Microelectronic Circuit Design Third Edition - Part I Solutions to Exercises
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1 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 V O mV or 5.1V 3.91 V V O V Page 1 The dc component is V A 4V. The signal consists of the remaining portion of v A : v a 5 sin 000πt + 3 cos 1000 πt) Volts. Page 19 i SC i 1 + βi 1 v s + β v s β +1) v s v OC R 1 R 1 R 1 R th v OC i SC β +1)R S R 1 β +1)R S + R 1 β +1 1 β +1) β +1)R S v S β +1)R S + R R S R th R S R 1 β +1 R 1 Page 3 5sin 000πt + 5 o 90 o ) v o 5cos 000πt + 5 o V o 5 65 o V s o A v [ ] 5 65o o o sin 000πt 65o) 1
2 Page 6 v s 0.5sin 000πt [ + sin 4000πt) +1.5sin 6000πt) ] The three spectral components are f Hz f 000 Hz f Hz a) The gain of the band - pass filter is zero at both f 1 and f 3. At f, V o 10 1V 10V, and v O 10.0sin 4000πt) volts. b) The gain of the low - pass filter is zero at both f and f 3. At f, V o 6 0.5V ) 3V, and v O 3.00sin 000πt) volts. Page 7 R 39kΩ ) or 35.1 kω R 4.9 kω R 3.6kΩ ) or 3.56 kω R 3.64 kω 39kΩ kΩ Page 9 V S P P nom mw R 1 + R 54kΩ P max 1.1x15 ) 0.95x54kΩ 5.31 mw P min 0.9x mw 1.05x54kΩ Page 33 R 10kΩ o C o C 9.0 kω R 10kΩ o C o C 10.6 kω
3 Page 47 n i.31x10 30 K 3 cm 6 ) 300K 3 exp CHAPTER 0.66eV 8.6x10 5 ev / K) 300K).7x1013 cm 3 n i n i 1.08x10 31 K 3 cm 6 ) 50K 3 exp 1.08x10 31 K 3 cm 6 ) 35K cm 3 m L x cm 3 exp 3 1.1eV 8.6x10 5 ev / K) 50K) 4.34x10 39 cm 3 L 6.13x10 10 m 1.1eV 8.6x10 5 ev / K) 35K) 4.01x1010 cm 3 Page 48 v p µ p E 500 cm V s 10 V cm 5.00x10 3 cm s E V L 1 x10 4 V cm 5.00x103 V cm v n µ n E 1350 cm V s 1000 V cm 1.35x10 6 cm s Page 48 µ n v n E 4.3x105 cm / s 100V / cm 4300 cm s µ n v n E 8.5x105 cm / s 100V / cm 8500 cm s µ p v p E.1x105 cm / s 100V / cm 100 cm s 3
4 Page 50 n i 1.08x exp ρ 1 σ x x10 1 n i 1.08x x ) 5.40x104 / cm 6 [ 1350) +.3x10 1 ) 500) ] 3 exp ρ 1 σ x x10 39 Page 54 n i 1.08x x ) 1.88x10 77 / cm Ω cm [ 6500) x10 39 ) 000) ] 1.69x1053 Ω cm 3 exp x ) 5.40x104 / cm 6 holes p N A N D x x10 15 n n i cm 3 p 5.40x104 8x x10 8 electrons cm 3 n N D x10 16 electrons n n i cm 3 p 100 x10 holes x103 n > p n - type silicon cm 3 Page 55 Reading from the graph for N T /cm 3, 150 cm /V-s and 400 cm /V-s. Reading from the graph for N T /cm 3, 800 cm /V-s and 30 cm /V-s. Page 57 σ x10 19 µ n n u n n 6.5x10 1 / cm 3 6.5x a) N T x10 16 / cm 3 b) N T 5x10 16 / cm 3 4
5 Page 58 holes p N A N D 4x10 18 n n i cm 3 p 100 5electrons 18 4x10 cm 3 N T 4x1018 and mobilities from Fig..8 cm 3 holes p N A N D 7x10 19 n n i cm 3 p 100 electrons x10 cm 3 N T 7x1019 cm 3 and mobilities from Fig..8 ρ 1 σ x Page 59 [ 100) + 7x10 19 ) 50) ] V T kt q 1.38x mv 1.60x10 19 V T 1.38x mv 1.60x10 19 V T 1.38x mV 1.60x mω cm Page 59 D n kt q µ cm n 5.8mV cm D p kt V s s q µ cm p 5.8mV cm V s s dn j n qd n dx 1.60x10 19 C 0 cm µm 30 A s cm 3 µm cm cm dp j p qd p dx 1.60x10 19 C 4 cm µm 64 A s cm 3 µm cm cm 5
