Microelectronic Circuit Design Fourth Edition - Part I Solutions to Exercises

Page Microelectronic Circuit esign Fourth Edition - Part I Solutions to Exercises CHAPTER V LSB 5.V 0 bits 5.V 04bits 5.00 mv V 5.V MSB.560V 000000 9 + 8 + 4 + 0 785 0 V O 785 5.00mV or ) 5.V 3.95 V V O + + 6 + 0 Page V LSB 5.0V 8 bits 5.00V 56bits ) 3.95 V.V 9.53 mv N 56bits 6.44bits 5.00V 6 3 +6 + 8 + 4 + 5 + 4 + 3 + + 0 000 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 0.00 0 o A v Page 5 [ ] A v R 5 R R 0kΩ R 50 kω 5 65o 5000 65o o 0.00 0 5sin 000πt 65o) 08/09/0

Page 6 v s 0.5sin 000πt [ ) + sin 4000πt) +.5sin 6000πt) ] The three spectral component frequencies are f 000 Hz f 000 Hz f 3 3000 Hz a) The gain of the band - pass filter is zero at both f and f 3. At f, V o 0 V ) 0V, and v o 0.0sin 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 + 0.) or 35. kω R 4.9 kω ) R 3.6kΩ + 0.0) or 3.56 kω R 3.64 kω 39kΩ 0. 3.6kΩ 0.0 Page 9 V I P P nom 5 4.7 mw R + R 54kΩ P max.x5 ) ) 0.95x54kΩ 5.3 mw P min 0.9x5 3. mw.05x54kω Page 33 R 0kΩ + 0 3 55 5 o C ) o C 9.0 kω R 0kΩ + 0 3 85 5 o C ) o C 0.6 kω 08/08/0

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 3 4.34x0 39 0 cm 3 ) 3 exp 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 49 v p µ p E 500 cm V s 0 V cm 5.00x0 3 cm s E V L x0 4 V cm 5.00x03 V cm v n µ n E 350 cm V s 000 V cm.35x0 6 cm s Page 49 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 4300 00V / 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 08/09/0

Page 5 n i.08x0 3 400 ) 3 exp ρ σ.60x0 9.3x0 n i.08x0 3 50. 8.6x0 5 400) 5.40x04 / cm 6 n i.3x0 / cm 3 [ ) 350) +.3x0 ) 500) ] ) 3 exp ρ σ.60x0 9 4.34x0 39 Page 55 n i.08x0 3 400 450 Ω cm. 8.6x0 5 50).88x0 77 / cm 6 n i 4.34x0 39 / cm 3 [ ) 6500) + 4.34x0 39 ) 000) ].69x053 Ω cm ) 3 exp. 8.6x0 5 400) 5.40x04 / cm 6 holes p N A N 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 6 5.00x03 n > p n - type silicon cm 3 Page 56 Reading from the graph for N T 0 6 /cm 3, 30 cm /V-s and 405 cm /V-s. Reading from the graph for N T 0 7 /cm 3, 800 cm /V-s and 30 cm /V-s. 6.75x0 8 electrons cm 3 n N x0 6 electrons cm 3 4 08/08/0

Page 58 ) 00 σ 000.60x0 9 µ n n u n n 6.5x0 / cm 3 6.5x0 9 a) N x0 6 / cm 3 N A 0 / cm 3 N T x0 6 / cm 3 ) ρ σ 0.00 Ω cm 70 µ n 9 + 70 447 0.9 cm /V s µ p 48 + 363 0.76 cm /V s + x06 + x06.3x0 7 6.3x0 6 b) N T N + N A x0 6 / cm 3 + 3x0 6 / cm 3 5x0 6 / cm 3 70 µ n 9 + 990 447 0.9 cm /V s µ p 48 + 9 0.76 cm /V s + 5x06 + 5x06.3x0 7 6.3x0 6 Page 59 Boron B) is a Column - III element, so it is an acceptor impurity. holes p N A N 4x0 8 n n i cm 3 p 00 5electrons p - type material 8 4x0 cm 3 N T 4x08 and mobilities from Fig..8 : µ cm 3 n 50 cm /V s µ p 70 cm /V s ρ qµ p p 0.0 Ω cm.60x0 9 4x0 8 - - - ) 70 ) Indium In) is a Column - III element, so it is an acceptor impurity. holes p N A N 7x0 9 n n i cm 3 p 00.43electrons p - type material 9 7x0 cm 3 N T 7x09 cm 3 and mobilities from Fig..8 : µ n 00 cm /V s µ p 50 cm /V s ρ σ.79 mω cm.60x0 9 7x0 9 ) 50) 5 08/09/0

