3. 串聯諧振器(Series Resonators)
• 串聯諧振電路
I R L
C
( )
1
inZ R j L
j C
ω ω
ω
= + +
CV
+
−
21
2
RP I R=
21
4
mW I L=
2 2 2
2 2 2
1 1 1 1
4 4 4
e C
C
W V C I I
C Cω ω
= = =
在諧振頻率時, :0 02 fω π= m eW W= 0
1
LC
ω =
( )0inZ Rω =
當電感與電容平均儲能相等時,串聯電路
的輸入阻抗將只剩下純電阻性的成分。
• 品質因素 (Quality Factor, Q):
average energy stored
energy loss/sec
m e
R
W W
Q
P
ω ω
+
=≜
當諧振發生時, 0ω ω=
0
0 0
0
2 2 1m e
R R
W W L
Q
P P R CR
ω
ω ω
ω
= = = =
X
Q
R
= 其中 0
0
1
orX L
C
ω
ω
=
此時,
Q是一種度量諧振器損耗多寡的參數,
電路損耗越低代表Q值越高。
Department of Electronic Engineering, NTUT3/70
4. 在諧振頻率時, :
此時,
並聯諧振器
2
2
R
V
P
R
=
21
4
eW V C=
2 2
2
2 2 2
1 1 1
4 4 4
m L
V V
W I L L
L Lω ω
= = =
0 02 fω π= m eW W= 0
1
LC
ω =
( )0
1
inY
R
ω =
LI
LR C
( )
1 1
inY j C
R j L
ω ω
ω
= + +
V
+
−
• 並聯諧振電路
當電感與電容平均儲能相等時,並聯電路
的輸入導納將只剩下純電導性的成分。
• 品質因素 (Quality Factor, Q):
average energy stored
energy loss/sec
m e
R
W W
Q
P
ω ω
+
=≜
B
Q
G
=
0
0
1
orB C
L
ω
ω
=
1
G
R
=
當諧振發生時, 0ω ω=
其中
Department of Electronic Engineering, NTUT4/70
5. 與串聯RL與RC電路比較
L R
C R
1X
Q
R CRω
=≜
雖然Q可以被定義,但是無法定義出
諧振頻率,因RC與RL電路並不會發
生諧振(沒有能量互丟的現象),因此
不屬於諧振電路。
• 串聯RL與RC電路可沿用Q之定義
• 並聯RL與RC電路
B R
Q
G Lω
=≜
B
Q CR
G
ω=≜
L
R
C
R
X L
Q
R R
ω
=≜
Department of Electronic Engineering, NTUT5/70
6. 定義諧振頻寬
• 以串聯諧振器為例,當 0ω ω→
( )
1
inZ R j L
j C
ω ω
ω
= + +
0ω ω ω= + ∆令 0ω∆ → 0
1
LC
ω =
( )
2 2 2
0 0
2 2 2
1
1 1inZ R j L R j L R j L
LC
ω ω ω
ω ω ω ω
ω ω ω
−
= + − = + − = +
( )( )0 0 0
2
0 0
2
2 2
QR
R j L R jL R j L R j
ω ω ω ω ω ω
ω ω ω
ω ω ω
+ − ⋅∆
= + + = + ∆ = + ∆
≃ 0L
Q
R
ω
=其中
其中 而
I L R
C
( )inZ ω
CV
+
−
• 定義頻寬為
0
2
BW
ω
ω
∆
≜
0
0
2
2
in
BW QR
Z R j R jBW QR
ω
ω
⋅
= + = + ⋅
inZ
0ω
ω
R
2R
2 ω∆
當 1
BW
Q
= 2inZ R jR R= + =
0
03-dB Bandwidth BW
Q
ω
ω= ⋅ =
1
3-dB Bandwidth in % BW
Q
= =
Department of Electronic Engineering, NTUT6/70
7. 並聯諧振器的3-dB頻寬
• 以並聯諧振電路為例,當 0ω ω→
( )
1 1 1
2inY j C jC
R j L R
ω ω ω
ω
= + + + ∆≃
0
2
BW
ω
ω
∆
≜定義
( ) 0
0
1 1
2
2
in
BW Q Q
Y j jBW
R R R R
ω
ω
ω
⋅
= + = +
當 1
BW
Q
=
1 1 2
inY j
R R R
= + =
0
03-dB Bandwidth BW
Q
ω
ω= ⋅ =
1
3-dB Bandwidth in % BW
