Microelectronic Circuit Design Third Edition - Part I Solutions to Exercises

Σχετικά έγγραφα
Microelectronic Circuit Design Fifth Edition - Part I Solutions to Exercises

Microelectronic Circuit Design Fourth Edition - Part II Solutions to Exercises

( )( ) ( ) ( )( ) ( )( ) β = Chapter 5 Exercise Problems EX α So 49 β 199 EX EX EX5.4 EX5.5. (a)

Chapter 5. Exercise Solutions. Microelectronics: Circuit Analysis and Design, 4 th edition Chapter 5 EX5.1 = 1 I. = βi EX EX5.3 = = I V EX5.

CHAPTER 25 SOLVING EQUATIONS BY ITERATIVE METHODS

Srednicki Chapter 55

Inverse trigonometric functions & General Solution of Trigonometric Equations

ANSWERSHEET (TOPIC = DIFFERENTIAL CALCULUS) COLLECTION #2. h 0 h h 0 h h 0 ( ) g k = g 0 + g 1 + g g 2009 =?

Matrices and Determinants

Section 8.3 Trigonometric Equations

Ηλεκτρονική ΙΙΙ Παύλος - Πέτρος Σωτηριάδης. Επανάληψη μέρος 2 ο. Εθνικό Μετσόβιο Πολυτεχνείο Σχολή Ηλεκτρολόγων Μηχανικών - Μηχανικών Υπολογιστών

Second Order Partial Differential Equations

Homework 8 Model Solution Section

( )( ) ( )( ) 2. Chapter 3 Exercise Solutions EX3.1. Transistor biased in the saturation region

IXBH42N170 IXBT42N170

Second Order RLC Filters

Math 6 SL Probability Distributions Practice Test Mark Scheme

Surface Mount Multilayer Chip Capacitors for Commodity Solutions

Capacitors - Capacitance, Charge and Potential Difference

ECE Spring Prof. David R. Jackson ECE Dept. Notes 2

Μικροηλεκτρονική - VLSI

Example Sheet 3 Solutions

Probability and Random Processes (Part II)

Potential Dividers. 46 minutes. 46 marks. Page 1 of 11

CHAPTER 101 FOURIER SERIES FOR PERIODIC FUNCTIONS OF PERIOD

Answers to practice exercises

CHAPTER 48 APPLICATIONS OF MATRICES AND DETERMINANTS

SCOPE OF ACCREDITATION TO ISO 17025:2005

NPN Silicon RF Transistor BFQ 74

the total number of electrons passing through the lamp.

Solutions to Exercise Sheet 5

Section 9.2 Polar Equations and Graphs

Strain gauge and rosettes

Mock Exam 7. 1 Hong Kong Educational Publishing Company. Section A 1. Reference: HKDSE Math M Q2 (a) (1 + kx) n 1M + 1A = (1) =

Pg The perimeter is P = 3x The area of a triangle is. where b is the base, h is the height. In our case b = x, then the area is

IXBK64N250 IXBX64N250

3.4 SUM AND DIFFERENCE FORMULAS. NOTE: cos(α+β) cos α + cos β cos(α-β) cos α -cos β

NPN SILICON OSCILLATOR AND MIXER TRANSISTOR

Applications. 100GΩ or 1000MΩ μf whichever is less. Rated Voltage Rated Voltage Rated Voltage

Smaller. 6.3 to 100 After 1 minute's application of rated voltage at 20 C, leakage current is. not more than 0.03CV or 4 (µa), whichever is greater.

EE512: Error Control Coding

GenX3 TM 300V IGBT IXGA42N30C3 IXGH42N30C3 IXGP42N30C3 V CES = 300V I C110. = 42A V CE(sat) 1.85V t fi typ. = 65ns

Aluminum Electrolytic Capacitors

Econ 2110: Fall 2008 Suggested Solutions to Problem Set 8 questions or comments to Dan Fetter 1

= 0.927rad, t = 1.16ms

Section 8.2 Graphs of Polar Equations

1. Ηλεκτρικό μαύρο κουτί: Αισθητήρας μετατόπισης με βάση τη χωρητικότητα

HOMEWORK 4 = G. In order to plot the stress versus the stretch we define a normalized stretch:

Thin Film Chip Resistors

Areas and Lengths in Polar Coordinates

Exercises 10. Find a fundamental matrix of the given system of equations. Also find the fundamental matrix Φ(t) satisfying Φ(0) = I. 1.

