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ELEctromagnetic DIAgnostics Lab.DIT Università di TrentoDIT - Università di Trento

Via Sommarive 14, I-38050 Trento ItaliaE-mail: [email protected] mail: [email protected]

MicrowaveMicrowaveMicrowave DevicesDevicesDevicesPowerPowerPower splitterssplitterssplitters///combinerscombinerscombiners

Master Master DegreeDegree Electronic and TelecommunicationElectronic and TelecommunicationA.A. A.A. 20122012--20132013

Power Splitters/Combiners

• Lossless, T-junction (waveguide/microstrip).

• Lossy Three resistors junction (microstrip).y j ( p)

• Wilkinson power splitter (microstrip)• Wilkinson power splitter (microstrip).

• Lumped element splitters (microstrip, also for low frequencies)

Lossless Junction – The T-Junction power splitter

The T-junction power divider is a simple Three port networkthat can be used for power division or power combiningthat can be used for power division or power combining.

You can develop a power splitter combiner with any kind ofYou can develop a power splitter combiner with any kind oftransmission line ( e. g. microstrip, waveguide).

Lossless dividers cannot be matched simultaneously at allportsports.

T junctionT-junction Waveguide


Z0

Z1

jBV0

+

ZYi

-

Z2Yin

The T-junction power divider as all dividers, can be modeled as a junctionof three transmission lines as shown above.In general at the junction there are fringing fields associated with thediscontinuity that can be accounted for by a lumped susceptance B.In practical applications (especially with waveguide) when B is notIn practical applications, (especially with waveguide) when B is notnegligible some types of reactive elements are added at the dividers tokeep B=0.


Z0

Z1

jBV0

+

ZYi

-

Z2Yin

In order to keep the characteristic impedance of T-junction ports thefollowing relation should be considered:

111ZZZ

jBYin =++=021 ZZZ

jin


If lossless lines are considered and if we also assume the term B=0 theprevious relation reduces to:p

111=+

021 ZZZ+

The above relation tell us that the microwave device cannot be matchedon all ports.The output line impedances Z1 and Z2 can be chosen to provide variouse output e peda ces 1 a d 2 ca be c ose to p o de a ouspower division ratios.To obtain the matching on all device ports quarter-waves transformerscan be used to bring the output line impedances back to the desiredcan be used to bring the output line impedances back to the desiredlevels.It is worth noticed that in such kind of device the output ports are notisolated!

Example 1: 3dB T-Junction power splitter designDesign a T-Junction 3dB power splitter. A device able to equally split thepower on the two ports. Design the splitter with microstrip technology.Th d i t b t d ith t d d 50 Oh d i t thThe device must be connected with standard 50 Ohm devices at theinput and output ports.

If we consider a 50 Ohm input line a 3 dB (equal split) power divider canbe made by using two output line of 100 Ohm

Z0=50Ohm line

Z1=100Ω line

Z0=50Ω line

0

Lambda/4 matchingtransformers

Input port

Z0=50Ohm line

Z2=100Ω line

Example 2: T-Junction power splitter designDesign a T-Junction power splitter. A device able to split the power on thetwo ports with a ratio 2:1. Moreover compute the refection coefficients

l ki i t th t t tseen looking into the output ports.

V 11 20

201 V

inPZV

P31

21

1

01 == 01 3ZZ =

0

0

21

ZV

Pin =

inPZV

P32

21 2

02 == 02 2

3 ZZ =inZ 32 22 02 2

Example 2: T-Junction power splitter designLayout of the T-Junction power splitter with a ratio 2:1 no the output ports.

Z1=75Ω line

Output port 1

Z0=50Ω line0Unmatched

Ports

Z2=150Ω lineInput port

Output port 2

Here we see an impedance of

Output port 2

Z1//Z2=50Ω line

Example 2: T-Junction power splitter designScattering matrix of the T-Junction power splitter with a ratio 2:1 no theoutput ports.

32

31

131211 0SSS

232231

33

232221

131211 0SSSSS

SSSS ==

333232

333231 SSSSS

In a good device these

In a good matched device thediagonal should be

zero In a good device theseTerms should be

zero

zero

Example 2: T-Junction power splitter designFrom the scattering matrix of the T-Junction power splitter with a ratio 2:1we observe two main problems:1) The output ports are coupled together so if you insert a signal in port 2

you can measure an amount of power at port 3.2) M th di l i t th S d th S t2) Moreover the diagonal is not zero the S22 and the S33 parameters arenot equal to zero, this means no matching conditions.

210SSS

232231

32

31

232221

131211 0SSSSS

SSSS ==

333232

333231 SSSSS

Example 2: T-Junction power splitter designEstimation of the reflection coefficients at the three ports. Layout of the T-

Junction power splitter with a ratio 2:1 no the output ports.