Q
= =
inY
0ω
ω
1
R
2
R
2 ω∆
關於關於關於關於Q的重要觀念的重要觀念的重要觀念的重要觀念::::
1. Q值越高、損耗越低、頻寬越窄
2. Q值越低、損耗越高、頻寬越寬
Department of Electronic Engineering, NTUT7/70
8. Loaded-Q與Unloaded Q
LR
Resonant
Circuit
Unloaded Q
• 串聯RLC電路
• 並聯RLC電路
( )
0
0
1
L
L L
X L
Q
R R R C R R
ω
ω
= = =
+ +
( )0
0
//
//L
L L
B R R
Q C R R
G L
ω
ω
= = =
• 定義Qe為外部Q (external Q)
0
0
1
e
L L
L
Q
R R C
ω
ω
= = for series RLC
0
0
L
e L
R
Q R C
L
ω
ω
= =
1 1 1
L eQ Q Q
= +
for parallel RLC
for both cases
負載效應
負載Q是無負載Q
與外部Q的並聯。
Department of Electronic Engineering, NTUT8/70
9. 阻抗匹配
Matching
Network
in sZ R=
+
−
sV
sR
LZo sZ R=
0inΓ =
Goal:
• 假設匹配網路為理想無損耗的情況下,為了達到最大功率傳輸的目
的,匹配網路是要設計來將ZL轉換為Z0 (matched with the transmission line)
或 Rs (matched with the source impedance when no line connected)。
• 當負載與傳輸線阻抗匹配時,將有最大功率傳輸至負載(assuming the
generator is matched)。
Department of Electronic Engineering, NTUT9/70
11. L型匹配 – Case (a) Rs < 1/GL
+
−
sV
sR
jX
jB LY
L L LY G jB= +
inZ
目標在於求匹配元件之電抗X與電受B,能使
Case (a) 1s LR G< in sZ R= 0Γ =
+
−
sV
sR
jX
( )Lj B B+ LG
L
L
B B
Q
G
+
=l
Series RC or RL
Parallel RC or RL
s
s
X
Q
R
=
( )
1
in s
L L
Z jX R
G jB jB
= + =
+ +
實部: ( ) 1s L LR G X B B+ + =
虛部: ( ) 0s L LR B B XG+ − =
阻抗匹配時,由匹配
網路往負載端視入的
輸入阻抗虛部為0。
L
s
s L
B BX
Q Q Q
R G
+
= = = =l
( )2
1 1s LR G Q + =
1
1
s L
Q
R G
= ± −
選擇
1
1s
s L
Q Q Q
R G
= = = + −l
1
1s s
s L
X R Q R
R G
= = −
1
1L L L L
s L
B G Q B G B
R G
= − = − − (>0, 電容)
(<0, 電感)
(>0, 電感)
當Q被決定了,X與B也就決定了。
或
Department of Electronic Engineering, NTUT11/70
12. L型匹配 – Case (b) Rs > RL
+
−
sV
sR
jX
jB LZ
L L LZ R jX= +
inY
Case (b) s LR R>
目標在於求匹配元件之電抗X與電受B,能使
1in sY R= 0Γ =或
s sQ BR= L
L
X X
Q
R
+
=l
+
−
sV
sR
( )Lj X X+
jB LR
Parallel
RC or RL
Series RC or RL
( )
1 1
in
L L s
Y jB
R jX jX R
= + =
+ +
( )s L s LBR X X R R+ = −
( ) 0L s LX X BR R+ − =虛部:
實部:
阻抗匹配時,由匹配
網路往負載端視入的
輸入導納虛部為0。
L
s s
L
X X
Q BR Q Q
R
+
= = = =l
2
L s LQ R R R= − 1s
L
R
Q
R
= ± −
選擇 1s
s
L
R
Q Q Q
R
= = = + −l
1s
L L L L
L
R
X R Q X R X
R
= − = − −
1
1s
s s L
RQ
B
R R R
= = −
(>0, 電感)