Page 60 V T kt q.38x0 3 50) 4.3 mv V.60x0 9 T.38x0 3 300) 5.8 mv.60x0 9 ) V T.38x0 3 400 34.5mV.60x0 9 n kt q µ cm n 5.8mV 36 35. cm V s s dn j n q n dx.60x0 9 C 0 cm 0 6 04 µm 30 A s cm 3 µm cm cm dp j p q p dx.60x0 9 C 4 cm 0 6 04 µm 64 A s cm 3 µm cm cm p kt q µ cm p 5.8mV 495.8 cm V s s 6 08/08/0

CHAPTER 3 Page 79 φ j V T ln N N A n i 0.05V ln x08 0 0 ) 0 0.05 V w do ε s q + φ j.7 N A.60x0 9 x0 + 8 N ) ) 8.85x0 4 0 0.05 ).63x0 6 cm 0.063 µm Page 80 E max ε s 0 x p qn A dx qn x A p E max.6x0 9 C 0 7 / cm 3 ε s E max ε s qn 0 ).3x0 5 cm) ).7 8.854x0 4 F / cm ) x n 75 kv cm dx qn x n ε s For the values in Ex.3. : E max.05v kv 798.63x0 6 cm cm x p 0.063µm + x08 0.058 µm x 0 0 n 0.063µm + 00 x0 8 Page 83 ) n kt.38x0 3 300 0.059 T 300 q.60x0 9.03 9 K 5.6x0 4 µm Page 85 i 40x0 5 A exp 0.55 43 µa i 40x0 5 A exp 0.70 57.9 ma 0.05 0.05 V 0.05V ) ln + 6x0 3 A 0.643 V 40x0 5 A i 5x0 5 A exp 0.04 3.99 fa i 5x0 5 A exp.0 5.00 fa 0.05 0.05 7 08/09/0

Page 87 a) V BE V T ln + I 0.05V ln + 40x0 6 A 0.593 V I S x0 5 A V BE V T ln + I 0.05V ln + 400x0 6 A 0.65 V I S x0 5 A b) V BE V T ln + I 0.058V ln + 40x0 6 A 0.6 V I S x0 5 A V BE V T ln + I 0.058V ln + 400x0 6 A 0.67 V I S x0 5 A ΔV BE 57.6 mv ΔV BE 59.4 mv Page 89 ln i ln i ) ln I S ) Assume i is constant, and E G E GO I S dv dt k q ln i kt di S I S q I S dt v d T V di S T I S Kn i KBT 3 exp E GO I S dt kt di S dt 3KBT exp E GO + KBT 3 exp E GO + E GO kt kt kt I S di S dt 3 T + E GO kt 3 T + qv GO kt 3 T + V GO V T T dv dt v d T V T I S di S dt v V 3V d GO T T Page 90 ) ) w d 0.3µm + 0V 0.979V 0.378 µm E V + φ j max 0.979V w d 0.378x0 4 cm I S 0 fa + 0V 0.8V 36.7 fa 58 kv cm Page 93 ) C jo.7 8.854x0 4 F / cm 0.3x0 4 cm.5 pf C j 5V ) 4.64 pf 5V + 0.979V 9.7 nf cm C j 0V ) 9.7 nf cm 0 cm).5x0 cm).5 pf C 0 5 A 0.05V 0 8 s 4.00 pf C 8x0 4 A 0.05V 0 8 s 30 pf C 5x0 A 0.05V 0 8 s 0.0 µf 8 08/08/0

