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 theyre 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 20201960 ( ) =14.5 x 1012=14.5 Tb/chip 1.3 (a) B2B1 =19.97x100.1977 Y21960( )19.97x100.1977 Y11960 ( ) =100.1977 Y2Y1 ( ) so 2 =100.1977 Y2Y1 ( )Y2 Y1 =log20.1977 =1.52 years (b) Y2 Y1 =log100.1977 = 5.06 years 1.4 N =1610x100.1548 20201970 ( ) = 8.85 x 1010 transistors/P 1.5 N2N1 =1610x100.1548 Y21970 ( )1610x100.1548 Y11970( ) =100.1548 Y2Y1 ( )(a) Y2 Y1 =log20.1548 =1.95 years(b) Y2 Y1 =log100.1548 = 6.46 years 1.6 . F = 8.00x100.05806 20201970 ( )m =10 nmNo, 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/P 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 = 75x106tubes( )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 =10.24V212bits =10.24V4096bits = 2.500 mV VMSB=10.24V2 = 5.120V1001001001102 = 211+ 28+ 25+ 22+ 2 = 234210 VO = 2342 2.500mV( )= 5.855 V 1.11 VLSB =5V28bits =5V256bits =19.53mVbit and 2.77V19.53mVbit=142 LSB14210 = 128 + 8 + 4 + 2( ) =100011102 10 1.12 VLSB =2.5V210 bits =2.5V1024 bits = 2.44mVbit01011011012 = 28+ 26+ 25+ 23+ 22+ 20( )10 = 36510 VO = 365 2.5V1024 = 0.891 V 1.13 VLSB =10V214bits = 0.6104mVbit and 6.83V10V214bits( )=11191 bits1119110 = 8192 + 2048 + 512 + 256 +128 + 32 +16 + 4 + 2 +1( )101119110 =101011101101112 1.14A 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 =5.12V4096 bits =1.25mVbit and VO = 1011101110112( )VLSB VLSB2VO = 211+ 29+ 28+ 27+ 25+ 24+ 23+ 2 +1( )101.25mV 0.0625VVO = 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 k, R2= 47 k and R3 = 180 k. V+-V1V2+-R1R2R3I3I2 V1 =10V22k22k+ 47k180k( )=10V22k22k+ 37.3k = 3.71 VV2 =10V37.3k22k+ 37.3k = 6.29 V Checking : 6.29 + 3.71 = 10.0 VI2 = I1180k47k+180k =10V22k+ 37.3k 180k47k+180k =134 AI3 = I147k47k+180k =10V22k+ 37.3k 47k47k+180k = 34.9 AChecking : I1 =10V22k+ 37.3k =169A 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 k, R2= 33 k and R3 = 11 k. V+-V1V2+-R1R2R3I3I2 V1 =18V56k56k+ 33k11k( )=15.7 V V2 =18V33k11k56k+ 33k11k( )= 2.31 VChecking : V1 +V2 =15.7 + 2.31=18.0 V which is correct.I1 =18V56k+ 33k11k( )= 280 A I2 = I111k33k+11k = 280 A( )11k33k+11k = 70.0 AI3 = I133k33k+11k = 280 A( )33k33k+11k = 210 A Checking : I2 + I3 = 280 A 1.22 I1 = 5mA5.6k+ 3.6k( )5.6k+ 3.6k( )+ 2.4k = 3.97 mA I2 = 5mA2.4k9.2k+ 2.4k =1.03 mAV3 = 5mA 2.4k 9.2k( )3.6k5.6k+ 3.6k = 3.72V Checking : I1 + I2 = 5.00 mA and I2R2 =1.03mA 3.6k( )= 3.71 V 1.23 I2 = 250A150k150k+150k =125 A I3 = 250A150k150k+150k =125 AV3 = 250A 150k150k( )82k68k+ 82k =10.3VChecking : I1 + I2 = 250 A and I2R2 =125A 82k( )=10.3 V 1.24 1-6 R. C. Jaeger & T. N. Blalock 6/9/06 R1+-vgmvvsvth+- Summing currents at the output node yields:v5x104+.002v = 0 so v = 0 and vth = vs v = vs R1vx+-vg vmix Summing currents at the output node :ix = v5x104 0.002v = 0 but v = vxix =vx5x104 + 0.002vx= 0 Rth =vxix=11R1 + gm= 495 Thvenin equivalent circuit: vs495 1-7 6/9/06 1.25 The Thvenin equivalent resistance is found using the same approach as Problem 1.24, and Rth =14k +.025 1= 39.6 R1vs+-vgmvin The short circuit current is:in =v4k + 0.025v and v = vs in =vs4k + 0.025vs = 0.0253vs Norton equivalent circuit: 39.6 0.0253vs 1.26 1-8 R. C. Jaeger & T. N. Blalock 6/9/06 (a) R1R2ivsi+-vth Vth =Voc = i R2 but i = vsR1 and Vth = vs R2R1 =120 vs 39k100k = 46.8 vs R1R2iiRthvxix Rth =vxix ; ix =vxR2 + i but i = 0 since VR1 = 0. Rth = R2 = 39 k. Thvenin equivalent circuit: 58.5vs39 k (b) R1R2iisi+-vth Vth =Voc = i R2 where i + bi + is = 0 and Vth = is +1 R2 = 38700 is 1-9 6/9/06 R1R2iiRthvx Rth =vxix ; ix =vxR2 + i but i + i = 0 so i = 0 and Rth= R2 = 39 k Thvenin equivalent circuit: 39 k 38700is 1.27iR1R2vsiin in = i but i = vsR1 and in = R1vs =10075kvs =1.33 x 103 vs From problem 1.26(a), Rth = R2 = 56 k. Norton equivalent circuit: 56 k 0.00133vs 1.28 1-10 R. C. Jaeger & T. N. Blalock 6/9/06 R1R2ivsiis is =vsR1 i =vsR1 + vsR1 = vs +1R1 R =vsis=R1 +1 =100k81 =1.24 1.29 The open circuit voltage is vth= gmv R2 and v = +isR1.vth= gmR1R2is = 0.0025( )105( )106( )is= 2.5 x 108isFor i = 0, v = 0 and R = R =1 Ms th 2 1.30 5 V3 V0f (Hz)500 1000 0 1.31 2 V0f (kHz)9 10 11 v = 4sin 20000t( )sin 2000t( )=42cos 20000t + 2000t( )+ cos 20000t 2000t( ) [ ]v = 2cos 22000t( )+ 2cos 18000t( ) 1.32 A=236o10500 = 2x10536o A = 2x105 A = 36o 1-11 6/9/06 1.33 (a) A=10245o2x1030o = 545o (b) A=10112o1030o =10012o 1.34 (a) Av = R2R1 = 620k14k = 44.3 (b) Av = 180k= 10.0 (c) Av = 62k1.6k = 38.8 18k 1.35 vot( )= R2R1vst( )= 90.1 sin 750t ( ) mVIS =VSR1 =0.01V910 =11.0A and is = 11.0 sin 750t( ) A 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 voR2 + i +vR1 = 0 or vs voR2 +vsR1 = 0 and Av =vovs=1+R2R1 1.38 Writing a nodal equation at the inverting input terminal of the op amp gives v1 vR1+v2 vR2 = i +v voR3 but v- = v+ = 0 and i- = 0vo = R3R1v1 R3R2v2 = 0.255sin3770t 0.255sin10000t volts 1.39vO = VREFb12 +b24 +b38 (a) vO = 502 +14 +18 = 1.875V (b) vO = 512 +04 +08 = 2.500V b1b2b3vO (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 Amplitudef106 kHz 1-12 R. C. Jaeger & T. N. Blalock 6/9/06 1-13 6/9/06 1.41 Band-pass amplifier f201 kHz 5 kHzAmplitude 1.42 High-pass amplifier f10 kHz16Amplitude 1.43 vOt( )=10x5sin 2000t( )+10x3cos 8000t( )+ 0x3cos 15000t( )vOt( )= 50sin 2000t( )+ 30cos 8000t( ) [ ]volts 1.44 vOt( )= 20x0.5sin 2500t( )+ 20x0.75cos 8000t( )+ 0x0.6cos 12000t( )vOt( )= 10.0sin 2500t( )+15.0cos 8000t( ) [ ]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)-2-10120 1 2 3 4 5x10-3 (b)-10-505100 1 2 3 4 5x10-3 1-14 R. C. Jaeger & T. N. Blalock 6/9/06 1-15 6/9/06 (c)-10-505100 1 2 3 4 5x10-3 (d)-10-505100 1 2 3 4 5x10-3 1.47 (a) 3000 1.01( ) R 3000 1+.01( ) or 2970 R 3030 (b) 3000 1.05( ) R 3000 1+.05( ) or 2850 R 3150(c) 3000 1.10( ) R 3000 1+.10( ) or 2700 R 3300 1.48 Vnom = 2.5V V 0.05V T =0.052.50 = 0.0200 or 2.00% 1.49 20000F 1.5( ) C 20000F 1+.2( ) or 10000F R 24000F 1.50 8200 10.1( ) R 8200 1+ 0.1( ) or 7380 R 9020 The resistor is within the allowable range of values. 1.51 (a) 5V 1.05( )V 5V 1+.05( ) or 5.75V V 5.25VV = 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 = RT =6562 6066100 0 = 4.96 oC 1-16 R. C. Jaeger & T. N. Blalock 6/9/06 Rnom = R0oC + TCR T( )= 6066 + 4.96 27( )= 6200 1-17 6/9/06 1.53 V2+-V+-V1R2I3I2 R1R3 Let RX = R2R3 then V1 =VR1R1 + RX=V11+RXR1RXmin=47k 0.9( )180k( )0.9( )47k 0.9( )+180k 0.9( )= 33.5k RXmax=47k1.1( )180k( )1.1( )47k1.1( )+180k1.1( )= 41.0kV1max=10 1.05( )1+33.5k22k1.1( )= 4.40V V1min=10 0.95( )1+41.0k22k 0.9( )= 3.09VI1 =VR1 + RX and I2 = I1R3R2 + R3 =VR1 + R2 +R1R2R3 I2max=10 1.05( )22000 0.9( )+ 47000 0.9( )+22000 0.9( )47000( )0.9( )180000 1.1( ) =158 A I2min=10 0.95( )22000 1.1( )+ 47000 1.1( )+22000 1.1( )47000( )1.1( )180000 0.9( ) =114 AI3 = I1R2R2 + R3 =VR1 + R3 +R1R3R2 I3max=10 1.05( )22000 0.9( )+180000 0.9( )+22000 0.9( )180000( )0.9( )47000 1.1( ) = 43.1 A I3min=10 0.95( )22000 1.1( )+180000 1.1( )+22000 1.1( )180000( )1.1( )47000 0.9( ) = 28.3 A 1.54 I1 = IR2 + R3R1 + R2 + R3 = I11+R1R2 + R3 and similarly I2 = I11+R2 + R3R1I1max=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 mAI2max=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 mAV3 = I2R3 =I1R1 +1R3 +R2R1R3V3max=5 1.02( )12400 1.05( )+13600 1.05( )+5600 0.95( )2400 1.05( )3600( )1.05( )= 4.18 VV3min=5 0.98( )12400 0.95( )+13600 0.95( )+5600 1.05( )2400 0.95( )3600( )0.95( )= 3.30 V 1.55 From Prob. 1.24 : Rth =1gm +1R1 Rthmax=10.002 0.8( )+15x1041.2( )= 619 Rthmin=10.002 1.2( )+15x1040.8( )= 412 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()( ) V1I2I3Min 3.23 V 116 A 29.9 A Max 3.71 V 151 A 40.9 A Average 3.71 V 133 A 35.1 A 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()( ) I1I2V3Min 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 k) = 36.5 V (b) 36 V (c) (0.1021 A)(97.80 k) = 9.99 V; 10 V CHAPTER 2 2.1 Based upon Table 2.1, a resistivity of 2.6 -cm < 1 m-cm, and aluminum is a conductor. 