Adapted from notes by ECE 5317-6351 Prof. Jeffery T. Williams Microwave Engineering Fall 2019 Prof. David R. Jackson Dept. of ECE Notes 18 Impedance Matching jX L Z =100 Ω Th [ ] jBc ZL =1000 [ Ω] 1 Impedance Matching Impedance matching is used to: We have ZLL= R + jX L . Maximize power from source to load We want Zin= R in + jX in . Minimize reflections (usually zero) Considerations: ⇓ Two constraints • Complexity • Implementation • Bandwidth • Adjustability A matching circuit typically requires at least 2 degrees of freedom. 2 Impedance Matching (cont.) Matching Methods: 1) Lumped element matching circuits 2) Transmission line matching circuits 3) Quarter-wave impedance transformers 3 Lumped-Element Matching Circuits Examples 4 Lumped-Element Matching Circuits (cont.) “tee” One extra degree of freedom “ladder” mm+2+2 elements mm extra degrees of freedom 5 Smith Chart Review Short-hand version Γ plane 6 Smith Charts Review (cont.) Γ plane 7 Smith Charts Review (cont.) ZY- Chart - + + - Γ plane 8 Series and Shunt Elements - + Γ plane Load Center of Smith chart: ZZYY=0, = 00 = 1/ Z + - ZZ0 ≡ in Note: The Smith chart is not actually being used as a transmission-line calculator but an impedance/admittance calculator. Hence, the normalizing impedance Z0 is arbitrary. (Usually we choose it to be the desired input resistance R .) in 9 High Impedance to Low Impedance Use when GL < Yin (The load is outside of the red G = 1 circle.) (High pass) Shunt-series “ell” C L Series Shunt Load ZZ0 ≡ in Shunt C Series L (Low pass) The shunt element puts us on the R = 1 Two possibilities circle; the series element is used to “tune out” the unwanted reactance. 10 Low Impedance to High Impedance Use when RL < Rin (The load is outside of the black R = 1 circle.) (Low pass) Series-shunt “ell” ZZ0 ≡ in Shunt C Series L Load Shunt L Series C (High pass) The series element puts us on the G = 1 Two possibilities circle; the shunt element is used to “tune out” the unwanted susceptance. 11 Example Use high impedance to low impedance matching. We want Zin =100 [ Ω ] jX L Z =100 Ω Th [ ] jBc ZL =1000 [ Ω] Z = 10 0.3 Ln, BB=0.3 ⇒= =3 [mS] Cn, C 100[Ω ] 1 XB=−=−Ω1/ 333 [ ] X C = − Cc ωC XXLn, =−=⇒= n 3 XL 3( 100[ Ω ]) X L =300 [ Ω ] ( XLL = ω ) Bn = 0.3 and X n = −3.0 12 Example (cont.) Here, the design example was repeated using a 50 [Ω] normalizing impedance. Note that the final normalized input impedance is 2.0. 5 GHz design frequency C = 0.096 [pF] L = 9.55 [nH] 13 Matching with a Pi Network Zin This works for low-high or high-low. ZL Note: This solution is not unique. Different ZL (load) Z (match) values for Bc2 could have been chosen. in Note: We could have also used parallel inductors and a series capacitor, or other combinations. 14 Pi Network Example (1000 Ω →100 Ω) Note that the final normalized input impedance is 2.0. 15 High Impedance to Low Impedance Exact Solution > RRL in GRLL= 1/ Shunt-series “ell” BX22= −1/ jX1 Rin jB2 RL 1 22+ GBL 2 B=±− GG G Re= Rin = G 2 L( in L ) G+ jB in L 2 GL Then we choose: 2 GL B = R 2 1 B2 22 in GGL +=in =−= GB+ X1 Im22 L 2 GL GLL++ jB22 G B 16 High Impedance to Low Impedance Exact Solution RR> L in GR= 1/ Summary LL BX22= −1/ jX1 Rin jB2 RL B2 =±− GGL( in G L ) B2 X1 = 22 GBL + 2 17 Transmission Line Matching Single-stub matching . This has already been discussed. Double-stub matching . This is an alternative matching method that gives more flexibility (details omitted – please see the Pozar book). 18 Quarter-Wave Transformer @ ff= 0 2ππλ β =g = λg 42 2 ZT ⇒=Zin ZL Γ=when = in 0 ZTL ZZ0 =λ Only true atf0 where g /4 Note: If ZL is not real, we can always add a reactive load in series or parallel to make it real (or add a length of transmission line between the load and the transformer to get a real impedance). 19 Quarter-Wave Transformer (cont.) At a general frequency: ZLT+ jZ tan β T ZZin= T ZTLT+ jZ tan β ZLT+ jZ t = ZtTT≡ tan β ZTL+ jZ t After some algebra (omitted): ZZ-- ZZ Γ=in 00 = L ZZin + 0 ZLL++ Z00 j2 t ZZ where we have used ZTL= ZZ0 20 Quarter-Wave Transformer (cont.) ZZ- Γ= L 0 ZLL++ Z00 j2 t ZZ After some more algebra (omitted): 1 Γ= θβ≡ 4ZZ T 1+ 0 L sec2 θ 2 (ZZL - 0 ) 2222 where we used 1+=+t 1tanββθTT ll = sec = sec 21 Quarter-Wave Transformer (cont.) The bandwidth is defined by the limit Γm (the maximum acceptable value of Γ). For example, using Γm = 1/3 corresponds to SWR = 2.0. This corresponds to Γm = - 9.54 dB. We set: Bandwidth region of transformer: 1 Γ=m π 4ZZ0 L 2 ∆=θββ( -) = 2- θ 1+ 2 sec θm 21TT 2 m (ZZL - 0 ) Solving for θm: Solve for θm Γ 2 ZZ θ = -1 m 0 L m cos -1 Γ2 ZZ- Note: cosx ∈( 0,π /2) 1- m L 0 22 Quarter-Wave Transformer (cont.) For TEM lines: π ff2 0 θβ=T = ⇒=f θ 2 f0 π 2 f0 ⇒=fmmθ π ∆ff2-( ff0 m ) 24mmθ ⇒==BW =2- = 2- (relative bandwidth) ff00 f 0π 4θm Note: BW= 2 - Multiply by 100 to get BW π in percent. Hence, using θm from the previous slide, 4 Γ 2 ZZ BW= 2 - cos-1 m 0 L π 2 |-|ZZ 1-Γm L 0 23 Quarter-Wave Transformer (cont.) Summary λg /4 @ f0 Z0 ZL 4 Γ 2 ZZ BW= 2 - cos-1 m 0 L π 2 |-|ZZ 1-Γm L 0 Smaller contrast between ZZ0 andL ⇒ larger BW Larger contrast between ZZ0 andL ⇒ smaller BW 24 Example λg /4 @ f0 Given: Z =100 [ Ω ] Z0 ZL L Z0 =50 [ Ω ] We’ll illustrate with two choices: dB Γ=mm1/ 3( Γ = -9.54 [dB]) ⇒ BW = 0.433( 43.3%) dB Γ=mm0.05( Γ = -26.0 [dB]) ⇒ BW = 0.060( 6.0%) 4 Γ 2 ZZ BW= 2 - cos-1 m 0 L π 2 |-|ZZ 1-Γm L 0 25 .