Microelectronic circuit design 3rd edition by r jaeger solutions mmzzhh

1.1 Answering machine Alarm clock Automatic door Automatic lights ATM Automobile: Engine controller Temperature control ABS Electronic dash Navigation system Automotive tune-up equipment Baggage scanner Bar code scanner Battery charger Cable/DSL Modems and routers Calculator Camcorder Carbon monoxide detector Cash register CD and DVD players Ceiling fan (remote) Cellular phones Coffee maker Compass Copy machine Cordless phone Depth finder Digital Camera Digital watch Digital voice recorder Digital scale Digital thermometer Electronic dart board Electric guitar Electronic door bell Electronic gas pump Elevator Exercise machine Fax machine Fish finder Garage door opener GPS Hearing aid Invisible dog fences Laser pointer LCD projector Light dimmer Keyboard synthesizer Keyless entry system Laboratory instruments Metal detector Microwave oven Model airplanes MP3 player Musical greeting cards Musical tuner Pagers Personal computer Personal planner/organizer (PDA) Radar detector Broadcast Radio (AM/FM/Shortwave) Razor Satellite radio receiver Security systems Sewing machine Smoke detector Sprinkler system Stereo system Amplifier CD/DVD player Receiver Tape player Stud sensor Talking toys Telephone Telescope controller Thermostats Toy robots Traffic light controller TV receiver & remote control Variable speed appliances Blender Drill Mixer Food processor Fan Vending machines Video game controllers Wireless headphones & speakers Wireless thermometer Workstations Electromechanical Appliances* Air conditioning and heating systems Clothes washer and dryer Dish washer Electrical timer Iron, vacuum cleaner, toaster Oven, refrigerator, stove, etc *These appliances are historically based only upon on-off (bang-bang) control However, many of the high end versions of these appliances have now added sophisticated electronic control 1-1 ©R C Jaeger & T N Blalock 6/9/06 1.2 B = 19.97 x 100.1997(2020−1960) = 14.5 x 1012 = 14.5 Tb/chip 1.3 (a) 0.1977(Y2 −1960) B2 19.97x10 0.1977(Y2 −Y1 ) 0.1977(Y2 −Y1 ) = = 10 so = 10 0.1977(Y1 −1960) B1 19.97x10 Y2 − Y1 = (b) log2 = 1.52 years 0.1977 Y2 − Y1 = log10 = 5.06 years 0.1977 1.4 0.1548(2020−1970) N = 1610x10 1.5 = 8.85 x 1010 transistors/μP (2 ) N 1610x10 0.1548(Y2 −Y1 ) = = 10 0.1548(Y1 −1970) N1 1610x10 log2 (a) Y2 − Y1 = = 1.95 years 0.1548 log10 (b) Y2 − Y1 = = 6.46 years 0.1548 0.1548 Y −1970 1.6 −0.05806(2020−1970) F = 8.00x10 μm = 10 nm No, this distance corresponds to the diameter of only a few atoms Also, the wavelength of the radiation needed to expose such patterns during fabrication is represents a serious problem 1.7 From Fig 1.4, there are approximately 600 million transistors on a complex Pentium IV microprocessor in 2004 From Prob 1.4, the number of transistors/μP will be 8.85 x 1010 in 2020 Thus there will be the equivalent of 8.85x1010/6x108 = 148 Pentium IV processors 1-2 ©R