2.2 Based upon Table 2.1, a resistivity of 1015 -cm > 105 -cm, and silicon dioxide is an insulator. 2.3 Imax= 107Acm2 5m( )1m( )108cm2m2 = 500 mA 2.4 =T xEBT nGi 5310 62 . 8exp For silicon, B = 1.08 x 1031 and EG = 1.12 eV: ni = 2.01 x10-10/cm3 6.73 x109/cm3 8.36 x 1013/cm3. For germanium, B = 2.31 x 1030 and EG = 0.66 eV: ni = 35.9/cm3 2.27 x1013/cm3 8.04 x 1015/cm3. 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 = 1014/cm3 for T = 506 K ni = 1016/cm3 for T = 739 K 2.6 ni= BT3exp EG8.62x105T with B = 1.27x1029K3cm6 T = 300 K and EG = 1.42 eV: ni = 2.21 x106/cm3T = 100 K: ni = 6.03 x 10-19/cm3 T = 500 K: ni = 2.79 x1011/cm3 20 2.7 vn= nE = 700cm2V s 2500Vcm = 1.75x106cmsvp= +pE = +250cm2V s 2500Vcm = +6.25x105cmsjn= qnvn= 1.60x1019C( )10171cm3 1.75x106cms = 2.80x104Acm2jp= qnvp= 1.60x1019C( )1031cm3 6.25x105cms =1.00x1010Acm2 2.8 ni2= BT3exp EGkT B = 1.08x10311010( )2=1.08x1031T3exp 1.128.62x105T 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 scmcm Ccm AQjv52210/ 01 . 0/ 1000 == = 2.10 2 26 734 10 4sec10 4 . 0cmMAcmAxcmcmCQv j = == = 21 2.11 vn= nE = 1000cm2V s 2000Vcm = +2.00x106cmsvp= +pE = +400cm2V s 2000Vcm = 8.00x105cmsjn= qnvn= 1.60x1019C( )1031cm3 +2.00x106cms = 3.20x1010Acm2jp= qnvp= 1.60x1019C( )10171cm3 8.00x105cms = 1.28x104Acm2 2.12 ( ) ( ) ( ) V 100 10 10 10 b 500010 105 4 54== = =cm xcmVVcmVcm xVE a 2.13 jp= qpvp= 1.60x1019C( )1019cm3 107cms =1.60x107Acm2ip= jpA= 1.60x107Acm2 1x104cm( )25x104cm( )= 4.00 A 2.14 For intrinsic silicon, = q nni+ pni ( )= qnin+ p ( ) 1000 cm( )1 for a conductorniq n+ p( )=1000 cm( )11.602x1019C 100 + 50( )cm2v sec=4.16x1019cm3ni2=1.73x1039cm6= BT3exp EGkT withB =1.08x1031K3cm6, 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, = q nni+ pni ( )= qnin+ p ( ) 105cm( )1 for an insulatorniq n+ p( )=105cm( )11.602x1019C( )2000 + 750( )cm2v sec =2.270x1010cm3ni2=5.152x1020cm6= BT3exp EGkT withB =1.08x1031K3cm6, 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 SiSiBSi Si SiPDonor 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 , 200 02 . 0 100002cmVcmcmAjjE = = = = 2.21 jndrift= qnnE = qnvn= 1.602x1019( )1016( )Ccm3 107cms =16000Acm2 2.22 N =1015atomscm3 1m( )10m( )0.5m( )104cmm 3= 5, 000 atoms 2.23 NA> ND: NA ND=10151014= 9x1014/cm3If we assume NA ND>> 2ni=1014/ cm3:p = NA ND= 9x1014/cm3 | n =ni2p=2510269x1014= 2.78x1012/cm3If we use Eq. 2.12: p =9x1014 9x1014( )2+ 4 5x1013( )22= 9.03x1014and n = 2.77x1012/cm3. The answers are essentially the same. 2.24 3 51622 23 143 11 3 16 16 1610 50 210 41010 410 2 2 10 4 10 10 5/cm x .x pn | n /cm x N N p/cm x n /cm x x N : N N NiD Ai D A D A= = = = == >> = = > 2.25 ND> NA: ND NA= 3x10172x1017=1x1017/cm32ni= 2x1017/cm3; Need to use Eq. (2.11)n =1017 1017( )2+ 4 1017( )22=1.62x1017/cm3p =ni2n=10341.62x1017= 6.18x1016/cm3 24 2.26 ND NA= 2.5x1018/ cm3Using Eq. 2.11: n =2.5x1018 2.5x1018( )2+ 4 1010( )22Evaluating this with a calculator yields n = 0, and n =ni2p= .No, the result is incorrect because of loss of significant digitswithin 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. 33 186 20 23 18i3 18 3 107 . 1610 610 and 10 6 So 2n >> / 10 6 and 10 re, temperatu room At /cm/cm x/cmpnn /cm x pcm x N N /cm niD A i= = = == = (b) At 200K, ni2=1.08x1031200( )3exp 1.128.62x105200( ) = 5.28x109/cm6ni= 7.27x104/cm3 NA ND>> 2ni, so p = 6x1018/cm3 and n =5.28x1096x1018= 8.80x1010/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. 33 176 20 23 17i3 17 3 10i33310 310 and 10 3 So 2n >> / 10 3 and / 10 n re, temperatu room At /cm/cm x/cmnnp /cm x ncm x N N cmiA D= = = == = (b) At 250K, ni2=1.08x1031250( )3exp 1.128.62x105250( ) = 4.53x1015/cm6ni= 6.73x107/cm3 ND NA>> 2ni, so n = 3x1017/ cm3 and n =4.53x10153x1017= 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 33 186 20 23 18i3 18 3 10i7 1610 610 and 10 6 So 2n >> / 10 6 and / 10 n re, temperatu room At (b)/cm ./cm x/cmpnn /cm x pcm x N N cmiD A= = = == = 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. 33 176 20 23 17i3 17 3 1033310 310 and 10 3 So 2n >> / 10 3 and 10 re, temperatu room At (b)/cm/cm x/cmpnn /cm x pcm x N N /cm niD A i= = = == = 2.31 ND = 4 x 1016/cm3. Assume NA = 0, since it is not specified. ND> NA: material is n - type | ND NA= 4x1016/ cm3>> 2ni= 2x1010/ cm3n = 4x1016/ cm3 | p =ni2n=10204x1016= 2.5x103/ cm3ND+ NA= 4x1016/ cm3 | Using Fig. 2.13, n=1030cm2V s and p= 310cm2V s =1qnn=11.602x1019C( )1030cm2V s 4x1016cm3 = 0.152 cm 26 2.32 NA = 1018/cm3. Assume ND = 0, since it is not specified. NA> ND: material is p - type | NA ND=1018/ cm3>> 2ni= 2x1010/ cm3p = 1018/ cm3 | n =ni2p=10201018=100/ cm3ND+ NA=1018/ cm3 | Using Fig. 2.13, n= 375cm2V s and p=100cm2V s =1qpp=11.602x1019C 100cm2V s 1018cm3 = 0.0624 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. NA> ND: material is p - type | NA ND= 7x1019/cm3>> 2ni= 2x1010/cm3p = 7x1019/cm3 | n=ni2p=10207x1019=1.43/cm3ND+ NA= 7x1019/ cm3 | Using Fig. 2.13, n=120cm2V s and p= 60cm2V s =1qpp=11.602x1019C 60cm2V s 7x1019cm3 =1.49 mcm 2.34 Phosphorus is a donor : ND= 5.5x1016/ cm3 | Boron is an acceptor : NA= 4.5x1016/ cm3ND> NA: material is n - type | ND NA=1016/ cm3>> 2ni= 2x1010/ cm3n =1016/cm3 | p =ni2p=10201016=104/cm3ND+ NA=1017/ cm3 | Using Fig. 2.13, n= 800cm2V s and p= 230cm2V s =1qnn=11.602x1019C 800cm2V s 1016cm3 = 0.781 cm 27 2.35 =1qpp | pp =11.602x1019C( )0.054cm( )=1.16x1020V cm s An iterative solution is required. Using the equations in Fig. 2.8: NA p p 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 =1qpp | pp =11.602x1019C( )0.75cm( )=8.32x1018V cm s An iterative solution is required. Using the equations in Fig. 2.8: NA p p 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 =1qnn | nn nND=11.602x1019C( )2cm( )=3.12x1018V cm s An iterative solution is required. Using the equations in Fig. 2.8: ND n nn 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) =1qnn | nn nND=11.602x1019C( )0.001cm( )=6.24x1021V cm s An iterative solution is required. Using the equations in Fig. 2.8: ND n nn 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) =1qpp | pp pNA=11.602x1019C( )0.001cm( )=6.24x1021V cm s An iterative solution is required using the equations in Fig. 2.8: NA p p 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: =1qpp | pp pNA=11.602x1019C( )1cm( )=6.25x1018V cm s An iterative solution is required. Using the equations in Fig. 2.8: NA p p 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: =1qpp | pp pNA=11.602x1019C( )0.25cm( )=2.5x1019V cm s NAp p 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 = 1.1x1017- 1.7x1016 = 9.3 x 1016/cm3(b) If donors are added: ND ND + NA n ND - NA nn 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 1016/cm3 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 n = 1250 cm2/V-s from Fig. 2.8. = qnn qnND= 1.602x1019C( )1250( )1016( )=2.00cm Now we add acceptors until = 5.0 (-cm) -1: = qpp | pp pNA ND ( )=5 cm( )11.602x1019C=3.12x1019V cm s NA ND + NA p NA - ND p 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 p = 405 cm2/V-s from Fig. 2.8. = qpp qpNA= 1.602x1019C( )405( )1016( )=0.649cm Now we add donors until = 5.5 (-cm) -1: = qnn | nn nND NA ( )=5.5 cm( )11.602x1019C=3.43x1019V cm s ND ND + NA n ND - NA p 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=1.38x1023T1.602x1019= 8.62x105T 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 = qDndndx = qVTndndxj = 1.602x1019C( )0.025V( )350cm2V s 101800 104 1cm4= 14.0kAcm2 2.46 j = qDpdpdx= 1.602x1019C( )15cm2s 1019/ cm32x104cm exp x2x104cm j =1.20x105exp 5000xcm Acm2I 0( )= j 0( )A= 1.20x105Acm2 10m2( )108cm2m2 =12.0 mA 32 2.47 