Baixe o app para aproveitar ainda mais
Esta é uma pré-visualização de arquivo. Entre para ver o arquivo original
Solutions Manual - Microelectronic Circuit Design - 4th Ed By Richard C. Jaeger, Travis N. Blalock - McGraw-Hill (2010) NOTE: these answers are for the International Edition (?) But they’re still very similar to the original (sometimes a, b, c, d answers will be switched around, and some numbers may be a little off. As a general rule, subtract 3 from the answer you are looking for and that should be the real one) Special thanks to Moser from NIU for the main files February 1, 2012 Go to this website for book updates / corrections by the publisher: http://www.jaegerblalock.com/ 1-1 ©R. C. Jaeger & T. N. Blalock 6/9/06 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. 1.2 B =19.97 x 100.1997 2020−1960( ) =14.5 x 1012 =14.5 Tb/chip 1.3 (a) B2 B1 = 19.97x10 0.1977 Y2 −1960( ) 19.97x100.1977 Y1 −1960( ) =100.1977 Y2 −Y1( ) so 2 =100.1977 Y2 −Y1( ) Y2 −Y1 = log20.1977 =1.52 years (b) Y2 −Y1 = log100.1977 = 5.06 years 1.4 N =1610x100.1548 2020−1970( ) = 8.85 x 1010 transistors/μP 1.5 N2 N1 = 1610x10 0.1548 Y2 −1970( ) 1610x100.1548 Y1 −1970( ) =100.1548 Y2 −Y1( ) (a) Y2 −Y1 = log20.1548 =1.95 years (b) Y2 −Y1 = log100.1548 = 6.46 years 1.6 . F = 8.00x10−0.05806 2020−1970( )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 1-3 6/9/06 1.8 P = 75x106 tubes( )1.5W tube( )=113 MW! I = 1.13 x 108W220V = 511 kA! 1.9 D, D, A, A, D, A, A, D, A, D, A 1.10 VLSB = 10.24V212 bits = 10.24V 4096bits = 2.500 mV VMSB = 10.24V2 = 5.120V 1001001001102 = 211 + 28 + 25 + 22 + 2 = 234210 VO = 2342 2.500mV( )= 5.855 V 1.11 VLSB = 5V28 bits = 5V 256bits =19.53 mV bit and 2.77V 19.53 mV bit =142 LSB 14210 = 128 + 8 + 4 + 2( ) =100011102 10 1.12 VLSB = 2.5V210 bits = 2.5V 1024 bits = 2.44 mV bit 01011011012 = 28 + 26 + 25 + 23 + 22 + 20( )10 = 36510 VO = 365 2.5V1024⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.891 V 1.13 VLSB = 10V214 bits = 0.6104 mV bit and 6.83V 10V 214 bits( )=11191 bits 1119110 = 8192 + 2048 + 512 + 256 +128 + 32 +16 + 4 + 2 +1( )10 1119110 =101011101101112 1.14 A 4 digit readout ranges from 0000 to 9999 and has a resolution of 1 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 VLSB = 5.12V212 bits = 5.12V 4096 bits =1.25 mV bit and VO = 1011101110112( )VLSB ± VLSB2 VO = 211 + 29 + 28 + 27 + 25 + 24 + 23 + 2 +1( )101.25mV ± 0.0625V VO = 3.754 ± 0.000625 or 3.753V ≤ VO ≤ 3.755V 1.16 IB = dc component = 0.002 A, ib = signal component = 0.002 cos (1000t) A 1.17 VGS = 4 V, vgs = 0.5u(t-1) + 0.2 cos 2000�t Volts 1.18 vCE = [5 + 2 cos (5000t)] V 1.19 vDS = [5 + 2 sin (2500t) + 4 sin (1000t)] V 1.20 V = 10 V, R1 = 22 kΩ, R2= 47 kΩ and R3 = 180 kΩ. V + - V 1 V 2 + - R 1 R 2 R3 I3 I2 V1 =10V 22kΩ22kΩ + 47kΩ 180kΩ( )=10V 22kΩ 22kΩ + 37.3kΩ = 3.71 V V2 =10V 37.3kΩ22kΩ + 37.3kΩ = 6.29 V Checking : 6.29 + 3.71 = 10.0 V I2 = I1 180kΩ47kΩ +180kΩ = 10V 22kΩ + 37.3kΩ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 180kΩ 47kΩ +180kΩ =134 μA I3 = I1 47kΩ47kΩ +180kΩ = 10V 22kΩ + 37.3kΩ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 47kΩ 47kΩ +180kΩ = 34.9 μA Checking : I1 = 10V22kΩ + 37.3kΩ =169μA and I1 = I2 + I3 1-4 ©R. C. Jaeger & T. N. Blalock 6/9/06 1-5 6/9/06 1.21 V = 18 V, R1 = 56 kΩ, R2= 33 kΩ and R3 = 11 kΩ. V + - V 1 V 2 + - R 1 R 2 R3 I3 I2 V1 =18V 56kΩ56kΩ + 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 = 18V56kΩ + 33kΩ 11kΩ( )= 280 μA I2 = I1 11kΩ 33kΩ +11kΩ = 280 μA( ) 11kΩ33kΩ +11kΩ = 70.0 μA I3 = I1 33kΩ33kΩ +11kΩ = 280 μA( ) 33kΩ33kΩ +11kΩ = 210 μA Checking : I2 + I3 = 280 μA 1.22 I1 = 5mA 5.6kΩ + 3.6kΩ( ) 5.6kΩ + 3.6kΩ( )+ 2.4kΩ = 3.97 mA I2 = 5mA 2.4kΩ 9.2kΩ + 2.4kΩ =1.03 mA V3 = 5mA 2.4kΩ 9.2kΩ( ) 3.6kΩ5.6kΩ + 3.6kΩ = 3.72V Checking : I1 + I2 = 5.00 mA and I2R2 =1.03mA 3.6kΩ( )= 3.71 V 1.23 I2 = 250μA 150kΩ150kΩ +150kΩ =125 μA I3 = 250μA 150kΩ 150kΩ +150kΩ =125 μA V3 = 250μA 150kΩ 150kΩ( ) 82kΩ68kΩ + 82kΩ =10.3V Checking : I1 + I2 = 250 μA and I2R2 =125μA 82kΩ( )=10.3 V 1.24 1-6 ©R. C. Jaeger & T. N. Blalock 6/9/06 R 1 + - v gmv vs vth + - Summing currents at the output node yields: v 5x104 + .002v = 0 so v = 0 and vth = vs − v = vs R 1 vx + - v g v m ix Summing currents at the output node : ix = − v5x104 − 0.002v = 0 but v = −vx ix = vx5x104 + 0.002vx = 0 Rth = vx ix = 1 1 R1 + gm = 495 Ω Thévenin equivalent circuit: vs 495 Ω 1-7 6/9/06 1.25 The Thévenin equivalent resistance is found using the same approach as Problem 1.24, and Rth = 14kΩ + .025 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −1 = 39.6 Ω R 1 vs + - v gmv in The short circuit current is : in = v4kΩ + 0.025v and v = vs in = vs4kΩ + 0.025vs = 0.0253vs Norton equivalent circuit: 39.6 Ω0.0253v s 1.26 1-8 ©R. C. Jaeger & T. N. Blalock 6/9/06 (a) R 1 R2 βi vs i + - vth Vth = Voc = −β i R2 but i = − vsR1 and Vth = β vs R2 R1 =120 vs 39kΩ100kΩ = 46.8 vs R 1 R2 βi i Rth vx ix Rth = vxix ; ix = vx R2 + βi but i = 0 since VR1 = 0. Rth = R2 = 39 kΩ. Thévenin equivalent circuit: 58.5v s 39 k Ω (b) R 1 R2 βi is i + - vth Vth = Voc = −β i R2 where i + bi + is = 0 and Vth = −β − isβ +1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ R2 = 38700 is 1-9 6/9/06 R 1 R2 βi i Rth vx Rth = vxix ; ix = vx R2 + βi but i + βi = 0 so i = 0 and Rth = R2 = 39 kΩ Thévenin equivalent circuit: 39 k Ω 38700i s 1.27 βi R 1 R2vs i in in = −β i but i = − vsR1 and in = β R1 vs = 10075kΩ vs =1.33 x 10 −3 vs From problem 1.26(a), Rth = R2 = 56 kΩ. Norton equivalent circuit: 56 k Ω0.00133v s 1.28 1-10 ©R. C. Jaeger & T. N. Blalock 6/9/06 R 1 R2 βi vs i is is = vsR1 − β i = vs R1 + β vs R1 = vs β +1 R1 R = vs is = R1β +1 = 100kΩ 81 =1.24 Ω 1.29 The open circuit voltage is vth = −gmv R2 and v = +isR1. vth = −gmR1R2is = − 0.0025( )105( )106( )i s= 2.5 x 108is For i = 0, v = 0 and R = R =1 MΩs th 2 1.30 5 V 3 V 0 f (Hz) 500 10000 1.31 2 V 0 f (kHz) 9 10 11 v = 4sin 20000πt( )sin 2000πt( )= 42 cos 20000πt + 2000πt( )+ cos 20000πt − 2000πt( )[ ] v = 2cos 22000πt( )+ 2cos 18000πt( ) 1.32 A = 2∠36 o 10−5∠00 = 2x10 5∠36o A = 2x105 ∠A = 36o 1-11 6/9/06 1.33 (a) A = 10 −2∠ − 45o 2x10−3∠0o = 5∠ − 45 o (b) A = 10 −1∠ −12o 10−3∠0o =100∠ −12 o 1.34 (a) Av = − R2R1 = − 620kΩ 14kΩ = −44.3 (b) Av = − 180k = −10.0 (c) Av = − 62kΩ1.6kΩ Ω = −38.8 18kΩ 1.35 vo t( )= − R2R1 vs t( )= −90.1 sin 750πt ( ) mV IS = VSR1 = 0.01V 910Ω =11.0μA and is = 11.0 sin 750πt( ) μA 1.36 Since the voltage across the op amp input terminals must be zero, v- = v+ and vo = vs. Therefore Av = 1. 1.37 Since the voltage across the op amp input terminals must be zero, v- = v+ = vs. Also, i- = 0. v− − vo R2 + i− + v−R1 = 0 or vs − vo R2 + vs R1 = 0 and Av = vovs =1+ R2 R1 1.38 Writing a nodal equation at the inverting input terminal of the op amp gives v1 − v− R1 + v2 − v− R2 = i− + v− − voR3 but v- = v+ = 0 and i- = 0 vo = − R3R1 v1 − − R3R2 v2 = −0.255sin3770t − 0.255sin10000t volts 1.39 vO = −VREF b12 + b2 4 + b3 8 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ (a) vO = −5 0 2 + 1 4 + 1 8 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = −1.875V (b) vO = −5 1 2 + 0 4 + 0 8 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = −2.500V b1b2b 3 vO (V) 000 0 001 -0.625 010 -1.250 011 -1.875 100 -2.500 101 -3.125 110 -3.750 111 -4.375 1.40 Low-pass amplifier Amplitude f 10 6 kHz 1-12 ©R. C. Jaeger & T. N. Blalock 6/9/06 1-13 6/9/06 1.41 Band-pass amplifier f 20 1 kHz 5 kHz Amplitude 1.42 High-pass amplifier f 10 kHz 16 Amplitude 1.43 vO t( )=10x5sin 2000πt( )+10x3cos 8000πt( )+ 0x3cos 15000πt( ) vO t( )= 50sin 2000πt( )+ 30cos 8000πt( )[ ] volts 1.44 vO t( )= 20x0.5sin 2500πt( )+ 20x0.75cos 8000πt( )+ 0x0.6cos 12000πt( ) vO t( )= 10.0sin 2500πt( )+15.0cos 8000πt( )[ ] volts 1.45 The gain is zero at each frequency: vo(t) = 0. 1.46 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) (a) -2 -1 0 1 2 0 1 2 3 4 5 x10-3 (b) -10 -5 0 5 10 0 1 2 3 4 5 x10-3 1-14 ©R. C. Jaeger & T. N. Blalock 6/9/06 1-15 6/9/06 (c) -10 -5 0 5 10 0 1 2 3 4 5 x10-3 (d) -10 -5 0 5 10 0 1 2 3 4 5 x10-3 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 1.48 Vnom = 2.5V ΔV ≤ 0.05V T = 0.052.50 = 0.0200 or 2.00% 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. 1.51 (a) 5V 1− .05( )≤ V ≤ 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 Vmeter =1.015Vact or Vact = 5.301.015 = 5.22V which is within specifications limits. 