C Jaeger & T N Blalock 6/9/06 1.8 ( ) P = 75x106 tubes (1.5W tube)= 113 MW! I= 1.13 x 108W = 511 kA! 220V D, D, A, A, D, A, A, D, A, D, A 1.9 1.10 10.24V 10.24V 10.24V = = 2.500 mV VMSB = = 5.120V 12 2 bits 4096bits 1001001001102 = 211 + 28 + 25 + 22 + = 234210 VO = 2342(2.500mV )= 5.855 V VLSB = 1.11 VLSB = 5V mV 5V = = 19.53 bit bits 256bits 2.77V = 142 LSB mV 19.53 bit and 14210 = (128 + + + 2) = 100011102 10 1.12 VLSB = 2.5V 2.5V mV = = 2.44 bit bits 1024 bits 10 ( 01011011012 = 28 + 26 + 25 + 23 + 22 + 20 ) 10 ⎛ 2.5V ⎞ VO = 365 ⎜ ⎟ = 0.891 V ⎝ 1024 ⎠ = 36510 1.13 ( ) mV 6.83V 14 10V = 0.6104 and bits = 11191 bits 14 10V bit bits 1119110 = (8192 + 2048 + 512 + 256 + 128 + 32 + 16 + + + 1) 10 VLSB = 1119110 = 101011101101112 1.14 A digit readout ranges from 0000 to 9999 and has a resolution of part in 10,000 The number of bits must satisfy 2B ≥ 10,000 where B is the number of bits Here B = 14 bits 1.15 5.12V mV V 5.12V = 1.25 and VO = (1011101110112 )VLSB ± LSB = 12 bit 2 bits 4096 bits 11 VO = + + + + + + + + 1.25mV ± 0.0625V VLSB = ( ) 10 VO = 3.754 ± 0.000625 or 3.753V ≤ VO ≤ 3.755V 1-3 6/9/06 1.16 IB = dc component = 0.002 A, ib = signal component = 0.002 cos (1000t) A 1.17 VGS = V, vgs = 0.5u(t-1) + 0.2 cos 2000 t Volts 1.18 vCE = [5 + cos (5000t)] V 1.19 vDS = [5 + sin (2500t) + sin (1000t)] V 1.20 V = 10 V, R1 = 22 kΩ, R2= 47 kΩ and R3 = 180 kΩ + V R I2 R2 V I + V R - V1 = 10V 22kΩ ( ) 22kΩ + 47kΩ 180kΩ = 10V 22kΩ = 3.71 V 22kΩ + 37.3kΩ 37.3kΩ = 6.29 V Checking : 6.29 + 3.71 = 10.0 V 22kΩ + 37.3kΩ ⎛ ⎞ 180kΩ 180kΩ 10V I2 = I1 =⎜ = 134 μA ⎟ 47kΩ + 180kΩ ⎝ 22kΩ + 37.3kΩ ⎠ 47kΩ + 180kΩ ⎛ ⎞ 47kΩ 47kΩ 10V I3 = I1 =⎜ = 34.9 μA ⎟ 47kΩ + 180kΩ ⎝ 22kΩ + 37.3kΩ ⎠ 47kΩ + 180kΩ V2 = 10V Checking : I1 = 10V = 169μA and I1 = I2 + I3 22kΩ + 37.3kΩ 1-4 ©R C Jaeger & T N Blalock 6/9/06 1.21 V = 18 V, R1 = 56 kΩ, R2= 33 kΩ and R3 = 11 kΩ V + R I R V I3 + V R - V1 = 18V 56kΩ ( ) 56kΩ + 33kΩ 11kΩ = 15.7 V V2 = 18V 33kΩ 11kΩ ( ) 56kΩ + 33kΩ 11kΩ = 2.31 V Checking :V1 + V2 = 15.7 + 2.31 = 18.0 V which is correct I1 = 18V ( ) 56kΩ + 33kΩ 11kΩ I3 = I1 = 280 μA I2 = I1 11kΩ 11kΩ = (280 μA) = 70.0 μA 33kΩ + 11kΩ 33kΩ + 11kΩ 33kΩ 33kΩ = (280 μA) = 210 μA 33kΩ + 11kΩ 33kΩ + 11kΩ 1.22 I1 = 5mA (5.6kΩ + 3.6kΩ) = 3.97 mA (5.6kΩ + 3.6kΩ)+ 2.4kΩ ( Checking : I2 + I3 = 280 μA I2 = 5mA 2.4kΩ = 1.03 mA 9.2kΩ + 2.4kΩ = 3.72V )5.6kΩ3.6kΩ + 3.6kΩ V3 = 5mA 2.4kΩ 9.2kΩ Checking : I1 + I2 = 5.00 mA and I2 R2 = 1.03mA(3.6kΩ)= 3.71 V 1.23 150kΩ 150kΩ = 125 μA I3 = 250μA = 125 μA 150kΩ + 150kΩ 150kΩ + 150kΩ 82kΩ V3 = 250μA 150kΩ 150kΩ = 10.3V 68kΩ + 82kΩ Checking : I1 + I2 = 250 μA and I2 R2 = 125μA(82kΩ)= 10.3 V I2 = 250μA ( ) 1-5 6/9/06 1.24 + R v v s + g v v m th - Summing currents at the output node yields: v + 002v = so v = and v th = vs − v = vs 5x10 + v - R ix g v m vx Summing currents at the output node : v − 0.002v = but v = −vx 5x10 vx v ix = + 0.002vx = Rth = x = = 495 Ω ix 5x10 + gm R1 ix = − Thévenin equivalent