射頻電子 - [第三章] 史密斯圖與阻抗匹配
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This document discusses Smith charts and impedance matching. It begins with an introduction to resonators, Q factor, and resonant bandwidth. It then covers basic impedance matching networks including L, T, and π networks. The document explains how to use Smith charts to represent LC circuits and perform impedance matching. It also discusses loaded Q versus unloaded Q and how to match impedances for different cases. Matching bandwidth is defined and conversions between series and parallel circuits are covered. The document provides an overview of important concepts regarding resonators, Q factor, impedance matching, and the use of Smith charts.
射頻電子 - [第三章] 史密斯圖與阻抗匹配
- 1. 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 Engineering National Taipei University of Technology
- 2. 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 Department of Electronic Engineering, NTUT2/70
- 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
- 10. 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 Department of Electronic Engineering, NTUT10/70
- 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
- 26. 範例 – π型匹配網路 (II) 2 k198.85 pF 12.42 nH + − sV 50 Ω 79.53 nH 31.83 pF 2 k198.85 pF 12.42 nH + − sV 79.58 nH 31.85 pF 2 k12.74 nH 203.95 pF + − sV 50 Ω 79.53 nH 31.83 pF 2 k12.74 nH 203.95 pF + − sV 50 Ω 31.85 pF 79.58 nH 50 Ω Department of Electronic Engineering, NTUT26/70
- 27. 範例 – π型匹配網路 (III) Department of Electronic Engineering, NTUT27/70
- 28. 範例 – T型匹配網路 (I) + − sV 50 Ω 1jX ( )100 50inZ f MHz= = Ω ( )100 0f MHzΓ = = 2ajB VR 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 R R = ≤ ⇒ ≥ Ω − 80050VR = Ω ( )100 80050inZ f MHz= = Ω ( )100 0f MHzΓ = = 80050VR = Ω LR2 k 1 80050 1 1 40 50 V s R Q R = ± − = ± − = ± 2 80050 1 1 6.247 2000 V L R Q 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 = − Department of Electronic Engineering, NTUT28/70
- 29. 範例 – T型匹配網路 (II) 2 k0.795 pF 3.183 µH + − sV 50 Ω 19.884 µH 0.124 pF 2 k3.185 µH 0.796 pF + − sV 50 Ω 19.884 µH 0.124 pF 2 k0.795 pF 3.183 µH + − sV 50 Ω 0.127 pF 20.394 µH 2 k3.185 µH 0.796 pF + − sV 50 Ω 0.127 pF 20.394 µH Department of Electronic Engineering, NTUT29/70
- 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
- 37. 範例 – 串接L型匹配 (III) 2 k11.61 pF 183.63 nH + − sV 50 1.161 µH 1.836 pF 2 k218.11 nH 13.79 pF + − sV 50 Ω 1.161 µH 1.836 pF 2 k11.61 pF 183.63 nH + − sV 50 Ω 2.181 pF 1.379 µH 2 k218.11 nH 13.79 pF + − sV 50 Ω 2.181 pF 1.379 µH Department of Electronic Engineering, NTUT37/70
- 38. 範例 – 串接L型匹配 (IV) Department of Electronic Engineering, NTUT38/70
- 39. 若需要更大的匹配頻寬 L-Shape Matching Network + − sV 50sR = Ω LR L-Shape Matching Network L-Shape Matching Network • Use multi-section L-shape matching networks to extend BW. Department of Electronic Engineering, NTUT39/70
- 40. 史密斯圖史密斯圖史密斯圖史密斯圖 Department of Electronic Engineering, NTUT40/70
- 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
- 43. 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 = ∠ − 1j 2j 1j− 2j− 0.707 45Γ = ∠ 0.707 45Γ = ∠ − Department of Electronic Engineering, NTUT43/70
- 44. 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 = Department of Electronic Engineering, NTUT44/70
- 45. 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 0.5j 1j 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 = ∠ Department of Electronic Engineering, NTUT45/70
- 46. 典型的史密斯圖 Short OpenLoad +jx -jx Inductive Capacitive Department of Electronic Engineering, NTUT46/70
- 47. 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∆ = Department of Electronic Engineering, NTUT47/70
- 48. 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 Department of Electronic Engineering, NTUT48/70
- 49. 找出阻抗所對應的反射係數 19.44∠ − Γ 1 3 3z j= − 1z 0.721 19.44Γ = ∠ − Department of Electronic Engineering, NTUT49/70
- 50. 找出反射係數 所對應的阻抗值 0.447 26.56Γ = ∠ 2 1z j= + 26.56∠ Γ Department of Electronic Engineering, NTUT50/70
- 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
- 52. ZY Chart U V Department of Electronic Engineering, NTUT52/70
- 53. 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 Department of Electronic Engineering, NTUT53/70
- 54. 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 Department of Electronic Engineering, NTUT54/70
- 55. 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 Department of Electronic Engineering, NTUT55/70
- 56. 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 Department of Electronic Engineering, NTUT56/70
- 57. 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 Department of Electronic Engineering, NTUT57/70
- 58. 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 Department of Electronic Engineering, NTUT58/70
- 59. 匹配到參考阻抗(史密斯圖中心點) • 大部分系統的參考阻抗 50refZ = Ω 1z2z 3z 4z 5z Goal Goal circle (r=1) Goal circle (g=1) Department of Electronic Engineering, NTUT59/70
- 60. 匹配範例 (I) ( )10 10LZ j= + Ω 0.2 0.2Lz j= + Goal 0.2j 0.4j 0.2x j∆ = 2j− 0j 2y j∆ = 0.2 0.4z j= + ( )50refZ = Ω C L 01@ 500 MHzinz f= = 0.2 0.2j 0.2j 0.5j− 02 0.2 50 10f Lπ = × Ω = 0 1 2 2 0.04 50 f Cπ = × = Ω 3.18 nHL = 12.74 pFC = C L 10 Ω 3.18 nH 3.18 nH 12.74 pF Department of Electronic Engineering, NTUT60/70
- 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= − Department of Electronic Engineering, NTUT62/70
- 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
- 69. 匹配頻寬與Q值的要求 (II) Department of Electronic Engineering, NTUT69/70
- 70. 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。 Department of Electronic Engineering, NTUT70/70