Z1=75Ω lineHere we see an impedance of

Z2//Z0=37.5Ω line

Z0=50Ω line

2 0

0

Input port

Here we see an impedance ofZ //Z 30Ω li

Z2=150Ω lineZ1//Z0=30Ω line

Example 2: T-Junction power splitter designEstimation of the reflection coefficients at the three ports. Layout of theT-Junction power splitter with a ratio 2:1 no the output ports.

05050)//( 0210 =

−=

−=Γ

ZZZ1005050)//( 021

0 ++ ZZZ

66.01503015030

)//()//( 120

1 −=+−

=−−

=ΓZZZZZZ

15030)//( 120 +ZZZ

75537)//( ZZZ33.0

755.37755.37

)//()//(

210

2102 −=

+−

=−−

=ΓZZZZZZ

Lossy Junction – The Three resistors power splitterIf a three ports divider contains lossy components it can be made to bematched at all ports, although the two output ports may not be isolated.

The circuit of this device is shown below

R RPort 2

R1R2

R1 R3

Port 1 R3

Port 3

Th i ti litt b l d id i th i it thThe resistive splitter can be analyzed considering the circuit theory.

Lossy Junction – The Three resistors power splitterIf we want to design a 3bB divider this means that at the two output portsthw power must be equally splitted.Let us consider the different arms of the device considering an

R

Let us consider the different arms of the device considering animpedance of Z0 on the ports. R2

R1Port 2

Port 1

Z0

Zt

R3Z0 Zt

Port 3Z0

Z0 4 ZZZ

Z

If h R R R Z /3 b i t i d t th

Zt00

0

34

3ZZZt =++=

If we choose R1=R2=R3=Z0/3 we can observe an input impedance at thecenter of the junction equal to Zt

Lossy Junction – The Three resistors power splitter

Z0/3 4/3Z0

ZiZin4/3Z0

Port 1 result matched and since the device is completely symmetricalalso the other ports result matched.

00012 ZZZZin =+= 000 33in

Lossy Junction – The Three resistors power splitter

If the voltage at Port 1 is V1, then the voltage Vt at the center of thejunction is given by:

32 20

VVVZ

== 1100 3

32

3

VVZZ

Vt =+

=

The voltages at Port 1 and two is half of the voltage V1 at Port 1,remember this is a 3dB power splitter

10

32 21

43 VVV

ZZ

VV tt ====

remember this is a 3dB power splitter.

10

0

32 243

ZZ

tt

+

Lossy Junction – The Three resistors power splitterThe scattering matrix of the three resistors 3dB power splitter is reportedin the following:1) The output ports are still coupled together so if you insert a signal inport 2 you can measure an amount of power at port 3.2) Now since this is a lossy device the diagonal of the scattering matrix is2) Now since this is a lossy device the diagonal of the scattering matrix iszero, this means that all the ports are matched.

00

11

21

21

131211

== SSSSSS

S0

0

21

21

21

21

333231

232221 ==SSSSSSS

Since S21= S31= S23=1/2 at Port 1, this means that we have -6dB belowthe input power level at port 1 since half of the input power is lostthe input power level at port 1, since half of the input power is lost(dissipated) in the resistors.

Lossy Junction – The Three resistors power splitterConcerning the power at the ports we can consider the following relations:

0

21

1 21

ZV

PPin == This is the power at Port 10

( )2( )inP

ZV

ZVPP

41

81

21

0

21

0

212

1

32 ==== This is the power at Port 2and Port 38 00

Th l t ti h th t h lf f th l i di i t d i thThe last equation shows that half of the supply power is dissipated in theresistors.

Lossy Junction – The Wilkinson power splitterT-junction splitter cannot be matched at all ports moreover the outputports are not isolated.

The three resistors splitters is a lossy device you can match all ports butthe two output ports are still coupledthe two output ports are still coupled.

We can use a lossy device with the output ports uncoupled this device isthe Wilkinson device, from the name of his inventor.

The Wilkinson device can be made to gives an arbitrary power divisionThe Wilkinson device can be made to gives an arbitrary power divisionbut at first we analyze the classical 3dB case.

The Wilkinson device can be realized with waveguide or microstrip, themost used is the microstrip splitter composed with two segments oflambda/4 transmission lines and a resistorlambda/4 transmission lines and a resistor.