(<0, 電容)
(>0, 電容)
當Q被決定了,X與B也就決定了。
Department of Electronic Engineering, NTUT12/70
13. (1) 已知串聯,要轉並聯:
串聯與並聯轉換
sR
sjX
pR
pjX
s
s
s
X
Q
R
=
s s sZ R jX= +
p
p
p
R
Q
X
=
p p
p
p p
R jX
Z
R jX
⋅
=
+
ps
s p
s p
RX
Q Q Q
R X
= = = =
p p
s p s s
p p
R jX
Z Z R jX
R jX
⋅
= = + =
+
( )2
1s pR Q R+ =
等效
彼此互轉
( )2
1p sR R Q= +
p
p
p
R
X
Q
=
(2) 已知並聯,要轉串聯:
2
1
p
s
R
R
Q
=
+
s s sX R Q=
Department of Electronic Engineering, NTUT13/70
14. 匹配頻寬 (I)
+
−
sV
sR
jX
jB LY
L L LY G jB= +
inZ
in s
in s
Z R
Z R
−
Γ =
+
+
−
sV
sR
jX
( )Lj B B+ LG
+
−
sV
sR
jX eqjB
eqR
2
1 1
1
eq
L
R
G Q
=
+
( )2
11 L
eq
eq
G Q
B
QR Q
+
= =
Case (a) 1s LR G<
把匹配後的完整網路
想辦法轉成「串聯」
或「並聯」RLC電路,
就可以直接使用諧振
器頻寬的定義來計算
匹配頻寬。
並聯準備轉成串聯
Department of Electronic Engineering, NTUT14/70
15. 匹配頻寬 (II)
阻抗匹配時的中心頻率 0
1
LC
ω ω= =
( )0
1
in eq s
eq
Z jX R R
jB
ω ω= = + + =
虛部虛部虛部虛部:
1
0
eq
jX
jB
+ = 1eqXB =
令 0X Lω= 0eqB Cω=
實部實部實部實部: eq sR R= 2
1 1
1
s
L
R
G Q
=
+
1
1
L s
Q
G R
= ± −
與
與
+
−
sV
sR
jX eqjB
eq sR R=
inQ1
2
L inQ Q=
• 定義RLC諧振器的 QL 與 Qin: ( )2
1in s L
eq
X
Q R Q G Q Q
R
= = ⋅ ⋅ + =
1
2
L inQ Q=而
• 找出 的 3-dB頻寬 :Γ
令 X Lω= eqB Cω=
當 0ω ω→ ,令 0ω ω ω= + ∆ 0
1
LC
ω =0ω∆ →
1
2
1 2 22
eqin s
in s s
s
eq
jX
jBZ R jL
Z R jL RjX R
jB
ω
ω
+
− ∆
Γ = =
+ ∆ ++ +
≃
與
其中 及
Department of Electronic Engineering, NTUT15/70
16. 匹配頻寬 (III)
( )2 ω= ∆
2 2 sL Rω∆ = 02 2
2 s sR R
L X
ω
ω∆ = =
3-dB Bandwidth in % BW=
0
22 2 1 2
1
1
s
L
s L
R
BW
X Q Q
R G
ω
ω
∆
= = = = =
−
1 2 2
1
1L
s L
BW
Q Q
R G
= = =
−
Γ
0ω
ω
1
2
1
2 ω∆
1
1
s L
Q
R G
= −
Case (a)
1
s
L
R
G
<
1s
L
R
Q
R
= −
Case (b)
s LR R>
• 找出 的 3-dB頻寬 :Γ
當
Case (b) s LR R>
同理可求得:
Department of Electronic Engineering, NTUT16/70
17. 範例 – L型匹配 (I)
• 下圖電路請以L型網路匹配之,並求出3-dB匹配頻寬。
1 2000
1 1
50s L
Q
R G
= ± − = ± −
1
, choose case (a)s
L
R
G
<
6.245Q = +
6
6
50 6.245 2 100 10
6.245 / 2000 0 2 100 10
s
L L
X R Q L
B G Q B C
π
π
= = ⋅ = ⋅ × ⋅
= − = − = ⋅ × ⋅
1st Solution: choose
Matching
Network
( )100in sZ f MHz R= =
+
−
sV
50sR = Ω 2000LR = Ω
( )100 0in f MHzΓ = =
Goal:
LR
