射頻電子 - [第三章] 史密斯圖與阻抗匹配
TRANSCRIPT
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高頻電子電路第三章史密斯圖與阻抗匹配
李健榮助理教授
Department of Electronic EngineeringNational Taipei University of Technology
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大綱
• 諧振器、Q值與諧振頻寬
• 阻抗匹配網路:L型、T型、π型與串接L型網路
• 史密斯圖• 串並聯LC於史密斯圖上的軌跡
• 使用史密斯圖進行阻抗匹配• 固定Q值軌跡與阻抗匹配
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串聯諧振器(Series Resonators)
• 串聯諧振電路
I R L
C
( ) 1inZ R j L
j Cω ω
ω= + +
CV
+
−
21
2RP I R=
21
4mW I L=
2 2 2
2 2 2
1 1 1 1
4 4 4e C
CW V C I I
C Cω ω= = =
在諧振頻率時, :0 02 fω π= m eW W= 0
1
LCω =
( )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/secm e
R
W WQ
Pω ω+ =≜
當諧振發生時, 0ω ω=
00 0
0
2 2 1m e
R R
W W LQ
P P R CR
ωω ωω
= = = =
XQ
R= 其中 0
0
1 or X L
Cω
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。
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在諧振頻率時, :
此時,
並聯諧振器
2
2R
VP
R=
21
4eW V C=2 2
2
2 2 2
1 1 1
4 4 4m L
V VW I L L
L Lω ω= = =
0 02 fω π= m eW W= 0
1
LCω =
( )0
1inY
Rω =
LI
LR C
( ) 1 1inY j C
R j Lω 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/secm e
R
W WQ
Pω ω+ =≜
BQ
G=
00
1 or B C
Lω
ω=
1G
R=
當諧振發生時, 0ω ω=
其中
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與串聯RL與RC電路比較
L R
C R1X
QR 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 RQ
G Lω=≜
BQ CR
Gω=≜
L
R
C
R
X LQ
R R
ω=≜
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定義諧振頻寬
• 以串聯諧振器為例,當 0ω ω→
( ) 1inZ R j L
j Cω ω
ω= + +
0ω ω ω= + ∆令 0ω∆ → 0
1
LCω =
( )2 2 20 0
2 2 2
11 1inZ R j L R j L R j L
LC
ω ω ωω ω ω ωω ω ω
− = + − = + − = +
( )( )0 0 02
0 0
22 2
QRR j L R jL R j L R j
ω ω ω ω ω ωω ω ωω ω ω
+ − ⋅∆= + + = + ∆ = + ∆
≃ 0LQ
R
ω=其中
其中 而
I L R
C
( )inZ ω
CV
+
−
• 定義頻寬為0
2BW
ωω∆
≜
0
0
22in
BW QRZ R j R jBW QR
ωω
⋅= + = + ⋅
inZ
0ωω
R
2R
2 ω∆
當 1BW
Q= 2inZ R jR R= + =
003-dB Bandwidth BW
Q
ωω= ⋅ = 13-dB Bandwidth in % BW
Q= =
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並聯諧振器的3-dB頻寬
• 以並聯諧振電路為例,當 0ω ω→
( ) 1 1 12inY j C jC
R j L Rω ω ω
ω= + + + ∆≃
0
2BW
ωω∆
≜定義
( ) 0
0
1 12
2in
BW Q QY j jBW
R R R R
ωωω
⋅= + = +
當 1BW
Q=
1 1 2inY j
R R R= + =
003-dB Bandwidth BW
Q
ωω= ⋅ =
13-dB Bandwidth in % BW
Q= =
inY
0ωω
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
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Loaded-Q與Unloaded Q
LRResonantCircuit
Unloaded Q
• 串聯RLC電路
• 並聯RLC電路
( )0
0
1L
L L
X LQ
R R R C R R
ωω
= = =+ +
( )00
////L
L L
B R RQ C R R
G Lω
ω= = =
• 定義Qe為外部Q (external Q)
0
0
1e
L L
LQ
R R C
ωω
= = for series RLC
00
Le L
RQ R C
Lω
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。
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阻抗匹配
MatchingNetwork
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 thegenerator is matched)。
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八種雙元件L型匹配網路