circuit: 495 Ω v s 1-6 ©R C Jaeger & T N Blalock 6/9/06 1.25 The Thévenin equivalent resistance is found using the same approach as Problem 1.24, and ⎛ ⎞−1 Rth = ⎜ + 025⎟ = 39.6 Ω ⎝ 4kΩ ⎠ + v R - vs g v m in The short circuit current is : v in = + 0.025v and v = vs 4kΩ v i n = s + 0.025vs = 0.0253vs 4kΩ Norton equivalent circuit: 0.0253v s 39.6 Ω 1-7 6/9/06 1.26 (a) + βi R vs v th R2 i Vth = Voc = −β i R2 - i =− but vs R1 and Vth = β vs R2 39kΩ = 120 vs = 46.8 vs R1 100kΩ ix βi R v Rth R2 x i Rth = vx ; ix ix = vx + βi R2 but i = since VR = Rth = R2 = 39 kΩ Thévenin equivalent circuit: 39 k Ω 58.5v s (b) + βi i s R R2 i v th - ⎛ i ⎞ Vth = Voc = −β i R2 where i + bi + is = and Vth = −β ⎜ − s ⎟ R2 = 38700 is ⎝ β + 1⎠ 1-8 ©R C Jaeger & T N Blalock 6/9/06 βi R Rth R2 v x i Rth = vx ; ix ix = vx + βi but R2 i + βi = so i = and Rth = R2 = 39 kΩ Thévenin equivalent circuit: 39 k Ω 38700i s 1.27 βi R vs R2 in i in = −β i but i = − vs R1 and in = β R1 vs = From problem 1.26(a), Rth = R2 = 56 kΩ 0.00133v s 100 vs = 1.33 x 10−3 vs 75kΩ Norton equivalent circuit: 56 k Ω 1-9 6/9/06 1.28 is v βi R s R2 i is = vs v v β +1 − β i = s + β s = vs R1 R1 R1 R1 R= vs R 100kΩ = = = 1.24 Ω is β + 81 1.29 The open circuit voltage is vth = −g mv R2 and v = +is R1 ( )( ) vth = −g m R1 R2is = −(0.0025) 105 106 i s= 2.5 x 108 is For is = 0, v = and Rth = R2 = MΩ 1.30 5V 3V f (Hz) 0 500 1000 1.31 2V f (kHz) 10 11 v = 4sin (20000πt )sin (2000πt )= [ ] cos(20000πt + 2000πt )+ cos(20000πt − 2000πt ) v = 2cos(22000πt )+ 2cos(18000πt ) 1.32 2∠36 o A = −5 = 2x105 ∠36 o 10 ∠0 A = 2x105 ∠A = 36o 1-10 ©R C Jaeger & T N Blalock 6/9/06 17.91 40(250µA) I1 | fT = = 159 MHz 2π (10 pF ) fT = gm 40IC1 = 2πCC 2πCC SR = MV V I1 500µA = = 50 = 50 since I2 > I1 The slew rate is symmetrical CC 10 pF µs s 17.92 SR = | IC1 = V I1 40µA V = = 8x10 = CC pF µs s *Problems 17.92 - Bipolar Op-amp VCC DC 10 VEE -10 I1 40U I2 400U I3 500U V1 DC PWL (0 5U 5.1U 10U 10.2U -5 15U -5 15.2U 20U 5) VF DC -0.0045 Q1 NBJT Q2 NBJT Q3 10 PBJT Q4 10 PBJT Q11 10 PBJT Q5 PBJT Q6 NBJT CC 5PF MODEL NBJT NPN BF=100 IS=1FA VAF=80 RB=250 TF=0.65NS CJC=2PF MODEL PBJT PNP BF=100 IS=1FA VAF=80 RB=250 TF=0.65NS CJC=2PF OP TRAN 05U 20U PROBE V(4) V(5) V(6) V(7) END Results: -8V/µs, +6V/µs 17.93 SR = V I1 100µA V = = 12.5x10 = 12.5 CC pF µs s 17-66 ©Richard C Jaeger and Travis N Blalock - 3/10/07 17.94 fT = gm 40IC1 = 2πCC 2πCC | IC = I1 | fT = 40(50µA) = 21.2 MHz 2π (15 pF ) *Problem 17.94 - Bipolar Op-amp VCC DC 10 VEE -10 I1 100U I2 500U I3 500U V1 DC 2.10M AC 0.5 V2 DC AC -0.5 Q1 NBJT Q2 NBJT Q3 10 PBJT Q4 10 PBJT Q11 10 PBJT Q5 PBJT Q6 NBJT CC 15PF MODEL NBJT NPN BF=100 IS=1FA VAF=80 RB=250 TF=0.65NS CJC=2PF MODEL PBJT PNP BF=100 IS=1FA VAF=80 RB=250 TF=0.65NS CJC=2PF OP TF V(7) V1 AC DEC 100 20MEG PRINT AC VM(7) VP(7) PROBE END Spice Results: (a) 16.2MHz (b) 16.3 MHz - 15 pF does not represent the effective value of CC ©Richard C Jaeger and Travis N Blalock - 