Solution Manual for Microelectronic Circuit Design 5th Edition by Jaeger Full file at https://TestbankDirect.eu/ CHAPTER 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 NiCad/Lithium Ion battery chargers 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, most high end versions of these appliances have now added sophisticated electronic control 1-1 Full file at https://TestbankDirect.eu/ ©R C Jaeger & T N Blalock 5/24/14 Solution Manual for Microelectronic Circuit Design 5th Edition by Jaeger Full file at https://TestbankDirect.eu/ 1.2 N = 1327x10(2020−1970)/6.52 = 61.9 x 10 transistors/chip 1.3 N = ( 2.233x10 ) x10 (2021−2014)/10.1 = 11.0 x 10 Transistors/Chip 1.4 B = 19.97 x 10 0.1997( 2021−1960 ) = 30.3 x 1012 = 30.3 Tb/chip 1.5 (a) (b) 1.6 N 1327x10 ( ) Y −Y /6.52 = = 10 ( 1) Y1 −1970) /6.52 ( N1 1327x10 (a) Y2 −Y1 = 6.52 log = 1.96 years (b) Y2 − Y1 = 6.52 log10 = 6.52 years Y −1970 /6.52 1.7 (2 ) N ( 2.233x10 ) x10 Y −Y /10.1 = = 10 ( 1) (Y1 −2014) /10.1 N1 ( 2.233x10 ) x10 Y −2014 /10.1 (a) Y2 −Y1 = 10.1log = 3.07 years (b) Y2 − Y1 = 10.1log10 = 10.1 years 1.8 Although this distance corresponds to the diameter of only a few atoms, ITRS projections are on track to produce feature sizes in this range See the Intel website for example 1-2 Full file at https://TestbankDirect.eu/ ©R C Jaeger & T N Blalock 3/23/15 Solution Manual for Microelectronic Circuit Design 5th Edition by Jaeger Full file at https://TestbankDirect.eu/ 1.9 4.02 x 108 W P = ( 268x10 tubes) (1.5W tube) = 402 MW! I= = 1.83 MA! 220V V = ( 268x10 tubes) (80cm / tube) = 21.4x10 cm3 = 21400 m 1.10 D, D, A, A, D, A, A, D, A, D, A 1.11 VLSB = 5V 5V mV = = 19.53 bits 256bits bit and 15710 = (128 +16 + + +1)10 = 100111012 3.06V = 156.7 bits →157 LSB mV 19.53 bit 1.12 VLSB = 2.5V 2.5V mV = = 2.441 bits 1024 bits bit 10 0101100110 = ( 28 + + + 2 + 21 ) = 35810 10 2.5V VO = 358 = 0.874 V 1024 1.13 10V 10V 10V = = 2.441 mV VMSB = = 5.000 V 12 bits 4096bits 1001001010012 = 211 + 28 + + + = 234510 VO = 2345 ( 2.441mV ) = 5.724 V VLSB = 1.14 10V mV 6.89V 15 = 0.3052 and ( bits) = 22577 bits 15 bit 10V bits 2257710 = (16384 + 4096 + 2048 + 32 +16 +1)10 VLSB = 2257710 = 1011000001100012 1.15 (a) A digit readout ranges from 0000 to 2000 and has a resolution of part in 2,000 The number of bits must satisfy 2B ≥ 2,000 where B is the number of bits Here B = 11 bits (b) 2B ≥ 106 yields B = 20 bits 1.16 5.12V 5.12V mV V = = 0.3125 and VO = (10101110111010 ) VLSB ± LSB 214 bits 16384 bits bit 