LZ1C
2C
LZL
C
LZ1L
2L
LZC
L
LZC
L
LZ2C
1C
LZL
C
LZ2L
1L
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L型匹配 – Case (a) Rs < 1/GL
+
−sV
sRjX
jBLY
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
sRjX
( )Lj B B+ LG
L
L
B BQ
G
+=l
Series RC or RL
Parallel RC or RL
ss
XQ
R=
( )1
in sL L
Z jX RG 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。
Ls
s L
B BXQ Q Q
R G
+= = = =l
( )2 1 1s LR G Q + = 11
s L
QR G
= ± −
選擇1
1ss L
Q Q QR G
= = = + −l
11s s
s L
X R Q RR G
= = −
11L L L L
s L
B G Q B G BR 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
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L型匹配 – Case (b) Rs > RL
+
−sV
sRjX
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 XQ
R
+=l
+
−sV
sR( )Lj X X+
jB LR
Parallel RC or RL
Series RC or RL
( )1 1
inL L s
Y jBR 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。
Ls s
L
X XQ BR Q Q
R
+ = = = =l
2L s LQ R R R= − 1s
L
RQ
R= ± −
選擇 1ss
L
RQ Q Q
R= = = + −
l
1sL L L L
L
RX R Q X R X
R= − = − −
11s
s s L
RQB
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。
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(1)已知串聯,要轉並聯:
串聯與並聯轉換
sRsjX
pR
pjX
ss
s
XQ
R=
s s sZ R jX= +
pp
p
RQ
X=
p pp
p p
R jXZ
R jX
⋅=
+
pss p
s p
RXQ Q Q
R X= = = =
p ps p s s
p p
R jXZ Z R jX
R jX
⋅= = + =
+
( )21s pR Q R+ =
等效
彼此互轉
( )21p sR R Q= +
pp
p
RX
Q=
(2)已知並聯,要轉串聯:
21p
s
RR
Q=
+
s s sX R Q=
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匹配頻寬 (I)
+
−sV
sRjX
jBLY
L L LY G jB= +inZ
in s
in s
Z R
Z R
−Γ =+
+
−sV
sRjX
( )Lj B B+ LG+
−sV
sRjX eqjB
eqR2
1 1
1eqL
RG Q
=+
( )211 L
eqeq
G QB
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
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匹配頻寬 (II)
阻抗匹配時的中心頻率 0
1
LCω ω= =
( )0
1in eq s
eq
Z jX R RjB
ω ω= = + + =
虛部虛部虛部虛部:1
0eq
jXjB
+ = 1eqXB =
令0X Lω= 0eqB Cω=
實部實部實部實部: eq sR R= 2
1 1
1 sL
RG Q
=+
11
L s
QG R
= ± −
與
與
+
−sV
sRjX eqjB
eq sR R=
inQ1
2L inQ Q=
• 定義RLC諧振器的 QL與 Qin: ( )21in s Leq
XQ R Q G Q Q
R= = ⋅ ⋅ + =
1
2L 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
LCω =0ω∆ →
1
21 2 22
eqin s
in s ss
eq
jXjBZ R jL
Z R jL RjX RjB
ωω
+− ∆Γ = =+ ∆ ++ +
≃
與
其中 及
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匹配頻寬 (III)
( )2 ω= ∆
2 2 sL Rω∆ = 02 22 s sR R
L X
ωω∆ = =
3-dB Bandwidth in % BW=
0
22 2 1 2
11
s
L
s L
RBW
X Q Q
R G
ωω∆= = = = =
−
1 2 2
11L
s L
BWQ Q
R G
= = =−
Γ
0ωω
1
2
12 ω∆
11
s L
QR G
= −
Case (a)1
sL
RG
<
1s
L
RQ
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:
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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 20001 1
50s L
QR G
= ± − = ± −
1, choose case (a)s
L
RG
<
6.245Q = +6
6
50 6.245 2 100 10
6.245 / 2000 0 2 100 10s
L L
X R Q L
B G Q B C
ππ
= = ⋅ = ⋅ × ⋅ = − = − = ⋅ × ⋅
1st Solution: choose
MatchingNetwork