3/10/07 17-67 17.95 (a) The SPICE results agree with those given in Ex 11.10: Ao = 67.2 dB, dB, fH = 4.5 kHz, fT = 10.5 MHz, and with a phase margin of 74o 100 -0 -100 -200 -300 1.0Hz DB(V(R3:2)) 10Hz P(V(R3:2)) 100Hz 1.0KHz 10KHz 100KHz 1.0MHz 10MHz 100MHz Frequency (b) Note, λ should be 0.02V-1 for the MOSFETs (b) The SPICE results give Ao = 86.5 dB, fT = 8.5 MHz and with a phase margin of 68.4o SR+ = 17.6 V/µsec, and SR- = -15.6 V/µsec 12V 8V 4V 0V -4V -8V -12V 0s 1us 2us 3us 4us 5us 6us 7us V(M3:d) Time 17-68 ©Richard C Jaeger and Travis N Blalock - 3/10/07 8us 9us 10us 17.96 ⎛ 2R ⎞ Vo1 =− Vo = ⎜1+ ⎟V = 2V+ ⎝ 2R ⎠ + Vo sRC G (V+ − Vo1) + sCV+ + (V+ − Vo2 )GF = Combining these yields V G and T (s) = Av1 Av = Av = o = ⎛G ⎞ ⎛ Vo1 R⎞ sC + ⎜ − GF ⎟ sRC⎜ sRC + − ⎟ ⎝2 ⎠ RF ⎠ ⎝ Av1 = ∠T ( jω o ) = → RF = 2R and T ( jω o ) = → ω o = RC 17.97 Define V1 as the output of the inverting amplifier and V2 as the output of the right - hand non - inverting amplifier ⎛ ⎛ ⎞ ⎞ ⎛ ⎞ ⎜ ⎜ ⎟ Z R sCR R R ⎟⎛ R2 ⎞ R V1 = −V2 = −V2 = −V2 | V2 = V1⎜ 1+ ⎟⎜ 1+ ⎜ ⎟ 1 ⎝ R1 ⎠ ⎟⎜⎝ R1 ⎟⎠ Z1 SCR + ⎜R+ ⎜R+ ⎟ ⎟ R+ ⎝ ⎝ sC sC ⎠ sC ⎠ 2 ⎡ ⎛ ⎛ jωCR ⎞ ⎛ R2 ⎞ ⎞ ⎛ R2 ⎞ ⎤ sCR V2⎢1+ ⎜ ⎟ ⎜1+ ⎟ ⎥ = | ⎜ ⎟ ⎜1+ ⎟ = −1 ⎝ jωCR + 1⎠ ⎝ R1 ⎠ ⎢⎣ ⎝ SCR + 1⎠ ⎝ R1 ⎠ ⎥⎦ 3[90 o − tan−1 (ωCR)]= 180 o | tan−1 (ωCR) = 30 o | ωCR = tan(30 o )= 0.5774 ⎤ 2⎡ ⎞3 ⎛ o ⎛ R2 ⎞ 2⎛ R2 ⎞ ⎢ tan(30 ) ⎥ R ωCR ⎜ ⎟ = ⎜1+ ⎟ = → = −1 = 1.83 ⎜1+ ⎟ 2 o ⎢ ⎥ ⎜ ⎟ R1 ⎝ R1 ⎠ ⎝ 1+ (ωCR) ⎠ ⎝ R1 ⎠ 1+ tan (30 ) ⎣ ⎦ 17.98 10 k Ω 10 k Ω 15 k Ω 6.2 k Ω fo = 2π (5kΩ)(500 pF ) = 63.7 kHz | vO = 3(0.7V ) = 6.85 V ⎛ 15kΩ ⎞⎛ 10kΩ ⎞ 10kΩ ⎜2 − ⎟⎜1+ ⎟− ⎝ 10kΩ ⎠⎝ 6.2kΩ ⎠ 10kΩ ©Richard C Jaeger and Travis N Blalock - 3/10/07 17-69 17.99 *Problem 17.99 - Wien-Bridge Oscillator C1 500PF IC=1 RA 5K C2 500PF RB 5K E1 1E6 R1 10K R2 15K R3 6.2K R4 10K D1 DMOD D2 DMOD MODEL DMOD D TRAN 10U 10M UIC PROBE V(1) V(2) V(3) V(4) V(5) V(6) END Results: f = 60.0 kHz, amplitude = 6.8 V 17.100 - + v 47 k Ω i 68 k Ω 0.7 V B 15 k Ω i v O + Using Eq (17.135), f o = ( ) 2π (5000) 10−9 = 18.4 kHz Using Eq (17.136), the total feedback resistance should be R1 = 12R = 60kΩ The current in R1 is I = VB = I (47kΩ) = VO V V = O = O The voltage at VB is R1 12R 60kΩ 47kΩ V V − VB VO − 0.7 − VB VO | In the diode network, I = O = O + 60kΩ 15kΩ 68kΩ 60kΩ 13 13 VO VO 60 + 60 − VO = 0.7 → V = 10.7 V O 15kΩ 68kΩ 60kΩ 68kΩ 17-70 ©Richard C Jaeger and Travis N Blalock - 3/10/07 17.101 *Problem 17.101 - Phase Shift Oscillator C1 1000PF IC=1 RA 5K C2 1000PF RB 5K C3 1000PF E1 0 1E6 R2 47K R3 15K R4 68K D1 DMOD D2 DMOD MODEL DMOD D TRAN 10U 20M UIC PROBE V(1) V(2) V(3) V(4) V(5) V(6) END Results: f = 17.5 kHz, amplitude = 11.5 V ©Richard C Jaeger and Travis N Blalock - 3/10/07 17-71 17.102 Note that the presence of rπ makes the analysis more complex than the FET case C4 is a coupling capacitor, and its impedance is neglected in the analysis C5 = C1 + Cπ However, the effect of rπ can usually be neglected in the fo calculation as shown below ⎡ ⎢s(C5 + Cµ )+ gπ + sL ⎢ − sC + g + g ⎣ ( m π) ⎤⎡ ⎤ ⎡ ⎤ ⎥⎢Vb ⎥ = ⎢0⎥ s(C2 + C5 ) + gm + gπ + GE ⎥⎦⎣Ve ⎦ ⎣0⎦ −(sC5 + gπ ) ∆ (s) = s2 [C5C2 + Cµ (C2 + C5 )]+ s[C2 gπ + Cµ (gm + gπ + GE ) + C5GE ] + gm + gπ + GE (C + C5 ) + gπ GE + sL L vb + g m (v b -v e ) v C Cπ - rπ L ve Cµ RE C2 ∆ ( jω o ) = | ω o2 = ⎤ ⎤ ⎡1 ⎡1 gm CC ⎢ + ⎥= ⎢ + ⎥ | CTC = Cµ + CTC ⎣ L rπ RE (C2 + C5 )⎦ CTC ⎣ L β o RE (C2 + C5 )⎦ C + C5 gm + gπ + GE ω oL ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ C2 C5 C2 C5 ⎥ = | ω o2 L⎢Cµ + ⎥ =1 ω o2 L⎢Cµ + + + rπ RE ⎥ gm RE ⎥ βo ⎢ ⎢ β o + 1+ 1+ gm RE + β o + 1+ 1+ gm RE + ⎢⎣ ⎢⎣ RE rπ ⎥⎦ gm RE β o ⎥⎦ ⎤ 100 pF (20 pF ) ⎡ 10mS = 16.7 pF | ω o2 = + ⎢ ⎥ (a) CTC = 16.7 pF ⎣5µH 100(1kΩ)(100 pF + 20 pF )⎦ 100 pF + 20 pF ω o2 = [2x105 + 833]→ f o = 17.5 MHz 16.7 pF 1 Note that the correction term is negligible : ω o ≅ (b) CTC = 1 LCTC + + C C C3 ω o [C2 gπ + Cµ (gm + gπ + GE ) + C5GE ]= 17-72 ©Richard C Jaeger and Travis N Blalock - 3/10/07 1 = 3.85 pF | f o ≅ = = 36.3 MHz 1 π LC π µ H 3.85 pF ( ) ( ) TC + + 100 pF 20 pF pF 1 max CTC = = 12.5 pF | f o ≅ = 20.1 MHz 1 π µ H 12.5 pF ( ) ( ) + + 100 pF 20 pF 50 pF ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ C2 C5 C5 C2 ⎥≅ ⎢ ⎥=1 + + (c ) ω o2 L⎢ gm RE ⎥ CTC ⎢ gm RE ⎥ βo ⎢β + 1+ β o 1+ g R + β + 1+ 1+ g R + m E m E ⎢⎣ o ⎢⎣ o gm RE β o ⎥⎦ gm RE β o ⎥⎦ ⎡ ⎤ ⎢ ⎥ 20 pF ⎢ 100 pF ⎥ = | MATLAB yields gm = 0.211 mS + gm (1kΩ) ⎥ 16.7 pF ⎢101+ 100 1+ gm (1kΩ) + ⎢⎣ gm (1kΩ) 100 ⎥⎦ = CTC IC = (0.211mS)(0.025V ) = 5.28 µA 17.103 Assuming the effect of rπ is negligible: (a) f o ≅ CEQ = (b) Cπ fo ≅ 2π CEQ L | CEQ = 1 + C4 C + µ | Cπ = 1 1 + C1 + Cπ C2 1 + 0.01uF 3pF + = 2π 40(5mA) − 3pF = 60.7 pF 10 π = 47.4 pF | f o ≅ 1 2π = 5.17MHz 47.4 pF (20µH ) 1 + (20 + 60.7)pF 100 pF 40(10mA) − 3pF = 124 pF | CEQ = 10 π 1 + 0.01uF 3pF + = 61.6 pF 1 1 + 20 pF + 124 pF 100 pF = 4.53MHz 61.6 pF (20µH ) ©Richard C Jaeger and Travis N Blalock - 3/10/07 17-73 17.104 CTC = 1 = = 21.1pF | CTC = 2 1 ω o L 4x10 π (3µH ) + C1 C2 ( ) | gmR ≥ C1 2I D R C → ≥ VGS − VTN C2 C2 2 Kn 1.25mA VGS − VTN ) = VGS + 4) | − ≤ VGS Suppose we pick VGS = -2 V to provide ( ( 2 2V good value of g m : VGS = −2V → I D = 0.625mA(−2 + 4) = 2.50mA | R = = 800Ω → 820 Ω 2.5mA 2(2.5mA)820Ω 2I D R C1 ≤ = = 2.05 | C1 ≤ 2.05C2 | Select C1 ≅ C2 ≅ 42 pF | Choosing C2 VGS − VTN −2 − (−4) ID = C1 = 47 pF from Appendix C, C2 = If 47pF and 39pF are used : f o = 1 − 21.1pF 47 pF = 38.3 pF which is close to 39 pF 2π (21.3 pF )(3µH ) = 19.9 MHz In order to obtain an exact frequency of oscillation, a 33-pF capacitor in parallel with a small variable capacitor could be used Note that including the FET capacitances would modify the design values 17.105 L C CGD C2 C3 C1 L C GS CGS C 1 = pF + = 31.27 pF 1 1 + + C2 C1 + C3 + CGS 50 pF 50 pF + + 10 pF 1 = = 9.00MHz fo = −5 2π LCTC 2π (10 H )(31.27x10−12 F ) CTC = CGD + gm ro ≥ 17-74 CGD C1 + C3 + CGS 50 pF + + 10 pF = = 1.20 which is easily met 50 pF C2 ©Richard