6 Page 79 E max 1 ε s 0 x p qn A dx qn x A p E max 1.6x10 19 C / cm 3 CHAPTER 3 ε s E max 1 ε s qn D x10 5 cm) 1.13x10 8 cm) x10 14 F / cm E max 1.6x10 19 C 10 0 / cm x10 14 F / cm x n 175 kv cm 175 kv cm dx qn D x n ε s For the values in Ex.3. : E max 1.05V kv x10 6 cm cm 1 x p 0.058µm 1+ x µm x 10 0 n 0.058µm x10 18 Page 8 T 1.60x mV x K x10 4 µm Page 84 i D 5x10 15 A exp fa i D 5x10 15 A exp fa Page 86 V T ln 1+ I D 0.05V ln 1+ 40x10 6 A V I S x10 15 A V T ln 1+ I D 0.05V ln x10 6 A V I S x10 15 A V T ln 1+ I D 0.058V ln 1+ 40x10 6 A 0.61 V I S x10 15 A V T ln 1+ I D 0.058V ln x10 6 A V I S x10 15 A Δ 57.6 mv Δ 59.4 mv Page 89 w d 0.113µm 1+ 10V 0.979V µm E V max 0.378x10 4 cm 581 kv cm 6
7 Page 89 I S 10 fa 1+ 10V 0.8V 36.7 fa Page 9 C jo x10 14 F / cm 0.113x10 4 cm 11.5 pf C j 5V ) 4.64 pf 5V V Page nf cm C j 0V ) 91.7 nf cm 10 cm) 1.5x10 cm) 11.5 pf C D 10 5 A 0.05V 10 8 s 4.00 pf C D 8x10 4 A 0.05V 10 8 s 30 pf C D 5x10 A 0.05V 10 8 s 0.0 µf Page 97 Two points on the load line : V D 0, 5V 5kΩ 1 ma; 0, V D 5V Page 108 From the answer, the diodes are on,on,off. 0.6V 10V ) 10V - V I I B D.5kΩ V B 0.6V 0V + V B 10kΩ 10kΩ V ) V 1.13 ma 10kΩ V D V 10kΩ 1.13 ma 0 V B 1.87 V Page 110 R min 5kΩ 1kΩ 1.67 kω V 0 O 0V 5 1 5kΩ +1kΩ 3.33 V V Z is off) V 5 V V O Z is conducting) Page 111 V L 0V 1kΩ + V 5V L 0.1kΩ + V L 5kΩ 0 V 6.5 V L I Z 6.5V 5V 0.1kΩ 1.5mA +100Ω 1.5mA) 78.1 mw P Z 5V 1.5mA Page 11 Line Regulation 0.1kΩ mv kΩ + 5kΩ V Load Regulation 0.1kΩ 5kΩ 98.0 Ω 7
8 Page 117 V dc V P V on V I dc 7.91V 0.5Ω 15.8 A V I dc r C T 15.8A 1 0.5F 60 s 0.57 V ΔT 1 ω V r 0.57V 0.91 ms θ c 10π 0.91ms) 180o V P 8.91V π 19.7o 1 π 60 Page 118 V dc V P V on V I dc 13.1V Ω 6.57 A C I dc T 6.57A 1 V r 0.1F 60 s 1.10 F ΔT 1 ω V r V P 1 π V ms θ c 10π 0.316ms 14.1V 180o Page 118 For V T 0.05V, V D kt q ln 1+ I D 0.05V ln A V A V D kt q ln 1+ I D 1.38x10 3 J / K 300K 1.60x10 19 C I S I S V D kt q ln 1+ I D 1.38x10 3 J / K 73K + 50K 1.60x10 19 C I S ln A V A ln A 1.07 V A π 6.8o 8
9 Page 151 i) K n ' µ n ε ox T ox 500 cm V s CHAPTER x10 14 F / cm) 69.1x10 6 C 5x10 7 cm V s 69.1µA V ' W µa 0µm ii) K n K n µa L V 1µm V K µa 60µm n µa V 3µm V K n 50 µa 10µm 000 µa V 0.5µm V iii) For V GS 0V and 1V, V GS < V TN and the transistor is off. Therefore 0. V GS - V TN V and V DS 0.1 V Triode region operation ' W K n L V V V DS GS TN V DS 5 µa 10µm V 11.3 µa V 1µm V GS - V TN V and V DS 0.1 V Triode region operation ' W K n L V V V DS GS TN V DS 5 µa 10µm V 36.3 µa V 1µm K n ' 5 µa 100µm 50 µa V 1µm V Page 15 R on V GS V TN + 1 ' W K n L V V GS TN ) 1 K n R on 1V µa V 50 µa V 1) 4.00 kω R on 1 000Ω) 3.00 V 1 50 µa V 1.00 kω 4 1) 9