( )100in sZ f MHz R= =
+
−sV
50 sR = Ω 2000 LR = Ω
( )100 0in f MHzΓ = =Goal:
LR
LR4.9696 pFC =
496.96 nHL =
+
−sV
50 sR = Ω
-9
12
496.96 10 (H) 496.96 (nH)
4.9696 10 (F) 4.9696 (pF)
L
C −
= × =
= × =
6.245= ±
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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
ππ
= = ⋅ − = − ⋅ × ⋅ = − = − − = − ⋅ × ⋅
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
50 sR = Ω
509.7 nHL =
5.097 pFC =
3dB
1 2 232%
| | 6.245L
BWQ Q
= = = =
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三元件匹配(High Q匹配、窄頻匹配)
Case (a) π型匹配
+
−sV
sR2jX
3jB L L LY G jB= +
, inZ Γ
1jB
+
−sV
sR1jX
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
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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= +
1V
L
RG
<V sR R<
• π型網路可拆分成兩個“L型”網路來進行分析:
+
−sV
sR2ajX
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)
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π型匹配 – Q值與匹配頻寬
1 1s
V
RQ
R= ± − 2
11
V L
QR G
= ± −11
2BW
Q= 2
2
2BW
Q=
( )1 2min ,BW BW BW≃
28/51
+
−sV
sR2ajX
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。
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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:
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T型匹配 – Q值與匹配頻寬
1 1V
s
RQ
R= ± − 2 1V
L
RQ
R= ± −1
1
2BW
Q= 2
2
2BW
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。
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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%。
MatchingNetwork
( )100in sZ f MHz R= =
+
−sV
50 sR = Ω 2000 LR = Ω
( )100 0in f MHzΓ = =Goal:
LR
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範例 – π型匹配網路 (I)
+
−sV
50sR = Ω2ajX
1jB 1.249Ω
( )100 50inZ f MHz= = Ω ( )100 0f MHzΓ = =
VR
1
501 1 6.247
1.2492s
V
RQ
R= ± − = ± − = ± 2
1 20001 1 40
1.2492V L
QR G
= ± − = ± − = ±
3jB
2bjX+
−sV
1.249VR = Ω
( )100 1.2492inZ f MHz= = Ω ( )100 0f MHzΓ = =
LG
1
2000LG =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 RR
= ≤ ⇒ ≤ 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 =
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範例 – π型匹配網路 (II)
2 kΩ198.85 pF
12.42 nH
+
−sV
50 Ω 79.53 nH
31.83 pF 2 kΩ198.85 pF
12.42 nH
+
−sV 79.58 nH
31.85 pF
2 kΩ12.74 nH
203.95 pF
+
−sV
50 Ω 79.53 nH
31.83 pF 2 kΩ12.74 nH
203.95 pF
+
−sV
50 Ω 31.85 pF
79.58 nH
50 Ω
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範例 – π型匹配網路 (III)
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範例 – T型匹配網路 (I)
+
−sV
50 Ω1jX
( )100 50 inZ f MHz= = Ω ( )100 0f MHzΓ = =
2ajBVR
3jX
2bjB+
−sV
1 2max(| |,| |) max( ( / 50) 1, ( / 2000) 1) ( / 50) 1v v vQ Q Q R R R= = − − = −2
5% 80050 ( / 50) 1
v
v
BW RR
= ≤ ⇒ ≥ Ω−
80050VR = Ω
( )100 80050 inZ f MHz= = Ω ( )100 0f MHzΓ = =
80050VR = Ω
LR2 kΩ
