Wilkinson Power Divider (1)

Chapter 7: The Wilkinson Power Divider 1) 2) 3) 4) 5) 6) 7) 8)

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Directional Coupler

Directional Coupler

 Z in ( β l ) =  Z oo o

 Z o +  jZ oo tan( β  l )  Z oo +  jZ o tan( β  l )

Odd mode input impedance.

 Z  ( β l ) =  Z oe e in

Circuit Diagram for Even and Odd Mode Analysis. Analysis.

 Z o +  jZ oe tan( β  l )  Z oe +  jZ o tan( β  l )

Even mode input impedance. 2

Directional Coupler Coupling Factor: C  =

 Z oe − Z oo  Z oe + Z oo

Voltage at port 3: V 3 = V 

 jC  tan ( β  l ) 1 − C 

2

+  j tan ( β  l ) 1 − C 

2

Voltage at port 2:

V 2 = V 

Voltage at port 4: V 4 = 0

1 − C  cos( β  l ) +  j sin ( β  l ) 2

 Z oo =  Z o  Z oe =  Z o

1 − C  1 + C  1 + C  1 − C  3

T Junction

Matching requirement for the T junction input:

1  Z 2

+

1  Z 3

=

1  Z 1

1) The Lossless T- Junction suffers from the problem of  not being matched at all ports. 2) It does not have any isolation between output ports.

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T Junction The resistive divider can be matched at all ports. Even though it is not lossless, isolation is still not achieved.

  Z o     Z o    Z in = +  + Z o   + Z o  3   3     3    Z o

⇒

 Z in =

 Z o 3

+

2 Z o 3

= Z o

Since the network is symmetric from all three ports, the output ports are also matched. Therefore, S 11=S22=S33=0

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Three Port Network  • A lossless three port network can be matched at all three ports, but

any matched lossless network must be nonreciprocal. • Alternatively, a lossless and reciprocal three port network can be physically matched at only two ports. • A Lossy three port network can be matched at all ports with isolation between ports.

 S11 [S ] =  S 21  S 31

S12 S 22 S32

S13 

 S 23  S33 

[S] matrix of a three port network.

0 [S ] =  S12  S13

S12 0

S 23

S13 

  S33 

S 23

[S] matrix of a lossless and nonreciprocal three port network.

0 [S ] =  S 21  S 31

S12

S13 

0

S 23

S32

0

  

[S] matrix of a lossless and nonreciprocal three port network. 6

The Wilkinson Power Divider/ Combiner

Four-way divider using 3 Wilkinson dividers in Microstrip form.

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The Wilkinson Power Divider/ Combiner • The Wilkinson power divider can be matched at all ports with isolation between the output ports. • The Wilkinson can be made using Microstrip or Stripline. • The Wilkinson power divider can be made to give arbitrary power division. • We will study the equal split (3 dB) case first.

λ /4 2 Z o Port 1

Zo

Port 2

2Zo

Zo 2 Z o

Zo

Port 3

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The Wilkinson Power Divider/ Combiner λ /4

Vg2

V2

2 1

1

V1 2

1 1

2

V3

Vg3

Normalized Wilkinson Power Divider/ Combiner

• We will analyze this circuit by reducing it to two simpler circuits driven by Symmetric (Even) and Anti-Symmetric (Odd) sources at the output ports. 9

The Wilkinson Power Divider/ Combiner λ /4 2 V1

Vg2

V2 1

2

1

1 1

2

V3

Vg3

• If voltages Vg2 and Vg3 are equal and are in phase, we have even excitation, and magnetic wall (open circuit) can be placed between the two arms • Alternatively, when Vg2= - Vg3 = 2 V we have odd excitation, electric wall (short circuit) can be placed in the plane of the symmetry 10

The Wilkinson Power Divider/ Combiner 2

λ /4 V2 V1

2

1 1

2

V3

1

Vg2 1

Vg3

Plan: • Consider odd and even excitations • Compute voltages at different ports • Compute reflection coefficient at ports • Create S-matrix We will consider odd and even excitation of ports 2 & 3. We know already that any other excitation can be built from that. Then we will excite port 1 - this is even with respect to ports 2 and 3.

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Even Mode Analysis For the even mode analysis, Vg2 = Vg3 = 2V and no current flows through the isolation resistor or through the connection at port 1. Hence we can treat the axis of symmetry as an open circuit and draw the circuit as shown below.

λ /4

V2

2 1

1

Vg2

V1

We can now examine the voltages at ports 1 and 2, (V 1 and V2) and determine the optimum value for Z. 12

Even Mode Analysis λ /4 V 2

2 1

2 Open Circuit  = 2

2

V1=? 2

=

Vg2

V1 2

Quarter wave  Z in transformer

1

 Z o

 Z  L

2

1

λ /4

V2 1

V3 2

V2=V Match!