jp= qppE qDpdpdx= qpp E VT1pdpdx = 0 E =VT1pdpdxE VT1NAdNAdx= 0.0251022exp 104x( )1014+1018exp 104x( )E 0( )= 0.02510221014+1018= 250VcmE 5x104cm( )= 0.0251022exp 5( )1014+1018exp 5( )= 246Vcm 2.48 jndrift= qnnE = 1.60x1019C( )350cm2V s 1016cm3 20Vcm = 11.2Acm2jpdrift= qppE = 1.60x1019C( )150cm2V s 1.01x1018cm3 20Vcm = 484Acm2jndiff= qDndndx= 1.60x1019C( )350 0.025cm2s 10410162x104cm4 = 70.0Acm2jpdiff= qDpdpdx= 1.60x1019C( )150 0.025cm2s 10181.01x10182x104cm4 = 30.0Acm2jT= 11.2 484 70.0 + 30.0 = 535Acm2 2.49 NA = 2ND ECEANAHolesEVNANDNANDNANDED 2.50 =hcE=6.626x1034J s( )3x108m/ s( )1.12eV( )1.602x1019J / eV( )=1.108 m 33 2.51 p-type siliconn-type siliconSi02Al - 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 siliconp-type siliconSi02PhotoresistStructure after exposure and development of photoresist layerMaskn-type siliconp-type siliconStructure following ion implantation of n-type impurity n+Ion implantationSide viewTop ViewMask for ion implantation 2.53 (a) N = 818 + 612 + 4 1()= 8 atoms(b) V = l3= 0.543x109m( )3= 0.543x107cm( )3=1.60x1022cm3(c) D =8 atoms1.60x1022cm3= 5.00x1022atomscm3(d ) m= 2.33gcm3 1.60x1022cm3= 3.73x1022g(e) From Table 2.2, silicon has a mass of 28.086 protons.mp=3.73x1022g28.082 8( )protons=1.66x1024gprotonYes, near the actual proton rest mass. 34 CHAPTER 3 3.1 j=VTlnNANDni2= 0.025V( )ln1019 cm3( )1018 cm3( )1020 cm6= 0.979Vwdo=2sq1NA+1ND j=2 11.7 8.854x1014F cm1( )1.602x1019C11019cm3+11018cm3 0.979V ( ) wdo= 3.73 x 106cm = 0.0373m xn=wdo1+NDNA=0.0373m1+1018cm31019cm3= 0.0339 m | xp=wdo1+NAND=0.0373m1+1019cm31018cm3= 3.39 x 10-3 mEMAX=qNAxps=1.60x1019C( )1019cm3( )3.39x107cm( )11.7 8.854x1014F /cm= 5.24 x 105Vcm 3.2 ppo= NA=1018cm3 | npo=ni2ppo=10201018=102cm3nno= ND=1015cm3 | pno=ni2nno=10201015=105cm3 j=VTlnNANDni2= 0.025V ( )ln1018cm3( )1015cm3( )1020cm6= 0.748 Vwdo=2sq1NA+1ND j=2 11.7 8.854x1014F cm1( )1.602x1019C11018cm3+11015cm3 0.748V( )wdo= 98.4 x 106cm = 0.984 m 3.3 ppo= NA=1018cm3 | npo=ni2ppo=10201018=102cm3nno= ND=1018cm3 | pno=ni2nno=10201018=102cm3 j=VTlnNANDni2= 0.025V( )ln1018 cm3( )1018 cm3( )1020 cm6= 0.921Vwdo=2sq1NA+1ND j=2 11.7 8.854x1014F cm1( )1.602x1019C11018cm3+11018cm3 0.921V ( ) wdo= 4.881x106cm = 0.0488 m 34 3.4 ppo= NA=1018cm3 | npo=ni2ppo=10201018=102cm3nno= ND=1018cm3 | pno=ni2nno=10201018=102cm3j=VTlnNANDni2= 0.025V( )ln1018 cm3( )1020 cm3( )1020 cm6=1.04Vwdo=2sq1NA+1ND j=2 11.7 8.854x1014F cm1( )1.602x1019C11018cm3+11020cm3 1.04V( ) wdo= 0.0369 m 3.5 ppo= NA=1016cm3 | npo=ni2ppo=10201016=104cm3nno= ND=1019cm3 | pno=ni2nno=10201019=10cm3 j=VTlnNANDni2= 0.025V( )ln1019 cm3( )1016 cm3( )1020 cm6= 0.864Vwdo=2sq1NA+1ND j=2 11.7 8.854x1014F cm1( )1.602x1019C11019cm3+11016cm3 0.864V( ) wdo= 0.334 m 3.6 wd= wdo1+VRj | (a) wd= 2wdo requires VR= 3j= 2.55 V | wd= 0.4m 1+50.85= 1.05 m 3.7 wd= wdo1+VRj | (a) wd= 3wdo requires VR= 8j= 4.80 V | wd=1m 1+100.6= 4.20 m 3.8 jn=E , =1=10.5 cm=2 cm | E =jn=1000A cm22 cm( )1= 500Vcm 353.9 jp=E | E =jn= jn = 5000A cm2( )2 cm( )=10.0 kVcm 3.10 j jn= qnv = 1.60x1019C( )4x1015cm3 107cms = 6400Acm2 3.11 j jp= qpv = 1.60x1019C( )5x1017cm3 107cms = 800kAcm2 3.12 jp= qppE qDpdpdx= 0 E = Dpp 1p= kTqdpdx 1pdpdx p( x) = Noexp xL | 1p | E = VTL= 0.025V104cm= 250Vcmdpdx=1LThe exponential doping results in a constant electric field. 3.13 jp= qDndndx= qnVTdndx | dndx=2000A/ cm21.60x1019C( )500cm2/ V s( )0.025V( )=1.00 x 1021cm4 3.14 10 =104 1016exp 40VD( )1[ ]+VD and the solver yields VD= 0.7464 V 3.15 f =10 104ID0.025lnID+ ISIS | f'= 1040.025ID+ IS | ID'= IDff' Starting the iteration process with ID = 100 A and IS = 10-13A: ID f f' 1.000E-04 8.482E+00 -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 10-15A: 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 =qVTk=1.60x1019C 0.025V( )1.38x1023J / K= 290 K 3.18 VT=kTq=1.38x1023J / K( )T1.60x1019C= 8.63x105TFor T = 218 K, 273 K and 358 K, VT = 18.8 mV, 23.6 mV and 30.9 mV 3.19 Graphing ID= ISexp40VDn 1 yields : 1.2 1.0 0.8 0.6 0.4 0.2 0.00123456(a)(b)(c) 373.20 nVT= nkTq=1.041.38x1023J / K( )300( )1.60x1019C= 26.88 mV T = 26.88mV1.602x10-191.38x10-23= 312 K 3.21 iD= ISexpvDnVT 1 or vDnVT= ln 1+iDIS For iD>> IS , vDnVT lniDIS 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= nVTln 1+IDIS | ID= ISexpVDnVT 1 (a) VD=1.05 0.025V( )ln 1+7x105A1018A = 0.837V | (b) VD=1.05 0.025V( )ln 1+5x106A1018A = 0.768V (c) ID=1018A exp01.05 0.025V 1 = 0 A(d) ID=1018A exp0.075V1.05 0.025V 1 = 0.943x1019 A(e) ID=1018A exp5V1.05 0.025V 1 = 1.00x1018 A 3.23 VD= nVTln 1+IDIS | ID= ISexpVDnVT 1 (a) VD= 0.025V ln 1+104A1017A = 0.748V | (b) VD= 0.025V ln 1+105A1017A = 0.691V 38 (c) ID=1017A exp00.025V 1 = 0 A(d) ID=1017A exp0.06V0.025V 1 = 0.909x1017 A(e) ID=1017A exp4V0.025V 1 = 1.00x1017 A 3.24 ID= ISexpVDVT 1 =1017A exp0.6750.025 1 = 5.32x106A = 5.32 AVD=VTlnIDIS+1 = 0.025V( )ln15.9x106A1017A+1 = 0.703 V 3.25 VD= nVTln 1+IDIS = 2 0.025V( )ln 1+40A1010A =1.34 VVD= 2 0.025V( )ln 1+100A1010A =1.38 V 3.26 (a) IS=IDexpVDnVT 1 =2mAexp0.820.025 1 =1.14x1017A(b) ID=1.14x1017A exp50.025 1 = 1.14x1017A 3.27 (a) IS=IDexpVDnVT 1 =300Aexp0.750.025 1 = 2.81x1017A(b) ID= 2.81x1017A exp30.025 1 = 2.81x1017A 393.28 VD= nVTln 1+IDIS | 10-14 IS1012 | VD= 0.025V( )ln 1+103A1012A = 0.518 VVD= 0.025V( )ln 1+103A1014A = 0.633 V | So, 0.518 V VD 0.633 V 3.29 VT=1.38x1023307( )1.60x1019= 0.0264V | ID= ISexpVD0.0264n 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=1.38x1023J / K( )T1.60x1019C= 8.63x105TFor T = 233 K, 273 K and 323 K, VT = 20.1 mV, 23.6 mV and 27.9 mV 3.31 kTq=1.38x1023303( )1.60x1019= 26.1 mV | VD= 0.0261V( )ln 1+1032.5x1016 = 0.757 VV = 1.8mV / K( )20K( )= 36.0 mV | VD= 0.757 0.036 = 0.721 V 40 3.32 kTq=1.38x1023298( )1.602x1019= 25.67 mV | (a) VD= 0.02567V( )ln 1+1041015 = 0.650 V V = 2.0mV / K( )25K( )= 50.0 mV(b) VD= 0.650 0.050 = 0.600 V 3.33 kTq=1.38x1023298( )1.602x1019= 25.67 mV | (a) VD= 0.02567V( )ln 1+2.5x1041014 = 0.615 V b( ) V = 1.8mV / K( )60K( )= 50.0 mV VD= 0.6150.108 = 0.507 V(c) V = 1.8mV / K( )80K( )= +144 mV VD= 0.615+ 0.144 = 0.758 V 3.34 dvDdT=vDVG3VTT=0.7 1.213 0.0259( )300= 1.96mVK 413.35 IS2IS1=T2T1 3exp EGk 1T21T1 =T2T1 3expEGkT1 1T1T2 f x( )= x( )3expEGkT1 11x x =T2T1 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= wdo1+VRj | (a) wd=1m 1+50.8= 2.69 m (b) wd=1m 1+100.8= 3.67 m 3.37 j=VTlnNANDni2= 0.025V( )ln1016cm3( )1015cm3( )1020cm6= 0.633 Vwdo=2sq1NA+1ND j=2 11.7 8.854x1014F cm1( )1.602x1019C11016cm3+11015cm3 0.633V( )wdo= 0.949 m | wd= wdo1+VRjwd= 0.949m 1+10V0.633V= 3.89 m | wd= 0.949m 1+100V0.633V= 12.0 m 42 3.38 j=VTlnNANDni2= 0.025V( )ln1018cm3( )1020cm3( )1020cm6=1.04 Vwdo=2sq1NA+1ND j=2 11.7 8.854x1014F cm1( )1.602x1019C11018cm3+11020cm3 1.04V( )wdo= 0.0368 m | wd= wdo1+VRjwd= 0.0368m 1+51.04= 0.0887 m | wd= 0.0368m 1+251.04= 0.184 m 3.39 Emax=2 j+VR( )wd=2 j+VR( )wdo1+VRj=2jwdo1+VRj3x105Vcm=2 0.6V( )104cm1+VR0.6VR= 374 V 3.40 E =2jwdo=2 0.748V( )0.984x104cm=15.2kVcm | j+VR=Emax2wdoj=3x105Vcm0.984x104cm( )2 0.748VVR= 291.30.748 = 291 V 3.41 VZ = 4 V; RZ = 0 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 >> j, so j can be neglected. Emax=qNAxpS=qNAwdS=qNAS2SqVRNANA=Emax2S2qVR=3x105( )211.7( )8.854x1014( )2 1.602x1019( )1000= 2.91 x 1014/ cm3 433.43 j=VTlnNANDni2= 0.025ln101510201020= 0.864Vwdo=2Sq1NA+1ND j=2 11.7( )8.854x1014( )1.602x101911015+11020 0.864 =1.057x104cmCjo"=Swdo=11.7 8.854x1014( )1.057x104= 9.80x10-9F / cm2 | Cj=Cjo"A1+VRj=9.80x10-90.05( )1+50.864=188 pF 3.44 j=VTlnNANDni2= 0.025ln101810151020= 0.748Vwdo=2Sq1NA+1ND j=2 11.7( )8.854x1014( )1.602x101911018+11015 0.748 = 0.984x104cmCjo"=Swdo=11.7 8.854x1014( )0.984x104= 10.5x10-9F / cm2 | Cj=Cjo"A1+VRj=10.5x10-90.02( )1+100.748= 55.4 pF 3.45 (a) CD=IDTVT=104A 1010s( )0.025V= 400 fF (b) Q = IDT=104A 1010s( )=10 fC(c) CD=25x103A 