1
800501 1 40
50V
s
RQ
R= ± − = ± − = ± 2
800501 1 6.247
2000V
L
RQ
R= ± − = ± − = ±
Case (b) T型匹配
1 40Q = + 1 1 0sX R Q Lω= = 3.183 µHL =
2 1 0a VB Q R Cω= = 0.795 pFC =
( )1 1 01sX R Q Cω= = − 0.796 pFC =
( )2 1 01a VB Q R Lω= = − 3.185 µHL =1 40Q = −
2 6.247Q = +3 2 0L LX R Q X Lω= − = 19.884 µHL =2 2 0b VB Q R Cω= = 0.124 pFC =
( )3 2 01L LX R Q X Cω= − = − 0.127 pFC =
( )2 2 01b VB Q R Lω= = − 20.394 µHL =2 6.247Q = −
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範例 – T型匹配網路 (II)
2 kΩ0.795 pF
3.183 µH
+
−sV
50 Ω 19.884 µH
0.124 pF
2 kΩ3.185 µH
0.796 pF
+
−sV
50 Ω 19.884 µH
0.124 pF
2 kΩ0.795 pF
3.183 µH
+
−sV
50 Ω 0.127 pF
20.394 µH
2 kΩ3.185 µH
0.796 pF
+
−sV
50 Ω 0.127 pF
20.394 µH
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範例 – T型匹配網路 (III)
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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
VR2jX
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:
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串接L型匹配 (Low Q)
1 1V
s
RQ
R= ± − 2
11
L V
QG R
= ± −
令 1 2Q Q Q= = sV
L
RR
G= 2 2 2
2 11
s L
BWQQ
R G
= =−
≃
+
−sV
sR
1jX
1jB VR
in sZ R= 0Γ =
+
−sV
VR2jX
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)。
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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
VRVR0Γ =
+
−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:
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串接L型匹配 (Low Q)
1 1s
V
RQ
R= ± − 2 1V
L
RQ
R= ± −
1 2Q Q Q= =V s LR R R= 2 2 2
2 11
s L
BWQQ
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
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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%。
MatchingNetwork
( )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= = ⋅ =
2 261.29%
1 2000 50 11s L
BW
R G
= = =−−
s V LR R R< <
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範例 –串接L型匹配 (II)
+
−sV
50 sR = Ω1jX
1jB
( )100 50 inZ f MHz= = Ω
2jX
2jB
( )100 0f MHzΓ = =
VR+
−sV
316.23 VR = Ω
( )100 0f MHzΓ = =( )100 316.23inZ f MHz= = Ω
LR
1
316.231 1 2.3075
50V
s
RQ
R= ± − = ± − = ± 2
20001 1 2.3075
316.23L
V
RQ
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 =
316.23 VR = Ω 2 kΩLR =
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範例 –串接L型匹配 (III)
2 kΩ11.61 pF
183.63 nH
+
−sV
50 Ω 1.161 µH
1.836 pF
2 kΩ218.11 nH
13.79 pF
+
−sV
50 Ω 1.161 µH
1.836 pF
2 kΩ11.61 pF
183.63 nH
+
−sV
50 Ω 2.181 pF
1.379 µH
2 kΩ218.11 nH
13.79 pF
+
−sV
50 Ω2.181 pF
1.379 µH
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範例 –串接L型匹配 (IV)
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若需要更大的匹配頻寬
L-ShapeMatchingNetwork
+
−sV
50 sR = Ω
LR
L-ShapeMatchingNetwork
L-ShapeMatchingNetwork
• Use multi-section L-shape matching networks to extend BW.
