第 2 章 通信电子线路分析基础
2-7 2-12 2-13 2-14 2-17 2-27 2-28
LOGO
仅为解题要点,而非完整解答
通信电子线路 第 2 章 通信电子线路分析基础 Page # 1
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2-7
f = 465 kHz p1 = 40/160 = 0.25, p2 = 0.25 CS = 200 + 10 · 0.252 + 16 · 0.252
= 201.625 pF , L = 581 mH
120T N2 N3 40T N1 40T
120T N2 N3 40T N1 40T
RS = 160 (10/0.252) (2.5/0.252)
= 80/3 kW
Rp
Q0 =100, Rp= Q0wL =169.75 kW, RS' = RSRp= 23.046 kW QL' =13.6, B' = 34.2 kHz
通信电子线路 第 2 章 通信电子线路分析基础 Page # 2
Copyright 2012 DyNE All rights reserved
2-12
串联谐振回路 f0 = 1.5 MHz, C0 = 100 pF, R = 5 W, Vs = 1 mV
, L0 = 113 mH
串联谐振时,回路电流达到最大值:
I0 = Vs /R = 0.2 mA
串联谐振时,电感和电容上的电压达到最大值:
VL0 = VC0 = I0wL = Q0Vs = 212 mV
通信电子线路 第 2 章 通信电子线路分析基础 Page # 3
Copyright 2012 DyNE All rights reserved
2-13
并联谐振回路 f0 = 5 MHz, C = 50 pF, B = 150 kHz
L = 20.264 mH
Is Ri C L Rp RL
当 f = 5.5 MHz 时, 若要求 B = 300 kHz:
QL' =QL/2, RS' =QL'w0L = RS/2
应该再并联一个阻值为 21.2 kW 的电阻
通信电子线路 第 2 章 通信电子线路分析基础 Page # 4
Copyright 2012 DyNE All rights reserved
2-14
并联谐振回路,抽头变换
L = 0.8 mH, Q0 = 100, C1 = C2 = 20 pF
Ci = 5 pF, Ri = 10 kW
Co = 2 pF, Ro = 5 kW
C1 Ci Ri L C2 Ro Co
C2' = C2 + Co = 22 pF CS = Ci + C1 & C2' = 15.5 pF
Rp= Q0wL = 22.7 kW Ro' = Ro/p2 = 22.05 kW, RS = Ri Rp Ro' = 5.28 kW
通信电子线路 第 2 章 通信电子线路分析基础 Page # 5
Copyright 2012 DyNE All rights reserved
2-17
最佳全谐振工作状态
f01 = f02 = 1 MHz, r1 = r2 = 1 kW, R1 = R2 = 20 W, h = 1
r = wL, L1 = L2 = 159 mH f ~ L1C1, C1 = C2 = 159 pF
M = 3.18 mH
a C1 b
M L1 R1 L2 C2 R2
初级等效
初级自身
通信电子线路 第 2 章 通信电子线路分析基础 Page # 6
Copyright 2012 DyNE All rights reserved
2-17
最佳全谐振工作状态
f01 = f02 = 1 MHz, r1 = r2 = 1 kW, R1 = R2 = 20 W, h = 1
r = wL, L1 = L2 = 159 mH f ~ L1C1, C1 = C2 = 159 pF
M = 3.18 mH
a C1 b
M L1 R1 L2 C2 R2
若次级谐振频率调为:f02 = 950 kHz , C2' = 177 pF
该反射阻抗呈容性
通信电子线路 第 2 章 通信电子线路分析基础 Page # 7
Copyright 2012 DyNE All rights reserved
2-27
v =Vmcoswt 根据二极管的伏安特性可知: • v 0 时,D 等效为一电阻
rD = 1/g
i g v v D R
i
半波整流 Im i
i(t) 为半波余弦波形 • 幅值为:
通信电子线路 第 2 章 通信
电子线路分析基础 Page # 8
Copyright 2012 DyNE All rights reserved
2-28
v =Vmcoswt 根据 2-27 的结论: v > 0 时,上 D 导通,下 D 截止 • i(t) 为半波余弦波形
1:2 v
N1 N2 N3 v
v
i R
D
D
v
全波整流
通信电子线路 第 2 章 通信电子线路分析基础 Page # 9
Copyright 2012 DyNE All rights reserved