Page 98 Two points on the load line : V 0, I 5V 5kΩ ma; I 0, V 5V The intersection of the two curves occurs at the Q- pt : 0.88 ma, 0.6 V ) Page 00 0 0 4 I + V V V T ln + I 0 0 4 I + 0.05ln + I Page 0 Page 0 fzero@id) 0-0000*id-0.05*log+id/e-3), 5e-4) ans 9.458e-004 I S fzero@id) 0-0000*e-4)*exp40*vd)-)-vd), 0.5) ans 0.636 Page 09 From the answer, the diodes are on,on,off. ) 0.6V 0V ) 0V - V I I + I B.5kΩ V B 0.6V 0V + V B 0 V B.87 V 0kΩ 0kΩ.87 0.6 0V ).87 0.6 0V ) I.3 ma I 0kΩ 0kΩ I > 0, I > 0, V 3 < 0. These results are consistent with the assumptions. I S.3 ma V 3.87 0.6).7 V Page R min 5kΩ kω.67 kω V 0 O 0V 5 5kΩ +kω 3.33 V V is off) V 5 V V Z O Z is conducting) 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 P Z 5V.5mA ) +00Ω.5mA) 78. mw Line Regulation 0.kΩ mv 9.6 0.kΩ + 5kΩ V.5mA Load Regulation 0.kΩ 5kΩ 98.0 Ω 9 08/09/0

Page 8 V dc V on 6.3 7.9 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 π 60 ) 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 4.V Page 0 At 300K : V kt q ln + I.38x0 3 J / K 300K.60x0 9 C I S θ c 0π 0.36ms) 80o π 6.8o ) ) ln + 48.6 A 0.994 V 0 5 A At 50 o C : V kt q ln + I I S.38x0 3 J / K 73K + 50K ln + 48.6A.07 V.60x0 9 C 0 5 A Note : For V T 0.05V, V kt q ln + I 0.05V ln + 48.6 A 0.96 V 0 5 A Page 7 V rms V dc + V on I SC ωc π 60Hz ΔT ω V r I S 5+.3 V C I dc V r T ) 0.F) 6V ) 340 A 0.0 ) 5) π 60) 6 A 0.0 5V ) s 0.F,000 µf 60 T 0.363 ms I P I dc ΔT A s 60 0.363ms ) 84 A 0 08/08/0

Page 53 K n ' µ n ε ox T ox 500 cm V s CHAPTER 4 3.9 8.854x0 4 F / cm) 69.x0 6 C 5x0 7 cm V s 69.µA V - - - ' W µa 0µm K n K n 50 000 µa L V µm V K µa 60µm n 50 000 µa V 3µm V K n 50 µa 0µm 000 µa V 0.5µm V - - - For 0V and V, < V TN and the transistor is off. Therefore I 0. - V TN -.5 0.5V and V S 0. V Triode region operation ' W I K n L V V V S GS TN V S 5 µa 0µm.5 0. 0.V.3 µa V µm - V TN 3-.5.5V and V S 0. V Triode region operation ' W I K n L V V V S GS TN V S 5 µa 0µm 3.5 0. 0.V 36.3 µa V µm K n ' 5 µa 00µm 50 µa V µm V Page 54 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 ) 08/09/0

Page 57 - V TN 5-4 V and V S 0 V Saturation region operation I K n V TN ) ma V ' W K n K n L W L ma 40µA 5 5 ) V 8.00 ma Assuming saturation region operation, W 5L 8.75 µm I K n V TN ) ma.5 ) V.3 ma V g m K n V TN ) ma )V.50 ms.5 V I Checking : g m.5ma V TN )V ).5.50 ms Page 58 - V TN 5-4 V and V S 0 V Saturation region operation I K n V TN ) + λv S ) ) + λv S ) I K n V TN - - - ma 5 ) V + 0.0 0 V ma V 5 ) V + 0 0 [ )] 9.60 ma [ )] 8.00 ma - V TN 4-3 V and V S 5 V Saturation region operation I K n V TN ) + λv S ) 5 I K n V TN ) + λv S ) 5 Page 59 µa 4 ) V + 0.0 5 V V [ )] 8 µa [ )] 0 µa µa 5 ) V + 0.0 0 Assuming pinchoff region operation, I K n 0 V TN ) 50 V TN + I K n V + 00 µa) 50µA/V 4.00 V Assuming pinchoff region operation, I K n V TN ) 50 µa V +V ) 00 µa µa V [ V )] 5 µa 08/08/0