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史密斯圖史密斯圖史密斯圖史密斯圖
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史密斯圖的建立
( ) o
o
Z ZZ
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50Ω。
• 定義正規化阻抗 z為
o o
Z R jXz r jx
Z Z
+= = = +
( )( )
11
1 1
r jxzU jV
z r jx
− +−Γ = = = ++ + + ( )
2 2
2 2
1
1
r xU
r x
− +=+ + ( )2 2
2
1
xV
r x=
+ +其中 及
• 反射係數
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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
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固定電阻圓(I)
r
x
( )U jVΓ = +Γ-plane
U
V
1 1z j= +
1 1z j= −0z =
0.447 63.4Γ = ∠
0.447 63.4Γ = ∠ −
( )z r jx= +z-plane
1 1z j= +1 1z j= −
0.447 63.43Γ = ∠
0.447 63.43Γ = ∠ −
1 2z j= +
1 2z j= −
1 2z j= +1 2z j= −
0.707 45Γ = ∠
0.707 45Γ = ∠ −
1j2j
1j−2j−
0.707 45Γ = ∠
0.707 45Γ = ∠ −
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固定電阻圓(II)
r
x( )z r jx= +z-plane
U
V
0z jx= +
0z r= =0.5r =
1r = 3r =
0.5z jx= +1z jx= + 3z jx= +
0r = 3r =1r =0.5r =
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固定電抗軌跡
r
x
( )z r jx= +z-plane
U
V
0.5z j=
0.5z j=
1z j=
3z j=
0.5z j= −1z j= −
3z j= −
0j
0.5j1j
3j
0.5j− 1j−
3j−
0.5 0.5z j= +1 0.5z j= +
1.5 0.5z j= +
1 126.87Γ = ∠
0.447 116.56Γ = ∠
0.243 75.97Γ = ∠
0.2773 33.69Γ = ∠
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典型的史密斯圖
Short OpenLoad
+jx
-jx
Inductive
Capacitive
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史密斯圖上的電抗軌跡
Short OpenLoad
+jx
-jx
Inductive
Capacitive
+j0.1
+j0.2
+j0.3
+j0.4
+j0.5+j0.6 +j1.6
+j1.7+j1.8
+j2.0
+j3.0
+j4.0
+j5.0+j6.0
0.4x∆ =
0.4x∆ =
0.4x∆ =
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由史密斯圖讀出阻抗值
1 1 1z j= +
2 0.4 0.5z j= +
3 3 3z j= −
4 0.2 0.6z j= −
5 0z = 1z2z
3z
4z
5z
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找出阻抗所對應的反射係數
19.44∠ −
Γ
1 3 3z j= −
1z
0.721 19.44Γ = ∠ −
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找出反射係數 所對應的阻抗值
0.447 26.56Γ = ∠
2 1z j= +
26.56∠
Γ
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考慮導納的史密斯圖
y g jb= +
U
V
U ′
V ′z r jx= +
1 1
1y g jb
z
− Γ= = = ++ Γ
1
1z
+ Γ=− Γ
Impedance Chart (Z-Chart) Admittance Chart (Y-Chart)
jx+
jx− jb+
jb−
Short Load Open Short Load Open
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ZY Chart
U
V
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串聯電感在史密斯圖上的軌跡
0.8Lz j=
0.3 0.3z j= −
0.3 0.5inz j= +
0.3 0.3z j= −
0.3 0.5inz j= +
0.8x∆ =
-j0.3
+j0.5
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串聯電容在史密斯圖上的軌跡
0.8Cz j= −
0.3 0.3z j= −
0.3 1.1inz j= −
0.3 0.3z j= −
0.3 1.1inz j= −0.8x∆ = −-j0.3
-j1.1
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並聯電感在史密斯圖上的軌跡
1.6 1.6y j= +
1.6 0.8iny j= −
2.4Ly j= −
1.6 1.6y j= +
1.6 0.8iny j= −
2.4y∆ = −
+j1.6
-j0.8
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並聯電容在史密斯圖上的軌跡
1.6 1.6y j= +
1.6 5iny j= +
3.4Cy j=
1.6 1.6y j= +
1.6 5iny j= +
3.4y∆ =
+j1.6
+j5