1010s( )0.025V=100 pF | Q = IDT= 5x103A 1010s( )= 0.50 pC 3.46 (a) CD=IDTVT=1A 108s( )0.025V= 0.400 F (b) Q = IDT=1A 108s( )=10.0 nC(c) CD=100mA 108s( )0.025V= 0.04 F | Q = IDT=100mA 108s( )=1.00 nC 44 3.47 j=VTlnNANDni2= 0.025ln101910171020= 0.921Vwdo=2Sq1NA+1ND j=2 11.7( )8.854x1014( )1.602x101911019+11017 0.921 = 0.110 mCjo=SAwdo=11.7 8.854x1014( )104( )0.110x104= 9.42 pF / cm2 | Cj=Cjo1+VRj=9.42 pF1+50.921= 3.72 pF 3.48 j=VTlnNANDni2= 0.025ln101910161020= 0.864Vwdo=2Sq1NA+1ND j=2 11.7( )8.854x1014( )1.602x101911019+11016 0.864 = 0.334mCjo=SAwdo=11.7 8.854x1014( )0.25cm2( )0.334x104= 7750 pF | Cj=Cjo1+VRj=7750 pF1+30.864= 3670 pF 3.49 VDCRFCC10 H10 HCL = C =Cjo1+VRj (a) C =39 pF1+1V0.75V= 25.5pF | fo=12 LC=12 105H( )25.5pF= 9.97MHz(b) C =39 pF1+10V0.75V=10.3pF | fo=12 LC=12 105H( )10.3pF=15.7MHz 3.50 (a) VD= 0.025V( )ln 1+50A107A = 0.501 V | (b) VD= 0.025V( )ln 1+50A1015A = 0.961 V 453.51 (a) VD= 0.025V( )ln 1+4x103A1011A = 0.495 V | (b) VD= 0.025V( )ln 1+4x103A1014A = 0.668 V 3.52 RS= Rp+ Rn Rp= pLpAp= 1cm( )0.025cm0.01cm2= 2.5Rn= nLnAn= 0.01cm( )0.025cm0.01cm2= 0.025 RS= 2.53 3.53 (a) VD'=VTln 1+IDIS = 0.025V( )ln 1+1035x1016 = 0.708VVD=VD'+ IDRS= 0.708V +103A 10( )= 0.718 V(b) VD=VD'+ IDRS= 0.708V +103A 100( )= 0.808 V 3.54 c=10m2 Ac=1m2 RC=cAc=10m21m2=10/ contact5 anode contacts and 14 cathode contactsResistance of anode contacts =105= 2Resistance of cathode contacts =1014= 0.71 3.55 (a) From Fig. 3.21a, the diode is approximately 10.5 m long x 8 m wide. Area = 84 m2. (b) Area = (10.5x0.13 m) x (8x0.13m) = 1.42 m2. 46 3.56 (a) 5 =104ID+ VD | VD= 0 ID= 0.500mA | ID= 0 VD= 5VForward biased - VD= 0.5 V ID=4.5V104= 0.450 mA(b) 6 = 3000ID+VD | VD= 0 ID= 2.00mA | ID= 0 VD= 6VIn reverse breakdown - VD= 4 V ID=2V3k= 0.667 mA(c) 3 = 3000ID+VD | VD= 0 ID= 1.00mA | ID= 0 VD= 3VReverse biased - VD= 3 V ID= 0 1 2 3 41 mA2 mA(a) Q-pointvD6 5-1 -2 -3 -4 -5 -6-1 mA-2 mA(b) Q-pointiD(c) Q-point 3.57 (a) 10 = 5000ID+VD | VD= 0 ID= 2.00 mA | VD= 5 V ID=1.00 mAForward biased - VD= 0.5V ID=9.5V5k=1.90 mA(b) -10 = 5000ID+VD | VD= 0 ID= 2.00 mA | VD= 5 V ID= 1.00 mAIn reverse breakdown - VD= 4V ID=6V5k= 1.20 mA(c) 2 = 2000ID+VD | VD= 0 ID= 1.00 mA | ID= 0 VD= 2 VReverse biased - VD= 2 V ID= 0 1 2 3 41 mA2 mA(a) Q-pointvD6 5-1 -2 -3 -4 -5 -6-1 mA-2 mA(b) Q-pointiD(c) Q-point 473.58 *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 A .OP .MODEL DIODE1 D IS=1E-15 .END 3.59 (a) 10 =104ID+VD | VD= 0 ID= 1.00 mA | VD= 5 V ID= 0.500 mAIn reverse breakdown - VD= 4 V ID=10 (4)V10k= 0.600 mA(b) 10 =104ID+VD | VD= 0 ID=1.00 mA | VD= 5 V ID= 0.500 mAForward biased - VD= 0.5 V ID=10 0.5V10k= 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 2 341 mA2 mA(b) Q-pointvD6 5-1 -2-3 -4 -5 -6-1 mA-2 mA(c) Q-pointiD(a) Q-point 48 1 2 3 4 5 6 70.0020.001-0.001-0.002iD (A)vD (V)-7 -6 -5 -4 -3 -2 -1 49 3.60 +-VRiDvD+- The load line equation: V = iD R + vD We need two points to plot the load line. (a) V = 6 V and R = 4k: For vD = 0, iD = 6V/4 k = 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 = 3k: For vD = 0, iD = -6V/3 k = -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 = 3k: Two points: (0V, -1mA), (-3V, 0mA); Q-pt: (-3 V, 0 mA) (d) V = +12 V and R = 8k: Two points: (0V, 1.5mA), (4V, 1mA); Q-pt: (0.5 V, 1.4 mA) (e) V = -25 V and R = 10k: 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-.002i (A)Dv (V)DQ-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 = iDR+VTlniDIS or 10 =104iD+ 0.025ln 1013iD ( ). We want to find the zero of the function f = 10 104iD0.025ln 1013iD ( ) iD f .001 -0.576 .0001 8.48 .0009 0.427 .00094 0.0259 - converged 3.63 f =10 104ID0.025ln 1+IDIS | f'= 1040.025ID+ 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 513.65 The one-volt source will forward bias the diode. Load line: 1=104ID+VD | ID= 0 VD=1V | VD= 0 ID= 0.1mA 50 A, 0.5 V( ) Mathematical model: f =1109exp 40VD( )1[ ]+VD 49.9 A, 0.501 V( ) Ideal diode model: ID = 1V/10k = 100A; (100A, 0 V) Constant voltage drop model: ID = (1-0.6)V/10k = 40.0A; (40.0A, 0.6 V) 3.66 Using Thvenin equivalent circuits yields and then combining the sources +-+ -+-VI1 k 1.2 k 1.5 V 1.2 V 0.3 V+-+-VI2.2 k (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 =2.2k0.3V= 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+ -+-VI2.2 k CVD: 0.3 V+-0.6 VI2.2 k + -onV (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 =2.2k0.3V 0.6V= 136 A. 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=02.2 k - +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=1015A exp0.3V0.025V 1 =163 pA, a very small current. 3.67 The nominal values are: VA= 3VR2R1+ R2 = 3V2k2k+ 3k =1.20V and RTHA=R1R2R1+ R2=2k 3k( )2k+ 3k=1.20kVC= 3VR4R3+ R4 = 3V2k2k+ 2k =1.50V and RTHC=R3R4R3+ R4=2k 2k( )2k+ 2k=1.00kIDnom=1.50 1.201.20 +1.00 Vk=136 A For maximum current, we make the Thvenin equivalent voltage at the diode anode as large as possible and that at the cathode as small as possible. VA=3V1+R1R2=3V1+2k 0.9( )2k1.1( )=1.65V and RTHA=R1R2R1+ R2=2k 0.9( )2k1.1( )2k 0.9( )+ 2k1.1( )= 0.990kVC=3V1+R3R4=3V1+3k1.1( )2k 0.9( )=1.06V and RTHC=R3R4R3+ R4=3k1.1( )2k 0.9( )3k1.1( )+ 2k 0.9( )=1.17kIDmax=1.651.060.990 +1.17 Vk= 274 A For minimum current, we make the Thvenin equivalent voltage at the diode anode as small as possible and that at the cathode as large as possible. VA=3V1+R1R2=3V1+2k1.1( )2k 0.9( )=1.350V and RTHA=R1R2R1+ R2=2k1.1( )2k 0.9( )2k1.1( )+ 2k 0.9( )= 0.990kVC=3V1+R3R4=3V1+3k 0.9( )2k1.1( )=1.347V and RTHC=R3R4R3+ R4=3k 0.9( )2k1.1( )3k 0.9( )+ 2k1.1( )=1.21kIDmin=1.350 1.3470.990 +1.21 Vk=1.39 A 0 533.68 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 = 30 = 3 V | I =3 7( )16k= 0.625 mA(b) Diode is forward biased : V = 5 + 0 = 5 V | I =5 5( )16k= 0.625 mA(c) Diode is reverse biased : I = 0 | V = 5+16k I( )= 5 V | VD= 10 V(d ) Diode is reverse biased : I = 0 | V = 7 16k I( )= 7 V | VD= 10 V (b) (a) Diode is forward biased : V = 30.7 = 2.3 V | I =2.3 7( )16k= 0.581 mA(b) Diode is forward biased : V = 5 + 0.7 = 4.3 V | I =5 4.3( )16k= 0.581 mA(c) Diode is reverse biased : I = 0 | V = 5+16k I( )= 5 V | VD= 10 V(d ) Diode is reverse biased : I = 0 | V = 7 16k I( )= 7 V | VD= 10 V 3.70 (a) (a) Diode is forward biased : V = 30 = 3 V | I =3 7( )100k=100 A(b) Diode is forward biased : V = 5 + 0 = 5 V | I =5 5( )100k=100 A(c) Diode is reverse biased : I = 0 A | V = 5+100k I( )= 5 V | VD= 10 V(d ) Diode is reverse biased : I = 0 A | V = 7 100k I( )= 7 V | VD= 10 V (b) 54 (a) Diode is forward biased : V = 30.6 = 2.4 V | I =2.4 7( )100k= 94.0 A(b) Diode is forward biased : V = 5 + 0.6 = 4.4 V | I =5 4.4( )100k= 94.0 A(c) Diode is reverse biased : I = 0 | V = 5+100k I( )= 5 V | VD= 10 V(d ) Diode is reverse biased : I = 0 | V = 7 100k I( )= 7 V | VD= 10 V 3.71 (a) (a) D1 on, D2 on : ID2=0 9( )22k= 409A | ID1= 409A6 043k= 270A D1: 409 A, 0 V( ) D2: 270 A, 0 V( )(b) D1 on, D2 off : ID2= 0 | ID1=6 043k=140 A | VD2= 9 0 = 9V D1: 140 A, 0 V( ) D2: 0 A, 9 V( ) (c) D1 off, D2 on : ID1= 0 | ID2=6 9( )65k= 230 A | VD1= 6 43x103ID2= 3.92 V D1: 0 A, 3.92 V( ) D2: 230 A, 0 V( )(d ) D1 on, D2 on : ID2=0 6( )43k=140 A | ID1=9 022k140 A= 270 A D1: 140 A, 0 V( ) D2: 270 A, 0 V( ) (b) (a) D1 on, D2 on :ID2=-0.750.75 9( )22k= 341A | ID1= 341A6 0.75( )43k=184AD1: 184 A, 0.75 V( ) D2: 341 A, 0.75 V( )(b) D1 on, D2 off :ID2= 0 | ID1=6 0.7543k=122A | VD2= 9 0.75 = 9.75VD1: 122 A, 0.75 V( ) D2: 0 A, 9.75 V( ) 55 (c) D1 off, D2 on :ID1= 0 | ID2=6 0.75 9( )65k= 219A | VD1= 6 43x103ID2= 3.43VD1: 0 A, 3.43 V( ) D2: 219 A, 0.75 V( )(d) D1 on, D2 on :ID2=0.750.75 6( )43k=140A | ID1=9 0.7522k400A = 235AD1: 235 A, 0.75 V( ) D2: 140 A, 0.75 V( ) 3.72 (a) (a) D1 and D2 forward biasedID2=0 9( )15Vk= 600A ID1= ID26 0( )15Vk= 200AD1: 