10 Page 155 V GS - V TN V and V DS 10 V Saturation region operation K n V V GS TN ) 1 ma V ' W K n K n L W L 1mA 40µA ) V 8.00 ma Assuming saturation region operation, W 5L 8.75 µm K n V V GS TN ) 1 ma.5 1) V 1.13 ma V g m K n V GS V TN ) 1 ma V 1.50 ms V.5 1 Checking : g m 1.15mA.5 1)V 1.50 ms Page 156 i) V GS - V TN V and V DS 10 V Saturation region operation K n V V GS TN ) 1+ λv DS ) 1 K n V V GS TN ) 1+ λv DS ) 1 ii) V GS - V TN ma 5 1) V V ma V 5 1) V [ ] 9.60 ma [ ] 8.00 ma V and V DS 5 V Saturation region operation K n V V GS TN ) 1+ λv DS ) 5 K n V V GS TN ) 1+ λv DS ) 5 Page 157 µa 4 1) V V V [ ] 118 µa [ ] 0 µa µa 5 1) V i) Assuming pinchoff region operation, K n 0 V TN V GS V TN + K n V µa) 50µA/V 4.00 V ii) Assuming pinchoff region operation, K n V V GS TN ) 50 µa V +V 100 µa µa V [ 1 V )] 5 µa 10
11 Page V V TN i) V TN V TO + γ V SB + 0.6V 0.6V V TN V 1.84 V ii a) V GS is less than the threshold voltage, so the transistor is cut off and 0. b) V GS - V TN -1 1 V and V DS 0.5 V Triode region operation K n V GS V TN V DS V DS 1 ma V 375 µa V c) V GS - V TN -1 1 V and V DS V Saturation region operation K n V GS V TN ) 1+ λv DS ) 0.5 ma V [ ] 50 µa 1) V Page 160 a V GS is greater than the threshold voltage, so the transistor is cut off and 0. b) V GS - V TN V and V DS 0.5 V Triode region operation K n V GS V TN V DS V DS 0.4 ma )V 150 µa V c) V GS - V TN +1 1 V and V DS V Saturation region operation K n V V GS TN ) 1+ λv DS ) 0.4 Page 165 Active area 10Λ µm ma V +1 Gate area 0Λ 0 0.5µm) 5 µm N [ ] 08 µa V ) 30 µm L Λ 1µm W 10Λ 5µm 10 4 µm) 14Λ) 16Λ 0.5µm) 1.79 x 106 transistors 11
12 Page 167 C GC 00 µf 5x10 6 m) 0.5x10 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 5x10 6 m) 1.50 ff m 5x10 6 m 5x10 6 m 1.75 ff 1.83 ff Page 169 KP K n 150U LAMBDA λ VTO V TN 1 PHI φ F 0.6 W W 1.5U L L 0.5U Page 173 i Assume saturation region operation and λ 0. Then is indpendent of V DS, and 50 µa. V DS V DD R D 10 50kΩ 50µA) 7.50 V. V DS V GS V TN, so our assumption was correct. Q - Point 50.0 µa, 7.50 V ) ii) V EQ 5x kΩ 10V.647 V 70kΩ + 750kΩ R A V EQ 70kΩ 750kΩ Assume saturation region ) V 33.9 µa V DS V DD R D kΩ 33.9µA) 6.61 V. V DS V GS V TN, so our assumption was correct. iii) V GS does not change : V GS 3.00 V. 30x10 6 V DS V DD R D kΩ 60.0µA Q - Point 33.9 µa, 6.61 V ) A V 3 1) V 60.0 µa 4.00 V. V DS V GS V TN, so our assumption was correct. Q - Point 60.0 µa, 4.00 V iv) V GS does not change : V GS 3.00 V. 5x10 6 A 3 1.5) V 8.1 µa V V DS V DD R D kΩ 8.1µA) 7.19 V. V DS V GS V TN, so our assumption was correct. Q - Point 8.1 µa, 7.19 V ) 1