第 2 章 通信电子线路分析基础
2-7 2-12 2-13 2-14 2-17 2-27 2-28
LOGO
仅为解题要点,而非完整解答
通信电子线路 第 2 章 通信电子线路分析基础 Page # 1
Copyright 2012 DyNE All rights reserved
2-7
f = 465 kHz p1 = 40/160 = 0.25, p2 = 0.25 CS = 200 + 10 · 0.252 + 16 · 0.252
= 201.625 pF , L = 581 mH
120T N2 N3 40T N1 40T
120T N2 N3 40T N1 40T
RS = 160 (10/0.252) (2.5/0.252)
= 80/3 kW
Rp
Q0 =100, Rp= Q0wL =169.75 kW, RS' = RSRp= 23.046 kW QL' =13.6, B' = 34.2 kHz
通信电子线路 第 2 章 通信电子线路分析基础 Page # 2
Copyright 2012 DyNE All rights reserved
2-12
串联谐振回路 f0 = 1.5 MHz, C0 = 100 pF, R = 5 W, Vs = 1 mV
, L0 = 113 mH
串联谐振时,回路电流达到最大值:
I0 = Vs /R = 0.2 mA
串联谐振时,电感和电容上的电压达到最大值:
VL0 = VC0 = I0wL = Q0Vs = 212 mV
通信电子线路 第 2 章 通信电子线路分析基础 Page # 3
Copyright 2012 DyNE All rights reserved
2-13
并联谐振回路 f0 = 5 MHz, C = 50 pF, B = 150 kHz
L = 20.264 mH
Is Ri C L Rp RL
当 f = 5.5 MHz 时, 若要求 B = 300 kHz:
QL' =QL/2, RS' =QL'w0L = RS/2
应该再并联一个阻值为 21.2 kW 的电阻
通信电子线路 第 2 章 通信电子线路分析基础 Page # 4
Copyright 2012 DyNE All rights reserved
2-14
并联谐振回路,抽头变换
L = 0.8 mH, Q0 = 100, C1 = C2 = 20 pF
Ci = 5 pF, Ri = 10 kW
Co = 2 pF, Ro = 5 kW
C1 Ci Ri L C2 Ro Co
C2' = C2 + Co = 22 pF CS = Ci + C1 & C2' = 15.5 pF
Rp= Q0wL = 22.7 kW Ro' = Ro/p2 = 22.05 kW, RS = Ri Rp Ro' = 5.28 kW
通信电子线路 第 2 章 通信电子线路分析基础 Page # 5
Copyright 2012 DyNE All rights reserved
2-17
最佳全谐振工作状态
f01 = f02 = 1 MHz, r1 = r2 = 1 kW, R1 = R2 = 20 W, h = 1
r = wL, L1 = L2 = 159 mH f ~ L1C1, C1 = C2 = 159 pF
M = 3.18 mH
a C1 b
M L1 R1 L2 C2 R2
初级等效
初级自身
通信电子线路 第 2 章 通信电子线路分析基础 Page # 6
Copyright 2012 DyNE All rights reserved
2-17
最佳全谐振工作状态
f01 = f02 = 1 MHz, r1 = r2 = 1 kW, R1 = R2 = 20 W, h = 1
r = wL, L1 = L2 = 159 mH f ~ L1C1, C1 = C2 = 159 pF
M = 3.18 mH
a C1 b
M L1 R1 L2 C2 R2
若次级谐振频率调为:f02 = 950 kHz , C2' = 177 pF
该反射阻抗呈容性
通信电子线路 第 2 章 通信电子线路分析基础 Page # 7
Copyright 2012 DyNE All rights reserved
2-27
v =Vmcoswt 根据二极管的伏安特性可知: • v 0 时,D 等效为一电阻
rD = 1/g
i g v v D R
i
半波整流 Im i
i(t) 为半波余弦波形 • 幅值为:
通信电子线路 第 2 章 通信
电子线路分析基础 Page # 8
Copyright 2012 DyNE All rights reserved
2-28
v =Vmcoswt 根据 2-27 的结论: v > 0 时,上 D 导通,下 D 截止 • i(t) 为半波余弦波形
1:2 v
N1 N2 N3 v
v
i R
D
D
v
全波整流
通信电子线路 第 2 章 通信电子线路分析基础 Page # 9
Copyright 2012 DyNE All rights reserved