LR4.9696 pFC =
496.96 nHL =
+
−
sV
50sR = Ω
-9
12
496.96 10 (H) 496.96 (nH)
4.9696 10 (F) 4.9696 (pF)
L
C −
= × =
= × =
6.245= ±
Department of Electronic Engineering, NTUT17/70
18. 範例 – L型匹配 (II)
6
6
50 ( 6.245) 1/ (2 100 10 )
( 6.245) / 2000 0 1/ (2 100 10 )
s
L L
X R Q C
B G Q B L
π
π
= = ⋅ − = − ⋅ × ⋅
= − = − − = − ⋅ × ⋅
6.245Q = −2nd Solution: choose
12
9
5.097 10 (F) 5.097 (pF)
509.7 10 (H) 509.7 (nH)
C
L
−
−
= × =
= × =
LR
+
−
sV
50sR = Ω
509.7 nHL =
5.097 pFC =
3dB
1 2 2
32%
| | 6.245L
BW
Q Q
= = = =
Department of Electronic Engineering, NTUT18/70
19. 三元件匹配(High Q匹配、窄頻匹配)
Case (a) π型匹配
+
−
sV
sR
2jX
3jB L L LY G jB= +
,inZ Γ
1jB
+
−
sV
sR
1jX
LZ L L LZ R jX= +2jB
3jX
Case (b) T型匹配
LY
,inZ Γ
目標在於求匹配元件之電抗X2與電受B1、B3,
能使 in sZ R= 0Γ =或
目標在於求匹配元件之電抗X1、X3 與電受B2,
能使 in sZ R= 0Γ =或
Department of Electronic Engineering, NTUT19/70
20. π型匹配 – 兩個L型的結合
+
−
sV
sR
2ajX
3jB LY L L LY G jB= +
VR
1jB
2bjX
VR
2 2 2a bX X X= +
1
V
L
R
G
<V sR R<
• π型網路可拆分成兩個“L型”網路來進行分析:
+
−
sV
sR
2ajX
1jB VR
in sZ R= 0Γ =
3jB LY
L L LY G jB= +
2bjX
+
−
sV
sR
in VZ R= 0Γ =
VR : 虛擬電阻(designed by yourself)
Department of Electronic Engineering, NTUT20/70
21. π型匹配 – Q值與匹配頻寬
1 1s
V
R
Q
R
= ± − 2
1
1
V L
Q
R G
= ± −1
1
2
BW
Q
= 2
2
2
BW
Q
=
( )1 2min ,BW BW BW≃
28/51
+
−
sV
sR
2ajX
1jB VR
in sZ R= 0Γ =
3jB LY
L L LY G jB= +
2bjX
+
−
sV
sR
in VZ R= 0Γ =
• 因RV同時小於Rs與(1/GL),因此我們可以透過設計RV值來得到高Q匹配(窄頻)。
• 我們也可以反過來說,如果要求匹配頻寬為BW,我們便可以反推虛擬電阻值RV
應該要設計為多少。
Department of Electronic Engineering, NTUT21/70
22. T型匹配 – 兩個L型的結合
+
−
sV
sR
1jX
LZ L L LZ R jX= +
VR
2ajB
3jX
2bjB
VR V LR R>V sR R>
sR
+
−
sV
1jX
in sZ R= 0Γ =
2ajB VR LZ
L L LZ R jX= +
3jX
2bjB
+
−
sV
VR
in VZ R= 0Γ =
VR : 虛擬電阻(designed by yourself)
• T型網路可拆分成兩個“L型”網路來進行分析:
Department of Electronic Engineering, NTUT22/70
23. T型匹配 – Q值與匹配頻寬
1 1V
s
R
Q
R
= ± − 2 1V
L
R
Q
R
= ± −1
1
2
BW
Q
= 2
2
2
BW
Q
=
( )1 2min ,BW BW BW≃
sR
+
−
sV
1jX
in sZ R= 0Γ =
2ajB VR LZ
L L LZ R jX= +
3jX
2bjB
+
−
sV
VR
in VZ R= 0Γ =
• 因RV同時大於Rs與(1/RL),因此我們可以透過設計RV值來得到高Q匹配(窄頻)。