Page 6 ) + 0.75 0 + 0.6 0.6) V ).5 V V TN + 0.75 3 + 0.6 0.6).84 V V TN V TO + γ V SB + 0.6V 0.6V V TN + 0.75.5 + 0.6 0.6 - - - a) is less than the threshold voltage, so the transistor is cut off and I 0. b) - V TN - V and V S 0.5 V Triode region operation I K n V TN V S V S ma 0.5 0.5V 375 µa V c) - V TN - V and V S V Saturation region operation I K n V TN ) + λv S ) 0.5 ma V [ )] 50 µa ) V + 0.0 Page 63 a ) is greater than the threshold voltage, so the transistor is cut off and I 0. b) - V TN - + V and V S 0.5 V Triode region operation I K n V TN V S V S 0.4 ma + 0.5 0.5)V 50 µa V c) - V TN + V and V S V Saturation region operation I K n V TN ) + λv S ) 0.4 ma V + [ ] 08 µa ) V + 0.0 ) 3 08/09/0

Page 67 C GC 00 µf 5x0 6 m) 0.5x0 6 m) 0.500 ff m Triode region : C G C GS C GC + C W 0.5 ff + 300 pf GSO m Saturation region : C GS 3 C + C W 0.333 ff + 300 pf GC GSO m C G C GSO W 300 pf 5x0 6 m).50 ff m 5x0 6 m 5x0 6 m ).75 ff ).83 ff Page 69 KP K n 50U LAMBA λ 0.033 VTO V TN PHI φ F 0.6 W W.5U L L 0.5U Page 70 µm Circuits/cm α 0.5µm 6 Power - elay Product α 3 4 64 times improvement 3 64 Page 7 i * ε µ ox W /α n T ox /α L/α v V v S GS TN v S α i P * V i * V α i P * A * αp W /α) L/α) α P 3 A a) f T π V L 05 500cm /V s V 0 4 cm) V cm L V 05 0 4 cm cm ) 7.96 GHz b) f T ) 0 V L 0 5 V ) α P 500cm /V s π 0.5x0 4 cm ) V 0 5 cm cm ) V ) 7 GHz 4 08/08/0

Page 7 a ) From the graph for 0.5 V, I 0 8 A b) From the graph for V TN 0.5 V, I 3x0 5 A c) I 0 9 transistors Page 76 ) 3x0-5 A/transistors) 3 µa Active area 0Λ Λ ) 0Λ 0 0.5µm).88 µm L Λ 0.50 µm W 0Λ.5 µm Gate area Λ 0Λ ) 0Λ 0 0.5µm) 0.33 µm Transistor area 0Λ + Λ + Λ) Λ + Λ + Λ N 04 µm) 3.50µm 8.6 x 06 transistors ) 4Λ 4 0.5µm) 3.50 µm Page 80 Assume saturation region operation and λ 0. Then I is indpendent of V S, and I 50 µa. ) 7.50 V. V S V TN, so our assumption was correct. ) V S V I R 0 50kΩ 50µA Q - Point 50.0 µa, 7.50 V - - - V EQ 70kΩ 70kΩ + 750kΩ 0V.647 V R EQ 70kΩ 750kΩ Assume saturation region. A I 5x0 6 V.647 ) V 33.9 µa V S V I R 0 00kΩ 33.9µA) 6.6 V. V S V TN, so our assumption was correct. - - - does not change : 3.00 V I 30x0 6 Q - Point 33.9 µa, 6.6 V ) A V 3 ) V 60.0 µa ) 4.00 V. V S V TN, so our assumption was correct. ) V S V I R 0 00kΩ 60.0µA Q - Point 60.0 µa, 4.00 V 5 08/09/0

Page 8 does not change : 3.00 V I 5x0 6 V S V I R 0 00kΩ 8.µA ) Q - Point 8. µa, 7.9 V - - - ) V S 0 5x0 6 0 5 A 3.5) V 8. µa V ) 7.9 V. V S V TN, so our assumption was correct. 3 ) + 0.0V S V S 0 5.05 V 4.76 V I 5x0 6 ) V S 0 5 + 0.0V S ) [ )] 5.4 µa 3 ) + 0.0 4.76 Page 8 For I 0, V S 0 V. For V S 0, I 0V 66.7kΩ 50µA. The load line intersects the 3 - V curve at I 50 µa, V S 6.7 V. 6 08/08/0

