电子学习题解答

电子学基础习题解答2021/3/261第1章 半导体器件的特性1.2

解:(1)

由

得:当

时(2)教材P502021/3/2621.5解:教材P51(a)(b)(c)(d)2021/3/2631.9解:(1)教材P52∴电路能正常工作(2)∴电路不能正常工作(3)∴稳压管可能被烧坏2021/3/2641.12解:2021/3/265第2章 放大器基础2021/3/266教材P130解:(a)能正常放大;不能正常放大,输入端交流短路;不能正常放大,C结反偏不能满足;不能正常放大,VGS＜0不能满足。2.2解:(a)(b)2021/3/2672.4解:由教材P132图P2.3(b)知:2021/3/2682.5解:①不带负载RL时IC

=

hfeIB

=

1.5mAVCE

=

VCC－ICRC

=

7.35V直流通路2021/3/2692021/3/2610②带负载RL时,IC

=

hfeIB

=

1.5mA直流通路2021/3/26112.7解:①教材P132VCEQ

=

VCC－ICQ(RC

+

RE)

≈

5.74V直流通路2021/3/2612②Ri

=

hie

//

RB1

//

RB2

≈

hie

=

1.1kΩRo

=

RC

=

5.1kΩ2021/3/26132.10解:①教材P134VCEQ

=

VCC－ICQ(RC

+

RE1+

RE2)

≈

3.72V②2021/3/26142021/3/26152.12解:教材P1352021/3/26162.17解:教材P1372021/3/26172021/3/26182021/3/2619第3章 集成运算放大器2021/3/2620VCCVEERC1RC2RS1RS2RET1T2+_vi+vi1_+vi2__+

vo静态工作点的计算:VCCRC1RS1RC2RS2T1T2+_voRE-

VEE直流通路2021/3/26213.2解:①教材P1752021/3/2622②2021/3/2623解毕2021/3/26243.3解:2021/3/2625考虑到有则2021/3/2626解毕3.4解:①教材P176②③2021/3/2627解毕2021/3/26283.7解:教材P1762021/3/2629第4章 反馈放大器2021/3/26304.2~4.3解:(a)2021/3/2631(b)反馈通路由R3、R2、R1构成;电压串联直流正反馈。反馈通路由R2

、R3

、R1构成;电压并联交直流负反馈。教材P231(c)2021/3/2632(d)由R6

、C3

、R4、C2构成级间电压串联交流正反馈通路;R4、R5构成局部电流串联直流负反馈通路;R4、C2构成

局部电流串联交流负反馈通路;R7构成局部电压串联交直流负反馈通路。由R6

构成级间电流串联交直流负反馈通路;R3构成局部电流串联直流负反馈;R4构成局部电流串联直流负反馈;R8构成局部电流串联交直流负反馈。4.8证明:教材P2332021/3/26334.9证明:教材P2332021/3/2634第5章 信号运算电路教材P275又∵RF

=100kΩ5.2解:由2021/3/2635得:5.3解:由于“虚断”,i+=0,因此i1

+i2

+i3=0,则有:又由于“虚短”,即v－=v＋,得:教材P2752021/3/26365.9解:教材P2772021/3/26376.3解:教材P333(c)

N2021/3/2638(d)

N(a)

N(b)

Y第6章 正弦波振荡器6.4解:教材P333⑴正反馈网络为RC串并联网络;负反馈由

R1、R2组成。⑵

A

=

1+R1/R2;⑶

。⑷fo

=1/2πRC。⑸R1≥3kΩ。⑹R1可使用负温度系数的热敏电阻。2021/3/26396.6解:教材P334(a)

N2021/3/2640(b)

N(c)

N(d)

N6.8解:教材P335并联型石英晶体正弦波振荡电路,fo在fs

和fp

之间,此时石英晶体的阻抗呈感性。谐振频率近似为2021/3/26416.12解:(1)教材P337当时当时2021/3/26428.1

解:(100101)=(45)=(25)2021/3/2643=

(37)2

8

16

10(1011)2

=

(13)8

=

(B)16

=

(11)10(1011.11)2

=

(13.6)8

=

(B.C)16

=

(11.75)10(11010.011)2

=

(32.3)8

=

(1A.6)16

=

(26.375)108.2

解:(29)10

=(11101)2

=(35)8

=(1D)16(219)10

=

(11011011)2

=

(333)8

=

(DB)16(0.65)10

=

(0.1010011)2

=

(0.514)8

=

(0.A6)16(17.84)10

=

(10001.11010111)2

=

(21.656)8

=

(11.D7)16教材P444第8章 数字电路基础8.3

解:教材P445ABCY000000100100011110011011110111112021/3/26448.4

解:(1)(2)(3)2021/3/2645(4)(5)2021/3/26468.5

证明:ABCBCA+BCA+BA+C(A+B)(A+C)0000000000100010010001000111111110001111101011111100111111111111由真值表得证:2021/3/26478.6

解:(1)(2)(3)(4)2021/3/26488.7

解:(1)(2)2021/3/26498.8

解:(1)CAB010001111010101010CDAB00

01

11

100011111111(2)化简后得：0111102021/3/2650(4)(3)CDAB00

01

11

10001111111CDAB00

01

11

1000011110111111110111102021/3/2651(6)(5)CDAB00

01

11

10000111101111××××1××CDAB00

01

11

10000111101111××××2021/3/26528.9

解:(1)(2)2021/3/26538.10

解:(a)(b)(a)(b)2021/3/26548.11

解:(1)三极管T可靠截止的条件为则有(2)三极管T可靠饱和的条件为①三极管饱和时,c、e结均正向偏置,则②由①、②式得:2021/3/2655第9章 组合逻辑电路教材P4909.3

解:ABCY1Y20000000101010010111010001101011100111101(锁)(报警)2021/3/26569.4

解:A2A1B2B1Y00000000100010000110010010101001100011101000110011101001011011001110111110111110B2B1A2A100

01

11

101

110001

11110

1

12021/3/26579.5

解:ABCDY00000000110010100110010010101001100011111000110010101001011111000110111110111110CDAB00

01

11

1000011110111111112021/3/26589.6

解:CY4Y3Y2Y10X4X3X2X112021/3/26599.7

解:假定采用共阳极数码管。真值表如下表2021/3/2660字形十六进制代码ASCII编码输

出A7A6A5A4A3A2A1YaYbYcYdYeYfYgA4110000010001000b6211000101100