• 我們也可以反過來說,如果要求匹配頻寬為BW,我們便可以反推虛擬電阻值RV
應該要設計為多少。
Department of Electronic Engineering, NTUT23/70
24. 範例 – π與T型匹配網路
• 使用π型與T型網路進行匹配,並要求匹配頻寬 BW < 5%。
Matching
Network
( )100in sZ f MHz R= =
+
−
sV
50sR = Ω 2000LR = Ω
( )100 0in f MHzΓ = =
Goal:
LR
Department of Electronic Engineering, NTUT24/70
25. 範例 – π型匹配網路 (I)
+
−
sV
50sR = Ω
2ajX
1jB 1.249Ω
( )100 50inZ f MHz= = Ω ( )100 0f MHzΓ = =
VR
1
50
1 1 6.247
1.2492
s
V
R
Q
R
= ± − = ± − = ± 2
1 2000
1 1 40
1.2492V L
Q
R G
= ± − = ± − = ±
3jB
2bjX
+
−
sV
1.249VR = Ω
( )100 1.2492inZ f MHz= = Ω ( )100 0f MHzΓ = =
LG
1
2000
LG =
Ω
1 2max(| |,| |) max( 50 / 1, 2000 / 1) 2000 / 1v v vQ Q Q R R R= = − − = −
2
5% 1.249
(2000 / ) 1
v
v
BW R
R
= ≤ ⇒ ≤ Ω
−
Case (a) π型匹配
1 6.247Q = +
1 1 0sB Q R Cω= = 198.85 pFC =
2 1 0a VX R Q Lω= = 12.42 nHL =
1 6.247Q = −
( )1 1 01sB Q R Lω= = −
203.95 pFC =( )2 1 01a VX R Q Cω= = −
12.74 nHL =
2 40Q = +
3 2 0L LB G Q B Cω= − = 31.83 pFC =
2 2 0b VX R Q Lω= = 79.53 nHL =
2 40Q = −
( )3 2 01L LB G Q B Lω= − = − 79.58 nHL =
( )2 2 01b VX R Q Cω= = − 31.85 pFC =
Department of Electronic Engineering, NTUT25/70
30. 範例 – T型匹配網路 (III)
Department of Electronic Engineering, NTUT30/70
31. 串接L型匹配 (Low Q匹配、寬頻匹配)
Case (a) 1s V LR R G< <
+
−
sV
sR
1jX
1jB LY L L LY G jB= +
in sZ R=
2jX
2jB
0Γ =
VRVR
+
−
sV
sR
1jX
1jB VR
in sZ R= 0Γ =
+
−
sV
VR
2jX
LY
L L LY G jB= +
2jB
in VZ R= 0Γ =
VR : 虛擬電阻(designed by yourself)
• 串接L型網路當然可以拆成兩個“L型”網路來進行分析:
Department of Electronic Engineering, NTUT31/70
32. 串接L型匹配 (Low Q)
1 1V
s
R
Q
R
= ± − 2
1
1
L V
Q
G R
= ± −
令 1 2Q Q Q= = s
V
L
R
R
G
=
2 2 2
2 1
1
s L
BW
QQ
R G
= =
−
≃
+
−
sV
sR
1jX
1jB VR
in sZ R= 0Γ =
+
−
sV
VR
2jX
LY
L L LY G jB= +
2jB
in VZ R= 0Γ =
• 我們想要最大頻寬的匹配,也就是要找到最小Q匹配(寬頻)。
Department of Electronic Engineering, NTUT32/70
33. 串接L型匹配 (Low Q)
Case (b) s V LR R R> >
+
−
sV
sR
1jX
1jB LZ L L LZ R jX= +
in sZ R=
2jX
2jB
VRVR
0Γ =
+
−
sV
sR
1jX
1jB
in sZ R= 0Γ =
VR LZ
L LR jX+
2jX
2jB
+
−
sV
VR
in VZ R= 0Γ =
VR : 虛擬電阻(designed by yourself)
