00mV or
Microelectronic Circuit Design Fourth Edition - Part II Solutions to Exercises
Page 9 Microelectronic Circuit Design Fourth Edition - Part II olutions to Exercises CHAPTER 6 NM 0.8V 0.4V 0.4 V NM H 3.6V.0V.6 V Page 94 V 0% V + 0. ΔV [ ].4 V [ ].4 V.6V + 0. 0.6 (.6) 0.6V 0.9 0.6 (.6)
( )( ) ( ) ( )( ) ( )( ) β = Chapter 5 Exercise Problems EX α So 49 β 199 EX EX EX5.4 EX5.5. (a)
hapter 5 xercise Problems X5. α β α 0.980 For α 0.980, β 49 0.980 0.995 For α 0.995, β 99 0.995 So 49 β 99 X5. O 00 O or n 3 O 40.5 β 0 X5.3 6.5 μ A 00 β ( 0)( 6.5 μa) 8 ma 5 ( 8)( 4 ) or.88 P on + 0.0065
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