Impedance Matching

A. Nassiri -ANL

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 Impedance Matching

• Impedance matching is important for the following reasons: – Maximum power is delivered when the load is matched to the line (assuming the generator is matched), and power loss in is minimized. – Impedance matching sensitive receiver components (antenna, low-noise amplifier, etc.) improves the signal-to- noise ratio of the system. – Impedance matching in a power distribution network (such as an antenna array feed network will reduce amplitude and phase errors.

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 2 Impedance Matching

• Factors that may be important in the selection of a particular matching network include the following: – Complexity: • A simpler matching network is usually cheaper, more reliable, and less lossy than a more complex design.

– Bandwidth: • Narrow or broadband.

– Implementation: • According to the technology used the matching network can be decided on.

– Adjustability: • Required if dealing with variable load. Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 3 Matching with lumped elements (L-Networks)

Network for zL inside the Network for zL outside the 1 + jx circle 1 + jx circle jX and jB Can be capacitors or inductors

1+ jx

8 possibilities for matching

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 4 Matching with lumped elements (L-Networks)

Analytic Solutions

Z L Z L = RL + jX L z = L Z  1  o   Zo = Z L //  + jX  jB  1 RL > Zo Zo = jX + jB +1/(RL + jX L )

RL + jX L RL + jX L Zo = jX + = jX + jB(RL + jX L ) +1 (1− BX L )+ jBRL

jX ((1− BX L )+ jBRL )+ RL + jX L Zo = (1− BX L )+ jBRL

(RL − XBRL )+ j(X L + X (1− BX L )) Zo = (1− BX L )+ jBRL

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 5 Matching with lumped elements (L-Networks)

ℜeLH = ℜeRH

Zo (1− BX L ) = (RL − XBRL ) B(XRL − X L Zo ) = RL − Zo

ℑmLH = ℑmRH

BRL Zo = X L + X (1− BX L ) X (1− BX L ) = BZo RL − X L

Solving the 2nd order equation

X ± R Z R2 + X 2 − Z R = L L o L L o L But R > Z B 2 2 L o RL + X L 1 X Z Z = + L o − o B(XRL − X L Zo ) = RL − Zo X B RL BRL

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 6 Matching with lumped elements (L-Networks)

Analytic Solutions

Z L = RL + jX L 1 1 = jB + Zo RL + j(X + X L ) RL < Zo 1 jB(R + j(X + X ))+1 = L L Zo RL + j(X + X L )

RL + j(X + X L ) = Zo ((1− B(X + X L ))+ jBRL )

ℜeLH = ℜeRH BZo (X + X L ) = Zo − RL X = ± RL (Zo − RL ) − X L

(Zo − RL ) RL ℑmLH = ℑmRH (X + X L ) = BZo RL B = ± Zo

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 7 Matching with lumped elements (L-Networks)

Smith Chart Solutions Example Design an L-section matching network to match a series RC load with an

impedance ZL=200-j100 Ω, to a 100 Ω, at a frequency of 500 MHz. Solution

Z L = 200 − j100 Ω RL > Zo

zL = 2 − j1 Ω

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 8 st 1 Step zL = 2 − j1 Ω

yl 180o

O zl

jb1 = j0.3

jb2 = − j0.7

y1 = 0.4 + j0.5

y2 = 0.4 − j0.5

Sol. 1 jb1 = j0.3

jx1 = + j1.2 y1 = 0.4 + j0.5

z1 =1− j1.2

Sol. 2 jb2 = − j0.7

jx2 = − j1.2 y = 0.4 + j0.5 1 z1 =1+ j1.2

jb1 = j0.3 jx1 = + j1.2 jb2 = − j0.7 jx2 = − j1.2

1 1 − j jB = jb× = jωC jB = jb× = Zo Zo ωL b 0.3 − Z −100 C = = = 0.955pF L = o = = 45.5 nH 6 ω π × 6 − ωZo 2π (500×10 )100 b 2 (500 10 )( 0.7) − j jX = jx× Z = jωL jX = jx× Z = o o ωC −1 −1 xZo 1.2×100 = = = = = C 6 L 6 38.2nH ω π × × − × ω 2π (500×10 ) zZo 2 (500 10 ) ( 1.2) 100 = 2.65 pF