• 串接L型網路當然可以拆成兩個“L型”網路來進行分析:
Department of Electronic Engineering, NTUT33/70
34. 串接L型匹配 (Low Q)
1 1s
V
R
Q
R
= ± − 2 1V
L
R
Q
R
= ± −
1 2Q Q Q= = V s LR R R=
2 2 2
2 1
1
s L
BW
QQ
R G
= =
−
≃
+
−
sV
sR
1jX
1jB
in sZ R= 0Γ =
VR LZ
L LR jX+
2jX
2jB
+
−
sV
VR
in VZ R= 0Γ =
• 串接L型網路當然可以拆成兩個“L型”網路來進行分析:
令
Department of Electronic Engineering, NTUT34/70
35. 範例 – 串接L型匹配 (I)
• 使用串接L型匹配使匹配頻寬能夠達到BW > 60%。
Matching
Network
( )100in sZ f MHz R= =
+
−
sV
50sR = Ω 2000LR = Ω
( )100 0in f MHzΓ = =
Goal:
LR
選擇RV : 50 2000 316.23V s LR R R= = ⋅ =
2 2
61.29%
1 2000 50 11
s L
BW
R G
= = =
−−
s V LR R R< <
Department of Electronic Engineering, NTUT35/70
36. 範例 – 串接L型匹配 (II)
+
−
sV
50sR = Ω
1jX
1jB
( )100 50inZ f MHz= = Ω
2jX
2jB
( )100 0f MHzΓ = =
VR
+
−
sV
316.23VR = Ω
( )100 0f MHzΓ = =( )100 316.23inZ f MHz= = Ω
LR
1
316.23
1 1 2.3075
50
V
s
R
Q
R
= ± − = ± − = ± 2
2000
1 1 2.3075
316.23
L
V
R
Q
R
= ± − = ± − = ±
1 2.3075Q = +
1 1 0sX R Q Lω= = 183.63 nHL =
1 1 0VB Q R Cω= = 11.61 pFC =
1 2.3075Q = −
( )1 1 01sX R Q Cω= = − 13.79 pFC =
( )1 1 01VB Q R Lω= = − 218.11 nHL =
2 2.3075Q = +
2 2 0VX R Q Lω= = 1.161 µHL =
( )2 2 0L LB Q R B Cω= − = 1.836 pFC =
2 2.3075Q = −
( )2 2 01VX R Q Cω= = − 2.181 pFC =
( ) ( )2 2 01L LB Q R B Lω= − = − 1.379 µHL =
316.23VR = Ω 2 kLR =
Department of Electronic Engineering, NTUT36/70
41. 史密斯圖的建立
( ) o
o
Z Z
Z
Z Z
−
Γ =
+
• 史密斯圖(Smith chart)也稱為反射係數圖( plane),某一阻抗所代表
的反射係數與其阻抗具有以下關係:
Γ
對所有的正實數Z都成立,而其中Zo是傳輸線特徵阻抗或系統參考阻
抗,一般為50 。
• 定義正規化阻抗 z 為
o o
Z R jX
z r jx
Z Z
+
= = = +
( )
( )
11
1 1
r jxz
U jV
z r jx
− +−
Γ = = = +
+ + + ( )
2 2
2 2
1
1
r x
U
r x
− +
=
+ + ( )
2 2
2
1
x
V
r x
=
+ +
其中 及
• 反射係數
Department of Electronic Engineering, NTUT41/70
42. 史密斯圖 (Smith Chart)
r
x
( )U jVΓ = +Γ-plane
U
V
1z j=
1z =
0z =
1
1
z
z
−
Γ =
+
1 1 1 90z j j= ⇒ = ∠
0 1 1 180z = ⇒ Γ = − = ∠
1 0z = ⇒ Γ =
1 90Γ = ∠
0Γ =
1Γ = −
( )z r jx= +z-plane
1 1 1 90z j j= − ⇒ Γ = − = ∠ −
1z j= −
Short Load Open
1z = ∞ ⇒ Γ =
1Γ =
Pure Imaginary: inductive
1 90Γ = ∠ −