C Jaeger and Travis N Blalock - 3/10/07 C 17.106 (a) CTC = CGD + fo = 1 + C2 C1 + C3 + CGS = 2π LCTC 2π max | CTC = pF + (10µH )(32.3pF ) = 32.3pF 1 + 50 pF 50 pF + pF + 10 pF = 8.87 MHz 1 = 38.4 pF | f o = = 8.12 MHz 1 2π (10µH )(38.4 pF ) + 50 pF 50 pF + 50 pF + 10 pF C +C +C 50 pF + pF + 10 pF = 1.30 and (b) gm ro ≥ GS | gm ro ≥ 50 pF C2 = pF + CTC gm ro ≥ 50 pF + 50 pF + 10 pF = 2.20 | ∴ gm ro ≥ 2.20 which is easily met 50 pF 17.107 (a) CD = C jo 1+ CD = f omin VTUNE | CD = φj 20 pF = 10.7 pF | CTC = = 40.0 pF 1 2V + 1+ 75 pF + 10.7 pF 75 pF 0.8V 20 pF = 3.92 pF | CTC = = 38.5 pF 1 20V + 1+ 75 pF + 3.92 pF 75 pF 0.8V 1 = = 7.96 MHz | f omax = = 8.11 MHz 2π (10µH )(40.0 pF ) 2π (10µH )(38.5 pF ) (b) In this circuit, R = ro : µ f = gm ro ≥ C1 + CD C2 | µf ≥ 78.9 pF = 1.05 75 pF ©Richard C Jaeger and Travis N Blalock - 3/10/07 17-75 17.108 CTC = 1 + 470 pF 220 pF The required g m R ≥ = 150 pF | f o = 2π (10µH )(150 pF ) = 4.11 MHz C2 220 pF = = 0.468 is met : C1 470 pF 2 Kn 1.25mA VGS − VTN ) = VGS + 4) and VGS = −820I D → VGS = −2.016V , I D = 2.46mA ( ( 2 2(2.46mA)(820Ω) 2I D R ≅ = 1.02 gm R = VGS − VTN −2.02 − (−4) This analysis is borne out by the SPICE simulation below VDD DC 10 R 820 C1 470PF IC=2 C2 220PF IC=0 *C1 220PF IC=2 *C2 470PF IC=0 L 10UH M1 NFET MODEL NFET NMOS VTO=-4 KP=1.25MA OP TRAN 10N 30U UIC PROBE END C 470 pF (b) For this case, the required g m R ≥ C2 = 220 pF = 2.14 is not met, ID = and the circuit fails to oscillate SPICE simulation confirms that the circuit does not oscillate 17-76 ©Richard C Jaeger and Travis N Blalock - 3/10/07 17.109 CTC = 3pF + = 30.3pF | f o = 1 2π + 50 pF + 10 pF 50 pF (10µH )(30.3pF ) = 9.15 MHz *Problem 17.109 NMOS Colpitts Oscillator VDD DC 12 LRFC 20MH C1 50PF C2 50PF L 10UH M1 0 NFET CGS 10PF CGD 4PF MODEL NFET NMOS VTO=1 KP=10MA LAMBDA=0.02 OP TRAN 50N 40U UIC PROBE END Results: f = 7.5 MHz, amplitude = 80 V peak-peak There is little to set the amplitude in this circuit, and the frequency of oscillation is significantly in error Also, µf of the transistor greatly exceeds the gain required for oscillation and the waveform at the drain is highly nonlinear The voltage at the gate is filtered by the LC network and is more sinusoidal in character A diode from ground to gate could be employed to help limit the amplitude of the oscillation 17.110 fo = 2π (10µH + 10µH )(20 pF ) = 7.96 MHz ©Richard C Jaeger and Travis N Blalock - 3/10/07 17-77 17.111 1 20 pF | CTC = | C = 220 pF | CD = | L = L1 + L2 = 20µH 1 2π LCTC V + 1+ TUNE C CD 0.8V 20 pF (a) CD = = 10.7 pF | CTC = = 10.2 pF 1 2V + 1+ 220 pF 10.7 pF 0.8V fo = fo = 2π 20µH (10.2 pF ) = 11.1 MHz 20 pF = 3.92 pF | CTC = = 3.85 pF 1 20V + 1+ 220PF 3.92 pF 0.8V L fo = = 18.1 MHz (b) µ f ≥ = 1.00 L2 2π 20µH (3.85 pF ) CD = 17.112 ωS = RQ | L= ωS LCS (a) L = 40(25000) 1 = 15.915 mH | CS = = = 15.916 fF 7 2x10 π ω S L (2x10 π ) 15.915mH 1 = 15.890 fF | f P = = 10.008 MHz 1 π 15.915mH 15.890 