SCHOOL OF MATHEMATICAL SCIENCES G11LMA Linear Mathematics Examination Solutions

Reminders: linear functions

(1) Describe the process by which mercury atoms become excited in a fluorescent tube (3)

Parametrized Surfaces

Phys460.nb Solution for the t-dependent Schrodinger s equation How did we find the solution? (not required)

EE101: Resonance in RLC circuits

Math221: HW# 1 solutions

2. THEORY OF EQUATIONS. PREVIOUS EAMCET Bits.

Aluminum Electrolytic Capacitors (Large Can Type)

CRASH COURSE IN PRECALCULUS

Section 7.6 Double and Half Angle Formulas

Areas and Lengths in Polar Coordinates

AREAS AND LENGTHS IN POLAR COORDINATES. 25. Find the area inside the larger loop and outside the smaller loop

ITU-R BT ITU-R BT ( ) ITU-T J.61 (

6.1. Dirac Equation. Hamiltonian. Dirac Eq.

w o = R 1 p. (1) R = p =. = 1

Homework 3 Solutions

Thick Film Chip Resistors

Example 1: THE ELECTRIC DIPOLE

Si Photo-transistor Chip TKA124PT

k A = [k, k]( )[a 1, a 2 ] = [ka 1,ka 2 ] 4For the division of two intervals of confidence in R +

MathCity.org Merging man and maths

Jesse Maassen and Mark Lundstrom Purdue University November 25, 2013

is like multiplying by the conversion factor of. Dividing by 2π gives you the

4.6 Autoregressive Moving Average Model ARMA(1,1)

B37631 K K 0 60

C.S. 430 Assignment 6, Sample Solutions

Answers - Worksheet A ALGEBRA PMT. 1 a = 7 b = 11 c = 1 3. e = 0.1 f = 0.3 g = 2 h = 10 i = 3 j = d = k = 3 1. = 1 or 0.5 l =

b. Use the parametrization from (a) to compute the area of S a as S a ds. Be sure to substitute for ds!

Lecture Stage Frequency Response (1/10/02) Page 210-1

Chapter 6 BLM Answers

SOLUTIONS TO MATH38181 EXTREME VALUES AND FINANCIAL RISK EXAM

9.09. # 1. Area inside the oval limaçon r = cos θ. To graph, start with θ = 0 so r = 6. Compute dr

ΕΓΚΟΛΠΙΟ e-ιπταµένου 322M e-haf 22/11/08 VERSION 1.0

Written Examination. Antennas and Propagation (AA ) April 26, 2017.

The Simply Typed Lambda Calculus

Instruction Execution Times

0CHIPSTAR MICROELECTRONICS 5.5W CS8571E CS8571E. Chipstar Micro-electronics. 470uF. 0.39uF 4 IN MODE: 0----AB CS8571 CS8571E FM AB D CS8571E

[1] P Q. Fig. 3.1

AT Surface Mount Package SOT-363 (SC-70) I I Y. Pin Connections B 1 C 1 E 1 E 2 C 2 B , 7:56 PM

ECE 308 SIGNALS AND SYSTEMS FALL 2017 Answers to selected problems on prior years examinations

SURFACE MOUNT NPN SILICON HIGH FREQUENCY TRANSISTOR

Risk! " #$%&'() *!'+,'''## -. / # $

Διπολικό Τρανζίστορ Bipolar Junction Transistor (BJT)

Operating Temperature Range ( C) ±1% (F) ± ~ 1M E-24 NRC /20 (0.05) W 25V 50V ±5% (J) Resistance Tolerance (Code)

MATH423 String Theory Solutions 4. = 0 τ = f(s). (1) dτ ds = dxµ dτ f (s) (2) dτ 2 [f (s)] 2 + dxµ. dτ f (s) (3)