0 V, 200 A( ) D2: 0 V, 600 A( ) (b) D1 forward biased, D2 reverse biased ID1=6 015Vk= 400A VD2= 9 0 = 9 VD1: 0 V, 400 A( ) D2: -9 V, 0 A( ) (c) D1 reverse biased, D2 forward biasedID2=6V 9V( )30k= 500A VD1= 6 15000ID2= 1.50VD1: 1.50 V, 0 A( ) D2: 0 V, 500 A( ) (d) D1 and D2 forward biasedID2=0 6( )15Vk= 400A ID1=9 0( )15Vk ID2= 200AD1: 0 V, 200 A( ) D2: 0 V, 400 A( ) (b) (a) D1 on, D2 on :ID2=-0.750.75 9( )15k= 500A | ID1= 500A6 0.75( )15k= 50.0AD1: 50.0 A, 0.75 V( ) D2: 500 A, 0.75 V( ) 56 (b) D1 on, D2 off :ID2= 0 | ID1=6 0.7515k= 350A | VD2= 9 0.75 = 9.75VD1: 350 A, 0.75 V( ) D2: 0 A, 9.75 V( ) (c) D1 off, D2 on :ID1= 0 | ID2=6 0.75 9( )30k= 475A | VD1= 6 15x103ID2= 1.13VD1: 0 A, 1.13 V( ) D2: 475 A, 0.75 V( ) (d) D1 on, D2 on :ID2=0.750.75 6( )15k= 400A | ID1=9 0.7515k400A=150AD1: 150 A, 0.75 V( ) D2: 400 A, 0.75 V( ) 3.73 Diodes are labeled from left to right (a) D1 on, D2 off, D3 on : ID2= 0 | ID1=10 03.3k+ 6.8k= 0.990mAID3+ 0.990mA=0 5( )2.4kID3=1.09mA | VD2= 5 10 3300ID1( )= 1.73VD1: 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= 0ID1=10 0( )V8.2k+12k= 0.495mA | VD2= 5 10 8200ID1( )= 0.941VID3=0 5V( )10k ID1= 0.005mAD1: 0.495 mA, 0 V( ) D2: 0 A, 0.941 V( ) D3: 0.005 mA, 0 V( ) 57 (c) D1 on, D2 on, D3 onID1=0 10( )8.2kV =1.22mA> 0 | I12K=0 2( )12kV = 0.167mA | ID2= ID1+ I12K=1.05mA> 0I10K=2 5( )10kV = 0.700mA | ID3= I10K I12K= 0.533mA> 0D1: 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= 0ID3=12 5( )4.7 + 4.7 + 4.7Vk=1.21mA> 0 | VD1= 0 5+ 4700ID3( )= 0.667V < 0VD2= 5 12 4700ID3( )= 1.33V < 0D1: 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( )3.3k+ 6.8k= 0.990mAID3+ 0.990mA=0.6 5( )2.4kID3= 0.843mA | VD2= 5 10 0.6 3300ID1( )= 1.13VD1: 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= 0ID1=10 0.6 5( )8.2k+12k+10kV = 0.477mA | VD2= 5 10 0.6 8200ID1( )= 0.490VVD3= 0 5+10000ID1( )= +0.230V < 0.6V so the diode is offD1: 0.477 mA, 0.600 V( ) D2: 0 A, 0.490 V( ) D3: 0 A, 0.230 V( ) (c) D1 on, D2 on, D3 onID1=0.6 9.4( )8.2Vk=1.07mA> 0 | I12K=0.6 1.4( )12Vk= 0.167mAID2= ID1+ I12K= 0.906mA> 0 | I10K=1.4 5( )10Vk= 0.640mA | ID3= I10K I12K= 0.807mA> 0D1: 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= 0ID3=11.4 5( )4.7 + 4.7 + 4.7Vk=1.16mA> 0 | VD1= 0 5+ 4700ID3( )= 0.452V < 0VD2= 5 11.4 4700ID3( )= 0.948V < 0D1: 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( )10k+10k=1.50mA | ID2= 0ID3=0 10( )10k=1.00mA | VD2=10 104ID10 = 5.00VD1: 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= 24V3.9k3.9k+11k= 6.28V | RTH=11k 3.9k= 2.88kIZ=6.28 42.88k= 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 mAQ-pointvDiD-6 -5 -4 -3 -2 -1 Q-Point: (-0.8 mA, -4 V) 3.80 IS=27 915k=1.20mAIL

9V1.2mA= 7.50 k 3.81 IS=27 915k=1.20mA | P = 9V( )1.20mA( )=10.8 mW 3.82 IZ=VSVZRSVZRL=VSRSVZ1RS+1RL | PZ=VZIZIZnom=30V15k9V115k+110k = 0.500 mA | PZnom= 9V 0.500mA( )= 4.5 mWIZmax=30V 1.05( )15k 0.95( )9V 0.95( )115k 0.95( )+110k(1.05) = 0.796 mAPZmax= 9V .95( )0.796mA( )= 6.81 mWIZmin=30V 0.95( )15k1.05( )9V 1.05( )115k1.05( )+110k(0.95) = 0.215 mAPZmin= 9V 1.05( )0.215mA( )= 2.03 mW 3.83 (a) VTH= 60V100150+100= 24.0V | RTH=150100= 60 | IZ=24 1560=150 mAP = 15IZ= 2.25 W | (b) IZ=60 15150= 300 mA | P = 15IZ= 4.50 W 613.84 IZ=VSVZRSVZRL=VSRSVZ1RS+1RL | PZ=VZIZIZnom=60 15( )V15015V100=150 mA | PZnom=15V 150mA( )= 2.25 WIZmax=60V 1.1( )150 0.90( )15V 0.90( )1150 0.90( )+1100(1.1) = 266 mAPZmax=15V 0.90( )266mA( )= 3.59 WIZmin=60V 0.90( )1501.1( )15V 1.1( )11501.1( )+1100(0.9) = 43.9 mAPZmin=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. T =11202VrVP | Vr=ITC=50.1 60( )= 0.8333VT =11202 0.8333( )6 = 1.40 ms 3.86 VD= nVTln 1+IDIS = 2 0.025V( )ln 1+48.6A109A =1.230 V 62 3.87 Von= nVTln 1+IDIS | VD=Von+ IDRSVD=1.6 0.025V( )ln 1+100A108A +100A 0.01( )=1.92 VPjunctionVonIDC=VonIPT2T= 0.92V100A2 1ms16.7ms = 2.75 WPR43TT IDC2RS=4316.7ms1ms 3A( )20.01= 2.00 WPtotal= 4.76 W 3.88 VDC=1Tv t( )dt0T=1TVPVon ( )T TVr2 = VPVon ( )0.05 VPVon( )2 = 0.975 VPVon ( )VDC= 0.975 18V( )=17.6 V 3.89 PD=1TiD2t( )RSdt0T=1TIP21tT 2RSdt0TPD=IP2RST12tT+t2T2 2dt0T=IP2RSTt t2T+t33T2 0TPD=IP2RSTT T +T3 =13IP2RSTT 3.90 Using SPICE with VP = 10 V. 0s 10ms 20ms 30ms 40ms 50mst15V10V5V0V-5V-10V-15VVoltage 633.91 (a) Vdc= VPVon( )= 6.3 2 1( )= 7.91V (b) C =I TVr=7.910.5510.5160=1.05F(c) PIV 2VP= 2 6.3 2 =17.8V (d) Isurge= CVP= 2 60( )1.05( )6.3 2( )= 3530A(e) T =12VrVP=12 60( )2 .25( )6.3 2= 0.628ms | IP= Idc2TT=7.91.52601.628ms= 841A 3.92 VOnom= VPVon ( )= 6.3 2 1( )= 7.91VVOmax= VPmaxVon ( )= 6.3 1.1( )2 1[ ]= 8.80VVOmin= VPminVon ( )= 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+10.000SPICE Graph Results: VDC = 9.29 V, Vr = 1.05 V, IP = 811 A, ISC = 1860 A Vdc= VPVon( )= 10 1( )= 9.00V | Vr=I TC=9.00V0.25160s10.5F=1.20VISC= CVP= 2 60( )0.5( )10( )=1890A | T =12VrVP=12 60( )2 1.2( )10=1.30msIP= Idc2TT=90.2526011.3ms= 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= VPVon( )= 6.3 2 1( )= 7.91V (b) C =I TVr=7.910.2510.51400= 0.158F(c) PIV 2VP= 2 6.3 2 =17.8V (d ) Isurge= CVP= 2 400( )0.158( )6.3 2( )= 3540A(e) T =12VrVP=12 400( )2 .25( )6.3 2= 94.3s | IP= Idc2TT=7.91.52400194.3s= 839A 3.95 (a) Vdc= VPVon ( )= 6.3 2 1( )= 7.91V (b) C =I TVr=7.910.2510.51105= 633F(c) PIV 2VP= 2 6.3 2 =17.8V (d ) Isurge= CVP= 2 105( )633F( )6.3 2( )= 3540A(e) T =12VrVP=12 105( )2 .25( )6.3 2= 0.377s | IP= Idc2TT=7.91.5210510.377s= 839A 3.96 65 (a) C =I TVr=13000 0.01( )160= 556 F (b) PIV 2VP= 2 3000 = 6000V(c) Vrms=30002= 2120 V (d ) T =12VrVP=12 60( )2 0.01( )= 0.375msIP= Idc2TT=1260 10.375ms = 88.9A (e) Isurge= CVP= 2 60( )556F( )3000( )= 629A 3.97 Assuming Von = 1 V: C =VPVonVrT1R=10.025160 303.3 = 6.06 F | PIV = 2VP= 2 3.3+1( )V = 8.6 V | Vrms=3.3+12= 3.04 VT =12TRCVPVonVP=12 60( )20.110 6.06F( )160s 3.3V4.3V = 0.520 msIP= Idc2TT= 30260s 10.520ms =1920 A | Isurge= CVP= 2 60/ s( )6.06F( )4.3V( )= 9820 A 3.98 0s 5ms 10ms 15ms 20ms 25ms 30ms40V20V0V-20Vv1vSvOTime 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 100msTime4.0kV3.0kV2.0kV1.0kV0V-1.0kV-2.0kVvOvS 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= VPVon ( )= 15 2 1( )= 20.2 V (b) C =I VrT2 =20.2V0.510.25V 1120s =1.35 F(c) PIV 2VP= 2 15 2 = 42.4 V (d ) Isurge= CVP= 2 60( )1.35( )15 2( )=10800 A(e) T =12VrVP=12 60( )2 .25( )15 2= 0.407 ms | IP= IdcTT=20.2V0.5160s 10.407ms=1650 A 3.101 (a) Vdc= VPVon ( )= 9 2 1( )= 11.7 V (b) C =I VrT2 =11.7V0.510.25V 1120s = 0.780 F(c) PIV 2VP= 2 9 2 = 25.5 V (d ) Isurge= CVP= 2 60( )0.780( )9 2( )= 3740 A(e) T =12VrVP=12 60( )2 .25( )9 2= 0.526 ms | IP= IdcTT=11.7V0.5160s 10.407ms = 958A 673.102 *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 5.0msTime20V10V0V-10V-20VvSvO Simulation Results: VDC = -13.4 V, Vr = 0.23 V, IP = 108 A VDC=VPVon=10 2 0.7 =13.4 V | Vr=13.431800122000F= 0.254 VT =11202VrVP=11202 0.254( )14.1= 0.504 msIP= IdcTT=13.4V3160s10.504 ms=150 A Simulation with RS = 0.02 . 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 =VPVonVrT1R=10.0251s120 30A3.3V = 3.03 F (b) PIV = 2VP= 2 3.3+1( )V = 8.6 V (c) Vrms=3.3+12= 3.04 V (d) T =12VrVP=12 60( )2 0.025( )3.3( )4.3= 0.520 ms(e) IP= IdcTT= 30A160s 10.520ms = 962 A | Isurge= CVP= 2 60/ s( )3.03F( )4.3V( )= 4910 A 3.104 (a) C =IVrT2=13000 0.01( )12 120=139 F (b) PIV 2VP= 6000 V(c) VS=30002= 2120 V (d ) T =12VrVP=12 60( )2 0.01( )= 0.375 msIP= IdcTT=1160s 10.375ms = 44.4 A (e) Isurge= CVP= 2 60/ s( )139F( )3000V( )=157 A 3.105 The circuit is behaving like a half-wave rectifier. The capacitor should charge during the first 1/2 cycle, but it is not. Therefore, diode D1 is not functioning properly. It behaves as an open circuit. 