13 Page 174 V DS 10 5x ) V DS V DS V 4.76 V I 5x10 6 D V DS V DS ) [ ] 5.4 µa 3 1) Page 176 For 0, V DS 10 V. For V DS 0, 10V 66.7kΩ 150µA. The load line intersects the V GS 3 - V curve at 50 µa, V DS 6.7 V. Page 178 i) 4 V GS + 30x x x Q - Point : 36.8 µa,5.81 V V GS 1) V GS 0.91V GS V GS.566V,.75V 36.8 µa V DS kΩ 36.8µA) 5.81 V ii) 4 V GS + 5x x10 4 5x Q - Point : 6.7 µa,6.96 V V GS 1.5) V GS 0.949V GS V GS.960V,.01V 6.7 µa V DS kΩ 6.7µA) 6.96 V ii) 4 V GS + 5x x10 4 5x Q - Point : 5.4 µa,6.5 V V GS 1) V GS 0.710V GS V GS.46V, 1.716V 5.4 µa V DS kΩ 5.4µA) 6.5 V 13
14 Page 180 i Assume saturation. For 99.5µA, V GS 4V µA 1.8kΩ 3.81V 99.5µA which agrees. V DS 10V 99.5µA 40.8kΩ) 5.94 V 5µA Saturation region operation is correct. 1.5MΩ ii) V EQ 10V 6.00 V 1.5MΩ +1MΩ R 6 V GS + 5x10 6 x10 3 EQ 1.5MΩ 1MΩ 600kΩ Assume saturation region. V GS 1) V GS V GS V GS 3.818V ) 99.3 µa V DS 10 40kΩ 99.3µA) 6.03 V 5x10 6 Saturation region operation is correct. Q- Point : 99.3 µa, 6.03 V iii) R 1 + R 10V µa 5MΩ R 1 10V 6V R 1 5MΩ 6V R 1 + R 10V 3 MΩ R MΩ R EQ 3MΩ MΩ 1. MΩ Page 18 5µA V GS V SB 000 V TN V SB Spreadsheet iteration yields 83. µa. Page 183 [ +.83 ] V 0. SB Equation 4.55 becomes V SB V SB 6.065V SB V SB 1.63V. R S 1.63V 10 4 A 16.3 kω 16 kω R )V D 3.7kΩ 4 kω 10 4 A Page 185 i) Using Eq. 4.58) V GS 3.3 x kΩ V GS x10 4 ii V GS 1) V GS V GS V TN ) V GS.3 0 V GS.097V.097 1) 10.3µA V DS V GS Q - Point : 10 µa,.10 V ) Since there was no voltage drop across R G in i), the equations and answers are identical to the first part. 14
15 Page V DS 1.8x10 6.5x V DS V DS V DS.96 mv Q - Point :. µa,.96 mv ) Checking; R on Page 189 i) x10 6 6x10 4 5x Q - Point : 5.4 µa, V 4 V DS. µa 6 1.8x10.µA.96 mv.5x ) V GS +1) V GS V GS V GS V GS.46V, V [ ] 6.5 V 5.4 µa V DS kΩ 5.4µA) Page 19 i) Circuits/cm α 1µm 0.5µm 16 Power - Delay Product 1 α times improvement 3 64 ε ii) i * D µ ox W /α n T ox /α L/α v V v DS GS TN v DS α i D P * V DD i * D V DD α i D P * A * αp W /α) L/α) α P 3 A iii) f T 1 π 500cm /V s 1V 10 4 cm) 7.96 GHz f T 1 500cm /V s π 0.5x10 4 cm 1V Page 193 The field in the first sentence at the top of the page should be 10 5 V / cm. V V L 104 cm L V V cm) 10 V L cm cm cm 1 V α P 17 GHz 15
16 CHAPTER 5 Page 13 α F α F β F β 0.50 F α F α 00 F α 3 F Page 15 i C A exp exp A exp ma i E A exp exp A exp ma i B A 100 exp A exp µa Page 17 i C A exp exp A exp µa i E A exp exp A exp µa i B A 75 exp A exp µa Page 17 i T A exp exp 10.7 ma i T A exp exp 1.07 ma