Aluminum Capacitors C, Tubular, Radial Lead

TRANSISTOR. POWER TRANSISTOR Power MOSFET Trench MOSFET Power Transistor Switching Power Transistor General Purpose Power Transistor

MATHEMATICS. 1. If A and B are square matrices of order 3 such that A = -1, B =3, then 3AB = 1) -9 2) -27 3) -81 4) 81

Transcript:

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 5.1V 3.91 V V O 1 + + 6 + 10 3.95 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 1 + 1 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 0.001 0 o A v [ ] 5 65o 5000 65o o 0.001 0 5sin 000πt 65o) 1

Page 6 v s 0.5sin 000πt [ + sin 4000πt) +1.5sin 6000πt) ] The three spectral components are f 1 1000 Hz f 000 Hz f 3 3000 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 1+ 0.1) or 35.1 kω R 4.9 kω R 3.6kΩ 1+ 0.01) or 3.56 kω R 3.64 kω 39kΩ 1 0.1 3.6kΩ 1 0.01 Page 9 V S P P nom 15 4.17 mw R 1 + R 54kΩ P max 1.1x15 ) 0.95x54kΩ 5.31 mw P min 0.9x15 3.1 mw 1.05x54kΩ Page 33 R 10kΩ 1+ 10 3 55 5 o C o C 9.0 kω R 10kΩ 1+ 10 3 85 5 o C o C 10.6 kω

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 3 4.34x10 39 10 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

Page 50 n i 1.08x10 31 400 3 exp ρ 1 σ 1 1.60x10 19.3x10 1 n i 1.08x10 31 50 1.1 8.6x10 5 400) 5.40x104 / cm 6 [ 1350) +.3x10 1 ) 500) ] 3 exp ρ 1 σ 1 1.60x10 19 4.34x10 39 Page 54 n i 1.08x10 31 400 1.1 8.6x10 5 50) 1.88x10 77 / cm 6 1450 Ω cm [ 6500) + 4.34x10 39 ) 000) ] 1.69x1053 Ω cm 3 exp 1.1 8.6x10 5 400) 5.40x104 / cm 6 holes p N A N D 10 16 x10 15 8x10 15 n n i cm 3 p 5.40x104 8x10 15 6.75x10 8 electrons cm 3 n N D x10 16 electrons n n i cm 3 p 100 x10 holes 16 5.00x103 n > p n - type silicon cm 3 Page 55 Reading from the graph for N T 10 16 /cm 3, 150 cm /V-s and 400 cm /V-s. Reading from the graph for N T 10 17 /cm 3, 800 cm /V-s and 30 cm /V-s. Page 57 σ 1000 1.60x10 19 µ n n u n n 6.5x10 1 / cm 3 6.5x10 19 100 a) N T x10 16 / cm 3 b) N T 5x10 16 / cm 3 4

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 1.4 19 7x10 cm 3 N T 7x1019 cm 3 and mobilities from Fig..8 ρ 1 σ 1 1.60x10 19 1.4 Page 59 [ 100) + 7x10 19 ) 50) ] V T kt q 1.38x10 3 50 4.31 mv 1.60x10 19 V T 1.38x10 3 300 5.8 mv 1.60x10 19 V T 1.38x10 3 400 34.5mV 1.60x10 19 1.79 mω cm Page 59 D n kt q µ cm n 5.8mV 136 35.1 cm D p kt V s s q µ cm p 5.8mV 49 1.7 cm V s s dn j n qd n dx 1.60x10 19 C 0 cm 10 16 104 µm 30 A s cm 3 µm cm cm dp j p qd p dx 1.60x10 19 C 4 cm 10 16 104 µm 64 A s cm 3 µm cm cm 5