Page 85 Upper group + V TN + V TN K n V TN K n ± K n V EQ K n Assume saturation region operation. I + K n V EQ V TN K n 0 V TN ± K n V TN + V TN V TN + V EQ V TN + K n K n V TN K n V TN + K n V EQ V TN K n + V EQ K n ) ) ) ) + 30µA) 39kΩ) 4 ) 30µA) 39kΩ) ) 5.8 V Saturation region is correct. Q - Point : 36.8 µa, 5.8 V ) V S 0 4kΩ 36.8µA Assume saturation region operation. I 5µA V S 0 4kΩ 6.7µA ) 36.8 µa ) 6.7 µa + 5µA) 39kΩ) 4.5) ) 39kΩ) ) 6.96 V Saturation region is correct. Q - Point : 6.7 µa, 6.96 V ) Assume saturation region operation. I 5µA V S 0 37kΩ 5.4µA ) 36.8 µa + 5µA) 6kΩ) 4 ) ) 6kΩ) ) 6.5 V Saturation region is correct. Q - Point : 5.4 µa, 6.5 V ) 7 08/09/0

Page 85 Lower group R V EQ MΩ V 0V 4 V Assume saturation region operation. R + R MΩ +.5MΩ ) ) + 5µA).8kΩ) 4 ) 5µA).8kΩ) ) 5.94 V Saturation region is correct. Q - Point : 99.5 µa, 5.94 V ) I + K n V EQ V TN K n V S 0 40.8kΩ 99.5µA ) 99.5 µa R V EQ.5MΩ V 0V 6 V Assume saturation region operation. R + R.5MΩ +MΩ I 5µA ) 99. µa ) kω) + 5µA) kω) 6 ) ) 6.03 V Saturation region is correct. Q - Point : 99. µa, 6.03 V ) V S 0 40kΩ 99.µA I Bias V EQ V R + R R + R V I Bias 0V µa 5 MΩ R V R R + R ) V EQ 5MΩ 4V R + R V 0V MΩ R 5MΩ R 3 MΩ R EQ R R. MΩ Page 87 ) I 5µA 6 000I V SB 000I V TN + 0.75 V SB + 0.6 0.6 Spreadsheet iteration yields I 83. µa. Page 83 [ ) +.83 ] V 0. SB Equation 4.55 becomes 6 - + 0.75 V SB + 0.6 0.6 V SB 6.065V SB + 7.3 0 V SB.63V..63V 0 4 A 6.3 kω 6 kω R 0 6.63)V 3.7kΩ 4 kω 0 4 A V TN ) 8 08/08/0

Page 89 Assume saturation region operation. ) 0 6 5x0 6 6x0 4 +) V GS + 0.70 4.6 0.46V, +.76V ) 5.4 µa V S 0 37kΩ 5.4µA) ) I 5x0 6.46 + Saturation region is correct. Q - Point : 5.4 µa, - 6.5 V [ ] 6.5 V Page 95 a ) V S 3 V, 3.5 ) +.5 V. The transistor is pinched off. I I SS V GS ma V 84 µa Pinchoff requires V S +.5 V 3.5V b) V S 6 V, 3.5) +.5 V. The transistor is pinched off. I I SS V GS ma V 50 µa Pinchoff requires V S +.5 V 3.5V c) V S 0.5 V, 3.5) +.5 V. The transistor is in the triode region. I I SS V S V S ma 3.5 Pinchoff requires V S +.5 V a) V S 0.5 V, 4 I I SS V S V S 0.mA b) V S 6 V, 4 I I SS V GS ) + 3.5 0.5 ) 0.5 5.0 µa ) + V. The transistor is in the triode region. ) + 4 0.5 0.5.9 µa 4) ) +3 V. The transistor is pinched off. 0.mA V 3 µa 4V 9 08/09/0

Page 97 a ) V S 3 V, 3 4 V. The transistor is pinched off. I I SS V GS.5mA 3V 56 µa Pinchoff requires V S V 4V b) V S 6 V, 4) 3 V. The transistor is pinched off. I I SS V GS.5mA V.4 ma Pinchoff requires V S 3 V 4V c) V S 0.5 V, 4) V. The transistor is in the triode region. I I SS ) V S V S.5mA 4 0.5 0.5 4 Pinchoff requires V S V ) 73 µa Page 98 BETA I SS BETA I SS.5mA ) 0.65 ma VTO V LAMBA λ 0.05 V - 5mA.5 ma VTO V LAMBA λ 0.0 V - 0 08/08/0