17 Page 0 V T ln I C V ln 10 4 A I S A V V T ln I C V ln 10 3 A I S A V Page 1 npn : > 0, V BC < 0 Forward Active Region pnp :V EB > 0, V CB > 0 Saturation Region Page 3 α F 1 α F β α R R α R , V BC << 0 I C I S A fa I E I S fa β R I B I S A fa β R << 0, V BC << 0 I E I S I C I S β R 3x10 16 A fa A aa I I S B I S A β R A fa 1/ 3 Page 5 a The currents do not depend upon V CC as long as the collector - base junction is reverse biased. Later when Early votlage V A is discussed, one should revisit this problem.) b) I E 100 µa, I B I E µA 51 V T ln I C V ln 98.0µA I S A V 1.96 µa, I C I B 50I B 98.0 µa 17
18 Page 6 a The currents do not depend upon V CC as long as the collector - base junction is reverse biased. Later when Early voltage V A is discussed, one should revisit this problem.) b) Forward - active region : I B 100 µa, I E +1)I B 5.10 ma, I C I B 5.00 ma V T ln I C V ln 5.00mA I S A V Checking : V BC V c) Forward - active region with V CB 0 requires V CC or V CC V Page 8 0.7V 9V I E 5.6kΩ V CE V C V E I C 1.48 ma I B I E µa, I β I C F B 1.45 ma 0.7) 3.47 V Q - Point : 1.45 ma, 3.47 V) I E β +1 F I C µA 10µA R 0.7V 9V 10µA Nearest 5% value is 8 kω. 8.3V 81.4 kω 10µA Page 9 0.7V 9V ) i) I E 1.48 ma I B I E 5.6kΩ µa, I β I C F B 1.45 ma ii) I E β +1 F I C I 1.0I V V ln I C C C BE T ln I S x10 15 I C +1) x10 16 exp V BE V using a calculator solver or spreadsheet I C 5x10 16 exp µa V CE I C 0.708) 5.44 V 0.05 Page 9 I SD I SBJT α F x10 14 A fa Page 3 0.7V 9V I C 5.6kΩ 1.48 ma I B I C ma, β R +1 - I β I E R B ma 18
19 Page 34 i) V CESAT 0.05V ) ln ii) 0.05V ln ln V BC 0.05V 1+ 1mA 1 40µA) 99.7 mv 0.5 1mA µA) 0.1mA )1mA A mv mA 1mA A mv V BE V BC 67.7mV Page cm / s D n kt q µ n 0.05V 500cm /V s I S qad n n i N AB W 1.6x10 19 C 50µm Page K i) V T 1.38x10 3 J / K 1.60x10 19 C ii) f β f T 300MHz cm / µm) 1.5cm / s) 10 0 / cm 6 ) / cm 3 ) 1µm).40 MHz 3.mV C D I C V T τ F A 10 A 0.03V 4x10 9 s) 1.4 µf Page 41 a) I C Aexp ma I B 1.74mA 19.3 µa b) I C Aexp ma 75 I B 1.45mA 19.3 µa Page 43 g m ms g m ) 40.0 ms C D 4.00mS 5 ps pf C D 40.0mS 5 ps) 1.00 pf 19
20 Page kΩ V EQ 180kΩ + 360kΩ 1V 4.00V R EQ 180kΩ 360kΩ 10kΩ V I B )16 kω.470µa I 75I 185.3µA I 76I 187.7µA C B E B V CE 1 000I C 16000I E 4.9V Q - point : 185 µa, 4.9 V Page 50 i) I I C 5 50I B 5 10I B 18kΩ ii) V EQ 18kΩ + 36kΩ 1V 4.00V R EQ 18kΩ 36kΩ 1kΩ V I B )16 kω µA I 500I 05.5µA I 76I 06.0µA C B E B V CE 1 000I C 16000I E 4.18V Q - point : 06 µa, 4.18 V Page 51 The voltages all remain the same, and the currents are reduced by a factor of 10. Hence all the resistors are just scaled up by a factor of kω 1. MΩ 8 kω 80 kω 6.8 kω 68 kω Page 53 i) V CE 0.7 V at the edge of saturation. 0.7V 1V - R C kΩ 05µA ii) SAT 4 1kΩ 4µA) 16kΩ 184µA) V V CESAT 1 56kΩ 160µA 16kΩ 184µA) V Page V I B )1 kω 95.4µA I 50I 4.77 ma I 76I 187.7µA C B E B V CE I C + I B 4.13V Q - po int : 4.77 ma, 4.13 V ) R C 38.9 kω 0
Microelectronic Circuit Design Fifth Edition - Part I Solutions to Exercises
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