1.52 TCR = ΔRΔT = 6562 − 6066 100 − 0 = 4.96 Ω oC 1-16 ©R. C. Jaeger & T. N. Blalock 6/9/06 Rnom = R 0oC + TCR ΔT( )= 6066 + 4.96 27( )= 6200Ω 1-17 6/9/06 1.53 V2 + - V + -V1 R2 I3I2R1 R3 Let RX = R2 R3 then V1 = V R1R1 + RX = V1 1+ RX R1 RX min = 47kΩ 0.9( )180kΩ( )0.9( ) 47kΩ 0.9( )+180kΩ 0.9( )= 33.5kΩ RXmax = 47kΩ 1.1( )180kΩ( )1.1( ) 47kΩ 1.1( )+180kΩ 1.1( )= 41.0kΩ V1 max = 10 1.05( ) 1+ 33.5kΩ 22kΩ 1.1( ) = 4.40V V1min = 10 0.95( ) 1+ 41.0kΩ 22kΩ 0.9( ) = 3.09V I1 = VR1 + RX and I2 = I1 R3R2 + R3 = V R1 + R2 + R1R2R3 I2 max = 10 1.05( ) 22000 0.9( )+ 47000 0.9( )+ 22000 0.9( )47000( )0.9( )180000 1.1( ) =158 μA I2 min = 10 0.95( ) 22000 1.1( )+ 47000 1.1( )+ 22000 1.1( )47000( )1.1( )180000 0.9( ) =114 μA I3 = I1 R2R2 + R3 = V R1 + R3 + R1R3R2 I3 max = 10 1.05( ) 22000 0.9( )+180000 0.9( )+ 22000 0.9( )180000( )0.9( )47000 1.1( ) = 43.1 μA I3 min = 10 0.95( ) 22000 1.1( )+180000 1.1( )+ 22000 1.1( )180000( )1.1( )47000 0.9( ) = 28.3 μA 1.54 I1 = I R2 + R3R1 + R2 + R3 = I 1 1+ R1 R2 + R3 and similarly I2 = I 1 1+ R2 + R3 R1 I1 max = 5 1.02( ) 1+ 2400 0.95( ) 5600 1.05( )+ 3600 1.05( ) mA = 4.12 mA I1min = 5 0.98( ) 1+ 2400 1.05( ) 5600 0.95( )+ 3600 0.95( ) mA = 3.80 mA I2 max = 5 1.02( ) 1+ 5600 0.95( )+ 3600 0.95( ) 2400 1.05( ) mA =1.14 mA I2min = 5 0.98( ) 1+ 5600 1.05( )+ 3600 1.05( ) 2400 0.95( ) mA = 0.936 mA V3 = I2R3 = I1 R1 + 1 R3 + R2 R1R3 V3 max = 5 1.02( ) 1 2400 1.05( )+ 1 3600 1.05( )+ 5600 0.95( ) 2400 1.05( )3600( )1.05( ) = 4.18 V V3 min = 5 0.98( ) 1 2400 0.95( )+ 1 3600 0.95( )+ 5600 1.05( ) 2400 0.95( )3600( )0.95( ) = 3.30 V 1.55 From Prob. 1.24 : Rth = 1 gm + 1R1 Rth max = 1 0.002 0.8( )+ 15x104 1.2( ) = 619 Ω Rthmin = 1 0.002 1.2( )+ 15x104 0.8( ) = 412 Ω 1-18 ©R. C. Jaeger & T. N. Blalock 6/9/06 1-19 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 μA Max 3.71 V 151 μA 40.9 μA Average 3.71 V 133 μA 35.1 μA 1.57 For one set of 200 cases using the Equations in Prob. 1.54: I = 0.005* 0.98 + 0.04* RAND()( ) R1 = 2400* 0.95+ 0.1* RAND()( ) R1 = 5600* 0.95+ 0.1* RAND()( ) R3 = 3600* 0.95+ 0.1* RAND()( ) I1 I2 V3 Min 3.82 mA 0.96 mA 3.46 V Max 4.09 mA 1.12 mA 4.08 V Average 3.97 mA 1.04 mA 3.73 V 1.58 3.29, 0.995, -6.16; 3.295, 0.9952, -6.155 1.59 (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 CHAPTER 2 2.1 Based upon Table 2.1, a resistivity of 2.6 μΩ-cm < 1 mΩ-cm, and aluminum is a conductor. 2.2 Based upon Table 2.1, a resistivity of 1015 Ω-cm > 105 Ω-cm, and silicon dioxide is an insulator. 2.3 Imax = 107 Acm2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 5μm( )1μm( ) 10 −8cm2 μm2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 500 mA 2.4 ⎟⎠ ⎞⎜⎝ ⎛−= − Tx EBTn Gi 5 3 1062.8 exp For silicon, B = 1.08 x 10 31 and EG = 1.12 eV: ni = 2.01 x10 -10 /cm 3 6.73 x10 9 /cm 3 8.36 x 10 13 /cm 3 . For germanium, B = 2.31 x 10 30 and EG = 0.66 eV: ni = 35.9/cm 3 2.27 x10 13 /cm 3 8.04 x 10 15 /cm 3 . 2.5 Define an M-File: function f=temp(T) ni=1E14; f=ni^2-1.08e31*T^3*exp(-1.12/(8.62e-5*T)); ni = 10 14 /cm 3 for T = 506 K ni = 10 16 /cm3 for T = 739 K 2.6 ni = BT 3 exp − EG8.62x10−5T ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ with B = 1.27x10 29 K−3cm−6 T = 300 K and EG = 1.42 eV: ni = 2.21 x10 6 /cm 3 T = 100 K: ni = 6.03 x 10 -19/cm 3 T = 500 K: ni = 2.79 x10 11 /cm 3 20 2.7 vn = −μnE = −700 cm 2 V − s ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2500 V cm ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = −1.75x10 6 cm s v p = +μ pE = +250 cm 2 V − s ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2500 V cm ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = +6.25x10 5 cm s jn = −qnvn = −1.60x10−19C( )1017 1cm3⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −1.75x106 cms⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 2.80x104 Acm2 jp = qnv p = 1.60x10−19C( )103 1cm3⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 6.25x105 cms⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =1.00x10−10 Acm2 2.8 ni 2 = BT 3 exp − EG kT ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ B = 1.08x10 31 1010( )2 =1.08x1031T 3 exp − 1.128.62x10−5T⎛ ⎝ ⎜ ⎞ ⎠ ⎟ Using a spreadsheet, solver, or MATLAB yields T = 305.22K Define an M-File: function f=temp(T) f=1e20-1.08e31*T^3*exp(-1.12/(8.62e-5*T)); Then: fzero('temp',300) | ans = 305.226 K 2.9 s cm cmC cmA Q jv 52 2 10 /01.0 /1000 −=−== 2.10 22 67 3 4104sec 104.0 cm MA cm Axcm cm CQvj ==⎟⎠ ⎞⎜⎝ ⎛⎟⎠ ⎞⎜⎝ ⎛== 21 2.11 vn = −μnE = −1000 cm 2 V − s ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −2000 V cm ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = +2.00x10 6 cm s v p = +μ pE = +400 cm 2 V − s ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −2000 V cm ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = −8.00x10 5 cm s jn = −qnvn = −1.60x10−19C( )103 1cm3⎛ ⎝ ⎜ ⎞ ⎠ ⎟ +2.00x106 cms⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = −3.20x10−10 Acm2 jp = qnv p = 1.60x10−19C( )1017 1cm3⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −8.00x105 cms⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = −1.28x104 Acm2 2.12 ( ) ( ) ( ) V 100101010 b 5000 1010 5 454 =⎟⎠ ⎞⎜⎝ ⎛=== −− cmxcm VV cm V cmx VEa 2.13 jp = qpv p = 1.60x10−19C( )1019cm3⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 10 7 cm s ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =1.60x10 7 A cm2 ip = j p A = 1.60x107 Acm2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1x10 −4cm( )25x10−4cm( )= 4.00 A 2.14 For intrinsic silicon, σ = q μnni + μ pni( )= qni μn + μ p( ) σ ≥1000 Ω − cm( )−1 for a conductor ni ≥ σq μn + μ p( )= 1000 Ω − cm( )−1 1.602x10−19C 100 + 50( ) cm2v − sec = 4.16x10 19 cm3 n i 2 = 1.73x10 39 cm6 = BT 3 exp − EG kT ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ with B =1.08x1031 K−3cm−6, k = 8.62x10-5eV/K and EG =1.12eV This is a transcendental equation and must be solved numerically by iteration. Using the HP solver routine or a spread sheet yields T = 2701 K. Note that this temperature is far above the melting temperature of silicon. 22 2.15 For intrinsic silicon, σ = q μnni + μ pni( )= qni μn + μ p( ) σ ≤10−5 Ω − cm( )−1 for an insulator ni ≥ σq μn + μ p( )= 10−5 Ω − cm( )−1 1.602x10−19C( )2000 + 750( ) cm2v − sec⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 2.270x10 10 cm3 n i 2 = 5.152x10 20 cm6 = BT 3 exp − EG kT ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ with B =1.08x1031 K−3cm−6, k = 8.62x10-5eV/K and EG =1.12eV Using MATLAB as in Problem 2.5 yields T = 316.6 K. 2.16 Si Si Si SiB Si Si Si P 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 will act 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 one less electron and will act as an acceptor impurity. 2.20 ( ) field. electric small a ,20002.010000 2 cm Vcm cm AjjE =−Ω⎟⎠ ⎞⎜⎝ ⎛=== ρσ 2.21 jn drift = qnμnE = qnvn = 1.602x10−19( )1016( ) Ccm3⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 107 cms⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =16000 Acm2 2.22 N = 10 15 atoms cm3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1μm( )10μm( )0.5μm( ) 10−4cm μm ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 3 = 5,000 atoms 2.23 N A > N D: N A − N D =1015 −1014 = 9x1014 /cm3 If we assume N A − N D >> 2ni =1014 / cm3 : p = N A − N D = 9x1014 /cm3 | n = ni 2 p = 2510 26 9x1014 = 2.78x1012 /cm3 If we use Eq. 2.12 : p = 9x10 14 ± 9x1014( )2 + 4 5x1013( )2 2 = 9.03x1014 and n = 2.77x1012 /cm3. The answers are essentially the same. 2.24 35 16 222 314 3113161616 10502 104 10104 102210410105 /cmx. xp n | n/cmxNNp /cmxn/cmxxN: NNN i DA iDADA ====−= =>>=−=−> 2.25 N D > N A: ND − N A = 3x1017 − 2x1017 =1x1017 /cm3 2ni = 2x1017 /cm3; Need to use Eq. (2.11) n = 10 17 ± 1017( )2 + 4 1017( )2 2 =1.62x1017 /cm3 p = ni 2 n = 10 34 1.62x1017 = 6.18x1016 /cm3 24 2.26 N D − N A = −2.5x1018 / cm3 Using Eq. 2.11: n = −2.5x10 18 ± −2.5x1018( )2 + 4 1010( )2 2 Evaluating this with a calculator yields n = 0, and n = ni 2 p = ∞. No, the result is incorrect because of loss of significant digits within the calculator. It does not have enough digits. 