Power Dividers and Directional Couplers - Arraytool · Power Dividers and Couplers Three-Port...

Power Dividers and Couplers Three-Port Networks Four-Port Networks A Few Other Power Dividers Problems

Power Dividers and Directional Couplers

S. R. [email protected]

Department of Electrical & Electronics EngineeringBITS Pilani, Hyderbad Campus

May 7, 2015

Course Outline RF & Microwave Engineering, Dept. of EEE, BITS Hyderabad

mailto:[email protected]


Outline

1 Power Dividers and Couplers

2 Three-Port Networks

3 Four-Port Networks

4 A Few Other Power Dividers

5 Problems



Power Dividers and Combiners

DividerP1

P1 = P2 + P3P2

P3P3=(1-αP1)

P2 = αP1

Combiner

Both the above shown devices are loss-less. Do you know why? Do they have to beloss-less always?

What happens if the above devices are reciprocal?



A Few Types of Dividers or Combiners

4

Z0

2Z0

2Z0

Z0

λ

Z0

Z0 Z0

Z0/ 2

Z0/ 2

Z0

Z0 Z0

12

3

12

3

Now, let’s discuss the differences between the above shown power dividers/couplers ..



3-port Networks

If all ports are matched, then Sii = 0, and if the network is reciprocal, [S] reduces to

[S] =

0 S12 S13S12 0 S23S13 S23 0

.

If the network is also loss-less,

|S12|2 + |S13|2 = 1

|S12|2 + |S23|2 = 1

|S13|2 + |S23|2 = 1

S∗13S23 = 0

S∗23S12 = 0

S∗12S13 = 0.

All the above equations can’t be satisfied simultaneously. So, three-port networks can-not be simultaneously loss-less, reciprocal, and matched at all ports. 4-port networksdo not have such kind of restriction.



Let’s Revisit the Previous Slide ...

4

Z0

2Z0

2Z0

Z0

λ

Z0

Z0 Z0

Z0/ 2

Z0/ 2

Z0

Z0 Z0

12

3

12

3



4-port Networks

The scattering matrix of a reciprocal four-port network matched at all ports has thefollowing form:

[S] =

0 S12 S13 S14

S12 0 S23 S24S13 S23 0 S34S14 S24 S34 0

If the network is loss-less too, then, it can be shown that

S∗13S23 + S∗14S24 = 0

S∗14S13 + S∗24S23 = 0

S∗12S23 + S∗14S34 = 0

S∗14S12 + S∗34S23 = 0

S∗12S24 + S∗13S34 = 0

S∗12S13 + S∗24S34 = 0

|S12|2 + |S13|2 + |S14|2 = 1

|S12|2 + |S23|2 + |S24|2 = 1

|S13|2 + |S23|2 + |S34|2 = 1

|S14|2 + |S24|2 + |S34|2 = 1



4-port Networks ... Cont’d

{S∗24(S∗13S23 + S∗14S24

)= 0

S∗13 (S∗14S13 + S∗24S23) = 0

=⇒ S∗14

(|S13|2 − |S24|2

)= 0{

S12 (S∗12S23 + S∗14S34) = 0S34(S∗14S12 + S∗34S23

)= 0

=⇒ S23

(|S12|2 − |S34|2

)= 0

If S14 = 0 and S23 = 0,

|S12|2 + |S13|2 = 1

|S12|2 + |S24|2 = 1

|S13|2 + |S34|2 = 1

|S24|2 + |S34|2 = 1,

which imply that |S13| = |S24| and |S12| = |S34|.



Directional Couplers

1 2

4 3

Input

Isolated

Through

Coupled

Coupling = C = 10 logP1

P3

Directivity = D = 10 logP3

P4

Isolation = I = 10 logP1

P4

Insertion loss = L = 10 logP1

P2



Outline





5 Problems



T-Junction Power Dividers


The above shown power dividers ... Are they lossy or loss-less?




Loss-less T-Junction Power Dividers ... EquivalentCircuit

Yin

jBV0Z0

Z1

Z2

+

–




Lossy T-Junction Power Divider

Zin

Port 1

Port 3

Port 2

Z0/3

V1Z0

Z0

Z0

P2

P3

P1

+

–V Z+

–

Z0/3

Z0/3

V 2+

–

V3

+

–

Z =4Z0

3

V = V12Z0/3

Z0/3 + 2Z0/3=

23

V1

V2 = V3 = VZ0

Z0 + Z0/3=

34

V =V1

2

So, it is a 6 dB power divider ... What happens to the rest of the power? And what isthe corresponding scattering matrix [S]?