Page 79 E max 1 ε s 0 x p qn A dx qn x A p E max 1.6x10 19 C 10 17 / cm 3 CHAPTER 3 ε s E max 1 ε s qn D 0 1.13x10 5 cm) 1.13x10 8 cm) 11.7 8.854x10 14 F / cm E max 1.6x10 19 C 10 0 / cm 3 11.7 8.854x10 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 799.63x10 6 cm cm 1 x p 0.058µm 1+ x1018 0.053 µm x 10 0 n 0.058µm 1+ 100 x10 18 Page 8 T 1.60x10 19 5.8mV 1.03 1.38x10 3 91K 1 5.06x10 4 µm Page 84 i D 5x10 15 A exp 0.04 1 3.99 fa i D 5x10 15 A exp.0 1 5.00 fa 0.05 0.05 Page 86 V T ln 1+ I D 0.05V ln 1+ 40x10 6 A 0.593 V I S x10 15 A V T ln 1+ I D 0.05V ln 1+ 400x10 6 A 0.651 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 1+ 400x10 6 A 0.671 V I S x10 15 A Δ 57.6 mv Δ 59.4 mv Page 89 w d 0.113µm 1+ 10V 0.979V 0.378 µm E 10.979V max 0.378x10 4 cm 581 kv cm 6

Page 89 I S 10 fa 1+ 10V 0.8V 36.7 fa Page 9 C jo 11.7 8.854x10 14 F / cm 0.113x10 4 cm 11.5 pf C j 5V ) 4.64 pf 5V 1+ 0.979V Page 9 91.7 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 1 1 + I B D.5kΩ V B 0.6V 0V + V B 10kΩ 10kΩ 1.87 0.6 0V ) 1.87 0.6 10V 1.13 ma 10kΩ V D3 1.87 0.6 1.7 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 19.6 0.1kΩ + 5kΩ V Load Regulation 0.1kΩ 5kΩ 98.0 Ω 7

Page 117 V dc V P V on 6.3 1 7.91 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 10 113.1 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 60 0.1V 0.316 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 1+ 48.6 A 0.961 V 10 15 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 1+ 48.6 A 0.994 V 10 15 A ln 1+ 48.6 A 1.07 V 10 15 A π 6.8o 8

Page 151 i) K n ' µ n ε ox T ox 500 cm V s CHAPTER 4 3.9 8.854x10 14 F / cm) 69.1x10 6 C 5x10 7 cm V s 69.1µA V ' W µa 0µm ii) K n K n 50 1000 µa L V 1µm V K µa 60µm n 50 1000 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 -1.5 0.5V and V DS 0.1 V Triode region operation ' W K n L V V V DS GS TN V DS 5 µa 10µm 1.5 0.1 0.1V 11.3 µa V 1µm V GS - V TN 3-1.5 1.5V and V DS 0.1 V Triode region operation ' W K n L V V V DS GS TN V DS 5 µa 10µm 3 1.5 0.1 0.1V 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 + 1 50 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

Page 155 V GS - V TN 5-1 4 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 5 1 5 1) 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 5-1 4 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 1+ 0.0 10 V ma V 5 1) V 1+ 0 10 [ ] 9.60 ma [ ] 8.00 ma 4-1 3 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 1+ 0.01 5 V V [ ] 118 µa [ ] 0 µa µa 5 1) V 1+ 0.01 10 50 i) Assuming pinchoff region operation, K n 0 V TN V GS V TN + K n V + 100 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

Page 159 1.51 V V TN i) V TN V TO + γ V SB + 0.6V 0.6V V TN 1+ 0.75 1.5+ 0.6 0.6 1 V 1.84 V 1+ 0.75 0 + 0.6 0.6 1+ 0.75 3+ 0.6 0.6 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 1 0.5 0.5V 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 1+ 0.0 Page 160 a V GS is greater 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 0.4 ma +1 0.5 0.5)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 10 0.5µm ma V +1 Gate area 0Λ 0 0.5µm) 5 µm N [ ] 08 µa V 1+ 0.0 ) 30 µm L Λ 1µm W 10Λ 5µm 10 4 µm) 14Λ) 16Λ 0.5µm) 1.79 x 106 transistors 11

Page 167 C GC 00 µf 5x10 6 m) 0.5x10 6 m) 0.500 ff m Triode region : C GD 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 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 0.0133 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 5x10 6 70kΩ 10V.647 V 70kΩ + 750kΩ R A V EQ 70kΩ 750kΩ Assume saturation region..647 1) V 33.9 µa V DS V DD R D 10 100kΩ 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 10 100kΩ 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 10 100kΩ 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