2.27 (a) Since boron is an acceptor, NA = 6 x 1018/cm3. Assume ND = 0, since it is not specified. The material is p-type. 3 318 6202 318 i 318310 7.16 106 10 and 106 So 2n >> /106 and 10 re, temperaturoomAt /cm /cmx /cm p nn/cmxp cmxNN/cmn i DAi ==== =−= (b) At 200K, ni 2 =1.08x1031 200( )3 exp − 1.128.62x10−5 200( ) ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ = 5.28x109 /cm6 ni = 7.27x104 /cm3 N A − N D >> 2ni, so p = 6x1018 /cm3 and n = 5.28x10 9 6x1018 = 8.80x10−10 /cm3 2.28 (a) Since arsenic is a donor, ND = 3 x 1017/cm3. Assume NA = 0, since it is not specified. The material is n-type. 3 317 6202 317 i 317310 i 333 103 10 and 103 So 2n >> /103 and /10n re, temperaturoomAt /cm /cmx /cm n np/cmxn cmxNNcm i AD ==== =−= (b) At 250K, ni 2 =1.08x1031 250( )3 exp − 1.128.62x10−5 250( ) ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ = 4.53x1015 /cm6 ni = 6.73x107 /cm3 N D − N A >> 2ni , so n = 3x1017 / cm3 and n = 4.53x10 15 3x1017 = 0.0151/ cm3 2.29 (a) Arsenic is a donor, and boron is an acceptor. ND = 2 x 1018/cm3, and NA = 8 x 1018/cm3. Since NA > ND, the material is p-type. 25 3 318 6202 318 i 318310 i 716 106 10 and 106 So 2n >> /106 and /10n re, temperaturoomAt (b) /cm. /cmx /cm p nn/cmxp cmxNNcm i DA ==== =−= 2.30 (a) Phosphorus is a donor, and boron is an acceptor. ND = 2 x 1017/cm3, and NA = 5 x 1017/cm3. Since NA > ND, the material is p-type. 3 317 6202 317 i 317310 333 103 10 and 103 So 2n >> /103 and 10 re, temperaturoomAt (b) /cm /cmx /cm p nn/cmxp cmxNN/cmn i DAi ==== =−= 2.31 ND = 4 x 1016/cm3. Assume NA = 0, since it is not specified. N D > N A : material is n - type | N D − N A = 4x1016 / cm3 >> 2ni = 2x1010 / cm3 n = 4x1016 / cm3 | p = ni 2 n = 10 20 4x1016 = 2.5x103 / cm3 N D + N A = 4x1016 / cm3 | Using Fig. 2.13, μn =1030 cm 2 V − s and μp = 310 cm2 V − s ρ = 1 qμnn = 1 1.602x10−19C( )1030 cm2V − s⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 4x1016 cm3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.152 Ω − cm 26 2.32 NA = 1018/cm3. Assume ND = 0, since it is not specified. N A > N D : material is p - type | N A − N D =1018 / cm3 >> 2ni = 2x1010 / cm3 p = 1018 / cm3 | n = ni 2 p = 10 20 1018 =100 / cm3 N D + N A =1018 / cm3 | Using Fig. 2.13, μn = 375 cm 2 V − s and μp =100 cm2 V − s ρ = 1 qμ p p = 1 1.602x10−19C 100 cm 2 V − s ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1018 cm3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.0624 Ω − cm 2.33 Indium is from column 3 and is an acceptor. NA = 7 x 1019/cm3. Assume ND = 0, since it is not specified. N A > N D : material is p - type | N A − N D = 7x1019 /cm3 >> 2ni = 2x1010 /cm3 p = 7x1019 /cm3 | n= ni 2 p = 10 20 7x1019 =1.43/cm3 N D + N A = 7x1019 / cm3 | Using Fig. 2.13, μn =120 cm 2 V − s and μp = 60 cm2 V − s ρ = 1 qμ p p = 1 1.602x10−19C 60 cm 2 V − s ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 7x1019 cm3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =1.49 mΩ − cm 2.34 Phosphorus is a donor : N D = 5.5x1016 / cm3 | Boron is an acceptor : N A = 4.5x1016 / cm3 N D > N A : material is n - type | N D − N A =1016 / cm3 >> 2ni = 2x1010 / cm3 n =1016 /cm3 | p = ni 2 p = 10 20 1016 =104 /cm3 N D + N A =1017 / cm3 | Using Fig. 2.13, μn = 800 cm 2 V − s and μp = 230 cm2 V − s ρ = 1 qμnn = 1 1.602x10−19C 800 cm 2 V − s ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1016 cm3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.781 Ω − cm 27 2.35 ρ = 1 qμ p p | μ p p = 1 1.602x10−19C( )0.054Ω − cm( )= 1.16x1020 V − cm − s An iterative solution is required. Using the equations in Fig. 2.8: NA μp μp p 1018 96.7 9.67 x 1020 1.1 x1018 93.7 1.03 x 1020 1.2 x 1017 91.0 1.09 x 1020 1.3 x 1019 88.7 1.15 x 1020 2.36 ρ = 1 qμ p p | μ p p = 1 1.602x10−19C( )0.75Ω − cm( )= 8.32x1018 V − cm − s An iterative solution is required. Using the equations in Fig. 2.8: NA μp μp p 1016 406 4.06 x 1018 2 x 1016 363 7.26 x 1018 3 x 1016 333 1.00 x 1019 2.4 x 1016 350 8.40 x 1018 2.37 Based upon the value of its resistivity, the material is an insulator. However, it is not intrinsic because it contains impurities. Addition of the impurities has increased the resistivity. 2.38 ρ = 1 qμnn | μnn ≈ μnN D = 1 1.602x10−19C( )2Ω − cm( )= 3.12x1018 V − cm − s An iterative solution is required. Using the equations in Fig. 2.8: ND μn μnn 1015 1350 1.35 x 1018 2 x 1015 1330 2.67 x 1018 2.5 x 1015 1330 3.32 x 1018 28 2.3 x 1015 1330 3.06 x 1018 29 2.39 (a) ρ = 1 qμnn | μnn ≈ μnN D = 1 1.602x10−19C( )0.001Ω − cm( )= 6.24x1021 V − cm − s An iterative solution is required. Using the equations in Fig. 2.8: ND μn μnn 1019 116 1.16 x 1021 7 x 1019 96.1 6.73 x 1021 6.5 x 1019 96.4 6.3 x 1021 (b) ρ = 1 qμp p | μp p ≈ μpNA = 1 1.602x10−19C( )0.001Ω − cm( ) = 6.24x1021 V − cm − s An iterative solution is required using the equations in Fig. 2.8: NA μp μp p 1.3 x 1020 49.3 6.4 x 1021 2.40 Yes, by adding equal amounts of donor and 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: ρ = 1 qμ p p | μ p p ≈ μ pN A = 1 1.602x10−19C( )1Ω − cm( )= 6.25x1018 V − cm − s An iterative solution is required. Using the equations in Fig. 2.8: NA μp μp p 1016 406 4.1 x 1018 1.5 x 1016 383 5.7 x 1018 1.7 x 1016 374 6.4 x 1019 30 To change the resistivity to 0.25 ohm-cm: ρ = 1 qμ p p | μ p p ≈ μ pN A = 1 1.602x10−19C( )0.25Ω − cm( )= 2.5x1019 V − cm − s NA μp μp p 6 x 1016 276 1.7 x 1019 8 x 1016 233 2.3 x 1019 1.1 x 1017 225 2.5 x 1019 Additional acceptor concentration = 1.1x10 17 - 1.7x10 16 = 9.3 x 10 16 /cm 3 (b) If donors are added: ND ND + NA μn ND - NA μnn 2 x 1016 3.7 x 1016 1060 3 x 1015 3.2 x 1018 1 x 1017 1.2 x 1017 757 8.3 x 1016 6.3 x 1019 8 x 1016 9.7 x 1016 811 6.3 x 1016 5.1 x 1019 4.1 x 1016 5.8 x 1016 950 2.4 x 1016 2.3 x 1019 So ND = 4.1 x 10 16 /cm 3 must be added to change achieve a resistivity of 0.25 ohm-cm. The silicon is converted to n-type material. 2.42 Phosphorus is a donor: ND = 1016/cm3 and μn = 1250 cm2/V-s from Fig. 2.8. σ = qμnn ≈ qμnN D = 1.602x10−19C( )1250( )1016( )= 2.00Ω − cm Now we add acceptors until σ = 5.0 (Ω-cm) -1: σ = qμ p p | μ p p ≈ μ p N A − N D( )= 5 Ω − cm( ) −1 1.602x10−19C = 3.12x10 19 V − cm − s NA ND + NA μp NA - ND μp p 1 x 1017 1.1 x 1017 250 9 x 1016 2.3 x 1019 2 x 1017 2.1 x 1017 176 1.9 x 1017 3.3 x 1019 1.8 x 1017 1.9 x 1017 183 1.7 x 1016 3.1 x 1019 31 2.43 Boron is an acceptor: NA = 1016/cm3 and μp = 405 cm2/V-s from Fig. 2.8. σ = qμ p p ≈ qμ pN A = 1.602x10−19C( )405( )1016( )= 0.649Ω − cm Now we add donors until σ = 5.5 (Ω-cm) -1: σ = qμnn | μnn ≈ μn N D − N A( )= 5.5 Ω − cm( ) −1 1.602x10−19C = 3.43x10 19 V − cm − s ND ND + NA μn ND - NA μp p 8 x 1016 9 x 1016 832 7 x 1016 5.8 x 1019 6 x 1016 7 x 1016 901 5 x 1016 4.5 x 1019 4.5 x 1016 5.5 x 1016 964 3.5 x 1016 3.4 x 1019 2.44 VT = kTq = 1.38x10−23T 1.602x10−19 = 8.62x10−5T T (K) 50 75 100 150 200 250 300 350 400 VT (mV) 4.31 6.46 8.61 12.9 17.2 21.5 25.8 30.1 34.5 2.45 j = −qDn − dndx ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = qVTμn dn dx j = 1.602x10−19C( )0.025V( ) 350 cm2V − s⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1018 − 0 0 −10−4 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 cm4 = −14.0 kA cm2 2.46 j = −qDp dpdx = −1.602x10 −19C( )15 cm2s⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − 1019 / cm3 2x10−4cm ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ exp − x 2x10−4cm ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ j =1.20x105 exp −5000 x cm ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ A cm2 I 0( )= j 0( )A = 1.20x105 Acm2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 10μm 2( )10−8cm2μm2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =12.0 mA 32 2.47 jp = qμp pE − qDp dpdx = qμp p E −VT 1 p dp dx ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0 → E = VT 1 p dp dx E ≈ VT 1N A dN A dx = 0.025 −10 22 exp −104 x( ) 1014 +1018 exp −104 x( ) E 0( )= −0.025 10221014 +1018 = −250 Vcm E 5x10−4cm( )= −0.025 1022 exp −5( )1014 +1018 exp −5( )= −246 Vcm 2.48 jn drift = qμnnE = 1.60x10−19C( )350 cm2V − s⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1016 cm3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −20 V cm ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = −11.2 A cm2 jp drift = qμ p pE = 1.60x10−19C( )150 cm2V − s⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1.01x1018 cm3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −20 V cm ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = −484 A cm2 jn diff = qDn dndx = 1.60x10 −19C( )350 ⋅ 0.025cm2s⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 104 −1016 2x10−4cm4 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = −70.0 A cm2 jp diff = −qDp dpdx = −1.60x10 −19C( )150 ⋅ 0.025cm2s⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1018 −1.01x1018 2x10−4cm4 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 30.0 A cm2 jT = −11.2 − 484 − 70.0 + 30.0 = −535 Acm2 2.49 NA = 2ND E C E A N A Holes E V NA N D N A N D N A N DE D 2.50 λ = hc E = 6.626x10 −34 J − s( )3x108 m / s( ) 1.12eV( )1.602x10−19 J / eV( ) =1.108 μm 33 2.51 p-type silicon n-type silicon Si02 Al - Anode Al - Cathode 2.52 An n-type ion implantation step could be used to 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. n-type silicon p-type silicon Si02 Photoresist Structure after exposure and development of photoresist layer Mask n-type silicon p-type silicon Structure following ion implantation of n-type impurity n+ Ion implantation Side view Top View Mask for ion implantation 2.53 (a) N = 8 1 8 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + 6 1 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + 4 1()= 8 atoms (b) V = l 3 = 0.543x10−9 m( )3 = 0.543x10−7cm( )3 =1.60x10−22cm3 (c) D = 8 atoms 1.60x1022cm3 = 5.00x1022 atoms cm3 (d ) m = 2.33 g cm3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1.60x10 22cm3 = 3.73x10−22 g (e) From Table 2.2, silicon has a mass of 28.086 protons. mp = 3.73x10 −22 g 28.082 8( )protons =1.66x10−24 g proton Yes, near the actual proton rest mass. 34 CHAPTER 3 3.1 φ j = VT ln NANDni2 = 0.025V( )ln 1019 ⋅ cm−3( )1018 ⋅ cm−3( ) 1020 ⋅ cm−6 = 0.979V wdo = 2εsq 1 NA + 1 ND ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ φ j = 2 11.7 ⋅ 8.854x10−14 F ⋅ cm−1( ) 1.602x10−19C 1 1019cm−3 + 1 1018cm−3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 0.979V( ) w do = 3.73 x 10−6cm = 0.0373μm xn = wdo 1+ ND NA = 0.0373μm 1+ 10 18cm−3 1019cm−3 = 0.0339 μm | x p = wdo 1+ NA ND = 0.0373μm 1+ 10 19cm−3 1018cm−3 = 3.39 x 10-3 μm E MAX = qNA xpεs = 1.60x10−19C( )1019cm−3( ) 3.39x10−7cm( ) 11.7 ⋅ 8.854 x10−14 F /cm = 5.24 x 10 5 V cm 3.2 ppo = NA = 10 18 cm3 | npo = ni 2 ppo = 10 20 1018 = 10 2 cm3 nno = ND = 10 15 cm3 | pno = ni 2 nno = 10 20 1015 = 10 5 cm3 φ j = VT ln NANDni2 = 0.025V( )ln 1018cm−3( )1015cm−3( ) 1020cm−6 = 0.748 V wdo = 2εsq 1 NA + 1 ND ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ φ j = 2 11.7 ⋅ 8.854x10−14 F ⋅ cm−1( ) 1.602x10−19C 1 1018cm−3 + 1 1015cm−3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 0.748V( ) wdo = 98.4 x 10−6cm = 0.984 μm 3.3 ppo = N A = 10 18 cm3 | npo = ni 2 p po = 10 20 1018 = 10 2 cm3 nno = N D = 10 18 cm3 | pno = ni 2 nno = 10 20 1018 = 10 2 cm3 φ j = VT ln NANDni2 = 0.025V( )ln 1018 ⋅ cm−3( )1018 ⋅ cm−3( ) 1020 ⋅ cm−6 = 0.921V wdo = 2εsq 1 NA + 1 ND ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ φ j = 2 11.7 ⋅ 8.854x10−14 F ⋅ cm−1( ) 1.602x10−19C 1 1018cm−3 + 1 1018cm−3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 0.921V( ) w do = 4.881x10−6cm = 0.0488 μm 34 3.4 ppo = N A = 10 18 cm3 | npo = ni 2 p po = 10 20 1018 = 10 2 cm3 nno = N D = 10 18 cm3 | pno = ni 2 nno = 10 20 1018 = 10 2 cm3 φ j = VT ln NANDni2 = 0.025V( )ln 1018 ⋅ cm−3( )1020 ⋅ cm−3( ) 1020 ⋅ cm−6 =1.04V wdo = 2εsq 1 NA + 1 ND ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ φ j = 2 11.7 ⋅ 8.854x10−14 F ⋅ cm−1( ) 1.602x10−19C 1 1018cm−3 + 1 1020cm−3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1.04V( ) w do = 0.0369 μm 3.5 ppo = N A = 10 16 cm3 | npo = ni 2 p po = 10 20 1016 = 10 4 cm3 nno = N D = 10 19 cm3 | pno = ni 2 nno = 10 20 1019 = 10 cm3 φ j = VT ln N A N Dni2 = 0.025V( )ln 1019 ⋅ cm−3( )1016 ⋅ cm−3( )1020 ⋅ cm−6 = 0.864V wdo = 2εsq 1 N A + 1 N D ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ φ j = 2 11.7 ⋅ 8.854x10−14 F ⋅ cm−1( ) 1.602x10−19C 1 1019cm−3 + 1 1016cm−3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 0.864V( ) wdo = 0.334 μm 3.6 wd = wdo 1+ VRφ j | (a) wd = 2wdo requires VR = 3φ j = 2.55 V | wd = 0.4μm 1+ 5 0.85 = 1.05 μm 3.7 wd = wdo 1+ VRφ j | (a) wd = 3wdo requires VR = 8φ j = 4.80 V | wd =1μm 1+ 10 0.6 = 4.20 μm 3.8 jn = σE , σ = 1ρ = 1 0.5 Ω⋅ cm = 2 Ω⋅ cm | E = jn σ = 1000A ⋅ cm−2 2 Ω⋅ cm( )−1 = 500 V cm 35 3.9 jp = σE | E = jnσ = jnρ = 5000A ⋅ cm −2( )2Ω⋅ cm( )=10.0 kVcm 3.10 j ≅ jn = qnv = 1.60x10−19C( ) 4x1015cm3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 107cm s ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 6400 A cm2 3.11 j ≅ jp = qpv = 1.60x10−19C( ) 5x1017cm3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 107cm s ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 800 kA cm2 3.12 jp = qμ p pE − qDp dpdx = 0 → E = − Dp μ p ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ 1p = − kT q dp dx ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 p dp dx p(x) = N o exp − xL ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ | 1 p | E = −VT L = − 0.025V 10−4cm = −250 V cm dp dx = 1 L The exponential doping results in a constant electric field. 3.13 jp = qDn dndx = qμnVT dn dx | dn dx = 2000A / cm 2 1.60x10−19C( )500cm2 /V − s( )0.025V( )= 1.00 x 1021 cm4 3.14 10 =10 4 ⋅10−16 exp 40VD( )−1[ ]+ VD and the solver yields VD = 0.7464 V 3.15 f =10 −104 ID − 0.025ln ID + ISIS | f ' = −104 − 0.025 ID + IS | ID ' = ID − ff ' Starting the iteration process with ID = 100 μA and IS = 10-13A: ID f f' 1.000E-04 8.482E+0 0 -1.025E+04 9.275E-04 1.512E- 01 -1.003E+04 9.426E-04 3.268E- 06 -1.003E+04 9.426E-04 9.992E- 16 -1.003E+04 36 3.16 (a) Create the following m-file: function fd=current(id) fd=10-1e4*id-0.025*log(1+id/1e-13); Then: fzero('current',1) yields ans = 9.4258e-04 (b) Changing IS to 10 -15 A: function fd=current(id) fd=10-1e4*id-0.025*log(1+id/1e-15); Then: fzero('current',1) yields ans = 9.3110e-04 3.17 T = qVT k = 1.60x10 −19C 0.025V( ) 1.38x10−23 J / K = 290 K 3.18 VT = kTq = 1.38x10−23 J / K( )T 1.60x10−19C = 8.63x10−5T For T = 218 K, 273 K and 358 K, VT = 18.8 mV, 23.6 mV and 30.9 mV 3.19 Graphing ID = IS exp 40VDn ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −1 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ yields : 1.21.00.80.60.40.20.0 0 1 2 3 4 5 6 (a) (b) (c) 37 3.20 nVT = n kTq =1.04 1.38x10−23 J / K( )300( ) 1.60x10−19C = 26.88 mV T = 26.88mV1.602x10 -19 1.38x10-23 = 312 K 3.21 iD = IS exp vDnVT ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −1 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ or vD nVT = ln 1+ iD IS ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ For iD >> IS , vDnVT ≅ ln iD IS ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ or ln ID( )= 1nVT ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ vD + ln IS( ) which is the equation of a straight line with slope 1/nVT and x-axis intercept at -ln (IS). The values of n and IS can be found from any two points on the line in the figure: e. g. iD = 10 -4 A for vD = 0.60 V and iD = 10 -9 A for vD = 0.20 V. Then there are two equations in two unknowns: ln 10-9( )= 40n⎛ ⎝ ⎜ ⎞ ⎠ ⎟ .20 + ln IS( ) or 9.21= 8n⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + ln IS( ) ln 10-4( )= 40n⎛ ⎝ ⎜ ⎞ ⎠ ⎟ .60 + ln IS( ) or 20.72 = 24n⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + ln IS( ) Solving for n and IS yields n = 1.39 and IS = 3.17 x 10 -12 A = 3.17 pA. 3.22 VD = nVT ln 1+ IDIS ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ | ID = IS exp VD nVT ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −1 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ (a) VD =1.05 0.025V( )ln 1+ 7x10−5 A10−18 A ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.837V | (b) VD =1.05 0.025V( )ln 1+ 5x10−6 A 10−18 A ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.768V (c) ID =10−18 A exp 01.05 ⋅ 0.025V ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −1 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = 0 A (d) ID =10−18 A exp −0.075V1.05 ⋅ 0.025V ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −1 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = −0.943x10 −19 A (e) ID =10−18 A exp −5V1.05 ⋅ 0.025V ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −1 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = −1.00x10 −18 A 3.23 VD = nVT ln 1+ IDIS ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ | ID = IS exp VD nVT ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −1 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ (a) VD = 0.025V ln 1+ 10 −4 A 10−17 A ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.748V | (b) VD = 0.025V ln 1+ 10−5 A 10−17 A ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.691V 38 (c) ID =10−17 A exp 00.025V ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −1 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = 0 A (d) ID =10−17 A exp −0.06V0.025V ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −1 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = −0.909x10 −17 A (e) ID =10−17 A exp −4V0.025V ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −1 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = −1.00x10 −17 A 3.24 ID = IS exp VDVT ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −1 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ =10 −17 A exp 0.675 0.025 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −1 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = 5.32x10 −6 A = 5.32 μA VD = VT ln IDIS +1⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.025V( )ln 15.9x10 −6 A 10−17 A +1⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.703 V 3.25 VD = nVT ln 1+ IDIS ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 2 0.025V( )ln 1+ 40A10−10 A ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =1.34 V VD = 2 0.025V( )ln 1+ 100A10−10 A ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =1.38 V 3.26 (a) IS = ID exp VD nVT ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −1 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = 2mA exp 0.82 0.025 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −1 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ =1.14x10−17 A (b) ID =1.14x10−17 A exp −50.025 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −1 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = −1.14x10 −17 A 3.27 (a) IS = ID exp VD nVT ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −1 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = 300μA exp 0.75 0.025 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −1 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = 2.81x10−17 A (b) ID = 2.81x10−17 A exp −30.025 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −1 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = −2.81x10 −17 A 39 3.28 VD = nVT ln 1+ IDIS ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ | 10 -14 ≤ IS ≤10−12 | VD = 0.025V( )ln 1+ 10−3 A10−12 A ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.518 V VD = 0.025V( )ln 1+ 10−3 A10−14 A ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.633 V | So, 0.518 V ≤ VD ≤ 0.633 V 3.29 VT = 1.38x10−23 307( ) 1.60x10−19 = 0.0264V | ID = IS exp VD0.0264n ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −1 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ Varying n and IS by trial-and-error with a spreadsheet: n IS 1.039 7.606E-15 VD ID-Measured ID-Calculated Error Squared 0.500 6.591E-07 6.276E-07 9.9198E-16 0.550 3.647E-06 3.885E-06 5.6422E-14 0.600 2.158E-05 2.404E-05 6.0672E-12 0.650 1.780E-04 1.488E-04 8.518E-10 0.675 3.601E-04 3.702E-04 1.0261E-10 0.700 8.963E-04 9.211E-04 6.1409E-10 0.725 2.335E-03 2.292E-03 1.8902E-09 0.750 6.035E-03 5.701E-03 1.1156E-07 0.775 1.316E-02 1.418E-02 1.0471E-06 Total Squared Error 1.1622E-06 3.30 VT = kTq = 1.38x10−23 J / K( )T 1.60x10−19C = 8.63x10−5T For T = 233 K, 273 K and 323 K, VT = 20.1 mV, 23.6 mV and 27.9 mV 3.31 kT q = 1.38x10 −23 303( ) 1.60x10−19 = 26.1 mV | VD = 0.0261V( )ln 1+ 10−32.5x10−16 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.757 V ΔV = −1.8mV / K( )20K( )= −36.0 mV | VD = 0.757 − 0.036 = 0.721 V 40 3.32 kT q = 1.38x10 −23 298( ) 1.602x10−19 = 25.67 mV | (a) VD = 0.02567V( )ln 1+ 10−410−15 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.650 V ΔV = −2.0mV / K( )25K( )= −50.0 mV (b) VD = 0.650 − 0.050 = 0.600 V 3.33 kT q = 1.38x10 −23 298( ) 1.602x10−19 = 25.67 mV | (a) VD = 0.02567V( )ln 1+ 2.5x10−410−14 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.615 V b( ) ΔV = −1.8mV / K( )60K( )= −50.0 mV VD = 0.615− 0.108 = 0.507 V (c) ΔV = −1.8mV / K( )−80K( )= +144 mV VD = 0.615+ 0.144 = 0.758 V 3.34 dvD dT = vD −VG − 3VT T = 0.7 −1.21− 3 0.0259( ) 300 = −1.96 mV K 41 3.35 IS2 IS1 = T2 T1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 3 exp − EG k ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 T2 − 1 T1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = T2 T1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 3 exp EG kT1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1− T1 T2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ f x( )= x( )3 exp EGkT1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1− 1 x ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ x = T2 T1 Using trial and error with a spreadsheet yields �T = 4.27 K, 14.6 K, and 30.7 K to increase the saturation current by 2X, 10X, and 100X respectively. x f(x) Delta T 1.00000 1.00000 0.00000 1.00500 1.27888 1.50000 1.01000 1.63167 3.00000 1.01500 2.07694 4.50000 1.01400 1.97945 4.20000 1.01422 2.00051 4.26600 1.01922 2.54151 5.76600 1.02422 3.22151 7.26600 1.02922 4.07433 8.76600 1.03422 5.14160 10.26600 1.03922 6.47438 11.76600 1.04422 8.13522 13.26600 1.04922 10.20058 14.76600 1.04880 10.00936 14.64000 1.10000 90.67434 30.00000 1.10239 100.0012 30.71610 3.36 wd = wdo 1+ VRφ j | (a) wd =1μm 1+ 5 0.8 = 2.69 μm (b) wd =1μm 1+ 100.8 = 3.67 μm 3.37 φ j = VT ln N A N Dni2 = 0.025V( )ln 1016cm−3( )1015cm−3( )1020cm−6 = 0.633 V wdo = 2εsq 1 N A + 1 N D ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ φ j = 2 11.7 ⋅ 8.854x10−14 F ⋅ cm−1( ) 1.602x10−19C 1 1016cm−3 + 1 1015cm−3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 0.633V( ) wdo = 0.949 μm | wd = wdo 1+ VRφ j wd = 0.949μm 1+ 10V0.633V = 3.89 μm | wd = 0.949μm 1+ 100V 0.633V = 12.0 μm 42 3.38 φ j = VT ln N A N Dni2 = 0.025V( )ln 1018cm−3( )1020cm−3( ) 1020cm−6 =1.04 V wdo = 2εsq 1 N A + 1 N D ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ φ j = 2 11.7 ⋅ 8.854x10−14 F ⋅ cm−1( ) 1.602x10−19C 1 1018cm−3 + 1 1020cm−3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1.04V( ) wdo = 0.0368 μm | wd = wdo 1+ VRφ j wd = 0.0368μm 1+ 51.04 = 0.0887 μm | wd = 0.0368μm 1+ 25 1.04 = 0.184 μm 3.39 Emax = 2 φ j + VR( ) wd = 2 φ j + VR( ) wdo 1+ VRφ j = 2φ j wdo 1+ VRφ j 3x105 V cm = 2 0.6V( ) 10−4cm 1+ VR 0.6 → VR = 374 V 3.40 E = 2φ j wdo = 2 0.748V( ) 0.984x10−4cm =15.2 kV cm | φ j + VR = Emax2 wdo φ j = 3x105 V cm 0.984x10−4cm( ) 2 0.748V VR = 291.3− 0.748 = 291 V 3.41 VZ = 4 V; RZ = 0 Ω since the reverse breakdown slope is infinite. 3.42 Since NA >> ND, the depletion layer is all on the lightly-doped side of the junction. Also, VR >> φj, so φj can be neglected. Emax = qN A x pεS = qN Awd εS = qN A εS 2εS q VR N A N A = Emax 2 εS 2qVR = 3x10 5( )2 11.7( )8.854x10−14( ) 2 1.602x10−19( )1000 = 2.91 x 1014 / cm3 43 3.43 φ j = VT ln N AN Dni2 = 0.025ln 10151020 1020 = 0.864V wdo = 2εSq 1 N A + 1 N D ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ φ j = 2 11.7( )8.854x10−14( ) 1.602x10−19 1 1015 + 1 1020 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 0.864 =1.057x10 −4cm Cjo " = εS wdo = 11.7 8.854x10−14( ) 1.057x10−4 = 9.80x10-9 F / cm2 | Cj = Cjo " A 1+ VRφ j = 9.80x10 -9 0.05( ) 1+ 5 0.864 =188 pF 3.44 φ j = VT ln N AN Dni2 = 0.025ln 10181015 1020 = 0.748V wdo = 2εSq 1 N A + 1 N D ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ φ j = 2 11.7( )8.854x10−14( ) 1.602x10−19 1 1018 + 1 1015 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 0.748 = 0.984x10 −4cm Cjo " = εS wdo = 11.7 8.854x10−14( ) 0.984x10−4 = 10.5x10-9 F / cm2 | Cj = Cjo " A 1+ VRφ j = 10.5x10 -9 0.02( ) 1+ 10 0.748 = 55.4 pF 3.45 (a) CD = IDτTVT = 10−4 A 10−10 s( ) 0.025V = 400 fF (b) Q = IDτT =10−4 A 10−10 s( )=10 fC (c) CD = 25x10−3 A 10−10 s( ) 0.025V =100 pF | Q = IDτT = 5x10−3 A 10−10 s( )= 0.50 pC 3.46 (a) CD = IDτTVT = 1A 10−8 s( ) 0.025V = 0.400 μF (b) Q = IDτT =1A 10−8 s( )=10.0 nC (c) CD = 100mA 10−8 s( ) 0.025V = 0.04 μF | Q = IDτT =100mA 10−8 s( )=1.00 nC 44 3.47 φ j = VT ln N AN Dni2 = 0.025ln 10191017 1020 = 0.921V wdo = 2εSq 1 N A + 1 N D ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ φ j = 2 11.7( )8.854x10−14( ) 1.602x10−19 1 1019 + 1 1017 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 0.921 = 0.110 μm Cjo = εS Awdo = 11.7 8.854x10−14( )10−4( ) 0.110x10−4 = 9.42 pF / cm2 | Cj = Cjo 1+ VRφ j = 9.42 pF 1+ 5 0.921 = 3.72 pF 3.48 φ j = VT ln N AN Dni2 = 0.025ln 10191016 1020 = 0.864V wdo = 2εSq 1 N A + 1 N D ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ φ j = 2 11.7( )8.854x10−14( ) 1.602x10−19 1 1019 + 1 1016 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 