The Wilkinson Power Divider


4

Z0

2Z0

2Z0

Z0

λZ0

Z0

2Z0

Z0

2Z 0

2Z0

λ/4

λ/4




The Wilkinson Power Divider ... Exploiting Symmetry

4

Z02Z0

2Z0

Z0

λPEC/PMC

2Z 0

Z0

2Z0

Z0

2Z 0

2Z0

λ/4

λ/4

2Z 0




The Wilkinson Power Divider ... Even Mode Analysis

2Z 0

Z0

2Z0

Z0

2Z 0

2Z0

λ/4

λ/4

2Z 0

2Z 0

Z0

2Z 0

λ/4 V0

+

V1

+e

V0

+

V0

+




The Wilkinson Power Divider ... Even Mode Analysis

Zein =

(√2Z0

)2

2Z0= Z0

Ve1 = −j

√2V0




The Wilkinson Power Divider ... Odd Mode Analysis

2Z 0

Z0

2Z0

Z0

2Z 0

2Z0

λ/4

λ/4

2Z 0

2Z 0

Z0

Z0

2Z 0

λ/4V0

+

-V0

+





2Z 0

Z0

2Z0

Z0

2Z 0

2Z0

λ/4

λ/4

2Z 0

2Z 0

Z0Z0

2Z 0

λ/4





Zoin = Z0

Vo1 = 0




Combining Even & Odd Mode Analyses ...

Z 0

Z0

2Z0

Z0

2Z 0

2Z0

λ/4

λ/4 V0 +V0

+

V0 -V0

+

-j√2V0

+

S22 = S33 = 0

S12 = S21 = S31 = S13 =−j√

2V0

2V0=−j√

2S23 = S32 = 0




What About S11?

/4

/41

1

1

Port 3

Port 2

2

Zin

2

1

1

1

Port 3

Port 2

2

Zin

2

2

Port 1

Port 1

S11 = 0




What About S11?

1

1

1

Port 3

Port 2

2

Zin

2

1

1

1

Port 3

Port 2

2

Zin

2

2

Port 1

Port 1

λ/4

λ/4

S11 = 0




So, S Paramters of a Wilkinson Power Divider are ...

0.5 f0 f0 1.5 f0–40

–30

–20dB

–10

0

|Sij|

|S12||S23|

|S11|

4

Z0

2Z0

2Z0

Z0

λ

13

2




The Wilkinson Unequal Power Divider ***

1

2

3

Z0Z03

Z02

R2 = Z0K

R3 = Z0/K

R

P3

P2= K2

Z03 = Z0

√1 + K2

K3

Z02 = Z0

√K (1 + K2)

R = Z0

(K +

1K

)




N−way, Equal-split Wilkinson Power Divider ***

Z0 N

Z0 N

Z0 NZ0

Z0

Z0

Z0

Z0Z0

Z0

Z0

Z0

Z0 N

.

.

.

λ/4



Outline





5 Problems



90◦ Hybrid Coupler

90◦ (Quadrature) Hybrid Coupler

Z0

Z0 Z0

Z0/ 2

Z0/ 2

Z0

Z0 Z0

1

4

(Input)

(Isolated)

(Output)

(Output)

2

3

4λ

4λ

[S] =−1√

2

0 j 1 0j 0 0 11 0 0 j0 1 j 0

................




Quadrature Hybrid Coupler - Normalized Form

4 3

1 2A1 = 1

B3B4

B1 B2

1

1 1

1

11 1

1

1

1/ 2

1/ 2




Even Mode Analysis

1 2

4 3

+1/2

+1/21 1

1

1 1 1

11

1/ 2

1/ 2

Line of symmetryI = 0V = max

+1/2

+1/2

Te

1 1

1

1

1

1

1

1

1/ 2

1/ 2

Open-circuited stubs(2 separate 2-ports)

Γe

[A BC D

]e=

[1 0j 1

] [0 j/

√2

j/√

2 0

] [1 0j 1

]=

1√2

[−1 j

j −1

]




Even Mode Analysis ... Cont’d

Γe =A + B− C−DA + B + C + D

=−1 + j− j + 1−1 + j + j− 1

×√

2√2= 0

Te =2

A + B + C + D=

2−1 + j + j− 1

×√

2 =−1√

2(1 + j)




Odd Mode Analysis

1 2

4 3

–1/2

+1/21 1

1

1 1 1

11

1/ 2

1/ 2

Line of antisymmetryV = 0I = max

–1/2

+1/2

To

1 1

1

1

1

1

1

1

11

1/ 2

1/ 2

Short-circuited stubs(2 separate 2-ports)

Γo

[A BC D

]o=

[1 0−j 1

] [0 j/

√2

j/√

2 0

] [1 0−j 1

]=

1√2

[1 jj 1

]




Odd Mode Analysis ... Cont’d

Γo =A + B− C−DA + B + C + D

= 0

To =2

A + B + C + D=

1√2(1− j)




Now, you calculate Sij values from the obtained Γe, Te, Γo, and Tovalues ...