Vg3

1

1 Vg2=2V 13

Even Mode Analysis V1

λ /4

V2

1 Vg3

2

2 x 0 +

V ( x) = V  (e

−  j β  x

+ Γe

 jβ x

Γ= )

+

Γ +1 V1 = j V  Γ −1

−

2

2

+

2

x=-λ /4

x=0 V 1 = V  (1 + Γ)

2

+

V 2 = V  (  j − Γ j ) = 1V e

−  j β 

λ  4

=e

−  j

π  2

= −  j 14

Even Mode Analysis

We had:

V 1 =  jV

Γ=

2−

2

2+

2

2−

2

2+

2

− 2+ 2

2 2

+1 =  jV −1

and

2− 2−

V 1 = j

2 +2+

2

2 − (2 +

2)

Γ +1 Γ −1

= −  j

4 2 2

V = − j 2V

V 1 = − j 2V

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Odd Mode Analysis 2

λ /4 V2 V1

2

1 1

2

V3

1

Vg2 1

Vg3

For the odd mode analysis Vg2 = -Vg3 = 2 V. So there is a voltage null along the axis of symmetry of the circuit. We can assume a virtual ground at that point and redraw the circuit. Again we are interested in the behavior at ports 1 and 2 and in determining the optimum value for Z. 16

Odd Mode Analysis V2

λ /4 2

2

1

1

Vg2

V1 2

1

2

V1=0

1

λ /4

1

V3

2 Short  Circuit  = Short  Circuit 

1

 Z o2  Z o2 Quarter wave = = Open Circuit   Z in = transformer  Z  L 0 Open Circuit  1 = 1

Match! V2=V

Vg3 Vg2

1

1 1

Vg2=2V 17

Input Impedance at Port 1 The final question is what is the input impedance at port 1, when ports 2 and 3 are terminated in their matched loads?

λ /4 1

V2 1

open

1

V1 1

2

Z in

1

V3

Excitation in port 1, when 2 & 3 are matched.

1

V2

2

 Z in =

1

Match!

2

2

2

1

1

=1

Therefore, the input to the divider is matched as long as the output ports are matched.

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Wilkinson Divider S Parameters

λ /4 V2

1 1

1

1

2

Matched, S11=0

1

V3 even

λ /4 V2

2

odd

2

1

λ /4 2

Vg2=2V

V2 1

Port 2 matched for even excitation

Port 2 matched for odd excitation

1 Port 2 matched for any excitation. By symmetry, port 3 is also matched S22 and S33=0

• Ports 2 and 3 are separated either by electric or magnetic wall. No power

goes between (ISOLATION). S23=S32=0 • Circuit is reciprocal -- matrix symmetric

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Wilkinson Divider S Parameters λ /4

V2

2

1

1

V1 2

1

Vg2 1

2

V3 −

S12 =

V 1

V 2+

V 1+ =V 3+ = 0

We know now that ports are matched. Thus:

S12 =

V 1 V 2

V 1 + V 1 e

=

o

e o V 2 + V 2

=

−  j 1V

2V

+0

+ 1V

= −j

1 2

  0  −  j S =  2  −  j   2

−  j 2 0 0

−  j   2  0    0   20

Wilkinson Power Divider Example Design an equal split Wilkinson power divider for a 50 Ω system impedance at frequency f o.

λ /4 2 Z o Port 1

Zo

Port 2

2Zo

Zo 2 Z o

Zo The quarter wave transmission lines in the divider should have a characteristic impedance of:  Z  = The shunt resistor has a value of:  R = 2 Z o = 100

Port 3

2  Z o = 70.7

Ω

Ω 21

Unequal Power Division It is possible to design the Wilkinson for an unequal dividing ratio. This can be done by different choices of impedances and isolation resistor as shown below. Note: Lines 2 and 3 are still λ / 4 long and that ports 2 and 3 are no longer matched to Z o.

For a given power ratio K  between ports 2 and 3: K 2 =

P3 P2

We have the following formulas to design the width of the arms. 1 + K 

2

 Z o 3 =  Z o

3

K 

 Z o 2 = K   Z o3 =  Z o 2

K 

(1 + K  ) 2

   

 R =  Z o  K  +

Note: These result will reduce to the equal-split case for K=1.

1  

 K   22

The Wilkinson Divider Summary • The Wilkinson Divider can also be generalized as an N-Way Divider or Combiner. • This circuit can be matched at all ports. • The isolation between ports can be achieved. • The Wilkinson divider can also be made with stepped multiple sections for increased bandwidth.

Photograph of a four-way power divider network using three microstrip Wilkinson power divider.

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