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 Z= R + jX . Minimize reflections in in in
(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