并联谐振电路

Is R0 = U0
.
1
.U U
∴并谐: U0
==
U0
1 Q2
0 2
0
θ(ω)= t g 1 Q 0 =φu-φis
0
.
1
.I I
串谐: = =
I0
I0
1 Q2
0 2
0
θ(ω)= t g 1 Q 0 =φi-φus
0
4.有载Q值 — QL 并：Q =ω0 C R0 =
R0 — 空载Q值(并谐回路Q值)
0L
R0q = R0‖Rs‖RL (有载谐振阻抗)
R0 q
QL=
=ω0CR0q=
0L
R0 q R0
Q — 有载Q值 (要求Rs、RL大)(适用电流源)
串：Q = 0 L = 1
R
0 CR
QL= 0 L
R Rs RL
(空载Q值)(LC回路Q值) (要求Rs、RL小)(适用电压源)
例：R=10KΩ，L=1mH，C=0.1uF，is= 10 cos 105 t 30 ° mA
.
I Rs
50
50
R0 = 0 C = 10 12 ×10 7 ×10 2 = 50KΩ
.+
Us
C
L
Req =R0‖Rs =
5025 50 25
×10
3
=
16.7KΩ
-
Is =
U=s
Rs
150
25 ×10 3= 6 mA
U0 = IsReq = 6 ×10 3 × 16.7×10 3 = 100 (V)
r
QQ
4 ×10 4
10 10 =20Ω
100
R= Q2 r
匹配： 1
m2wenku.baidu.com
Rs
=
Q2
r
m2
=
Rs Q2 r
=
8 ×10 3 1002 ×20
= 4 ×10
2
m= 2 ×10
1 = N1
N
∴N1 = 2 ×10 1 ×100 =20T
. . . . 例：谐振时IRL=15A、I=9A，求IC
..
. . . 解：谐振时U和I同相，IRL滞后U角φL，IC超前U90°角
I=IRL+IC
.
. Y(jω)= Is= G + j( C
1
)
U
L
=|Y|∠θ(ω)
|Y| = G2 C 1 2
L
θ(ω) =
C 1
tg 1
L
G
.
Is
1
. . G j C
IG
IC
Iｌ
.
Is=Is∠φis固定、ω可变
G—C、L损耗电导
.jωL
IL
. . 1.谐振条件 总B=0 → ω0 C – 1 = 0 → Y(jω)=G → Is和U同相 0L
I0 =
2 25
550×Is
=
2
mA
IL = QL Is = Q I0=100 mA
.
Is
. I.
.
.
IC
IL
I0
Rs
L
R0
C
例：串谐的等效电路
jωL
Z=jX= j L 1
C
Z→
1
1
X=0 → ω0=
，Z0=0Ω
j C
LC
并谐的等效电路
Y= j C 1 =j C 1 =jB
j L
L
Z→
1
jωL
求：①ω0、Q、Z0、ρ、BW ②u、iR、iL、iC
解：
1
ω0 = =
1 = 105rad/s
LC 10 3 ×10 7
.
Q =ω0CR= 105 ×10 7 ×10 4=100
Is
ρ= L=ω0L =
1
=100
C
0C
Z0 = R =10KΩ
，BW =
0
Q=
103rad/s
.R C.
IR
IC
+. .L U
j C
1
1
B=0 → ω0= LC ，Z0=Y0 =∞
例：Q=100 (空载)，N=100T，L=400uF，C=100pF， Rs=8KΩ，使并谐电路获最大功率，抽头N1=？
N
L
is
Rs
N1 C
is’
Rs’ R
L
C
r
解：Rs’=
1 m2
Rs
，m= N1
N
Q= 0 L → r= 0 L =
L
C=
.
Is
jωL
Req =R0‖RL =
25 ×100 25 100
×10=3
20KΩ
R
+
.
1 RL U0
j L
-
U0 = IsReq = 10 ×10 3 ×20×10 3= 200 (V)
QL = Qeq =ω0CReq = Req Q= 20 ×50 = 40
R0
25
BW
=
0 QL
=
5
×10 40
1
∴ω0 =
— 并联谐振频率
LC
2.ω= ω0 时电压、电流、阻抗
11
Z0= Y( j ) = G =R0 — 谐振阻抗(最大、呈电阻性)
. . . . . . U0= Is = Is =R0Is (最大、且U0和I0同相) Y0 G0
. . . . IL0= 1 U0 = j R0 Is = -jQIs
. . . . j 0 L
0L
. . . IC0= jω0C U0 = jω0C R0 Is = jQIs
IR0= G U0 = Is
R0
Q=
0L
=ω0 C R0 =
R0
C — 并谐电路的品质因数
L
IC0 = IL0= Q Is (电流谐振) IL= Q Is — 环流
BW=ωC2 -ωC1 = 0
6
=1.25×10
5 rad/s
IL = QL Is = 40 ×10 2 = 0.4 A
.
Is
.
R jωL IL
+
. RL
1
U0
j L
-
例：L=100uH、C=100pF、Q=50、Us =150V、Rs =25KΩ，
求ω0、I0、U0、IL
解：
1
ω0 = =
1 = 107rad/s
LC 10 4 ×10 10
IL -
. . ②ω0 =ω0 =
10r5ad/s
(并谐) ，U0 = Is Z0 =
100
∠30°(V)
2
. . . . IR0 =Is = 10 ∠30°(mA) ，IC0 = jQIs = 1 ∠120°(A)
.
.2 1
2
IL0 = -jQIs = ∠-60°(A)
2
∴u0(t) = 100cos 105 t 30 ° (V) iR0(t) = 10 cos 105 t 30 ° (mA) iC0(t) = cos 105 t 120° (A) iL0(t) = cos 105 t 60 ° (A)
Q
Q↑ → BW↓ Q↓ → BW↑
3.频率特性
.
.U
H(jω)= =
1
= 1
1
Is G j C 1
G 1
j C
1
L
G LG
C G
1 LG
=
0C 0 G
0 1 0 LG
=
0
0
CR0
0 R0 0L
=Q 0
0
.
.U
H(jω)= =
1
1
Is
G 1
jQ
0
0
1. .
R0 =
,
G
例：Is=10mA、L=100uH、R=10Ω、C=400pF、
RL=100KΩ，求谐振时R0、U0、IL、BW
解：
1
1
ω0 =
=
LC
10 4 ×4×10 10= 5 × 106rad/s
L
10 4
Q= 0 L = C = 4 ×10 10 =50 (空载Q值)
RR
10
Z0 = Q2 R = 502 ×10 = 25KΩ