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串並聯LC於史密斯圖上的軌跡
Higher impedanceLower impedance
Series L
Series C
Shunt L
Shunt C
+jx
-jx
Inductive
Capacitive
Short
Open
Lower admittanceHigher admittance
-jb
+jb
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八種雙元件L型匹配網路
LZ1C
2C
LZL
C
LZ1L
2L
LZC
L
LZC
L
LZ2C
1C
LZL
C
LZ2L
1L
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匹配到參考阻抗(史密斯圖中心點)
• 大部分系統的參考阻抗 50 refZ = Ω
1z2z
3z
4z
5z
Goal
Goal circle (r=1)
Goal circle (g=1)
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匹配範例 (I)
( )10 10 LZ j= + Ω
0.2 0.2Lz j= +
Goal
0.2j
0.4j
0.2x j∆ =
2j−
0j
2y j∆ =
0.2 0.4z j= +
( )50 refZ = Ω
C
L
01@ 500 MHzinz f= =
0.2
0.2j
0.2j
0.5j−
02 0.2 50 10f Lπ = × Ω =
0
12 2 0.04
50f Cπ = × =
Ω
3.18 nHL =
12.74 pFC =
C
L 10 Ω
3.18 nH
3.18 nH
12.74 pF
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匹配範例 (II)
( )10 10 LZ j= + Ω
0.2 0.2Lz j= +
Goal
0.2j
0.4j−
0.6x j∆ = −
2j
0j
2y j∆ = −
0.2 0.4z j= −
LC 0.2
0.2j
01@ 500 MHzinz f= =
0.6j−
( ) 1
02 0.6 50 30f Cπ − = × Ω =
( ) 1
0
12 2 0.04
50f Lπ − = × =
Ω
10.6 pFC =
7.95 nHL =
LC
10.6 pF
7.95 nH
10 Ω
3.18 nH
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匹配範例 (III)
1 L
C
( )8 12 mSoutY j= −
Goal( )50 Ω
0.4 0.6outy j= −
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匹配至任意阻抗
LZC
L
50 20 inZ j= + Ω
100 100 LZ j= + Ω
Goal
100 refZ = Ω
LZ
C
L
0.5 0.2 inZ j= + Ω
1 1 Lz j= + Ω
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頻率變高時的阻抗變化軌跡 (I)
L
R
C
R
L
RC
LR
C
( )1inZ R j Lω ω= +
( ) ( )11 50
inin
Zz r jx
ωω = = +
Ω( )1in aZ ω
( )1in bZ ω
( )2inZ ω
( )2in aZ ω
( )2in bZ ω
( )3inZ ω
( )3in aZ ω
( )1
1inZ R j
Cω
ω= −
( )3in bZ ω ( )4inZ ω( )4in bZ ω
( )4in aZ ω
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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
CL R
RCRC
L
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固定Q軌跡 (I)
n
X xQ
R r= =
1nQ =
2nQ =
Short Open
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固定Q軌跡 (II)
Short Open
very intensivevery intensive
intensive
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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
nL
QQ =
• 在特定的匹配頻寬BW要求下,QL的值應設計為: 0
1
L
f BWQ
=
• 範例:設計一個T型匹配網路,使其能將負載阻抗 轉到50 LZ = Ω10 15 inZ j= − Ω
10.4
LQ= 1
2.50.4LQ = =
在阻抗匹配時: 2.52
nL
QQ = =
5nQ =所以匹配網路本身的節點Q值應該要為
0L
fQ
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。
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匹配頻寬與Q值的要求 (II)
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本章總結
• 阻抗匹配即是在於將負載阻抗ZL透過阻抗變換將其轉至與源阻抗Zs成共軛,以達最大功率傳輸之目的。換言之,我們也可以說,阻抗匹配是將源阻抗Zs透過阻抗變換將其轉至與負載阻抗ZL成共軛,以達最大功率傳輸之目的。
• 常用的匹配網路包含有L型、π型、T型與串接L型網路。 L型匹配頻寬中等,且頻寬無法調整。π型與T型具有可設計頻寬的優點,且匹配頻寬比L型還窄。串接L型匹配網路可增加頻寬,但缺點是需要以電路尺寸來換取頻寬。
• 在現今電腦輔助設計發達的時代,雖然史密斯圖已經很少被用來計算反射係數或阻抗,但它對於微波工程師仍然是一種非常有幫助的設計工具。
• 高Q值的電路具有較窄的頻寬,反之,低Q值的電路則具有較高的頻寬。因此,Q值越高的電路,對於頻率飄移或元件變異會更敏感。
• 越低的Q值雖然表示了電路的頻寬越寬,但也暗示了損耗的增加。
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