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 15 Example

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 16 The Quarter Wave Transformer

Z L + jZo tan βl Zin = Zo Zo + jZ L tan βl

Z L + jZ1 Z L π + jZ1 tan 2 tan βl 2 Z1 For only Zin = Z1 = Z1 = = Zo real loads Z1 Z1 Z + jZ + jZ L β L π L tan l tan 2 Z1 = ZoZ L Need to drive the mismatch versus frequency? π Z + jZ t Let = L 1 t = tan βl = tanθ at f = fo θ = Zin Z1 2 Z1 + jZ Lt  Z + jZ t   L 1  − Z1  Zo Z − Z Z + jZ t Z (Z + jZ t)− Z (Z + jZ t) Γ = in o =  1 L  = 1 L 1 o 1 L Z + Z  Z + jZ t  Z (Z + jZ t)+ Z (Z + jZ t) in o  L 1  + 1 L 1 o 1 L Z1  Zo  Z1 + jZ Lt 

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 17 The Quarter Wave Transformer

Z (Z − Z )+ jt(Z 2 − Z Z ) Γ = 1 L o 1 o L 2 2 & Z1 = ZoZ L Z1(Z L + Zo )+ jt(Z1 + ZoZ L ) Z (Z − Z ) Z − Z Γ = 1 L o Γ = L o 2 Γ Z1(Z L + Zo )+ j2tZ1 Z L + Zo + j2t ZoZ L

Z − Z 1 Γ = L o = 2 2 1/ 2 2 2 2 2 1/ 2 ((Z L + Zo ) + 4t ZoZ L ) ((Z L + Zo ) (Z L − Zo ) + (4t ZoZ L (Z L − Zo ) ))

2 (Z L + Zo ) 4ZoZ L 2 =1+ 2 (Z L − Zo ) (Z L − Zo ) 1 1 Γ = = 2 2 2 1/ 2 2 2 1/ 2 (1+ 4ZoZ L (Z L − Zo ) + (4t ZoZ L (Z L − Zo ) )) (1+ (4ZoZ L (Z L − Zo ) )sec θ )

2 2 2 2 2 1+ t =1+ tan θ = sec θ (4ZoZ L (Z L − Zo ) )sec θ >1

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 18 The Quarter Wave Transformer

λ o π 2 If we are operating close to fo l ≈ θ ≈ sec θ >>1 4 2 Z − Z Γ ≈ L o cosθ 2 ZoZ L  π  ∆θ = 2 −θ   2 m  2 1  2 Z Z  =1+  o L secθ  Γ2  − m  m  Z L Zo  Γ 2 Z Z cosθ = m o L m 2 Z − Z 1− Γm L o

2πf v p πf θ = βl = = v p 4 fo 2 fo Γm= Maximum reflection coefficient 2θ f Quarter magnitude that can be tolerated m o Wavelength at fm = π fo

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 19 The Quarter Wave Transformer

∆f 2( f − f ) 2 f 4θ 4  Γ 2 Z Z  = o m = 2 − m = 2 − m = 2 − cos−1  m o L  f f f π π  2 Z − Z  o o o  1− Γm L o 

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 20 Example

Design a single-section quarter-wave matching transformer to match a 10 Ω load to a 50 Ω line, at fo = 3 GHz. Determine the percent bandwidth for which the SWR ≤ 1.5. Solution

Z1 = ZoZ L = (50)(10) = 22.36 Ω SWR −1 1.5 −1 Length λ / 4 @3 GHz Γm = = = 0.2 SWR +1 1.5 +1   ∆f 4 − Γ 2 Z Z = 2 − cos 1  m o L  f π  2 Z − Z  o  1− Γm L o 

∆f 4  0.2 2 (50)(10) = 2 − cos−1   = 0.29, or 29% f π 2 10 − 50 o  1− (0.2)  Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 21 The Theory of Small Reflections

• Single-Section Transformer Γ?

Z2 − Z1 Γ1 = Z2 + Z1 Z L − Z2 Γ3 = Z L + Z2 Γ2 = −Γ1

2Z2 T21 =1+ Γ1 = Z1 + Z2

2Z1 T12 =1+ Γ2 = Z1 + Z2

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 22 The Theory of Small Reflections

∞ Γ = Γ +T T Γ e−2 jθ +T T Γ2Γ e−4 jθ + −2 jθ n n −2 jnθ 1 12 21 3 12 21 3 2 = Γ1 +T12T21Γ3e ∑Γ2 Γ3 e n=0 ∞ − θ n 1 T T Γ e 2 j Γ = −Γ ∑ x = for x <1 Γ = Γ + 12 21 3 2 1 1 −2 jθ 2Z n=0 1− x − Γ Γ 2 1 2 3e T21 =1+ Γ1 = Z1 + Z2 2Z1 T12 =1+ Γ2 = Z1 + Z2 (1+ Γ )(1− Γ )Γ e−2 jθ (1− Γ2 )Γ e−2 jθ Γ + Γ2Γ e−2 jθ + (1− Γ2 )Γ e−2 jθ Γ = Γ + 1 1 3 = Γ + 1 3 = 1 1 3 1 3 1 −2 jθ 1 −2 jθ −2 jθ 1+ Γ1Γ3e 1+ Γ1Γ3e 1+ Γ1Γ3e

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 23 The Theory of Small Reflections

Γ + Γ e−2 jθ Γ = 1 3 −2 jθ 1+ Γ1Γ3e

If the discontinuities between the impedances Z1, Z2, and Z2, ZL are small, then |Γ1 Γ3| <<1.

−2 jθ Γ ≅ Γ1 + Γ3e

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 24 Multi-section Transformer

Z1 − Zo Zn+1 − Zn Z L − Z N Γ0 = Γn = ΓN = Z1 + Zo Zn+1 + Zo Z L + Z N −2 jθ Γ ≅ Γ1 + Γ3e −2 jθ −4 jθ −2 j(N −2)θ −2 j(N −1)θ −2 jNθ Γ = Γ(θ ) ≅ Γ0 + Γ1e + Γ2e ++ ΓN −2e + ΓN −1e + ΓN e Assume that the transformer is symmetrical

Γ0 = ΓN Γ1 = ΓN −1 Γ2 = ΓN −2 etc. −2 jθ −4 jθ −2 j(N −2)θ −2 j(N −1)θ −2 jNθ Γ(θ ) = Γ0 + Γ1e + Γ2e ++ Γ2e + Γ1e + Γ0e −2 jNθ −2 jθ −2 j(N −1)θ −4 jθ −2 j(N −2)θ Γ(θ ) = Γ0 (1+ e )+ Γ1(e + e )+ Γ2 (e + e )+

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 25 Multi-section Transformer

− jNθ jNθ − jNθ j(N −2)θ − j(N −2)θ j(N −4)θ − j(N −4)θ Γ(θ ) = e [Γ0 (e + e )+ Γ1(e + e )+ Γ2 (e + e )+]

 1  Γ θ = − jNθ Γ θ + Γ − θ ++ Γ − θ ++ Γ ( ) 2e 0 cos N 1 cos(N 2) n cos(N 2n) N / 2   2  For N even

 1  Γ θ = − jNθ Γ θ + Γ − θ ++ Γ − θ ++ Γ θ ( ) 2e 0 cos N 1 cos(N 2) n cos(N 2n) (N −1)/ 2 cos   2  For N odd Finite Fourier Cosine Series

By choosing the Γns and enough sections (N) we can achieve the required response.

Binomial Multisection Chebyshev Multisection Matching Transformers Matching Transformers

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 26