0.864 = 0.334μm Cjo = εS Awdo = 11.7 8.854x10−14( )0.25cm2( ) 0.334x10−4 = 7750 pF | Cj = Cjo 1+ VRφ j = 7750 pF 1+ 3 0.864 = 3670 pF 3.49 VDC RFC C 10 μH 10 μH C L = C = Cjo 1+ VRφ j (a) C = 39 pF 1+ 1V 0.75V = 25.5 pF | fo = 1 2π LC = 1 2π 10−5 H( )25.5 pF = 9.97MHz (b) C = 39 pF 1+ 10V 0.75V =10.3pF | fo = 1 2π LC = 1 2π 10−5 H( )10.3pF =15.7 MHz 3.50 (a) VD = 0.025V( )ln 1+ 50A10−7 A ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.501 V | (b) VD = 0.025V( )ln 1+ 50A10−15 A ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.961 V 45 3.51 (a) VD = 0.025V( )ln 1+ 4x10−3 A10−11 A ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.495 V | (b) VD = 0.025V( )ln 1+ 4x10−3 A 10−14 A ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.668 V 3.52 RS = Rp + Rn Rp = ρ p LpAp = 1Ω − cm( ) 0.025cm 0.01cm2 = 2.5Ω Rn = ρn LnAn = 0.01Ω − cm( ) 0.025cm 0.01cm2 = 0.025Ω RS = 2.53 Ω 3.53 (a) VD ' = VT ln 1+ IDIS ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.025V( )ln 1+ 10 −3 5x10−16 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.708V VD = VD' + ID RS = 0.708V +10−3 A 10Ω( )= 0.718 V (b) VD = VD' + ID RS = 0.708V +10−3 A 100Ω( )= 0.808 V 3.54 ρc =10Ω − μm2 Ac =1μm2 RC = ρcAc = 10Ω − μm2 1μm2 =10Ω / contact 5 anode contacts and 14 cathode contacts Resistance of anode contacts = 10Ω 5 = 2Ω Resistance of cathode contacts = 10Ω 14 = 0.71Ω 3.55 (a) From Fig. 3.21a, the diode is approximately 10.5 μm long x 8 μm wide. Area = 84 μm2. (b) Area = (10.5x0.13 μm) x (8x0.13μm) = 1.42 μm2. 46 3.56 (a) 5 =104 ID + VD | VD = 0 ID = 0.500mA | ID = 0 VD = 5V Forward biased - VD = 0.5 V ID = 4.5V104Ω = 0.450 mA (b) − 6 = 3000ID + VD | VD = 0 ID = −2.00mA | ID = 0 VD = −6V In reverse breakdown - VD = −4 V ID = −2V3kΩ = −0.667 mA (c) − 3 = −3000ID + VD | VD = 0 ID = −1.00mA | ID = 0 VD = −3V Reverse biased - VD = −3 V ID = 0 1 2 3 4 1 mA 2 mA (a) Q-point vD 65 -1-2-3-4-5-6 -1 mA -2 mA (b) Q-point iD (c) Q-point 3.57 (a) 10 = 5000ID + VD | VD = 0 ID = 2.00 mA | VD = 5 V ID =1.00 mA Forward biased - VD = 0.5V ID = 9.5V5kΩ =1.90 mA (b) -10 = 5000ID + VD | VD = 0 ID = −2.00 mA | VD = −5 V ID = −1.00 mA In reverse breakdown - VD = −4V ID = −6V5kΩ = −1.20 mA (c) − 2 = 2000ID + VD | VD = 0 ID = −1.00 mA | ID = 0 VD = −2 V Reverse biased - VD = −2 V ID = 0 1 2 3 4 1 mA 2 mA (a) Q-point vD 65 -1-2-3-4-5-6 -1 mA -2 mA (b) Q-point iD (c) Q-point 47 3.58 *Problem 3.58 - Diode Circuit SPICE Results V 1 0 DC 5 R 1 2 10K VD = 0.693 V D1 2 0 DIODE1 ID = 0.431 μA .OP .MODEL DIODE1 D IS=1E-15 .END 3.59 (a) −10 =104 ID + VD | VD = 0 ID = −1.00 mA | VD = −5 V ID = −0.500 mA In reverse breakdown - VD = −4 V ID = −10 − (−4)V10kΩ = −0.600 mA (b) 10 =104 ID + VD | VD = 0 ID =1.00 mA | VD = 5 V ID = 0.500 mA Forward biased - VD = 0.5 V ID = 10 − 0.5V10kΩ = 0.950 mA (c) − 4 = 2000ID + VD | VD = 0 ID = −2.00 mA | ID = 0 VD = − 4 V Reverse biased - VD = −4 V ID = 0 1 2 3 4 1 mA 2 mA (b) Q-point vD 65 -1-2-3-4-5-6 -1 mA -2 mA (c) Q-point iD (a) Q-point 48 1 2 3 4 5 6 7 0.002 0.001 -0.001 -0.002 iD (A) vD (V)-7 -6 -5 -4 -3 -2 -1 49 3.60 + -V R iD vD + - The load line equation: V = iD R + vD We need two points to plot the load line. (a) V = 6 V and R = 4 kΩ: For vD = 0, iD = 6V/4 kΩ = 1.5 mA and for iD = 0, vD = 6V. Plotting this line on the graph yields the Q-pt: (0.5 V, 1.4 mA). (b) V = -6 V and R = 3 kΩ: For vD = 0, iD = -6V/3 kΩ = -2 mA and for iD = 0, vD = -6V. Plotting this line on the graph yields the Q-pt: (-4 V, -0.67 mA). (c) V = -3 V and R = 3 kΩ: Two points: (0V, -1mA), (-3V, 0mA); Q-pt: (-3 V, 0 mA) (d) V = +12 V and R = 8 kΩ: Two points: (0V, 1.5mA), (4V, 1mA); Q-pt: (0.5 V, 1.4 mA) (e) V = -25 V and R = 10 kΩ: Two points: (0V, -2.5mA), (-5V, -2mA); Q-pt: (-4 V, -2.1 mA) 1 2 3 4 5 6 7-1-2-3-4-5 -6 -7 .001 .002 -.001 -.002 i (A) D v (V) D Q-Point (-4V,-0.67 mA) Q-Point (0.5V,1.4 mA) Load line for (a) Load line for (b) Q-Point (-4V,-2.1 mA) (e) (c) Q-Point (-3V,0 mA) (d) Q-Point (0.5V,1.45 mA) 50 3.61 Using the equations from Table 3.1, (f = 10-10 -9 exp ..., etc.) VD = 0.7 V requires 12 iterations, VD = 0.5 V requires 22 iterations, VD = 0.2 V requires 384 iterations - very poor convergence because the second iteration (VD = 9.9988 V) is very bad. 3.62 Using Eqn. (3.28), V = iD R + VT ln iDIS ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ or 10 =10 4 iD + 0.025ln 1013iD( ). We want to find the zero of the function f = 10 −10 4iD − 0.025ln 1013iD( ) iD f .001 -0.576 .0001 8.48 .0009 0.427 .00094 0.0259 - converged 3.63 f =10 −104 ID − 0.025ln 1+ IDIS ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ | f ' = −104 − 0.025 ID + IS x f(x) f'(x) 1.0000E+00 -9.991E+03 -1.000E+04 9.2766E-04 1.496E-01 -1.003E+04 9.4258E-04 3.199E-06 -1.003E+04 9.4258E-04 9.992E-16 -1.003E+04 9.4258E-04 9.992E-16 -1.003E+04 3.64 Create the following m-file: function fd=current(id) fd=10-1e4*id-0.025*log(1+id/1e-13); Then: fzero('current',1) yields ans = 9.4258e-04 + 1.0216e-21i 51 3.65 The one-volt source will forward bias the diode. Load line: 1 =10 4 ID + VD | ID = 0 VD =1V | VD = 0 ID = 0.1mA → 50 μA, 0.5 V( ) Mathematical model: f =1−10 −9 exp 40VD( )−1[ ]+ VD → 49.9 μA, 0.501 V( ) Ideal diode model: ID = 1V/10kΩ = 100 μA; (100 μA, 0 V) Constant voltage drop model: ID = (1-0.6)V/10kΩ = 40.0μA; (40.0 μA, 0.6 V) 3.66 Using Thévenin equivalent circuits yields and then combining the sources + - +- + - V I 1 k Ω1.2 k Ω 1.5 V1.2 V 0.3 V +- + - V I 2.2 k Ω (a) Ideal diode model: The 0.3 V source appears to be forward biasing the diode, so we will assume it is "on". Substituting the ideal diode model for the forward region yields I = 2.2kΩ 0.3V = 0.136 mA. This current is greater than zero, which is consistent with the diode being "on". Thus the Q-pt is (0 V, +0.136 mA). Ideal Diode: 0.3 V +- + - V I 2.2 k Ω CVD: 0.3 V + - 0.6 V I 2.2 k Ω +- on V (b) CVD model: The 0.3 V source appears to be forward biasing the diode so we will assume it is "on". Substituting the CVD model with Von = 0.6 V yields I = 2.2kΩ 0.3V − 0.6V = −136 μA. This current is negative which is not consistent with the assumption that the diode is "on". Thus the diode must be off. The resulting Q-pt is: (0 mA, -0.3 V). 0.3 V+ - I=0 2.2 k Ω- +V 52 (c) The second estimate is more realistic. 0.3 V is not sufficient to forward bias the diode into significant conduction. For example, let us assume that IS = 10 -15 A, and assume that the full 0.3 V appears across the diode. Then iD =10−15 A exp 0.3V0.025V ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −1 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ =163 pA, a very small current. 3.67 The nominal values are: VA = 3V R2R1 + R2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 3V 2kΩ 2kΩ + 3kΩ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =1.20V and RTHA = R1R2 R1 + R2 = 2kΩ 3kΩ( ) 2kΩ + 3kΩ =1.20kΩ VC = 3V R4R3 + R4 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 3V 2kΩ 2kΩ + 2kΩ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =1.50V and RTHC = R3R4 R3 + R4 = 2kΩ 2kΩ( ) 2kΩ + 2kΩ =1.00kΩ ID nom = 1.50 −1.20 1.20 +1.00 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ V kΩ =136 μA For maximum current, we make the Thévenin equivalent voltage at the diode anode as large as possible and that at the cathode as small as possible. VA = 3V 1+ R1 R2 = 3V 1+ 2kΩ 0.9( ) 2kΩ 1.1( ) =1.65V and RTHA = R1R2R1 + R2 = 2kΩ 0.9( )2kΩ 1.1( ) 2kΩ 0.9( )+ 2kΩ 1.1( )= 0.990kΩ VC = 3V 1+ R3 R4 = 3V 1+ 3kΩ 1.1( ) 2kΩ 0.9( ) =1.06V and RTHC = R3R4R3 + R4 = 3kΩ 1.1( )2kΩ 0.9( ) 3kΩ 1.1( )+ 2kΩ 0.9( )=1.17kΩ ID max = 1.65−1.06 0.990 +1.17 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ V kΩ = 274 μA For minimum current, we make the Thévenin equivalent voltage at the diode anode as small as possible and that at the cathode as large as possible. VA = 3V 1+ R1 R2 = 3V 1+ 2kΩ 1.1( ) 2kΩ 0.9( ) =1.350V and RTHA = R1R2R1 + R2 = 2kΩ 1.1( )2kΩ 0.9( ) 2kΩ 1.1( )+ 2kΩ 0.9( )= 0.990kΩ VC = 3V 1+ R3 R4 = 3V 1+ 3kΩ 0.9( ) 2kΩ 1.1( ) =1.347V and RTHC = R3R4R3 + R4 = 3kΩ 0.9( )2kΩ 1.1( ) 3kΩ 0.9( )+ 2kΩ 1.1( )=1.21kΩ ID min = 1.350 −1.347 0.990 +1.21 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ V kΩ =1.39 μA ≅ 0 53 3.68 SPICE Input Results *Problem 3.68 NAME D1 V1 1 0 DC 4 MODEL DIODE R1 1 2 2K ID 1.09E-10 R2 2 0 2K VD 3.00E-01 R3 1 3 3K R4 3 0 2K D1 2 3 DIODE .MODEL DIODE D IS=1E-15 RS=0 .OP .END The diode is essentially off - VD = 0.3 V and ID = 0.109 nA. This result agrees with the CVD model results. 