3.106 (a) Vdc= VP2Von ( )= 15 2 2( )= 19.2 V (b) C =I VrT2 =19.2V0.510.25V 1120s =1.28 F(c) PIV VP=15 2 = 21.2 V (d ) Isurge= CVP= 2 60/ s( )1.28F( )15 2( )=10200 A(e) T =12VrVP=12 60( )2 .25( )15 2= 0.407 ms | IP= IdcTT=19.2V0.51s60 10.407ms =1570 A 3.107 (a) C =I VrT2 =1A3000V 0.01( )1120s = 278 F (b) PIV VP= 3000 V(c) VS=30002= 2120 V (d ) T =12VrVP=12 60( )2 0.01( )= 0.375 msIP= IdcTT=1A160s 10.375ms= 44.4 A (e) Isurge= CVP= 2 60( )278F( )3000( )= 314 A 693.108 (a) C =I VrT2 =30A0.025( )3.3V( )1120s = 3.03 F (b) PIV Vdc+ 2Von= 3.3+ 2( )= 5.3 V(c) Vrms=5.32= 3.75 V (d ) T =12VrVP=12 60( )20.025 3.3( )5.3 = 0.468 msIP= IdcTT= 30A160s 10.468ms=1070 A (e) Isurge= CVP= 2 60/ s( )3.03F( )3.3V( )= 3770 A 3.109 V1 = VP - Von = 49.3 V and V2 = -(VP -Von) = -49.3V. 3.110 *Problem 3.110 VS1 1 0 DC 0 AC 0 SIN(0 35 60) VS2 0 2 DC 0 AC 0 SIN(0 35 60) D1 1 3 DIODE D4 2 3 DIODE D2 4 1 DIODE D3 4 2 DIODE C1 3 0 0.1 C2 4 0 0.1 R1 3 0 500 R2 4 0 500 .MODEL DIODE D IS=1E-10 RS=0 .OPTIONS RELTOL=1E-6 .TRAN 10US 50MS .PRINT TRAN V(3) V(4) .PROBE V(3) V(4) .END 0s 10ms 20ms 30ms 40ms 50msTime40V20V0V-20V-40Vv1v2 3.111 (a) Vdc= VP2Von ( )= 15 2 2( )= 19.2 V (b) C =I VrT2 =19.2V0.510.25 1120 =1.28 F(c) PIV VP=15 2 = 21.2 V (d ) Isurge= CVP= 2 60/ s( )1.28F( )15V 2( )=10200 A(e) T =12VrVP=12 60( )2 .25( )15 2= 0.407 ms | IP= IdcTT=19.2V0.5160s 10.407ms =1570 A 70 3.112 3.3-V, 15-A power supply with Vr 10 mV. Assume Von = 1 V. Rectifier Type Half Wave Full Wave Full Wave Bridge Peak Current 533 A 266 A 266 A PIV 8.6 V 8. 6 V 5.3 V Filter Capacitor 25 F 12.5 F 12.5 F (i) The large value of C suggests we avoid the half-wave rectifier. This will reduce the cost and size of the circuit. (ii) The PIV ratings are all low and do not indicate a preference for one circuit over another. (iii) The peak current values are lower for the full-wave and full-wave bridge rectifiers and also indicate an advantage for these circuits. (iv) We must choose between use of a center-tapped transformer (full-wave) or two extra diodes (bridge). At a current of 15 A, the diodes are not expensive and a four-diode bridge should be easily found. The final choice would be made based upon cost of available components. 3.113 200-V, 3-A power supply with Vr 4 V. Assume Von = 1 V. Rectifier Type Half Wave Full Wave Full Wave Bridge Peak Current 189 A 94.3 A 94.3 A PIV 402 V 402 V 202 V Filter Capacitor 12,500 F 6250 F 6250 F (i) The the half-wave rectifier requires a larger value of C which may lead to more cost. (ii) The PIV ratings are all low enough that they do not indicate a preference for one circuit over another. (iii) The peak current values are lower for the full-wave and full-wave bridge rectifiers and also indicate an advantage for these circuits. (iv) We must choose between use of a center-tapped transformer (full-wave) or two extra diodes (bridge). At a current of 3 A, the diodes are not expensive and a four-diode bridge should be easily found. The final choice would be made based upon cost of available components. 71 3.114 3000-V, 1-A power supply with Vr 120 V. Assume Von = 1 V. Rectifier Type Half Wave Full Wave Full Wave Bridge Peak Current 133 A 66.6 A 66.6 A PIV 6000 V 6000 V 3000 V Filter Capacitor 41.7 F 20.8 F 20.8 F (i) A series string of multiple capacitors will normally be required to achieve the voltage rating. (ii) The PIV ratings are high, and the bridge circuit offers an advantage here. (iii) The peak current values are lower for the full-wave and full-wave bridge rectifiers but neither is prohibitively large. (iv) We must choose between use of a center-tapped transformer (full wave) or extra diodes (bridge). With a PIV of 3000 or 6000 volts, multiple diodes may be required to achieve the require PIV rating. 3.115 iD0+( )=5V1k= 5 mA | IF=5VD1k=50.61k= 4.4 mA Ir=30.61k= 3.6 mA | S= 7ns( ) ln 14.4mA3.6mA = 5.59 ns 3.116 *Problem 3.143 - Diode Switching Delay V1 1 0 PWL(0 0 0.01N 5 10N 5 10.02N -3 20N -3) R1 1 2 1K D1 2 0 DIODE .TRAN .01NS 20NS .MODEL DIODE D TT=7NS IS=1E-15 .PROBE V(1) V(2) I(V1) .OPTIONS RELTOL=1E-6 .OP .END 0s 5ns 10ns 15ns 20nsTime1050-5-10v1vD Simulation results give S = 4.4 ns. 72 3.117 iD0+( )=5V5=1 A | IF=5Von5=50.61= 0.880 A IR=30.65= 0.720 A | S= 250ns( ) ln 10.880A0.720A = 200 ns 3.118 *Problem 3.145(a) - Diode Switching Delay V1 1 0 DC 1.5 PWL(0 0 .01N 1.5 7.5N 1.5 7.52N -1.5 15N -1.5) R1 1 2 0.75K D1 2 0 DIODE .TRAN .02NS 100NS .MODEL DIODE D TT=50NS IS=1E-15 CJO=0.5PF .PROBE V(1) V(2) I(V1) .OPTIONS RELTOL=1E-6 .OP .END 0s 5ns 10ns 15ns 20ns 25nsTime2.01.00-1.0-2.0v1vD For this case, simulation yields S = 3 ns. *Problem 3.145(b) - Diode Switching Delay V1 1 0 DC 1.5 PWL(0 1.5 7.5N 1.5 7.52N -1.5 15N -1.5) R1 1 2 0.75K D1 2 0 DIODE .TRAN .02NS 100NS .MODEL DIODE D TT=50NS IS=1E-15 CJO=0.5PF .PROBE V(1) V(2) I(V1) .OPTIONS RELTOL=1E-6 .OP .END 0s 10ns 20ns 30ns 40nsTime2.01.00-1.0-2.0v1vD For this case, simulation yields S = 15.5 ns. 73 In case (a), the charge in the diode does not have time to reach the steady-state value given by Q = (1mA)(50ns) = 50 pC. At most, only 1mA(7.5ns) = 7.5 pC can be stored in the diode. Thus is turns off more rapidly than predicted by the storage time formula. It should turn off in approximately t = 7.5pC/3mA = 2.5 ns which agrees with the simulation results. In (b), the diode charge has had time to reach its steady-state value. Eq. (3.103) gives: (50 ns) ln (1-1mA/(-3mA)) = 14.4 ns which is close to the simulation result. 3.119 IC=11015exp 40VC ( )1[ ]A | For VC= 0, ISC=1AVOC=140ln 1+11015 = 0.864 VP =VCIC=VC11015exp 40VC( )1[ ] [ ]dPdVC=11015exp 40VC( )1[ ]40x1015VCexp 40VC( )= 0 Using the computer to find VC yields VC = 0.7768 V, IC = 0.9688 A, and Pmax = 7.53 Watts 3.120 (a) For VOC, each of the three diode teminal currents must be zero, andVOC=VC1+VC2+VC3=140V ln 1.05x1015( )+ ln 1.00x1015( )+ ln 0.95x1015( ) [ ]= 2.59 V(b) For ISC, the external currents cannot exceed the smallest of the short circuit currentof the individual diodes. Thus, ISC= min 1.05A,1.00A,0.95A [ ]= 0.95 ANote that diode three will be reversed biased in part (b). Using the computer to find VC yields VC = 0.7768 V, IC = 0.9688 A, and Pmax = 7.53 Watts 3.121 =hcE(a) =6.625x1034J s 3x108m/ s( )1.12eV 1.602x1019j / eV( )=1.11 m - far infrared(b) =6.625x1034J s 3x108m/ s( )1.42eV 1.602x1019j / eV( )= 0.875 m - near infrared 74 CHAPTER 4 4.1 (a) VG > VTN corresponds to the inversion region (b) VG VGSVTN( ) so the saturation region is correct ID=3752AV25m0.5um 2 1 ( )2V2=1.88 mA | Kn= Kn' WL= 375AV25m0.5um = 3.75 mAV2d( ) VGSVTN= 3-1 = 2V and VDS= 3.3 | VDS> VGSVTN( ) so the saturation region is correct ID=3752AV25m0.5m 31 ( )2V2= 7.50 mA 4.21 a ( ) For VGS= 0, VGS VGS - VTN so the transistor is operating in the saturation region. ID= Kn'2 WL VGSVTN( )2=2002 AV2 101 2 0.75( )2=1.56 mA (c) VGS < VTN so the transistor is cutoff with ID = 0. (d) IDKn' so (a) ID=300200 460A = 690A (b) ID= 2.34 mA (c) ID= 0 4.23 (a) VGS - VTN = 4 V, VDS = 6 V. VDS > VGS - VTN --> Saturation region (b) VGS < VTN --> Cutoff region (c) VGS - VTN = 1 V, VDS = 2 V. VDS > VGS - VTN --> Saturation region (d) VGS - VTN = 0.5 V, VDS = 0.5 V. VDS = VGS - VTN --> Boundary between triode and saturation regions (e) The source and drain of the transistor are now reversed because of the sign change in VDS. Assuming the voltages are defined relative to the original S and D terminals as in Fig. P4.11(b), VGS = 2 - (-0.5) = 2.5 V, VGS - VTN = 2.5 - 1 = 1.5 V, and VDS = 0.5 V --> triode region (f) The source and drain of the transistor are again reversed because of the sign change in VDS. Assuming the voltages are defined relative to the original S and D terminals as in Fig. P4.11(b), VGS = 3 - (-6) = 9 V, VGS - VTN = 9 - 1 = 8.0 V, and VDS = 6 V --> triode region G2 V+-+-0.5 VD --> 'S'S --> 'D'VG'S' = +2.5 V VD'S' = +0.5 V(e)G3 V+-+-6.0 VD --> 'S'S --> 'D'VG'S' = +9.0 V VD'S' = +6.0 V(f) 80 4.24 (a) VGS - VTN = 2.6 V, VDS = 3.3 V. VDS > VGS - VTN --> Saturation region (b) VGS < VTN --> Cutoff region (c) VGS - VTN = 1.3 V, VDS = 2 V. VDS > VGS - VTN --> Saturation region (d) VGS - VTN = 0.8 V, VDS = 0.5 V. VDS < VGS - VTN --> triode region (e) The source and drain of the transistor are now reversed because of the sign change in VDS. Assuming the voltages are defined relative to the original S and D terminals as in Fig. 4.54(b), VGS = 2 - (-0.5) = 2.5 V, VGS - VTN = 2.5 0.7 = 1.8 V, and VDS = 0.5 V --> triode region (f) The source and drain of the transistor are again reversed because of the sign change in VDS. Assuming the voltages are defined relative to the original S and D terminals as in Fig. 4.54(b), VGS = 3 - (-3) = 6 V, VGS - VTN = 6 0.7 = 5.3 V, and VDS = 3 V --> triode region 4.25 R2R1R4R3VDD+-DSBG 4.26 (a)I+VDDDGSBBBDDGGSS (b) +VDDIDGSBBDGS 4.27 81 VDS = 3.3V, VGS VTN = 1.3 V; VDS > VGS - VTN so the transistor is saturated. a ( ) gm= Kn VGSVTN( )= 250AV220m1m 2 0.7 ( )= 6.50 mS b( ) gm= Kn VGSVTN( )= 250AV220m1m 3.30.7( )=13.0 mS 4.28 a( ) gm=iDvGS=760 1405 3AV= 310 S | As a check, we can use the results from Problem 4.22. gm= Kn VGSVTN( )=125AV24 1.5( )V = 313 Sb( ) gm=iDvGS=390 154 2AV=188 S | Checking: gm=125AV231.5( )V =188 S 4.29 VDS > VGS - VTN so the transistor is saturated. (a) ID= Kn2 VGSVTN( )21+ VDS( )=2502AV25 0.75( )21+ 0.025 6( ) ( )= 2.60 mA(b) ID= Kn2 VGSVTN( )2=2502AV25 0.75( )2= 2.26 mA 4.30 VDS > VGS - VTN so the transistor is saturated. (a) ID= Kn2 VGSVTN( )21+ VDS( )=5002AV24 1( )21+ 0.02 5( ) ( )= 2.48 mA(b) ID= Kn2 VGSVTN( )2=5002AV24 1( )2= 2.25 mA 4.31 (a) The transistor is saturated by connection. ID=12V VGS105=100x1062101 AV2 VGS0.75V( )212.5VGS217.8VGS 4.97 = 0VGS= 0.266V, 1.214V VGS=1.214 V since it must exceed 0.75VID=12 1.214105=108 A Checking : 100x1062101 AV21.214 0.75V ( )2=108 A (b) ID=12V VGS105=1000x1062 AV2 VGS0.75V( )21+ 0.025VGS( ) 82 Starting with the solution from part (a) and solving iteratively yields VGS = 1.20772 V and ID = 108 A, essentially no change. (c) ID=12V VGS105=100x1062251 AV2 VGS0.75V( )262.5VGS291.75VGS+11.16 = 0VGS= 0.446V, 1.046V VGS=1.046 V since VGS must exceed the threshold voltage.ID=12 1.046105=110 A Checking : ID=100x1062251 AV21.046 0.75V ( )2=110 A 4.32 (a) The transistor is saturated by connection. ID=12V VGS5x104=100x1062101 AV2 VGS0.75V( )231.25VGS2 45.88VGS+ 5.58 = 0VGS= 0.0588V, 1.401V VGS=1.401 V since VGS must exceed the threshold voltage.ID=12 1.4015x104= 212 A Checking : ID=100x1062101 AV21.4010.75V ( )2= 212 A (b) ID=12V VGS5x104=1000x1062 AV2 VGS0.75V( )21+ 0.02VGS( ) Starting with the solution from part (a) and solving iteratively yields VGS = 1.3925 V and IDS = 212 A, essentially no change 4.33 (a) Since VDS = VGS and VTN > 0 for both transistors, both devices are saturated. Therefore ID1= Kn'2 WL VGS1VTN( )2 and ID2= Kn'2 WL VGS2VTN( )2. From the circuit, however, ID2 must equal ID1 since IG = 0 for the MOSFET: I = ID1= ID2 or Kn'2 WL VGS1VTN( )2= Kn'2 WL VGS2VTN( )2 which requires VGS1 = VGS2. Using KVL: VDD=VDS1+VDS2=VGS1+VGS2= 2VGS2VGS1=VGS2= VDD2= 5VI = Kn'2 WL VGS1VTN( )2=1002AV21015 0.75( )2V2= 9.03 mA (b) The current simply scales by a factor of two (see last equation above), and ID = 18.1 mA. (c) For this case, 83 ID1= Kn'2 WL VGS1VTN( )21+VDS1( ) and ID2= Kn'2 WL VGS2VTN( )21+VDS2( ). Since VGS = VDS for both transistors ID1= Kn'2 WL VGS1VTN( )21+VGS1( ) and ID2= Kn'2 WL VGS2VTN( )21+VGS2( ) and ID1 = ID2 = I Kn'2 WL VGS1VTN( )21+VGS1( )= Kn'2 WL VGS 2VTN( )21+VGS2( ) which again requires VGS1 = VGS2 = VDD/2 = 5V. I = Kn'2 WL VGS1VTN( )21+ VDS( )=1002AV21015 0.75( )2V21+ .04( )5( )=10.8 mA 4.34 (a) Since VDS = VGS and VTN > 0 for both transistors, both devices are saturated (by connection). Therefore ID1= Kn'2 WL 1 VGS1VTN( )2 and ID2= Kn'2 WL 2 VGS2VTN( )2. From the circuit, however, ID2 must equal ID1 since IG = 0 for the MOSFET: I = ID1= ID2 or Kn'2101 VGS1VTN( )2= Kn'2401 VGS2VTN( )2 which requires VGS1 = 2VGS2 - VTN. Using KVL: VDD=VDS1+VDS2=VGS2+VGS1= 3VGS2VTNVGS2= VDD+VTN3=10 + 0.753= 3.583V VGS1= 6.417I = Kn'2 WL VGS1VTN( )2=1002AV21016.417 0.75( )2V2=16.1 mAChecking : I =1002AV24013.5830.75( )2V2=16.1 mA which agrees. (b) For this case with VGS = VDS for both transistors and ID1 = ID2, ID1= Kn'2 WL 1 VGS1VTN( )21+VGS1( ) and ID2= Kn'2 WL 2 VGS2VTN( )21+VGS2( ) where VGS2 = VDD VGS1. Therefore, Kn'2101 VGS1VTN( )21+VGS1( )= Kn'2401 10 VGS1VTN( )21+ 10 VGS1( ) ( ) VGS1 = 6.3163, VGS2 = 3.6837, ID1 = 20.4 mA, Checking: ID2 = 20.4 mA which agrees. 84 4.35 (a) Since VDS = VGS and VTN > 0 for both transistors, both devices are saturated (by connection). Therefore ID1= Kn'2 WL 1 VGS1VTN( )2 and ID2= Kn'2 WL 2 VGS2VTN( )2. From the circuit, however, ID2 must equal ID1 since IG = 0 for the MOSFET: I = ID1= ID2 or Kn'2251 VGS1VTN( )2= Kn'212.51 VGS2VTN( )2 Solving for VGS2 yields:VGS2= 2VGS1 2 1( )VTN Also, VDD=VDS1+VDS2 or VGS1=10 VGS2VGS1=10 + 2 1( )VTN1+ 2= 4.271V VGS2= 5.729VI = Kn'2 WL VGS1VTN( )2=1002AV2251 4.2710.75( )2V2=15.5 mAChecking: I =1002AV212.51 5.729 0.75( )2V2=15.5 mA - agrees. (b) For this case with VGS = VDS for both transistors and ID1 = ID2, ID1= Kn'2 WL 1 VGS1VTN( )21+VGS1( ) and ID2= Kn'2 WL 2 VGS2VTN( )21+VGS2( ) where VGS2 = VDD VGS1. Therefore, Kn'2101 VGS1VTN( )21+VGS1( )= Kn'2401 10 VGS1VTN( )21+ 10 VGS1( ) ( ) VGS1 = 4.3265 V, VGS2 = 5.6735 V, ID1 = 19.4 mA, Checking: ID2 = 19.4 mA both agree 4.36 VGS - VTN = 5 - (-2) = 7 V > VDS = 6 V so the transistor is operating in the triode region. (a) ID= 250x1065 2( )62 6 = 6.00 mA (b) Our triode region model is independent of , so ID = 6.00 mA. 4.37 Since VDS = VGS, and VTN < 0 for an NMOS depletion mode device, VGS - VTN will be greater than VDS and the transistor will be operating in the triode region. 85 4.38 (a) VDS= 6V | VGSVTN= 0 3( )= 3V so the transistor is saturatedID= Kn2 VGS1VTN( )2=2502AV20 3V ( ) [ ]2=1.13 mA(b) ID=2502AV20 3V ( ) [ ]21+ 0.025 6 ( ) ( )=1.29 mA 4.39 +10 V100 k WL=101DGSIDS(a) -10 V100 k WL=101DGS(b)IDS (a) If the transistor were saturated, then ID=100x1062101 2( )2= 2.00 mA but this would require a power supply of greater than 200 V (2 mA x 100 k). Thus the transistor must be operating in the triode region. 10V VDS105=1030 2( )VDS2 VDS10 VDS= 50VDS4 VDS( ) and VDS= 0.0504V using the quadratic equation.ID=1032 0.05042 0.0504 = 99.5 A Checking : 10V VDS105= 99.5 A (b) For R = 50 k and W/L = 20/1, 10V VDS5x104= 2x1030 2( )VDS2 VDS10 VDS= 50VDS4 VDS( ), the same as part (a).ID= 2x1032 0.05042 0.0504 =199 A Checking : 10V VDS5x104=199 A (c) In this circuit, the drain and source terminals of the transistor are reversed because of the power supply voltage, and the current direction is also reversed. However, now VDS = VGS and since the transistor is a depletion-mode device, it is still operating in the triode region. 86 10V VDS105=1000x106VDS 2( )VDS2 VDS10 VDS= 50VDS4 +VDS( ) and VDS= 0.04915V using the quadratic equation.ID=1030.04915 2( )0.049152 0.04915 = 99.5 A Checking: 10V VDS105= 99.5 A (d) In this circuit, the drain and source terminals of the transistor are reversed because of the power supply voltage, and the current direction is also reversed. However, now VDS = VGS and since the transistor is a depletion-mode device, it is still operating in the triode region. 10V VDS5x104= 2000x106VDS 2( )VDS2 VDS10 VDS= 50VDS4 +VDS( ) Same as part (c). VDS= 0.04915V using the quadratic equation.ID=1030.04915 2( )0.049152 0.04915 = 99.5 A Checking: 10V VDS105= 99.5 A 4.40 See figures in previous problem but use W/L = 20/1. (a) If the transistor were saturated, then ID=25x1062201 1( )2= 250 A but this would require a power supply of greater than 25 V. Thus the transistor must be operating in the triode region. 