S Paramters of a Quadrature Hybrid

0.5 f0 f0 1.5 f0–40

–30

–20

–10

0

4

Z0

Z0 Z0

Z0/ 2

Z0/ 2

Z0

Z0 Z0

1

4

(Input)

(Isolated)

(Output)

(Output)

2

3

λ

4λ |Sij|

|S12|

|S13|

|S11|

|S14|



180◦ Rat-race or Ring Hybrid Coupler

Rat-race Coupler

2

4

3

1

2Z0

Z0

Z0

Z0

Z0(Σ)

(Δ)

λ/4

λ/4

λ/4

3λ/4

[S] =−1√

2

0 j 1 0j 0 0 11 0 0 j0 1 j 0

................




Rat-race Coupler

2

4

3

1

2Z0

Z0

Z0

Z0

Z0(Σ)

(Δ)

λ/4

λ/4

λ/4

3λ/4 [S] =−1√

2

0 j 1 0j 0 0 11 0 0 j0 1 j 0

................




Even Mode Analysis

4

1 22

1

3

+1/2

+1/21

1

O.C.

O.C.

O.C.

Te

√2√2

+1/2

Γe

λ/4

3λ/8√2λ/8

[A BC D

]e=

[1 0

j/√

2 1

] [0 j

j/2 0

] [1 0

−j/√

2 1

]=

[1/√

2 jj −1/

√2

]




Even Mode Analysis ... Cont’d

Γe =A + B− C−DA + B + C + D

=−j√

2

Te =2

A + B + C + D=−j√

2




Odd Mode Analysis

4

1 22

1

3

+1/2

–1/21

1

S.C.

S.C.

S.C.

To

√2√2

+1/2

Γe

λ/4

3λ/8√2λ/8

[A BC D

]e=

[1 0

−j/√

2 1

] [0 j

j/2 0

] [1 0

j/√

2 1

]=

[−1/√

2 jj 1/

√2

]




Odd Mode Analysis ... Cont’d

Γo =A + B− C−DA + B + C + D

=j√2

To =2

A + B + C + D=−j√

2




Now, you calculate Sij values from the obtained Γe, Te, Γo, and Tovalues ...




S Paramters of a Rat-race Coupler

0.5 f0 f0 1.5 f0–40

–30

–20

–10

0

dB

2

4

3

1

2Z0

Z0

Z0

Z0

Z0

|Sij|

|S12|

|S13|

|S11|

(Σ)

(Δ)

λ/4

λ/4

λ/4

3λ/4



Outline





5 Problems



Bethe Hole Directional coupler

a3

4

2

(Isolated)

(Through)

1(Input)

3(Coupled)

2(Through)

4 (Isolated)

1S b

b

y

x

z

(Coupled)

(Input)



Multi-Hole Couplers

4 3

1 2

g /4

0° –90°

(Isolated)

(Out of phase) (In phase)

(Input)

(Coupled)

(Through)

d

B0A

B F

A A

F0A

n = 0

B1A F1A

n = 1

B2A F2A

n = 2

B3A F3A

. . .

BNA FN A

n = N



Coupled Line Couplers

1

3 4

2

Coupled Isolated

ThroughInput

Z0e, Z0o

I3Z0 Z0

+V3 +V4

I4

I1Z0 Z0

+V1 +V2

I2

2V0

1 2

43



Multisection Coupled Line Couplers

V3 V4

V1 V2

C1 C2 CN – 1 CN

Coupled Isolated

Input Through

. . .



Lange Couplers

4

1

W

Isolated

Input

2

3

Coupled

Through

Z0

S

Z0 Z0

Z0

/4

2

Through

Input

4

1 3

Coupled

Isolated



Tapered Coupled Line Hybrid

Differenceinput

Output

Suminput

Output



Magic T

4

2

1

3

(Σ)

(Δ)



Outline





5 Problems


Power Dividers and Directional Couplers - Arraytool · Power Dividers and Couplers Three-Port...

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