13 11 VO = ( + + + + + + + + ) 0.3125mV ± 0.1625mV VLSB = 10 VO = 3.49813± 0.0001625 or 3.49798V ≤ VO ≤ 3.49829V 1-3 Full file at https://TestbankDirect.eu/ ©R C Jaeger & T N Blalock 3/23/15 Solution Manual for Microelectronic Circuit Design 5th Edition by Jaeger Full file at https://TestbankDirect.eu/ 1.17 IB = dc component = 7.50 mA, ib = signal component = 0.003 cos (1000t) A 1.18 VGS = 2.5 V, vgs = 0.5u(t-1) + 0.1 cos 2000πt Volts 1.19 vCE = [5 + cos (5000t)] V 1.20 vDS = [5 + sin (2500t) + sin (1000t)] V 1.21 V = V, R1 = 24 kΩ, R2= 30 kΩ and R3 = 11 kΩ 30kΩ 11kΩ 24kΩ V1 = 1V = 0.749 V V2 = 1V = 0.251 V 24kΩ + ( 30kΩ 11kΩ) 24kΩ + ( 30kΩ 11kΩ) Checking:V1 +V2 = 0.749 + 0.251 = 1.00 V which is correct I1 = 1V = 31.2 μ A 24kΩ + ( 30kΩ 11kΩ) I = I1 I = I1 R3 11kΩ = ( 31.2μ A) = 8.37 μ A R2 + R3 30kΩ +11kΩ R2 30kΩ = (31.2μ A ) = 22.8 μ A R2 + R3 30kΩ +11kΩ 1-4 Full file at https://TestbankDirect.eu/ Checking: I + I = 31.2 μ A ©R C Jaeger & T N Blalock 3/23/15 Solution Manual for Microelectronic Circuit Design 5th Edition by Jaeger Full file at https://TestbankDirect.eu/ 1.22 V = V, R1 = 30 kΩ, R2= 24 kΩ and R3 = 15 kΩ 30kΩ 30kΩ V1 = 8V = 8V = 6.12 V 30kΩ + ( 24kΩ 15kΩ) 30kΩ + 9.23kΩ V2 = 8V 24kΩ 15kΩ 30kΩ + ( 24kΩ 15kΩ) = 1.88 V Checking: 6.12+1.88 = 8.00 V 15kΩ R3 8V = = 78.4 μ A R2 + R3 30kΩ + 9.23kΩ 24kΩ +15kΩ R2 8V 24kΩ = = 125 μ A I = I1 R2 + R3 30kΩ + 9.23kΩ 24kΩ +15kΩ I = I1 Checking: I1 = 8V = 204 μ A and I1 =I + I 30kΩ + 9.23kΩ 1.23 150kΩ 150kΩ I = 200μ A I = 200μ A = 100 μ A = 100 μ A 150kΩ +150kΩ 150kΩ +150kΩ 82kΩ V3 = 200μ A (150kΩ 150kΩ) = 8.2V 68kΩ + 82kΩ Checking: I1 + I = 200 μ A 1.24 I1 = 4mA and I R2 = 100μ A (82kΩ) = 8.2 V (3.9kΩ + 5.6kΩ) = 3.19 mA (3.9kΩ + 5.6kΩ) + 2.4kΩ I = 4mA 2.4kΩ = 0.807 mA 9.5kΩ + 2.4kΩ 5.6kΩ = 4.52 V 3.9kΩ + 5.6kΩ Checking: I1 + I = 4.00 mA and I R3 = 0.807mA ( 5.6kΩ) = 4.52 V V3 = 4mA ( 2.4kΩ 9.5kΩ) 1-5 Full file at https://TestbankDirect.eu/ ©R C Jaeger & T N Blalock 3/23/15 Solution Manual for Microelectronic Circuit Design 5th Edition by Jaeger Full file at https://TestbankDirect.eu/ 1.25 Summing currents at the open circuited output node yields: v +.025v = so v = and v th = vi − v = vi 10 To find the Thévenin equivalent resistance, we apply a test source to the output with vi set to zero: 39.8 Ω vi Thévenin equivalent circuit: Summing currents at the output node: v ix = − − gm v = but v = −vx R1 v v 1 ix = x + gm vx = Rth = x = = = 39.8 Ω 1 R1 ix + gm + 0.025S R1 10kΩ 1-6 Full file at https://TestbankDirect.eu/ ©R C Jaeger & T N Blalock 3/23/15 Solution Manual for Microelectronic Circuit Design 5th Edition by Jaeger Full file at https://TestbankDirect.eu/ 1.26 -3 2.01 x 10 vi 467 Ω Norton equivalent circuit: The short circuit current is: v v + 0.002v and v = vi → in = i + 0.002vi = 2.01x10−3 vi 75kΩ 75kΩ To find the Thévenin equivalent resistance, we apply a test source to the output with vi set to zero: in = Summing currents at the output node: v ix = − − gm v = but v = −vx R1 v v 1 ix = x + gm vx = Rth = x = = = 467 Ω 1 R1 ix + gm + 0.002S R1 75kΩ 1-7 Full file at https://TestbankDirect.eu/ ©R C Jaeger & T N Blalock 3/23/15 Solution Manual for Microelectronic Circuit Design 5th Edition by Jaeger Full file at https://TestbankDirect.eu/ 1.27 (a) vi R1 and in = vx + βi R2 but in = −β i but i = − Rth = vx ; ix ix = β R1 vi = 150 vi = 3.85 x 10 −3 vi 39kΩ i = since vR1 = -3 3.85 x 10 vi Rth = R2 = 100 kΩ 100 kΩ Noton equivalent circuit: 100 ii 99300 (b) β 150 vth = voc = −β iR2 where i + β i + ii = and vth = R2 ii = 99300ii ii = 100kΩ 151 β +1 Rth is found in part (a) 1-8 Full file at https://TestbankDirect.eu/ ©R C Jaeger & T N Blalock 3/23/15 Solution Manual for Microelectronic Circuit Design 5th Edition by Jaeger Full file at https://TestbankDirect.eu/ 1.28 (a) vth = voc = −β i R2 Rth = vx ; ix ix = but i =− vx + βi R2 but vi R1 and vth = β vi i = since v R1 = R2 56kΩ = 120 vi = 89.6 vi R1 75kΩ Rth = R2 = 75 kΩ 75 kΩ 89.6 vi Thévenin equivalent circuit: (b) ii β 120 vth = voc = −β i R2 where i + β i + ii = and vth = R2 ii = 74400 ii ii = 75kΩ 121 β +1 1-9 Full file at https://TestbankDirect.eu/ ©R C Jaeger & T N Blalock 3/23/15 Solution Manual for Microelectronic Circuit Design 5th Edition by Jaeger Full file at https://TestbankDirect.eu/ Rth = vx ; ix ix = vx + βi R2 but i + β i = so i = and Rth = R2 = 75 kΩ 75 kΩ 74400 ii Thévenin equivalent circuit: 1.29 (a) vi v v β +1 − β i = i + β i = vi R1 R1 R1 R1 (b) Source is ii in part (b) (b) (a) ii = R= vi R 100kΩ = = = 1.32 kΩ ii β +1 76 vi = −iR1 and ii = −i − β i = − ( β +1) i R= vi −1 100kΩ =− R1 = = 1.32 kΩ ii β +1 76 1.30 The open circuit voltage is vth = −gm v R2 where v = +ii R1 vth = −gm R1 R2ii = − ( 0.0025) ( 2x10 ) ( 2x10 ) ii = 1.0 x 10 ii For ii = 0, v = 0, and Rth = R2 = MΩ 1-10 Full file at https://TestbankDirect.eu/ ©R C Jaeger & T N Blalock 3/23/15 Solution Manual for Microelectronic Circuit Design 5th Edition by Jaeger Full file at https://TestbankDirect.eu/ 1.31 ( a) RAB = 10kΩ +10kΩ 