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射頻電子 - [第三章] 史密斯圖與阻抗匹配

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史密斯圖與阻抗匹配

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射頻電子 - [第三章] 史密斯圖與阻抗匹配

  1. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 27. 範例 – π型匹配網路 (III) Department of Electronic Engineering, NTUT27/70
  28. 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. 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. 30. 範例 – T型匹配網路 (III) Department of Electronic Engineering, NTUT30/70
  31. 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. 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. 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. 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. 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. 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. 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. 38. 範例 – 串接L型匹配 (IV) Department of Electronic Engineering, NTUT38/70
  39. 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. 40. 史密斯圖史密斯圖史密斯圖史密斯圖 Department of Electronic Engineering, NTUT40/70
  41. 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. 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. 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. 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. 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. 46. 典型的史密斯圖 Short OpenLoad +jx -jx Inductive Capacitive Department of Electronic Engineering, NTUT46/70
  47. 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. 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. 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. 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. 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. 52. ZY Chart U V Department of Electronic Engineering, NTUT52/70
  53. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 66. 固定Q軌跡 (I) n X x Q R r = = 1nQ = 2nQ = Short Open Department of Electronic Engineering, NTUT66/70
  67. 67. 固定Q軌跡 (II) Short Open very intensive very intensive intensive Department of Electronic Engineering, NTUT67/70
  68. 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. 69. 匹配頻寬與Q值的要求 (II) Department of Electronic Engineering, NTUT69/70
  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

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