Page 00 VTO 5 V BETA I SS 5mA - 0. ma LAMBA λ 0.0 V 5 ) V G V S I G R G I S 0 I I K n V TN ) K n V TN 0.4mA - - - 5) kω 5 ) V R GS S V TN 5 + V GS 5 I SS V GS Assuming pinchoff, I.9 ma and V S 9.9mA kω +kω 3.09 V, V S >, and pinchoff is correct. - - - Assuming pinchoff, I SS V GS.5 V I I SS V GS V S.5mA kω + kω.5 5 - - - a) V G I G R G 0nA 680kΩ 5mA.5 5 for I SS K n V TN and V TN V GS 5 + 5 0 and the rest is identical. ) 3.7 V. 5x0 3 ) x0 3 ) +.5 ma 5 V GS +.5 + 5 0 ) 7.00 V. ) +.5 V, V S >, and pinchoff is correct. Q - Point :.5 ma, 7.00 V ) ) 6.80 mv. V G V S I G R G I SS V GS.00680 5x0 3 ) 0 3 ) + 5 V GS 5 + 5 0 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 I SS V GS.6 V I I SS V GS V S.54mA kω +kω ) 0.680 V and now cannot be neglected. 0.680 5x0 3 ) 0 3 ) + 5mA.6 5.54 ma ) 7.38 V Q - Point :.54 ma, 7.38 V) 5 V GS 5 + 8.4 0 08/09/0

CHAPTER 5 Page 3 a) β F α F 0.970 α F 0.970 3.3 β 0.993 F 0.993 4 β 0.50 F.50 0.333 b) α F β F β F + 40 4 0.976 α 00 F 0 0.995 α 3 F 4 0.750 Page 5 i C 0 5 A exp 0.700 exp 9.30 0 5 A exp 9.30.45 ma 0.05 0.05 0.5 0.05 i E 0 5 A exp 0.700 exp 9.30 + 0 5 A exp 0.700.46 ma 0.05 0.05 00 0.05 i B 0 5 A 00 exp 0.700 + 0 5 A exp 9.30 4.5 µa 0.05 0.5 0.05 Page 7 i C 0 6 A exp 0.750 exp 0.700 0 6 A exp 0.700 563 µa 0.05 0.05 0.4 0.05 i E 0 6 A exp 0.750 exp 0.700 + 0 6 A exp 0.750 938 µa 0.05 0.05 75 0.05 i B 0 6 A 75 exp 0.750 + 0 6 A exp 0.700 376 µa 0.05 0.4 0.05 i T 0 5 A exp 0.750 exp 0.7 ma 0.05 0.05 i T 0 6 A exp 0.750 exp 4.5.07 ma 0.05 0.05 08/08/0

Page 30 V BE V T ln I C + 0.05V ln 0 4 A I S 0 6 A + 0.69 V V BE V T ln I C + 0.05V ln 0 3 A I S 0 6 A + 0.748 V Page 3 npn :V BE > 0, V BC < 0 Forward Active Region pnp :V EB > 0, V CB > 0 Saturation Region Page 33 β F α F α F 0.95 0.05 9 β α R R 0.5 α R 0.75 3 V BE 0, V BC << 0 : I C I S + 0 6 A + 0.400 fa 0.333 I E I S 0.00 fa β R I B I S 0 6 A 0.300 fa β R 0.333 V BE << 0, V BC << 0 : I C I S β R 3x0 6 A 0.300 fa I E I S β F 0 6 A 9.0 5.6 aa I I S B I S 0 6 A β F β R 9.0 0 6 A 0.305 fa / 3 Page 35 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 + 0.05V ln 98.0µA I S 0 6 A + 0.690 V 3 08/09/0