Lossy Junction – The Wilkinson power splitterThe structure of the device is reported below:

02Z 0Z

0Z

02Z0Z

To analyze this device we must normalize4/λ

0

To analyze this device we must normalize considering Z0 the impedance of the

transmission lines and considering the different simmetry of the devicedifferent simmetry of the device

This device could be analyzed considering an anlogy with electronic circuits

Even/Odd analysis of the Wilkinson power splitterThis device could be analyzed considering an anlogy with electronic circuits

If we want to know the response of the devices at Port 2, we could insert a generator Vg2=4V withp g g2Port 3 (Vg3) closed on a short. We used a similar techniques for the estimation of the [Y] matrix.

Or we can use more efficient analysis is considering Vg2=Vg3=2V and then Vg2=-Vg3=2V: thesum of the two different effects produce the deired result.

The technique that considers these two different excitations (symmetric and anti-symmetric) is more simple if the device is symmetrical. The device simmetry permits toconsider the two output port as isolated and to study the device as a simple two portsp p y p pdevice.

This kind of analisys is called even/odd analisys and we will use it for other

If the two generators Vg2 and Vg3 presents the same voltages V2 = V3 The current

Even/Odd analysis of the Wilkinson power splitter

If the two generators Vg2, and Vg3 presents the same voltages V2 = V3. The currentcan’t flow through the resistor. The resistor could be removed, and also the junction atthe input node could be seen like an open circuit

2V

2V2V2V2V

Along the horizontal axis the current can’t flow because the two generators impose the same potential: there is a so called magnetic wall


2V

2V2V2V

The upper/lower segment of the device are simple quarter wave lenght linesclosed on a resistor (R=2). The impedance at Port 2 is given by:

2Z2

ZZ ein =

This lead to a matching condition on port 2, and thanks to the simmetry also

00 1122 == ee SSon port 3.


2V

2V2V2V

The upper/lower segment of the device are simple quarter wave lenght linesclosed on a resistor (R=2). The impedance at Port 2 is given by:

)()( zjzjzjzj eeVeVeVxV ββββ Γ++ −+−−+ )()( jjjj eeVeVeVxV ββββ Γ+=+=Now we need V1 and V2 and we know that the impedance at node 2 is 1, sowe obtain a resistive divider and V2 is given by:

VjVVV e =Γ+=−= + )1()( 42λ

we obtain a resistive divider and V2 is given by:

11)1()0(1 −Γ

+Γ=Γ+== + jVVVV e


2V

2V2V2V

22− 1VS e22222

1 jVV e −=→+

=Γ2

12

21 jjVVS e =

−=

Even/Odd analysis of the Wilkinson power splitterIf the two generators Vg2, and Vg3 presents the same voltages but V2 = -V3. we are ing g g gthe odd mode. There is a voltage null in the middle of the circuit. Thus we can bisect thecircuit by grounding it at two points on its midplane and obtain:

2V

-2VThis is an electric wall


L ki i t P t 2 i d f /2 i th ll lLooking into Port 2, we seen an impedance of r/2, since the parallelconnected transmission line is quarter wave long shorted at port 1 so itlooks like an open circuit at port 2

Thus port 2 will be matched only if we choose r equal to at this point weobtain that the return loss for the odd mode is given by:g y

00 1122 == oo SS

Concerning the insertion loss parameters we have short all power isdelivered toward the resistor place in the middle so the contributedelivered toward the resistor place in the middle so the contribute

00 1312 == oo SS 1312

Fi ll i t fi d th i t i d t t 1 f th Wilki di id


Finally, we muist find the input impedance at port 1 of the Wilkinson dividerwhen ports 2 and 3 are terminated with matched load. It is similar to theprevious even mode we can remove the central resistor

The input impedance is given by this simple formula since it is the parallelbetween two quarter wave transformers:

( ) 1221 2

==inZ

Even/Odd analysis of the Wilkinson power splitterSummarizing the scattering matrix of the 3dB Wilkinson power splitter isSummarizing the scattering matrix of the 3dB Wilkinson power splitter isreported in the following:

000

2

22

232221

131211j

jj

SSSSSS

S −

−−

==0000

2

2

333231

232221jSSS

SSSS−

Note that when the divider is driven at port 1 and the output ports aret h d Th d i i l l l fl t d f t 2 3 imatched. The device is lossless, only reflected power from ports 2 or 3 is

dissipated in the resistor. Moreover the two output ports are isolated.

Wilkinson power splitter with lumped elementsThe wilkinson is realized with two quarter wave transformers but if youThe wilkinson is realized with two quarter wave transformers but if youneed compactness you can simply remove the two quarterwave linesand simulate they with a Pi-Greca network realized with two capacitorsand one inductor:and one inductor:

1C 0ZL002 Zf

Ceq π=

0

0

2 fLeq π

=

This technique could be used also for low frequencies and high power.

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