3.69 (a) (a) Diode is forward biased :V = 3− 0 = 3 V | I = 3− −7( ) 16kΩ = 0.625 mA (b) Diode is forward biased :V = −5 + 0 = −5 V | I = 5− −5( ) 16kΩ = 0.625 mA (c) Diode is reverse biased : I = 0 | V = −5+16kΩ I( )= −5 V | VD = −10 V (d ) Diode is reverse biased : I = 0 | V = 7 −16kΩ I( )= 7 V | VD = −10 V (b) (a) Diode is forward biased :V = 3 − 0.7 = 2.3 V | I = 2.3 − −7( ) 16kΩ = 0.581 mA (b) Diode is forward biased :V = −5 + 0.7 = −4.3 V | I = 5− −4.3( ) 16kΩ = 0.581 mA (c) Diode is reverse biased : I = 0 | V = −5+16kΩ I( )= −5 V | VD = −10 V (d ) Diode is reverse biased : I = 0 | V = 7 −16kΩ I( )= 7 V | VD = −10 V 3.70 (a) (a) Diode is forward biased :V = 3 − 0 = 3 V | I = 3− −7( ) 100kΩ =100 μA (b) Diode is forward biased :V = −5 + 0 = −5 V | I = 5− −5( ) 100kΩ =100 μA (c) Diode is reverse biased : I = 0 A | V = −5+100kΩ I( )= −5 V | VD = −10 V (d ) Diode is reverse biased : I = 0 A | V = 7 −100kΩ I( )= 7 V | VD = −10 V (b) 54 (a) Diode is forward biased :V = 3 − 0.6 = 2.4 V | I = 2.4 − −7( ) 100kΩ = 94.0 μA (b) Diode is forward biased :V = −5 + 0.6 = −4.4 V | I = 5− −4.4( ) 100kΩ = 94.0 μA (c) Diode is reverse biased : I = 0 | V = −5+100kΩ I( )= −5 V | VD = −10 V (d ) Diode is reverse biased : I = 0 | V = 7 −100kΩ I( )= 7 V | VD = −10 V 3.71 (a) (a) D1 on, D2 on : ID2 = 0 − −9( ) 22kΩ = 409μA | ID1 = 409μA − 6 − 0 43kΩ = 270μA D1 : 409 μA, 0 V( ) D2 : 270 μA, 0 V( ) (b) D1 on, D2 off : ID2 = 0 | ID1 = 6 − 043kΩ =140 μA | VD2 = −9 − 0 = −9V D1 : 140 μA, 0 V( ) D2 : 0 A, − 9 V( ) (c) D1 off, D2 on : ID1 = 0 | ID2 = 6 − −9( ) 65kΩ = 230 μA | VD1 = 6 − 43x10 3 ID2 = −3.92 V D1 : 0 A,−3.92 V( ) D2 : 230 μA,0 V( ) (d ) D1 on, D2 on : ID2 = 0 − −6( ) 43kΩ =140 μA | ID1 = 9 − 0 22kΩ −140 μA = 270 μA D1 : 140 μA,0 V( ) D2 : 270 μA,0 V( ) (b) (a) D1 on, D2 on : ID2 = -0.75− 0.75− −9( ) 22kΩ = 341μA | ID1 = 341μA − 6 − −0.75( ) 43kΩ =184μA D1 : 184 μA, 0.75 V( ) D2 : 341 μA, 0.75 V( ) (b) D1 on, D2 off : ID2 = 0 | ID1 = 6 − 0.7543kΩ =122μA | VD2 = −9 − 0.75 = −9.75V D1 : 122 μA, 0.75 V( ) D2 : 0 A, − 9.75 V( ) 55 (c) D1 off, D2 on : ID1 = 0 | ID2 = 6 − 0.75− −9( ) 65kΩ = 219μA | VD1 = 6 − 43x10 3 ID2 = −3.43V D1 : 0 A, − 3.43 V( ) D2 : 219 μA, 0.75 V( ) (d) D1 on, D2 on : ID2 = 0.75− 0.75− −6( ) 43kΩ =140μA | ID1 = 9 − 0.75 22kΩ − 400μA = 235μA D1 : 235 μA, 0.75 V( ) D2 : 140 μA, 0.75 V( ) 3.72 (a) (a) D1 and D2 forward biased ID2 = 0 − −9( ) 15 V kΩ = 600μA ID1 = ID2 − 6 − 0( ) 15 V kΩ = 200μA D1 : 0 V, 200 μA( ) D2 : 0 V, 600 μA( ) (b) D1 forward biased, D2 reverse biased ID1 = 6 − 015 V kΩ = 400μA VD2 = −9 − 0 = −9 V D1 : 0 V, 400 μA( ) D2 : -9 V, 0 A( ) (c) D1 reverse biased, D2 forward biased ID2 = 6V − −9V( ) 30kΩ = 500μA VD1 = 6 −15000ID2 = −1.50V D1 : −1.50 V, 0 A( ) D2 : 0 V, 500 μA( ) (d) D1 and D2 forward biased ID2 = 0 − −6( ) 15 V kΩ = 400μA ID1 = 9 − 0( ) 15 V kΩ − ID2 = 200μA D1 : 0 V, 200 μA( ) D2 : 0 V, 400 μA( ) (b) (a) D1 on, D2 on : ID2 = -0.75 − 0.75 − −9( ) 15kΩ = 500μA | ID1 = 500μA − 6 − −0.75( ) 15kΩ = 50.0μA D1 : 50.0 μA, 0.75 V( ) D2 : 500 μA, 0.75 V( ) 56 (b) D1 on, D2 off : ID2 = 0 | ID1 = 6 − 0.7515kΩ = 350μA | VD2 = −9 − 0.75 = −9.75V D1 : 350 μA, 0.75 V( ) D2 : 0 A, − 9.75 V( ) (c) D1 off, D2 on : ID1 = 0 | ID2 = 6 − 0.75 − −9( ) 30kΩ = 475μA | VD1 = 6 −15x10 3 ID2 = −1.13V D1 : 0 A, −1.13 V( ) D2 : 475 μA, 0.75 V( ) (d) D1 on, D2 on : ID2 = 0.75 − 0.75 − −6( ) 15kΩ = 400μA | ID1 = 9 − 0.75 15kΩ − 400μA =150μA D1 : 150 μA, 0.75 V( ) D2 : 400 μA, 0.75 V( ) 3.73 Diodes are labeled from left to right (a) D1 on, D2 off, D3 on : ID2 = 0 | ID1 = 10 − 03.3kΩ + 6.8kΩ = 0.990mA ID3 + 0.990mA = 0 − −5( ) 2.4kΩ → ID3 =1.09mA | VD2 = 5− 10 − 3300ID1( )= −1.73V D1 : 0.990 mA, 0 V( ) D2 : 0 mA, −1.73 V( ) D3 : 1.09 mA, 0 V( ) (b) D1 on, D2 off, D3 on : ID2 = 0 | ID3 = 0 ID1 = 10 − 0( )V 8.2kΩ +12kΩ = 0.495mA | VD2 = 5− 10 − 8200ID1( )= −0.941V ID3 = 0 − −5V( ) 10kΩ − ID1 = 0.005mA D1 : 0.495 mA, 0 V( ) D2 : 0 A, − 0.941 V( ) D3 : 0.005 mA, 0 V( ) 57 (c) D1 on, D2 on, D3 on ID1 = 0 − −10( ) 8.2kΩ V =1.22mA > 0 | I12K = 0 − 2( ) 12kΩ V = −0.167mA | ID2 = ID1 + I12K =1.05mA > 0 I10K = 2 − −5( ) 10kΩ V = 0.700mA | ID3 = I10K − I12K = 0.533mA > 0 D1 : 1.22 mA, 0 V( ) D2 : 1.05 mA, 0 V( ) D3 : 0.533 mA, 0 V( ) (d) D1 off, D2 off, D3 on : ID1 = 0, ID2 = 0 ID3 = 12 − −5( ) 4.7 + 4.7 + 4.7 V kΩ =1.21mA > 0 | VD1 = 0 − −5+ 4700ID3( )= −0.667V < 0 VD2 = 5− 12 − 4700ID3( )= −1.33V < 0 D1 : 0 A, − 0.667 V( ) D2 : 0 A, −1.33 V( ) D3 : 1.21 mA, 0 V( ) 3.74 Diodes are labeled from left to right (a) D1 on, D2 off, D3 on : ID2 = 0 | ID1 = 10 − 0.6 − −0.6( ) 3.3kΩ + 6.8kΩ = 0.990mA ID3 + 0.990mA = −0.6 − −5( ) 2.4kΩ → ID3 = 0.843mA | VD2 = 5− 10 − 0.6 − 3300ID1( )= −1.13V D1 : 0.990 mA, 0.600 V( ) D2 : 0 A, −1.13 V( ) D3 : 0.843 mA, 0.600V( ) (b) D1 on, D2 off, D3 off : ID2 = 0 | ID3 = 0 ID1 = 10 − 0.6 − −5( ) 8.2kΩ +12kΩ +10kΩV = 0.477mA | VD2 = 5− 10 − 0.6 − 8200ID1( )= −0.490V VD3 = 0 − −5+10000ID1( )= +0.230V < 0.6V so the diode is off D1 : 0.477 mA, 0.600 V( ) D2 : 0 A, − 0.490 V( ) D3 : 0 A, 0.230 V( ) (c) D1 on, D2 on, D3 on ID1 = −0.6 − −9.4( ) 8.2 V kΩ =1.07mA > 0 | I12K = −0.6 − 1.4( ) 12 V kΩ = −0.167mA ID2 = ID1 + I12K = 0.906mA > 0 | I10K = 1.4 − −5( ) 10 V kΩ = 0.640mA | ID3 = I10K − I12K = 0.807mA > 0 D1 : 1.07 mA, 0.600 V( ) D2 : 0.906 mA, 0.600 V( ) D3 : 0.807 mA, 0.600 V( ) 58 (d) D1 off, D2 off, D3 on : ID1 = 0, ID2 = 0 ID3 = 11.4 − −5( ) 4.7 + 4.7 + 4.7 V kΩ =1.16mA > 0 | VD1 = 0 − −5+ 4700ID3( )= −0.452V < 0 VD2 = 5− 11.4 − 4700ID3( )= −0.948V < 0 D1 : 0 A, − 0.452 V( ) D2 : 0 A, − 0.948 V( ) D3 : 1.16 mA, 0.600 V( ) 3.75 *Problem 3.75(a) (Similar circuits are used for the other three cases.) V1 1 0 DC 10 V2 4 0 DC 5 V3 6 0 DC -5 R1 2 3 3.3K R2 3 5 6.8K R3 5 6 2.4K D1 1 2 DIODE D2 4 3 DIODE D3 0 5 DIODE .MODEL DIODE D IS=1E-15 RS=0 .OP .END NAME D1 D2 D3 MODEL DIODE DIODE DIODE ID 9.90E-04 -1.92E-12 7.98E-04 VD 7.14E-01 -1.02E+00 7.09E-01 NAME D1 D2 D3 MODEL DIODE DIODE DIODE ID 4.74E-04 -4.22E-13 2.67E-11 VD 6.95E-01 -4.21E-01 2.63E-01 NAME D1 D2 D3 MODEL DIODE DIODE DIODE ID 8.79E-03 1.05E-03 7.96E-04 VD 7.11E-01 7.16E-01 7.09E-01 NAME D1 D2 D3 MODEL DIODE DIODE DIODE ID -4.28E-13 -8.55E-13 1.15E-03 VD -4.27E-01 -8.54E-01 7.18E-01 For all cases, the results are very similar to the hand analysis. 3.76 59 ID1 = 10 − −20( ) 10kΩ +10kΩ =1.50mA | ID2 = 0 ID3 = 0 − −10( ) 10kΩ =1.00mA | VD2 =10 −10 4 ID1 − 0 = −5.00V D1 : 1.50 mA, 0 V( ) D2 : 0 A,−5.00 V( ) D3 : 1.00 mA, 0 V( ) 3.77 *Problem 3.77 V1 1 0 DC -20 V2 4 0 DC 10 V3 6 0 DC -10 R1 1 2 10K R2 4 3 10K R3 5 6 10K D1 3 2 DIODE D2 3 5 DIODE D3 0 5 DIODE .MODEL DIODE D IS=1E-14 RS=0 .OP .END NAME D1 D2 D3 MODEL DIODE DIODE DIODE ID 1.47E-03 -4.02E-12 9.35E-04 VD 6.65E-01 -4.01E+00 6.53E-01 The simulation results are very close to those given in Ex. 3.8. 3.78 VTH = 24V 3.9kΩ3.9kΩ +11kΩ = 6.28V | RTH =11kΩ 3.9kΩ = 2.88kΩ IZ = 6.28 − 42.88kΩ = 0.792mA > 0 | IZ ,VZ( )= 0.792 mA,4 V( ) 60 3.79 −6.28 = 2880ID + VD | ID = 0,VD = −6.28V | VD = 0, ID = −6.282880 = −2.18mA -1 mA -2 mA Q-point vD i D -6 -5 -4 -3 -2 -1 Q-Point: (-0.8 mA, -4 V) 3.80 IS = 27 − 915kΩ =1.20mA → IL <1.20 mA | RL > 9V 1.2mA = 7.50 kΩ 3.81 IS = 27 − 915kΩ =1.20mA | P = 9V( )1.20mA( )=10.8 mW 3.82 IZ = VS −VZRS − VZ RL = VS RS −VZ 1RS + 1 RL ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ | PZ = VZ IZ IZ nom = 30V 15kΩ − 9V 1 15kΩ + 1 10kΩ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.500 mA | PZ nom = 9V 0.500mA( )= 4.5 mW IZ max = 30V 1.05( ) 15kΩ 0.95( )− 9V 0.95( ) 1 15kΩ 0.95( )+ 1 10kΩ(1.05) ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ = 0.796 mA PZ max = 9V .95( )0.796mA( )= 6.81 mW IZ min = 30V 0.95( ) 15kΩ 1.05( )− 9V 1.05( ) 1 15kΩ 1.05( )+ 1 10kΩ(0.95) ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ = 0.215 mA PZ min = 9V 1.05( )0.215mA( )= 2.03 mW 3.83 (a) VTH = 60V 100Ω150Ω +100Ω = 24.0V | RTH =150Ω 100Ω = 60Ω | IZ = 24 −15 60 =150 mA P = 15IZ = 2.25 W | (b) IZ = 60 −15150 = 300 mA | P = 15IZ = 4.50 W 61 3.84 IZ = VS −VZRS − VZ RL = VS RS −VZ 1RS + 1 RL ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ | PZ = VZ IZ IZ nom = 60 −15( )V 150Ω − 15V 100Ω =150 mA | PZ nom =15V 150mA( )= 2.25 W IZ max = 60V 1.1( ) 150Ω 0.90( )−15V 0.90( ) 1 150Ω 0.90( )+ 1 100Ω(1.1) ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ = 266 mA PZ max =15V 0.90( )266mA( )= 3.59 W IZ min = 60V 0.90( ) 150Ω 1.1( ) −15V 1.1( ) 1 150Ω 1.1( )+ 1 100Ω(0.9) ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ = 43.9 mA PZ min =15V 1.1( )43.9mA( )= 0.724 W 3.85 Using MATLAB, create the following m-file with f = 60 Hz: function f=ctime(t) f=5*exp(-10*t)-6*cos(2*pi*60*t)+1; Then: fzero('ctime',1/60) yields ans = 0.01536129698461 and �T = (1/60)-0.0153613 = 1.305 ms. ΔT = 1 120π 2Vr VP | Vr = ITC = 5 0.1 60( )= 0.8333V ΔT = 1 120π 2 0.8333( ) 6 = 1.40 ms 3.86 VD = nVT ln 1+ IDIS ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 2 0.025V( )ln 1+ 48.6A10−9 A ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =1.230 V 62 3.87 Von = nVT ln 1+ IDIS ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ | VD = Von + ID RS VD =1.6 0.025V( )ln 1+ 100A10−8 A ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ +100A 0.01Ω( )=1.92 V Pjunction ≅ VonIDC = Von IPΔT2T = 0.92V 100A 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1ms 16.7ms ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 2.75 W PR ≅ 43 T ΔT ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ IDC 2 RS = 43 16.7ms 1ms ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 3A( ) 2 0.01Ω = 2.00 W Ptotal = 4.76 W 3.88 VDC = 1T v t( )dt0 T∫ = 1T VP −Von( )T − TVr2 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = VP −Von( )− 0.05 VP −Von( ) 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = 0.975 VP −Von( ) VDC = 0.975 18V( )=17.6 V 3.89 PD = 1T iD 2 t( )RS dt 0 T∫ = 1T IP2 1− tΔT ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 RS dt 0 ΔT∫ PD = IP 2 RS T 1− 2tΔT + t 2 ΔT 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 dt 0 ΔT∫ = IP2 RST t − t 2 ΔT + t 3 3ΔT 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 0 ΔT PD = IP 2 RS T ΔT − ΔT + ΔT 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 1 3 IP 2 RS ΔT T ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 3.90 Using SPICE with VP = 10 V. 0s 10ms 20ms 30ms 40ms 50ms t 15V 10V 5V 0V -5V -10V -15V Voltage 63 3.91 (a) Vdc = − VP −Von( )= − 6.3 2 −1( )= −7.91V (b) C = I TVr = 7.910.55 10.5 160 =1.05F (c) PIV ≥ 2VP = 2 ⋅ 6.3 2 =17.8V (d) Isurge = ωCVP = 2π 60( )1.05( )6.3 2( )= 3530 A (e) ΔT = 1ω 2Vr VP = 1 2π 60( ) 2 .25( ) 6.3 2 = 0.628ms | IP = Idc 2TΔT = 7.91 .5 2 60 1 .628ms = 841A 3.92 VO nom = − VP −Von( )= − 6.3 2 −1( )= −7.91V VO max = − VPmax −Von( )= − 6.3 1.1( ) 2 −1[ ]= −8.80V VO min = − VPmin −Von( )= − 6.3 0.9( ) 2 −1[ ]= −7.02V 3.93 *Problem 3.93 VS 1 0 DC 0 AC 0 SIN(0 10 60) D1 2 1 DIODE R 2 0 0.25 C 2 0 0.5 .MODEL DIODE D IS=1E-10 RS=0 .OPTIONS RELTOL=1E-6 .TRAN 1US 80MS .PRINT TRAN V(1) V(2) I(VS) .PROBE V(1) V(2) I(VS) .END V(2) *REAL(Rectifier)* Time (s)Circuit3_93b-Transient-8 +0.000e+000 +10.000m +20.000m +30.000m +40.000m +50.000m +60.000m +70.000m -10.000 -5.000 +0.000e+000 +5.000 +10.000 SPICE Graph Results: VDC = 9.29 V, Vr = 1.05 V, IP = 811 A, ISC = 1860 A Vdc = − VP −Von( )= − 10 −1( )= −9.00V | Vr = I TC = 9.00V0.25Ω 160s 10.5F =1.20V ISC = ωCVP = 2π 60( )0.5( )10( )=1890A | ΔT = 1ω 2VrVP = 1 2π 60( ) 2 1.2( ) 10 =1.30ms IP = Idc 2TΔT = 9 0.25 2 60 1 1.3ms = 923A 64 V(1) V(2) *REAL(Rectifier)* Time (s)Circuit3_93b-Transient-11 (Amp) -10.000 -5.000 +0.000e+000 +5.000 +10.000 +0.000e+000 +20.000m +40.000m +60.000m +80.000m +100.000m +120.000m +140.000m SPICE Graph Results: VDC = -6.55 V, Vr = 0.58 V, IP = 150 A, ISC = 370 A Note that a significant difference is caused by the diode series resistance. 3.94 (a) Vdc = − VP −Von( )= − 6.3 2 −1( )= −7.91V (b) C = I TVr = 7.910.25 10.5 1400 = 0.158F (c) PIV ≥ 2VP = 2 ⋅ 6.3 2 =17.8V (d ) Isurge = ωCVP = 2π 400( )0.158( )6.3 2( )= 3540 A (e) ΔT = 1ω 2Vr VP = 1 2π 400( ) 2 .25( ) 6.3 2 = 94.3μs | IP = Idc 2TΔT = 7.91 .5 2 400 1 94.3μs = 839A 3.95 (a) Vdc = − VP −Von( )= − 6.3 2 −1( )= −7.91V (b) C = I TVr = 7.910.25 10.5 1105 = 633μF (c) PIV ≥ 2VP = 2 ⋅ 6.3 2 =17.8V (d ) Isurge = ωCVP = 2π 105( )633μF( )6.3 2( )= 3540A (e) ΔT = 1ω 2Vr VP = 1 2π 105( ) 2 .25( ) 6.3 2 = 0.377μs | IP = Idc 2TΔT = 7.91 .5 2 105 1 0.377μs = 839A 3.96 65 (a) C = I T Vr = 1 3000 0.01( ) 1 60 = 556 μF (b) PIV ≥ 2VP = 2 ⋅ 3000 = 6000V (c) Vrms = 3000 2 = 2120 V (d ) ΔT = 1ω 2Vr VP = 1 2π 60( ) 2 0.01( )= 0.375ms IP = Idc 2TΔT =1 2 60 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 0.375ms ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 88.9A (e) Isurge = ωCVP = 2π 60( )556μF( )3000( )= 629A 3.97 Assuming Von = 1 V: C = VP −Von Vr T 1 R = 1 0.025 1 60 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 30 3.3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 6.06 F | PIV = 2VP = 2 3.3 +1( )V = 8.6 V | Vrms = 3.3 +12 = 3.04 V ΔT = 1ω 2T RC VP −Von VP = 1 2π 60( ) 2 0.110Ω 6.06F( ) 1 60 s ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 3.3V 4.3V ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.520 ms IP = Idc 2TΔT = 30 2 60 s ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 0.520ms ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =1920 A | Isurge = ωCVP = 2π 60 / s( )6.06F( )4.3V( )= 9820 A 3.98 0s 5ms 10ms 15ms 20ms 25ms 30ms 40V 20V 0V -20V v1 vS vO Time VDC = 2(VP - Von) = 2(17 - 1) = 32 V. 3.99 *Problem 3.99 VS 2 1 DC 0 AC 0 SIN(0 1500 60) D1 2 3 DIODE D2 0 2 DIODE C1 1 0 500U C2 3 1 500U RL 3 0 3K .MODEL DIODE D IS=1E-15 RS=0 .OPTIONS RELTOL=1E-6 .TRAN 0.1MS 100MS .PRINT TRAN V(2,1) V(3) I(VS) 66 .PROBE V(3) V(2,1) I(VS) .END 0s 20ms 40ms 60ms 80ms 100ms Time 4.0kV 3.0kV 2.0kV 1.0kV 0V -1.0kV -2.0kV vO vS Simulation Results: VDC = 2981 V, Vr = 63 V The doubler circuit is effectively two half-wave rectifiers connected in series. Each capacitor is discharged by I = 3000V/3000� = 1 A for 1/60 second. The ripple voltage on each capacitor is 33.3 V. With two capacitors in series, the output ripple should be 66.6 V, which is close to the simulation result. 3.100 (a) Vdc = − VP −Von( )= − 15 2 −1( )= −20.2 V (b) C = I Vr T2⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 20.2V0.5Ω 10.25V⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1120s⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =1.35 F (c) PIV ≥ 2VP = 2 ⋅15 2 = 42.4 V (d ) Isurge = ωCVP = 2π 60( )1.35( )15 2( )=10800 A (e) ΔT = 1ω 2Vr VP = 1 2π 60( ) 2 .25( ) 15 2 = 0.407 ms | IP = Idc TΔT = 20.2V 0.5Ω 1 60 s ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 0.407ms =1650 A 3.101 (a) Vdc = − VP −Von( )= − 9 2 −1( )= −11.7 V (b) C = I Vr T2⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 11.7V0.5Ω 10.25V⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1120s⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.780 F (c) PIV ≥ 2VP = 2 ⋅ 9 2 = 25.5 V (d ) Isurge = ωCVP = 2π 60( )0.780( )9 2( )= 3740 A (e) ΔT = 1ω 2Vr VP = 1 2π 60( ) 2 .25( ) 9 2 = 0.526 ms | IP = Idc TΔT = 11.7V 0.5Ω 1 60 s ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 0.407ms ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 958A 67 3.102 *Problem 3.102 VS1 1 0 DC 0 AC 0 SIN(0 14.14 400) VS2 0 2 DC 0 AC 0 SIN(0 14.14 400) D1 3 1 DIODE D2 3 2 DIODE C 3 0 22000U R 3 0 3 .MODEL DIODE D IS=1E-10 RS=0 .OPTIONS RELTOL=1E-6 .TRAN 1US 5MS .PRINT TRAN V(1) V(2) V(3) I(VS1) .PROBE V(1) V(2) V(3) I(VS1) .END 0s 1.0ms 2.0ms 3.0ms 4.0ms 5.0ms Time 20V 10V 0V -10V -20V vS vO Simulation Results: VDC = -13.4 V, Vr = 0.23 V, IP = 108 A VDC = VP −Von =10 2 − 0.7 =13.4 V | Vr = 13.43 1 800 1 22000μF = 0.254 V ΔT = 1 120π 2Vr VP = 1 120π 2 0.254( ) 14.1 = 0.504 ms IP = Idc TΔT = 13.4V 3Ω 1 60 s 1 0.504 ms =150 A Simulation with RS = 0.02 Ω. V(1) V(2) *REAL(Rectifier)* Time (s)Circuit3_102-Transient-15 -15.000 -10.000 -5.000 +0.000e+000 +5.000 +10.000 +15.000 +0.000e+000 +2.000m +4.000m +6.000m +8.000m +10.000m +12.000m +14.000m Simulation Results: VDC = -12.9 V, Vr = 0.20 V, IP = 33.3 A, ISC = 362 A. RS results in a significant reduction in the values of IP and ISC. 68 3.103 (a) C = VP −Von Vr T 1 R = 1 0.025 1s 120 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 30A 3.3V ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 3.03 F (b) PIV = 2VP = 2 3.3+1( )V = 8.6 V (c) Vrms = 3.3+1 2 = 3.04 V (d) ΔT = 1ω 2Vr VP = 1 2π 60( ) 2 0.025( )3.3( ) 4.3 = 0.520 ms (e) IP = Idc TΔT = 30A 1 60 s ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 0.520ms ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 962 A | Isurge = ωCVP = 2π 60 / s( )3.03F( )4.3V( )= 4910 A
Compartilhar