Page 36 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 + 0.05V ln 5.00mA I S 0 6 A + 0.789 V Checking : V BC 5 + 0.789 4. Forward - active region with V CB 0 requires V CC V BE or V CC 0.764 V Page 38 a ) Resistor R is changed. 0.7V 9V ) I E 5.6kΩ V CE V C V E 9 4300I C b) I E β + F.48 ma I B I E β F + I E 5 9. µa I β I 50I C F B B.45 ma ) 0.7) 3.47 V Q - Point :.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 39 0.7V 9V I E ) 5.6kΩ 8.3V 8.4 kω 0µA.48 ma I B I E β F + I E 50 9. 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 I S x0 5 I C +) V BE + 800.0 5x0 6 )exp V BE 9 V BE 0.7079 V using a calculator solver 0.05 or spreadsheet. I C 5x0 6 exp 0.7079 99 µa V CE 9 4300I C 0.708 0.05 I S I SBJT α F x0 4 A 0.95.0 fa ) 5.44 V 4 08/08/0

Page 4 Resistor R is changed : ) 5.6kΩ 0.7V 9V I C Page 44 ) ln V CESAT 0.05V ) ln V BESAT 0.05V ) ln V BCSAT 0.05V.48 ma I B I C β R + I C 0.74 ma I β I E R B )I B 0.74 ma 0.5 + ma 40µA ) ma 50 40µA ) 99.7 mv 0.mA + 0.5)mA 0 5 A 0.694 mv + 0.5 50 0.mA ma 50 0 5 A 0.67 mv V CESAT V BESAT V BCSAT 67.7mV + 0.5 0.5 50 Page 47 a) n kt q µ 0.05V n 500cm /V s b) I S qa n n i N AB W.6x0 9 C 50µm Page 50 V T.38x0 3 J / K.60x0 9 C f β f T β F ) 373K 300 MHz 5 ).5cm / s ).40 MHz ) 0 4 cm / µm).5cm / s) 0 0 / cm 6 ) 0 8 / cm 3 ) µm) 3.mV C I C V T τ F 0 8 A aa 0 A 0.03V 4x0 9 s).4 µf 5 08/09/0

Page 5 I C 0 5 Aexp 0.7 + 0.74 ma β F 75 + 0 90.0 I B.74mA 9.3 µa 0.05 50 50 90.0 I C 0 5 Aexp 0.7.45 ma β F 75 I B.45mA 9.3 µa 0.05 75 Page 53 g m 40 V 0 4 A C 4.00mS 5ps Page 56 ) 4.00 ms g m 40 ) 40.0 ms V 0 3 A ) 0.00 pf C 40.0mS 5ps).00 pf ) V T kt q.38x0 3 300 5.9 mv IS.60x0 9 BF 80 VAF 70 V I C 350µA exp V BE exp 0.68.39 fa 0.059 Page 60 8kΩ V EQ 8kΩ + 36kΩ V 4.00 V R EQ 8kΩ 36kΩ kω 4.00 0.7 V I B + 75+)6 kω.687 µa I 75I 0 µa I 76I C B E B 04 µa V CE 000I C 6000I E 4.9 V Q - point : 0 µa, 4.9 V V T ) 80kΩ V EQ 80kΩ + 360kΩ V 4.00 V R EQ 80kΩ 360kΩ 0 kω 4.00 0.7 V I B 0 + 75+)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 - point : 85 µa, 4.93 V 6 08/08/0

Page 6 I I C 5 50I B 0I B 5 8kΩ V EQ 8kΩ + 36kΩ V 4.00 V R EQ 8kΩ 36kΩ kω 4.00 0.7 V I B + 500 +)6 kω 0.4 µa I 500I 05.6 µa I 76I C B E B 06.0 µa ) V CE 000I C 6000I E 4.8 V Q - point : 06 µa, 4.8 V Page 6 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 64 V CE 0.7 V at the edge of saturation. V - R C + 76 75 6kΩ 05µA V BESAT 4 kω 4µA V CESAT 56kΩ 60µA ) 6kΩ 84µA) 0.768 V ) 6kΩ 84µA) 0.096 V ) 0.7V R C 38.9 kω Page 65 9 0.7 V I B 36 + 50 +) kω 95.4 µa I 50I 4.77 ma I 5I C B E B 4.87 ma V CE 9 000 I C + I B ) 4.3 V Q - point : 4.77 ma, 4.3 V ) Page 66 VBE V) IC A) V'BE V) 0.70000.055E-04 0.6756 0.6756.038E-04 0.6778 0.6778.037E-04 0.6778 0.6778.037E-04 0.6778 7 08/09/0