10kΩ + (10kΩ 10kΩ) = 16 kΩ ( b) RCD = 10kΩ 10kΩ + (10kΩ 10kΩ) = kΩ ( c) REF = 10kΩ 10kΩ (10kΩ +10kΩ) = kΩ ( d ) Terminals B & D are the same as E & F RBD = kΩ 1.32 1.33 1.34 (a) If the 36 kΩ resistor was shorted, or the 82 kΩ resistor was open, then the output voltage would be If the 82 kΩ resistor was shorted, the output would be 18 V (unless the 36 kΩ resistor was also shorted (b) If the 68 kΩ resistor was shorted, or the 27 kΩ resistor was open, then the output voltage would be +9 V If the 27 kΩ resistor was shorted, or the 68 1-11 Full file at https://TestbankDirect.eu/ ©R C Jaeger & T N Blalock 3/23/15 Solution Manual for Microelectronic Circuit Design 5th Edition by Jaeger Full file at https://TestbankDirect.eu/ kΩ resistor was open, the output would be -9 V Otherwise the voltage would be between –9 and +9 volts 1.35 v 5V 3V f (Hz) 1000 2000 1.36 1.37 4∠56 o = 4x10 ∠56 o −4 10 ∠0 A= A = 4x10 ∠A = 56 o 1.38 10 −1∠ −12 o (a) A = = 50∠ −12 o A = 50 ∠A = -12 o −3 o 2x10 ∠0 (b) 1.39 (a) Av = − R2 560kΩ =− = −46.7 R1 12kΩ (b) Av = − 360kΩ = −20.0 18kΩ 1-12 Full file at https://TestbankDirect.eu/ (c) Av = − 62kΩ = −31.0 2kΩ ©R C Jaeger & T N Blalock 3/23/15 Solution Manual for Microelectronic Circuit Design 5th Edition by Jaeger Full file at https://TestbankDirect.eu/ 1.40 R2 7500 vi ( t ) = − ( 0.01sin 750π t ) = (−82.4 sin 750π t ) mV 910 R1 v 0.01V = 11.0μ A and ii ( t ) = (11.0 sin 750π t ) μ A ii = i = R1 910Ω vo ( t ) = − 1.41 Since the voltage across the op amp input terminals must be zero, v- = v+ and vo = vi Therefore Av = 1.42 Since the voltage across the op amp input terminals must be zero, v- = v+ = vi Also, i- = 1.43 Writing a nodal equation at the inverting input terminal of the op amp gives v1 − v− v2 − v− v −v + = i− + − o but v- = v+ = and i- = R1 R2 R3 R R vo = − v1 − − v2 = (−0.255sin3770t − 0.250sin10000t ) volts R1 R2 1-13 Full file at https://TestbankDirect.eu/ ©R C Jaeger & T N Blalock 3/23/15 Solution Manual for Microelectronic Circuit Design 5th Edition by Jaeger Full file at https://TestbankDirect.eu/ 1.44 b1b2b3 vO (V) 000 001 -0.625 010 -1.250 011 -1.875 100 -2.500 101 -3.125 110 -3.750 111 -4.375 1.45 Low-pass amplifier 1.46 Band-pass amplifier 1-14 Full file at https://TestbankDirect.eu/ ©R C Jaeger & T N Blalock 3/23/15 Solution Manual for Microelectronic Circuit Design 5th Edition by Jaeger Full file at https://TestbankDirect.eu/ 1.47 High-pass amplifier 1.48 Refers to Prob 1.45 1.49 1.50 The gain is zero at each