Page 174 V DS 10 5x10 6 10 5 3 1) 1+ 0.01V DS V DS 10 5 1.05 V 4.76 V I 5x10 6 D V DS 10 5 1+ 0.01V DS ) [ ] 5.4 µa 3 1) 1+ 0.01 4.76 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 + 30x10 6 3.9x10 4 30x10 6.566 1 Q - Point : 36.8 µa,5.81 V V GS 1) V GS 0.91V GS 5.838 0 V GS.566V,.75V 36.8 µa V DS 10 114kΩ 36.8µA) 5.81 V ii) 4 V GS + 5x10 6 3.9x10 4 5x10 6.960 1.5 Q - Point : 6.7 µa,6.96 V V GS 1.5) V GS 0.949V GS 5.955 0 V GS.960V,.01V 6.7 µa V DS 10 114kΩ 6.7µA) 6.96 V ii) 4 V GS + 5x10 6 6.x10 4 5x10 6.46 1 Q - Point : 5.4 µa,6.5 V V GS 1) V GS 0.710V GS 4.161 0 V GS.46V, 1.716V 5.4 µa V DS 10 137kΩ 5.4µA) 6.5 V 13

Page 180 i Assume saturation. For 99.5µA, V GS 4V - 99.5µA 1.8kΩ 3.81V 99.5µA which agrees. V DS 10V 99.5µA 40.8kΩ) 5.94 V 5µA 3.81 1 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 +1.636V GS 0.8 0 V GS 3.818V 3.818 1) 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 6 000 V SB 000 V TN 1+ 0.75 V SB + 0.6 0.6 Spreadsheet iteration yields 83. µa. Page 183 [ +.83 ] V 0. SB Equation 4.55 becomes 6-1+ 0.75 V SB + 0.6 0.6 V SB 6.065V SB + 7.31 0 V SB 1.63V. R S 1.63V 10 4 A 16.3 kω 16 kω R 10 6 1.63)V D 3.7kΩ 4 kω 10 4 A Page 185 i) Using Eq. 4.58) V GS 3.3 x10 4 10kΩ 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

Page 186 4 V DS 1.8x10 6.5x10 4 4 1 V DS V DS V DS.96 mv Q - Point :. µa,.96 mv ) Checking; R on Page 189 i) 10 6 5x10 6 6x10 4 5x10 6.46 +1 Q - Point : 5.4 µa, - 6.5 V 4 V DS. µa 6 1.8x10.µA.96 mv.5x10 4 4 1) V GS +1) V GS V GS + 0.710V GS 4.161 0 V GS.46V, +1.716V [ ] 6.5 V 5.4 µa V DS 10 137kΩ 5.4µA) Page 19 i) Circuits/cm α 1µm 0.5µm 16 Power - Delay Product 1 α 1 3 4 1 64 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 105 10 4 cm) 10 V L 10 5 10 5 cm cm cm 1 V α P 17 GHz 15

CHAPTER 5 Page 13 α F 0.970 1 α F 1 0.970 3.3 β 0.993 F 1 0.993 14 β 0.50 F 1.50 0.333 α F +1 40 41 0.976 α 00 F 01 0.995 α 3 F 4 0.750 Page 15 i C 10 15 A exp 0.700 exp 9.30 10 15 A exp 9.30 1 1.45 ma 0.05 0.05 0.5 0.05 i E 10 15 A exp 0.700 exp 9.30 + 10 15 A exp 0.700 1 1.46 ma 0.05 0.05 100 0.05 i B 10 15 A 100 exp 0.700 1 + 10 15 A exp 9.30 1 14.5 µa 0.05 0.5 0.05 Page 17 i C 10 16 A exp 0.750 exp 0.700 10 16 A exp 0.700 1 563 µa 0.05 0.05 0.4 0.05 i E 10 16 A exp 0.750 exp 0.700 + 10 16 A exp 0.750 1 938 µa 0.05 0.05 75 0.05 i B 10 16 A 75 exp 0.750 1 + 10 16 A exp 0.700 1 376 µa 0.05 0.4 0.05 Page 17 i T 10 15 A exp 0.750 exp 10.7 ma 0.05 0.05 i T 10 16 A exp 0.750 5 exp 1.07 ma 0.05 0.05 16

