微电子电路设计第四版课后习题答案

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Richard C. Jaeger所著的《微电子电路设计》第四版的课后习题答案
19.9710019c010614.510 612=14.5 Tb/chip 19.9710 0.1977(,-1960 =10 0.1977( 0.197 199710 0.1977(1-1960) log 2 0.1977 log10 2 5.06 0.1977 161010 0.1548(2020-1970) 88510° transistors/P 16101→154(-90)101548(:-) 161010 0.1548(2-1970) log 2 2 1.95 0.1548 log10 6.46 0.1548 =8.0010 0.05806(2020-1970 =10 No, this distance corresponds to the diameter of only a few atoms. also, the wavclength of the radiation needed to expose such patterns during fabrication is represents a serious problem 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/Hp will be 8 85X 10 2020. Thus there will be the equivalent of 8.85x10 6x10=148 Pentium Iv processors 1-2 CR C. Jaeger &t.N. blalock 6/9/06 1.13103 (75 10 tubes 1.5W/tube=113 MW! 220 DD..ADA 10.2410.24 10.24 =2.500 MSB =5.120 2 4096 100101012+2+2+2=23420=2342(2500)=5855 m V 2.77V 19.53 and 142 LSB 256 mv 19.53 bit 142=(28+8+4+2)0=1000110 2.5 2.5 =2.44 1024 01010104-(2+2+2+2+2+2=35 365 =0.891 1024 10 6.83V 0.6104 and 1119l bits bit ()= 11l0=8192+2048+512+256+128+32+16+4+2+1) 111910=10101110110111 a4 digit readout ranges from 0000 to 9999 and has a resolution of 1 part in 10,000 e number of bits must satisfy 2>10,000 where b is thc number of bits. Hcrc b=14 bits 5.12 5.12 125—and 101110111011 4096 (2+2”+28+27+25+24+23+2+1)1.25±0.0625 3.754±0.000625or3.753V≤V≤3.755 6/9/06 bdc component =0.002 A, 1b=signal component =0.002 cos(1000t)A VGs=4V,Vgs=0.5u(t-1)+0.2 cos 2000 t Volts V=[5+2cos(50000v Ds=[5+2sin(2500+4sin(10000V V=10V,R1=22k,R2=47 kQ and r3=180kg2. R CR 229 229 10 10 229+147g1809 222+37.3g=371 37.3g 6.29 Checking:6.29+3.71=10.0 22g+37.3g2 180g 10 180g 134 47Ω+180g(22g+37.39)479+1809 47g 47g 34.9 47Ω+180g(22g+37.39479+1809 10 Checking:=229+373-169an,=2+3 CR C. Jaeger & t N blalock 6/9/06 V=18V,R1=56kΩ2,R,=33 kQ and r3=11k R R 18 56Ω 33 15.7 2.31 56g+(3391g 569+(3919) Checking: +,=15.7+231=18.0 which is correct 18 =2801 33g+11g =(280)A1 70.0 56g+(33g1lg 33g+11g 33Ω 801 33Ω 31339+119 33Ω+11g =2104 Checking:2+3=2801 56g+369 2.4g 3.97 5 =1.03 5.69+369)+2.49 9.29+2.49 3.6 4g29.2g2 3.72 5.6Ω+3.6g Checking: +2=5.00 an 22=103(3692)=31 150g 150Ω 250L 125 3=2501 125 150g+150g 1509+1509 =250(509509 82g 10.3 689+82g Checking:1+2=250 uar ,=125(829)=10.3 6/9/06 R th Summing currents at the output node yields + 002v=0 so v =0 and 5x10 R Summing currents at the output node 0.002=0 510 +0.002=0 =—=4959 10 Thevenin equivalent circuit 495g 1-6 CR C. Jaeger & t N blalock 6/9/06 The Thevenin equivalent resistance is found using the same approach as Problem 1. 24, and +.025=3969 49 The short circuit current is +0.025and 4kQ +0.025=0.0253 4kQ Norton cquivalent circuit 0.0253v 39.6g 1-7 6/9/06 R R 39g B ano 120 468 1009 R R B but =0 since VR=0 39Ω. Thevenin equivalent circuit 39k 8.5 (b) R B 2 whe 38700 B+1 1-8 CR C. Jaeger &t.N. blalock 6/9/06 βi R +b but +B=0 So =0 and 39g Thevenin equivalent circuit 39k Q 27 R R B but i and B_100 1.3310 From problem 1. 26(a),Rh-R2-56kQ2. Norton equivalent circuit 0.00133v 56k Q 6/9/06

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