frequency: vo(t) = 1.51 t=linspace(0,.005,1000); w=2*pi*1000; v=(4/pi)*(sin(w*t)+sin(3*w*t)/3+sin(5*w*t)/5); v1=5*v; v2=5*(4/pi)*sin(w*t); v3=(4/pi)*(5*sin(w*t)+3*sin(3*w*t)/3+sin(5*w*t)/5); plot(t,v) plot(t,v1) plot(t,v2) plot(t,v3) 1-15 Full file at https://TestbankDirect.eu/ ©R C Jaeger & T N Blalock 3/23/15 Solution Manual for Microelectronic Circuit Design 5th Edition by Jaeger Full file at https://TestbankDirect.eu/ -1 -2 x10-3 (a) 10 -5 -10 x10-3 (b) 10 -5 -10 (c) 1-16 Full file at https://TestbankDirect.eu/ x10-3 ©R C Jaeger & T N Blalock 3/23/15 Solution Manual for Microelectronic Circuit Design 5th Edition by Jaeger Full file at https://TestbankDirect.eu/ 10 -5 -10 x10-3 (d) 1.52 (a) 4700 (1−.01) ≤ R ≤ 4700 (1+.01) or 4650Ω ≤ R ≤ 4750Ω (b) 4700 (1−.05) ≤ R ≤ 4700 (1+.05) or 4460Ω ≤ R ≤ 4940Ω (c) 4700 (1−.10) ≤ R ≤ 4700 (1+.10) or 4230Ω ≤ R ≤ 5170Ω 1.53 1.54 1.55 Yes, the resistor is within the allowable range of values 1.56 (a) 5V (1−.05) ≤ V ≤ 5V (1+.05) or 4.75V ≤ V ≤ 5.25V V = 5.30 V exceeds the maximum range, so it is out of the specification limits (b) However, if the meter is reading 1.5% high, then the actual voltage would be 1.57 1-17 Full file at https://TestbankDirect.eu/ ©R C Jaeger & T N Blalock 3/23/15 Solution Manual for Microelectronic Circuit Design 5th Edition by Jaeger Full file at https://TestbankDirect.eu/ 1.58 At 30 o C, 7500Ω (1− 0.05) ≤ R ≤ 7500Ω (1+ 0.05) or 7120 Ω ≤ R ≤ 7880 Ω Adding the effect of TC for ΔT = 45o C: 2200 R = 7120Ω 1+ 45 = 7820 Ω R max = 7780Ω1+ 45 ( 2.2x10 −3 ) = 8550 Ω 10 7820 Ω ≤ R ≤ 8550 Ω with accumulated rounding 7830 Ω ≤ R ≤ 8650 Ω more exact calculation 1.59 I = 200 μA, R1 = 150 kΩ, R2 = 68 kΩ and R3 = 82 kΩ I1 = I R2 + R3 =I R1 R1 + R2 + R3 1+ R2 + R3 I1max = I 2max = 200 (1.02 ) μ A = 112 μ A 150kΩ ( 0.90 ) 1+ 68kΩ (1.1) + 82kΩ (1.1) I1min = 200 (1.02) μ A = 112 μ A 68kΩ ( 0.90) + 82kΩ ( 0.90 ) 1+ 150kΩ (1.1) V3 = I R3 = V3max = V3min = and similarly I = I R + R3 1+ R1 200 ( 0.98) μ A = 88.2 μ A 150kΩ (1.1) 1+ 68kΩ ( 0.90 ) + 82kΩ ( 0.90 ) I 2min = 200 ( 0.98) μ A = 88.2 μ A 68kΩ (1.1) + 82kΩ (1.1) 1+ 150kΩ ( 0.90) I 1 R + + R1 R3 R1 R3 200μ A (1.02) 68kΩ ( 0.9 ) 1 + + 150kΩ (1.1) 82kΩ (1.1) 150kΩ (1.1) (82kΩ) (1.1) 200μ A ( 0.98) 9.60 V 68kΩ (1.1) 1 + + 150kΩ ( 0.9) 82kΩ ( 0.9 ) 150kΩ ( 0.9 ) (82kΩ) ( 0.9 ) 1-18 Full file at https://TestbankDirect.eu/ = 6.89 V ©R C Jaeger & T N Blalock 3/23/15 Solution Manual for Microelectronic Circuit Design 5th Edition by Jaeger Full file at https://TestbankDirect.eu/ V = V, R1 = 24 kΩ, R2 = 30 kΩ and R3 = 11 kΩ R1 V1 Let RX = R2 R3 then V1 = V = R1 + RX 1+ RX R1 1.60 R X = V1max = I1 = 30kΩ ( 0.9) (11kΩ) ( 0.9) = 7.24kΩ 30kΩ ( 0.9) +11kΩ ( 0.9) 1(1.05) = 0.824 V 7.24kΩ 1+ 24kΩ (1.1) V R1 + RX I 2max = I 2min = I = I1 I 3max = I 3min = and I = I1 1(1.05) 24kΩ ( 0.9 ) + 7.24kΩ V1min = R max X = 30kΩ (1.1) (11kΩ) (1.1) = 8.85kΩ 30kΩ (1.1) +11kΩ (1.1) 1( 0.95) = 0.674 V 8.85kΩ 1+ 24kΩ ( 0.9) V R3 = R2 + R3 R1 + RX 1+ R2 R3 = 11.3 μ A 30kΩ) ( 0.9 ) ( 1+ 11kΩ (1.1) 1( 0.95) = 6.22 μ A 24kΩ (1.1) + 8.85kΩ ( 30kΩ) (1.1) 1+ 11kΩ ( 0.9 ) R2 I = R2 + R3 1+ R3 R2 1(1.05) 24kΩ ( 0.9 ) + 7.24kΩ = 28.0 μ A 11kΩ) ( 0.9 ) ( 1+ 30kΩ (1.1) 1( 0.95) = 18.6 μ A 24kΩ (1.1) + 8.85kΩ (11kΩ) (1.1) 1+ 30kΩ ( 0.9 ) 1.61 Rth = From Prob 1.24: Rthmax = gm + R1 1 0.002 ( 0.9 ) + 7.5x10 (1.2 ) = 552 Ω Rthmin = 1-19 Full file at https://TestbankDirect.eu/ 1 0.002 (1.1) + 7.5x10 ( 0.8) = 451 Ω ©R C Jaeger & T N Blalock 3/23/15 Solution Manual for Microelectronic Circuit Design 5th Edition by Jaeger Full file at https://TestbankDirect.eu/ 1.62 For one set of 200 cases using the Equations in Prob 1.59 (mA & kΩ): I = 0.200 * ( 0.98 + 0.04 * RAND()) R2 = 68* ( 0.9 + 0.2 * RAND()) R1 = 150 * ( 0.9 + 0.2 * RAND()) R3 = 82 * ( 0.9 + 0.2 * RAND()) I1 I2 V3 Min 89.9 μA 91.2 μA 7.34 V Max 110 μA 109 μA 9.23 V Average 100 μA 99.8 μA 8.23 V 1.63 For one set of 200 cases using the equations in Prob 1.60 V = 1* ( 0.95 + 0.1* RAND()) R1 = 24000 * ( 0.9 + 0.2 * RAND()) R2 = 30000 * ( 0.9 + 0.2 * RAND()) R3 = 11000 * ( 0.9 + 0.2 * RAND()) V1 I2 I3 Min 0.685 V 6.70 μA 19.7 μA Max 0.814V 10.1 μA 27.1 μA Average 0.754 V 8.49 μA 22.9 μA 1.64 3.29, 0.995, -6.16; 3.295, 0.9952, -6.155 1.65 (a) (1.763 mA)(20.70 kΩ) = 36.5 V (b) 36 V (c) (0.1021 A)(97.80 kΩ) = 9.99 V; 10 V 1-20 Full file at https://TestbankDirect.eu/ ©R C Jaeger & T N Blalock 3/23/15 ... the Intel website for example 1-2 Full file at https://TestbankDirect.eu/ ©R C Jaeger & T N Blalock 3/23/15 Solution Manual for Microelectronic Circuit Design 5th Edition by Jaeger Full file at... https://TestbankDirect.eu/ ©R C Jaeger & T N Blalock 3/23/15 Solution Manual for Microelectronic Circuit Design 5th Edition by Jaeger Full file at https://TestbankDirect.eu/ 1.25 Summing currents at the open circuited... C Jaeger & T N Blalock 3/23/15 Solution Manual for Microelectronic Circuit Design 5th Edition by Jaeger Full file at https://TestbankDirect.eu/ 1.26 -3 2.01 x 10 vi 467 Ω Norton equivalent circuit: