Circuiti a Microonde: Introduzione

IntroductionA microwave circuit is an interconnection of components whose size is comparable with the wavelength at the operation frequency

Type of Components:• Interconnection: it is not an ideal connection (zero resistance) as

in the case of lumped-element networks• Pseudo-lumped components (they present, in a first

approximation, the same frequency behavior of the lumped components, having however a finite size)

• Distributed components (transmission line sections or stubs arbitrarily terminated)

In microwave circuits the junction between two (or more) component do not correspond to an ideal node (as in lumped circuits). The junction must be suitably represented (typically by means of its scattering parameters) in order to take into account the phenomena which always arise at a discontinuity in a transmission line.

Discontinuity between two transmission lines

Zc Zc

Discontinuity

Zc ZcS Matrix

Coax 1: R1=5 cm, r1=2.17 cm

Coax 2: R2=3 cm, r1=1.3 cm

1 2

1 2

60 ln 60 ln 50 cR RZr r

= ⋅ = ⋅ = Ω

No good!

Correct model. Why?

L1=10 cm

L2=10 cm

Higher modes excitation

TEM wave

Transverse E, H fields

At the discontinuity the E field cannot be transverse

Higher modes are then generated, allowing the fulfillment of the continuity conditions. These modes however cannot propagate (they are below cutoff)

Incident TEM wave

Transmitted TEM wave

Amplitude of higher modes excited

The energy of the em field associated to the higher modes is confined closely to the discontinuity, and it affects the incident wave like a lumped reactance connected at the junction of the two coaxial lines

Scattering Matrix of the discontinuityThe scattering parameters of the following configuration can be evaluated by means of an em simulator (numerical solution of the em fields). It has (f0=1 GHz, Zc=50 Ω):

11 22 12 210.1792 20 , 0.9838 109.61 S S S S= = ∠ ° = = ∠ °

Discontinuity50 Ω 50 Ω

10 cm10 cm

Microwave network

Port 1 Port 2

To get the S parameters of discontinuity we need to move the reference sections (ports 1 and 2) by 10 cm inwards:

( )210 10 2.094 120fcm cm radπφ βν

= − ⋅ = ⋅ = − − °

( ) ( )11 22 11 12 21 12exp 2 0.1792 100 , exp 2 0.9838 10.39 S S S j S S S jφ φ′ ′ ′ ′= = − = ∠− ° = = − = ∠− °

L1=10 cm

L2=10 cm

Port 1

Port 2

Frequency dependance

500 600 700 800 900 1000 1100 1200 1300 1400 1500Frequency (MHz)

Return Loss vs. Frequency

-22

-20

-18

-16

-14

-12

-10

DB(|S(1,1)|)CoaxStep

The effects of the discontinuities are generally frequency-dependent

List of some discontinuities and components in the library of MWOffice

Example: microstrip implementation of a double stub matching network

Ideal scheme

Zc, L1

ZS1, LS1 ZS2, LS2

1 2Zc, L0Zc, L0

T junction

Open

Discontinuities inclusion

MLINID=TL3W=2.2 mmL=10 mm

MLINID=TL6W=0.622 mmL=79.68 mm

MOPEN$ID=TL9

1 2

3

MTEE$ID=TL5

MLINID=TL7W=1.33 mmL=32.92 mm

PORTP=1Z=50 Ohm

MOPEN$ID=TL2

1 2

3

MTEE$ID=TL4



PORTP=2Z=50 Ohm

MSUB

Tand=Rho=

T=H=Er=

0 1 .035 mm0.8 mm2.55

Comparison

0.9 0.95 1 1.05 1.1Frequency (GHz)

-50

-40

-30

-20

-10

0

1 2

Z

IMPEDID=Z1R=17 OhmX=66 Ohm

DoppioStub

Microtrip

Ideal

Computing the models of the discontinuities

• Equivalent circuits (lumped and/or distributed elements, numerically evaluated)

• Analytical formulas (simplest cases, low accuracy)

• Electromagnetic analysis (by means of em simulators, very time consuming)

• Different models for the same discontinuity are often available (high accuracy=long computation time)

Microstrip discontinuities: junctions

Step (2-port)Tee (3-port) Cross

(4-port)

S1-S2

S2S1

S3S3S1

S4

S2

MLINID=TL1W=1 mmL=4 mm

1

2

3

4

MCROSS$ID=TL2






1 2

3

MTEE$ID=TL2


MSTEPX$ID=MS1Offset=0 mm



1 2

Bend (2-port)



MBENDAID=TL2W=2 mmANG=90 Deg



MCURVEID=TL2W=2 mmANG=45 DegR=2 mm



MUBEND$ID=TL4S=2 mmM=0.5

S2

S1

S2

S1

S2

S1

Terminations (1-port)


MOPENX$ID=MO1

Open end Via hole


VIA1PID=V1D=1.5 mmH=1 mmT=0.05 mmRHO=1


MRSTUB2WID=TL2W=2 mmRo=7 mmTheta=50 Deg

Radial Stub

S1

S1

S1

Pseudo-lumped componentsSeries inductor

Ls

1 1

1 1

s sL

s s

j L j LY

j L j L

ω ω

ω ω

−

=−

( ) ( )

( ) ( )

1 1tan sin

1 1

sin tan

c s c sS

c s c s

jZ l jZ lY

jZ l jZ l

β β

β β

−

=−

( ) ( ) 0 tan sin

L S s s s s

s c s

c ss

Y Y l l l lL Z l

Z lLv

β β β βω β

≅ ⇒ ≅ ⇒

≅

≅

Zc

ls

Cp

1 1

1 1

p pC

p p

j C j CZ

j C j C

ω ω

ω ω

= ( ) ( )

( ) ( )

1 1tan sin

1 1

sin tan

c s c sS

c s c s

jY l jY lZ

jY l jY l

β β

β β

=

( ) ( ) 0 tan sin

L S s s s s

p c s

c sp

Z Z l l l lC Y l

Y lCv

β β β βω β

≅ ⇒ ≅ ⇒

≅

≅

Zc

ls

Shunt capacitor:

Other componentsInterdigital capacitor


MICAP$ID=MI1W=1 mmS=1 mmG=1 mmL=10 mmN=4WP=1 mm


S2S1

Spiral Inductors

EPSB=1TDB=0TB=0.001 mmRhoB=1



Two-port circuits

• S matrix for reciprocal, lossless networks:

11 22

2 212 11

11 22 21

12

S S

S Sφ φ φ π

=

= −

+ − = ±⋅ =*S S U

Only three real parameters are needed for defining S (for instance |S11|, φ11, φ22)

How to get a transformer at microwaves?

• The transformer is a 2-port widely used in many circuital applications

• The ideal component is reciprocal and lossless.• At microwave frequencies it is however very difficult to

be realized. In those application where the goal is the impedance (admittance) scaling, the component employed is the impedance (admittance) inverter:

n:1

ZL

2LZ inZ n=

Ideal Transformer

K ZL

2

L

ZinKZ =

Ideal Impedance inverter

S parameters of the impedance inverter

( )( )

2 2 20 0 0

11 22 2 2200 0

Z Z=

Z Z

K K ZS SK ZK

− −= =

++

Z0 K Z0

S12 has been obtained by imposing the lossless conditions:

22 2 0

12 11 2 20

11 22 12 12 11

1 2

2 2 2

K ZS SK Z

π πφ φ φ π φ φ

⋅= − = +

+ − = ± ⇒ = ± = ±

012 21 2 2

0

2 K ZS S jK Z

⋅= = ±

+

( )11 22 0 φ φ π= = 12 21 2πφ φ= = ±

Practical implementation of invertersNote: the parameter K of the ideal inverter is independent on frequency. The physical implementation however can only approximate this condition in a limited frequency band

Zc=K

λ0/4

-jK -jK

jK

jJ

-jJ-jJ

J=1/K

φ φ

XZc Zc

φ φ

Yc Yc

jB

( )1

21 2tan , 2 1c c

X KXZ K Z

φ − = − =

− ( )1

21 2tan , 2 1c c

B JBY J Y

φ − = − =

−

Note: for X (B) positive, φ is <0 and K/Zc (J/Yc) is <1.

Equivalent model for a 2-port lossless circuit

S Kφ1 φ2

Zc Zc

11

11

1

1c

SK Z

S−

=+

11 221 2,

2 2S Sπ πφ φ−∠ −∠

= =

11

11

1

1c

SK Z

S+

=−

11 221 2,

2 2S Sφ φ∠ ∠

= − = −

Choice 1 (|K|<Zc):

Choice 2 (|K|>Zc):

12 1 22K S π φ φ∠ = ∠ − + + (must be 0 or π)

Circuiti a Microonde: Introduzione

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