fF ( ) + 15.915 fF 10 pF 1 (c) CP = = 15.907 fF | f P = = 10.003 MHz 1 2π 15.915mH (15.907 fF ) + 15.915 fF 32 pF (b) CP = 17-78 ©Richard C Jaeger and Travis N Blalock - 3/10/07 17.113 C2 15 mH Cπ 470pF Cµ 5pF C1 100pF 20fF (a) CTC = 1 = 19.995 fF | f P = = 9.190 MHz 1 2π 15mH (19.995 fF ) + + 20 fF 470 pF 100 pF (b) IC = 100 CTC = fP = 40(2.14mA) − 0.7 = 2.14mA | Cπ = − pF = 49.5 pF 100kΩ + 101(1kΩ) 2π (2.5x10 Hz) 1 + 20 fF pF + = 19.996 fF 1 + 100 pF 470 pF + 49.5 pF 2π 15mH (19.996 fF ) = 9.190 MHz 17.114 max CTC = max CTC = 19.60 fF | f P = = 9.28 MHz 1 1 π 15mH 19.60 fF ( ) + + + 20 fF 1pF 100 pF 470 pF 1 = = 19.98 fF | f P = = 9.19 MHz 1 1 π 15mH 19.98 fF ( ) + + + 20 fF 35 pF 100 pF 470 pF ©Richard C Jaeger and Travis N Blalock - 3/10/07 17-79 17.115 *Problem 17.115 BJT Colpitts Crystal Oscillator VCC DC VEE DC -5 Q1 NBJT RE 1K RB 100K C1 100PF C2 470PF LC 15M CC 20FF IC=5 RC 50 MODEL NBJT NPN BF=100 VA=50 TF=1N CJC=5PF OP TRAN 2N 20U UIC PROBE END In the period of time used in the simulation results, node at the interior of the crystal oscillates vigorously, but the oscillation is not coupled well to the other nodes 17-80 ©Richard C Jaeger and Travis N Blalock - 3/10/07 [...]... 1.53 V1 + I2 R 1 R2 V I3 + R V2 3 - Let RX = R2 R3 R min X = V1max = I1 = I2min = I3 = I1 I3max = I3min = R1 = R1 + RX 47kΩ(0.9)(180kΩ)(0.9) 47kΩ(0.9)+ 180kΩ(0.9) 10(1.05) 33.5kΩ 1+ 22kΩ(1.1) V R1 + RX I2max = then V1 = V = 4.40V and I2 = I1 V1 R 1+ X R1 = 33.5kΩ R max = X V1min = R3 = R2 + R3 47kΩ(1.1)(180kΩ)(1.1) 47kΩ(1.1)+ 180kΩ(1.1) 10(0.95) 41.0kΩ 1+ 22kΩ(0.9) = 3.09V V R1 + R2 + R1 R2 R3 10(1.05)... as an acceptor impurity 2.18 Since Ge is from column IV, acceptors come from column III and donors come from column V (a) Acceptors: B, Al, Ga, In, Tl (b) Donors: N, P, As, Sb, Bi 23 2.19 (a) Germanium is from column IV and indium is from column III Thus germanium has one extra electron and will act as a donor impurity (b) Germanium is from column IV and phosphorus is from column V Thus germanium has... 47000(1.1)+ R2 = R2 + R3 = 41.0kΩ = 158 μA 180000(1.1) 22000(1.1)(47000)(1.1) = 114 μA 180000(0.9) V R1 + R3 + R1 R3 R2 10(1.05) 22000(0.9)+ 180000(0.9)+ 22000(0.9)(180000)(0.9) 10(0.95) 22000(1.1)+ 180000(1.1)+ = 43.1 μA 47000(1.1) 22000(1.1)(180000)(1.1) = 28.3 μA 47000(0.9) 1-17 6/9/06 1.54 I1 = I R2 + R3 =I R1 + R2 + R3 1+ and similarly I2 = I R1 R2 + R3 mA = 4.12 mA 2400(0.95) 1+ 1 R + R3 1+ 2 R1 I1min... as in Problem 2.5 yields T = 316.6 K 2.16 Si Si Si P B Si Si Si Si Donor electron fills acceptor vacancy No free electrons or holes (except those corresponding to ni) 2.17 (a) Gallium is from column 3 and silicon is from column 4 Thus silicon has an extra electron and will act as a donor impurity (b) Arsenic is from column 5 and silicon is from column 4 Thus silicon is deficient in one electron and... Also, i- = 0 v− − vo v + i− + − = 0 R2 R1 or vs − v o vs + =0 R2 R1 and A v = vo R = 1+ 2 vs R1 1.38 Writing a nodal equation at the inverting input terminal of the op amp gives v −v v1 − v− v2 − v− + = i− + − o R1 R2 R3 vo = − but v- = v+ = 0 and i- = 0 R3 R v1 − − 3 v2 = −0.255sin 3770t − 0.255sin10000t volts R1 R2 1-11 6/9/06 1.39 ⎛b b b ⎞ ⎛ 0 1 1⎞ ⎛ 1 0 0⎞ vO = −VREF ⎜ 1 + 2 + 3 ⎟ (a) vO = −5⎜ + +... 1.47 (a) 3000(1− 01)≤ R ≤ 3000(1+ 01) or 2970Ω ≤ R ≤ 3030Ω (b) 3000(1− 05)≤ R ≤ 3000(1+ 05) or 2850Ω ≤ R ≤ 3150Ω (c) 3000(1− 10) ≤ R ≤ 3000(1+ 10) or 2700Ω ≤ R ≤ 3300Ω ΔV ≤ 0.05V 0.05 = 0.0200 or 2.00% 2.50 1.48 Vnom = 2.5V 1.49 20000μF (1− 5)≤ C ≤ 20000μF (1+ 2) or 10000μF ≤ R ≤ 24000μF 1.50 8200(1− 0.1)≤ R ≤ 8200(1+ 0.1) or 7380Ω ≤ R ≤ 9020Ω The resistor is within the allowable range of values T= 1-15... form the n+ region following step (f) in Fig 2.17 A mask would be used to cover up the opening over the p-type region and leave the opening over the n-type silicon The masking layer for the implantation could just be photoresist Mask Ion implantation Photoresist Si02 n+ p-type silicon n-type silicon p-type silicon n-type silicon Structure following ion implantation of n-type impurity Structure after... acceptor impurities the mobilities are reduced, but the hole and electron concentrations remain unchanged See Problem 2.37 for example However, it is physically impossible to add exactly equal amounts of the two impurities 2.41 (a) For the 1 ohm-cm starting material: 1 1 6.25x1018 ρ= | μp p ≈ μpN A = = qμ p p 1.602x10−19 C (1Ω − cm) V − cm − s ( ) An iterative solution is required Using the equations in Fig... = 3.30 V 1.55 Rth = From Prob 1.24 : Rthmax = 1 gm + 1 1 0.002(0.8)+ 5x10 4 (1.2) 1 R1 = 619 Ω Rthmin = 1-18 1 1 0.002(1.2)+ 5x10 4 (0.8) = 412 Ω R C Jaeger & T N Blalock 6/9/06 1.56 For one set of 200 cases using the equations in Prob 1.53 V = 10 * (0.95 + 0.1* RAND()) R1 = 22000 * (0.9 + 0.2 * RAND()) R1 = 4700 * (0.9 + 0.2 * RAND()) R3 = 180000 * (0.9 + 0.2 * RAND()) V1 I2 I3 Min 3.23 V 116 μA 29.9... ≤ 5V (1+ 05) or 5.75V ≤ V ≤ 5.25V V = 5.30 V exceeds the maximum range, so it is out of the specification limits (b) If the meter is reading 1.5% high, then the actual voltage would be 5.30 = 5.22V which is within specifications limits Vmeter = 1.015Vact or Vact = 1.015 1.52 R 6562 − 6066 Ω = = 4.96 o ΔT 100 − 0 C = R o + TCR (ΔT)= 6066 + 4.96(27)= 6200Ω TCR = R nom 0 C 1-16 R C Jaeger & T N Blalock ... vx v ix = + 0.002vx = Rth = x = = 495 Ω ix 5x10 + gm R1 ix = − Thévenin equivalent circuit: 495 Ω v s 1-6 ©R C Jaeger & T N Blalock 6/9/06 1.25 The Thévenin equivalent resistance is found using... 39.6 Ω ⎝ 4kΩ ⎠ + v R - vs g v m in The short circuit current is : v in = + 0.025v and v = vs 4kΩ v i n = s + 0.025vs = 0.0253vs 4kΩ Norton equivalent circuit: 0.0253v s 39.6 Ω 1-7 6/9/06 1.26 (a)... Thévenin equivalent circuit: 39 k Ω 58.5v s (b) + βi i s R R2 i v th - ⎛ i ⎞ Vth = Voc = −β i R2 where i + bi + is = and Vth = −β ⎜ − s ⎟ R2 = 38700 is ⎝ β + 1⎠ 1-8 ©R C Jaeger & T N Blalock