10V VDS105=100x106201 0 1( )VDS2 VDS10 VDS=100VDS2 VDS( ) and VDS= 0.05105V using the quadratic equation.ID= 2.00x10310.051052 0.05105 = 99.5 A Checking: 10 0.0510105 V = 99.5 A (b) In this circuit, the drain and source terminals of the transistor are reversed because of the power supply voltage, and the current direction is also reversed. However, now VDS = VGS and since the transistor is a depletion-mode device, it is still operating in the triode region. VDS=10 105( )100x106( )201 VDS 1 ( )VDS2 VDSVDS=10 200VDS1+ VDS2 and VDS= 0.04858V using the quadratic equation.ID= 2000x1061+0.048582 0.04858 = 99.5 A Checking : 10 - 0.04858105 V= 99.5 A 87 4.41 (a) VTN= 0.75 + 0.75 1.5 + 0.6 0.6( )=1.26VVGSVTN= 2 1.26 = 0.74V > VDS= 0.2V Triode regionID= 200x106101 2 1.26 0.22 0.2 = 256 A (compared to 460 A)(b) VGSVTN= 2 1.26 = 0.74V < VDS= 2.5V Saturation regionID=200x1062101 2 1.26( )2= 548 A (compared to 1.56 mA)(c) VGS 0, vBC = 0, forward-active region; vBE = 0, vBC > 0, reverse-active region; vBE > 0, vBC = 0, forward-active region (b) vEB < 0, vCB < 0, cutoff region (c) vEB > 0, vCB < 0, forward-active region (d) vBE > 0, vBC < 0, forward-active region; vBE > 0, vBC > 0, saturation region 5.40 (a) vBE = 0, vBC < 0 cutoff region (b) vBC < 0, IE = 0, cutoff region 5.41 (a) vBE > 0, vBC > 0 saturation region (b) vBE > 0, vBC = 0, forward-active region (c) vBE = 0, vBC > 0, reverse-active region 135 5.42 Emitter-Base Voltage Collector-Base Voltage 0.7 V -0.65 V 0.7 V Saturation Forward Active -0.65 V Reverse Active Cutoff 5.43 (a) vBE > 0, vBC = 0, forward-active region (b) vBE = 0, vBC > 0, reverse-active region 5.44 (a) vEB = 0, vCB > 0, reverse-active region (b) vEB > 0, vCB = 0, forward-active region 5.45 (a) vEB > 0, vCB > 0, saturation region (b) vEB > 0, vCB = 0, forward-active region (c) vEB = 0, vCB > 0, reverse-active region 5.46 a( ) pnp transistor with VEB= 3V and VCB= 3V Cutoff | Using Eq. (5.17) :IC= + ISR=1015A2= 0.5x10-15= 0.5 fA | IE= ISF=1015A75= 13.3x10-18= 13.3 aA IB= IS1F+1R =1015A175+14 = 0.263x1015= 0.263 fAb( ) npn transistor with VBE= 5V and VBC= 5V Cutoff | The currents are the same as in part (a). 5.47 iC=1016exp0.30.025 exp50.025 10161exp50.025 1 =16.3 pAiE=1016exp0.30.025 exp50.025 +101619exp0.30.025 1 =17.1 pAiB=101619exp0.30.025 1 +10161exp50.025 1 = 0.857 pA 136 These currents are all very small - for most practical purposes it still appears to be cutoff. Since VBE > 0 and VBC < 0, the transistor is actually operating in the forward-active region. Note that IC = FIB. 5.48 An npn transistor with VBE= 0.7V and VBC= 0.7V Forward - active regionUsing Eq. (5.45) : IE= F+1( )IB | F= IEIB1=10mA0.15mA1= 65.7 IE= IS1+1F exp VBEVT | IS=0.01A1+165.7 exp0.70.025 = 6.81x1015A= 6.81 fA 5.49 A pnp transistor with VEB= 0.7V and VCB= 0.7V Forward - active regionUsing Eq. (5.44) : F= ICIB=2.5mA0.04mA= 62.5 | IC= ISexp VEBVT | IS=2.5mAexp0.7V0.025V =1.73 fA 5.50 IE=0.7V 3.3V( )47k= 55.3A | IB= IEF+1=55.3A81= 0.683AIC= FIB= 80 0.683A( )= 54.6A | Check : IB+ IC= IE is ok 5.51 (a) f= fTF=500MHz759= 6.67 MHz(b) The graph represents the Bode magnitude plot. Thus s( )=F1+ s=Fs +=Ts + s( )= s( ) s( )+1=Ts +Ts ++1=Ts +T+=Fs + F+1( )=FF+11+ sF+1( )F1+ sT j( )=F1+T 2 137 5.52 vEB> 0 vCB< 4VTiC= ISexp VEBVT + ISR ISexp VEBVT iE= ISexp VEBVT + ISFexp VEBVT = ISFexp VEBVT iB= ISFexp VEBVT ISR ISFexp VEBVT iC= FiB | iC=FiE BiBiEECi = iC F B+-0.7 VvEBiC 5.53 An npn transistor with VBE= 0.7V and VBC= +0.7V Reverse - active regionUsing Eq. (5.51) : IC= R+1( )IB | R= ICIB1= 75A40A1= 0.875 IE= ISexp VBCVT | IE= 35A | IS= 35Aexp0.70.025 = 2.42x1017A= 0.0242 fA= 24.2 aA 5.54 A pnp transistor with VEB= 0.7 V and VCB= +0.7 V Reverse active regioniC= ISexp VCBVT ISRexp VCBVT = ISRexp VCBVT iE= ISexp VCBVT ISF ISexp VCBVT iB= ISF+ ISRexp VCBVT ISRexp VCBVT R= iEiB= 0.1mA( )0.15mA= 0.667 | IS= iEexp VCBVT = 104Aexp0.70.025 = 6.91x1017A 5.55 IC= 0.7V 3.3V( )56k= 46.4 A | IB= ICR+1= 46.4A1.75= 26.5 AIE= IC+ IB= 46.4A+ 26.5A = 19.9 A 138 5.56 FOR= ICIB=1mA1mA=1 | VCESAT=VTln1R 1+FORR+1( )1FORF | R=RR+1=23VCESAT= 0.025ln32 1+12 +1( )1150 =17.8 mVVBE=VTln IB+ 1R( )ICIS1F+1R = 0.025V( )ln1mA+ 10.667( )1mA1015A 0.02 +1.0.667( ) = 0.724 V 5.57 iC= IS exp vEBVT ISR exp vCBVT | iB= ISF exp vEBVT + ISR exp vCBVT | Simultaneoussolution yields : vEB=VTln iB+ 1R( )iCIS 1F+ 1R( ) | vCB=VTln iB iCFIS1R 1F+ 1R( ) vECSAT= vEBvCB=VTln1R 1+ iCR+1( ) iB1 iCF iB for iB> iCF 5.58 (a) Substituting iC = 0 in Eq. 5.30 gives VCESAT=VTln1R = 0.025V( )ln10.5 = 0.0173 V =17.3 mV (b) By symmetry VECSAT=VTln1F or by using iE = 0 and iC = -iB, 139 VCESAT=VTln1R 11R+11+1F=VTln1R RR+1F+1F=VTln1R R1FVCESAT=VTln F( ) and VECSAT=VTln1F VECSAT=VTln1F = 0.025V( )ln10.99 = 0.000251 V = 0.251 mV 5.59 (a) Substituting iC = 0 in Eq. 5.30 gives VCESAT=VTln1R = 0.025V( )ln10.33 = 27.7 mV (b) By symmetry VECSAT=VTln1F = 0.025V( )ln10.95 =1.28 mV 5.60 (a) VCESAT=VTln1R 1+FORR+1( )1FORF | R=RR+1=0.90.9 +1= 0.47370.1= 0.025ln10.4737 1+FOR0.9 +1( )1FOR15 FOR=11.050.4737exp(4) =1+FOR0.9 +1( )1FOR15 FOR=11.05 | IB= ICFOR=20A11.05=1.81A (b) 0.04 = 0.025ln10.4737 1+FOR0.9 +1 ( )1FOR15 FOR=1.97 | IB= ICFOR=20A1.97=10.1A 140 5.61 With VBE = 0.7 and VBC = 0.5, the transistor is technically in the saturation region, but calculating the currents using the transport model in Eq. (5.13) yields iC=1016exp0.70.025 exp0.50.025 10161exp0.50.025 1 =144.5 AiE=1016exp0.70.025 exp0.50.025 +101639exp0.70.025 1 =148.3 AiB=101639exp0.70.025 1 +10161exp0.50.025 1 = 3.757 A At 0.5 V, the collector-base junction is not heavily forward biased compared to the base-emitter junction, and IC= 38.5IB FIB. The transistor still acts as if it is operating in the forward-active region. 5.62 (a) The current source will forward bias the base- emitter junction (VBE 0.7V) andthe collector - base junction will then be reverse biased (VBC 2.3V ). Therefore, thenpn transistor is in the forward - active region.IC= FIB= ISexp VBEVT | VBE= 0.025ln50 175x106A( )1016A = 0.803 V (b) Since IB=175A and IC= 0, IC< FIB, and the transistor is saturated.Using Eq. (5.53) : VBE= 0.025ln175x106+ 01016150+ 1.51.5 = 0.714 V | Using Eq. (5.54) with iC= 0,VCESAT= 0.025ln1R = 0.025lnR+1R = 0.025ln1.50.5 = 27.5 mV 141 5.63 iC= ISexp vBEVT exp vBCVT + ISRexp vBCVT 1 iE= ISexp vBEVT exp vBCVT + ISFexp vBEVT 1 vBE> 4VT and vBC< 4VTiC ISexp vBEVT and iE= IS1+1F exp vBEVT = ISFexp vBEVT iCFiEvBEVTln iCIS =VTlnFiEIS 5.64 ISD= ISF=1fA0.98=1.02 fA 5.65 Both transistors are in the forward - active region. For simplicity, assume VA= .I = IC1+ IB1+ IB2 | Since the transistors are identical and have the same VBE,IC2= IC1 and IB1= IB2 | I = IC1+ 2IB1= F+ 2( )IB1 | IC2= FIB2= FIB1IC2=FF+ 2 I =2525+ 225A | IC2= 23.2 A | See the Current Mirror in Chapter 15. 5.66 CD= ICVTF=50x10120.025 IC= 2x109IC (F) (a) 4 fF (b) 0.4 pF (c) 40 pF 5.67 Using Fig. 2.8 with N =1018cm3, n= 260cm2v - s and p=100cm2v - s(a) npn : F= WB22Dn= WB22VTn=1x104cm( )22 0.025V( )260cm2v - s = 0.769 ns(b) pnp : F= WB22Dp= WB22VTp=1x104cm( )22 0.025V( )100cm2v - s = 2.00 ns 142 5.68 For f >> f, fT= f =10 75MHz( )= 750 MHz | f= fTF=750MHz200= 3.75 MHz 5.69 F= fTf=900MHz5MHz=180 | For f >> 5 MHz, f( )= fTf=900MHz50MHz=18 5.70 NA=6x1018cm3n=130 cm2v s using Fig. 2.8. Dn= nVT=130 cm2v s0.025V( )= 3.25cm2sIS= qADnni2NAWB=1.60x1019C 25x108cm2( )3.25cm2s 1020cm6 6x1018cm30.4x104cm( )= 5.42x1020A 5.71 WB= 2DnF | F12f=12 5x109( )= 31.8 psNA=5x1018cm3n=135 cm2v s using Fig. 2.8.Dn= nVT=135 cm2v s0.025V( )= 3.38cm2sWB 2 3.38cm2s 31.8x1012s = 0.147 m 5.72 IC= FIB= FO1+ VCEVA IB | FO1+5VA =240A3A and FO1+10VA =265A3A1+