Pure Imaginary: capacitive
Department of Electronic Engineering, NTUT42/70
51. 考慮導納的史密斯圖
y g jb= +
U
V
U′
V′z r jx= +
1 1
1
y g jb
z
− Γ
= = = +
+ Γ
1
1
z
+ Γ
=
− Γ
Impedance Chart (Z-Chart) Admittance Chart (Y-Chart)
jx+
jx− jb+
jb−
Short Load Open Short Load Open
Department of Electronic Engineering, NTUT51/70
61. 匹配範例 (II)
( )10 10LZ j= + Ω
0.2 0.2Lz j= +
Goal
0.2j
0.4j−
0.6x j∆ = −
2j
0j
2y j∆ = −
0.2 0.4z j= −
L
C 0.2
0.2j
01@ 500 MHzinz f= =
0.6j−
( )
1
02 0.6 50 30f Cπ
−
= × Ω =
( )
1
0
1
2 2 0.04
50
f Lπ
−
= × =
Ω
10.6 pFC =
7.95 nHL =
L
C
10.6 pF
7.95 nH
10 Ω
3.18 nH
Department of Electronic Engineering, NTUT61/70
62. 匹配範例 (III)
1 L
C
( )8 12 mSoutY j= −
Goal
( )50 Ω
0.4 0.6outy j= −
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63. 匹配至任意阻抗
LZC
L
50 20inZ j= + Ω
100 100LZ j= + Ω
Goal
100refZ = Ω
LZ
C
L
0.5 0.2inZ j= + Ω
1 1Lz j= + Ω
Department of Electronic Engineering, NTUT63/70
64. 頻率變高時的阻抗變化軌跡 (I)
L
R
C
R
L
R
C
L
R
C
( )1inZ R j Lω ω= +
( )
( )1
1
50
in
in
Z
z r jx
ω
ω = = +
Ω
( )1in aZ ω
( )1in bZ ω
( )2inZ ω
( )2in aZ ω
( )2in bZ ω
( )3inZ ω
( )3in aZ ω
( )1
1
inZ R j
C
ω
ω
= −
( )3in bZ ω ( )4inZ ω
( )4in bZ ω
( )4in aZ ω
Department of Electronic Engineering, NTUT64/70
65. 頻率變高時的阻抗變化軌跡 (II)
( )2inZ ω ( )1inZ ω
( )4inZ ω ( )3inZ ω
( )1in bZ ω
( )1in aZ ω
( )2in bZ ω
( )2in aZ ω
( )4in bZ ω
( )3in bZ ω
( )3in aZ ω
( )4in aZ ω
L R
C
L R
RCRC
L
Department of Electronic Engineering, NTUT65/70
66. 固定Q軌跡 (I)
n
X x
Q
R r
= =
1nQ =
2nQ =
Short Open
Department of Electronic Engineering, NTUT66/70
67. 固定Q軌跡 (II)
Short Open
very intensive
very intensive
intensive
Department of Electronic Engineering, NTUT67/70
68. 匹配頻寬與Q值的要求 (I)
• 前面我們已經知道,在阻抗匹配時:
2
n
L
Q
Q =
• 在特定的匹配頻寬BW要求下,QL 的值應設計為: 0
1
L
f BW
Q
=
• 範例:設計一個T型匹配網路,使其能將負載阻抗 轉到50LZ = Ω
10 15inZ j= − Ω
1
0.4
LQ
=
1
2.5
0.4
LQ = =
在阻抗匹配時: 2.5
2
n
L
Q
Q = =
5nQ =所以匹配網路本身的節點Q值應該要為
0
L
f
Q
BW
=
並且能達到匹配頻寬40%的要求。
在下一頁我們會看到如何利用Smith Chart與固定Q軌跡來完成
這個匹配條件。
Department of Electronic Engineering, NTUT68/70