Page 0 V T ln I C +1 0.05V ln 10 4 A I S 10 16 A +1 0.691 V V T ln I C +1 0.05V ln 10 3 A I S 10 16 A +1 0.748 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 0.95 0.05 19 β α R R 0.5 1 α R 0.75 1 3 0, V BC << 0 I C I S 1+ 1 10 16 1 A 1+ 0.400 fa 0.333 I E I S 0.100 fa β R I B I S 10 16 A 0.300 fa β R 0.333 << 0, V BC << 0 I E I S I C I S β R 3x10 16 A 0.300 fa 10 16 A 19.0 5.6 aa I I S B I S 10 16 A β R 19.0 10 16 A 0.305 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 +1 100µA 51 V T ln I C +1 0.05V ln 98.0µA I S 10 16 A +1 0.690 V 1.96 µa, I C I B 50I B 98.0 µa 17

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 +1 0.05V ln 5.00mA I S 10 16 A +1 0.789 V Checking : V BC 5 + 0.789 4.1 V c) Forward - active region with V CB 0 requires V CC or V CC 0.764 V Page 8 0.7V 9V I E 5.6kΩ V CE V C V E 9 4300I C 1.48 ma I B I E +1 9.1 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 51 50 100µA 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 +1 9.1 µa, I β I C F B 1.45 ma ii) I E β +1 F I C 51 50 I 1.0I V V ln I C C C BE T +1 0.05ln I S x10 15 I C +1) + 800 5.10x10 16 exp V BE 1 9 0.7079 V using a calculator solver or spreadsheet. 0.05 I C 5x10 16 exp 0.7079 99 µa V CE 9 4300I C 0.708) 5.44 V 0.05 Page 9 I SD I SBJT α F x10 14 A 0.95 1.0 fa Page 3 0.7V 9V I C 5.6kΩ 1.48 ma I B I C 0.741 ma, β R +1 - I β I E R B 0.741 ma 18

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 1 50 40µA) 0.1mA + 1 0.5)1mA 10 15 A 1 0.694 mv +1 0.5 50 0.1mA 1mA 50 10 15 A 1 1 0.67 mv V BE V BC 67.7mV +1 0.5 0.5 50 Page 37 1.5cm / 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 40 373K i) V T 1.38x10 3 J / K 1.60x10 19 C ii) f β f T 300MHz 15 10 4 cm / µm) 1.5cm / s) 10 0 / cm 6 ) 10 18 / cm 3 ) 1µm).40 MHz 3.mV C D I C V T τ F 10 18 A 10 A 0.03V 4x10 9 s) 1.4 µf Page 41 a) I C 10 15 Aexp 0.7 1+ 10 1.74 ma 75 1+ 10 90.0 I B 1.74mA 19.3 µa 0.05 50 50 90.0 b) I C 10 15 Aexp 0.7 1.45 ma 75 I B 1.45mA 19.3 µa 0.05 75 Page 43 g m 40 10 4 4.00 ms g m 40 10 3 ) 40.0 ms C D 4.00mS 5 ps 0.100 pf C D 40.0mS 5 ps) 1.00 pf 19

Page 49 180kΩ V EQ 180kΩ + 360kΩ 1V 4.00V R EQ 180kΩ 360kΩ 10kΩ 4.00 0.7 V I B 10 + 75+1)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 4.00 0.7 V I B 1 + 500 +1)16 kω 0.4111µA 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 10. 10 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 + 76 75 16kΩ 05µA ii) SAT 4 1kΩ 4µA) 16kΩ 184µA) 0.768 V V CESAT 1 56kΩ 160µA 16kΩ 184µA) 0.096 V Page 54 9 0.7 V I B 36 + 50 +1)1 kω 95.4µA I 50I 4.77 ma I 76I 187.7µA C B E B V CE 9 1000 I C + I B 4.13V Q - po int : 4.77 ma, 4.13 V ) R C 38.9 kω 0