7 3 The Wilkinson Power Divider - University Of Kansas

4/14/20097 3 The Wilkinson Power Divider.doc1/17.3 – The Wilkinson Power DividerReading Assignment: pp. 318-323The Wilkinson power divider is the most popular power dividerdesigns.It is very similar to a lossless 3dB divider, but has oneadditional component!HO: THE WILKINSON POWER DIVIDERQ: I don’t see how the Wilkinson power divider designprovides the scattering matrix you claim. Is there any way toanalyze this structure to verify its performance?A: Yes! We simply need to apply an odd/even mode analysis.HO: WILKINSON DIVIDER EVEN/ODD MODE ANALYSISJim StilesThe Univ. of KansasDept. of EECS

4/14/2009The Wilkinson Power Divider 723.doc1/3The WilkinsonPower DividerThe Wilkinson power divider is a 3-port device with ascattering matrix of: 0 S j 2 j 2 j002 jb2 0 0 j2a1 jb322a2 j jb12a32Note this device is matched at port 1 (S11 0 ), and we find thatmagnitude of column 1 is:222S11 S21 S31 1Thus, just like the lossless divider, the incident power on port 1is evenly and efficiently divided between the outputs of port 2and port 3:P2 2 S21 P1 P1 2P3 2 S31 P1 P1 2But now look closer at the scattering matrix. We also note thatthe ports 2 and 3 of this device are matched !S22 S33 0Likewise, we note that ports 2 and ports 3 are isolated:Jim StilesThe Univ. of KansasDept. of EECS

4/14/2009The Wilkinson Power Divider 723.doc2/3S23 S32 0Æ It’s the (nearly) ideal 3dB power divider!!!Q: So just how do we make this Wilkinson power divider?It looks a lot like a lossless 3dB divider, only with an additionalresistor of value 2Z 0 between ports 2 and 3:This resistor is the secret to the Wilkinson power divider, andis the reason that it is matched at ports 2 and 3, and the reasonthat ports 2 and 3 are isolated.Note however, that the quarter-wave transmission line sectionsmake the Wilkinson power divider a narrow-band device.Jim StilesThe Univ. of KansasDept. of EECS

4/14/2009The Wilkinson Power Divider 723.doc3/3Figure 7.12 (p. 322)Frequency response of an equal-split Wilkinson power divider. Port 1 is the inputport; ports 2 and 3 are the output ports.www.seefunkschule.at/privat/studienzeit/3dB0 g010.htmJim StilesThe Univ. of KansasDept. of EECS

4/14/2009Wilkinson Divider Even and Odd Mode Analysis.doc1/14Even/Odd Mode Analysisof the Wilkinson DividerConsider a matched Wilkinson power divider, with a source atport 2:Port 2 Z0λ Vs-42 Z0Port 12 Z0Z02 Z0λPort 3Z04Too simplify this schematic, we remove the ground plane, whichincludes the bottom conductor of the transmission lines:Port 2λZ042 Z0Port 1Z0 Vs-2 Z02 Z0λ4Port 3Z0Jim StilesThe Univ. of KansasDept. of EECS

4/14/2009Wilkinson Divider Even and Odd Mode Analysis.doc2/14Q: How do we analyze this circuit ?A: Use Even-Odd mode analysis!Remember, even-odd mode analysis uses two importantprinciples:a) superpositionb) circuit symmetryTo see how we apply these principles, let’s first rewrite thecircuit with four voltage sources:VsVsV2λZ02 Z02 Z0λ2Z042 Z0V12 Vs 2V34Vs2Z0Turning off one positive source at each port, we are left withVso2an odd mode circuit:V2λV1 o2 Z0Z0Jim Stiles2 Z02 Z0λOdd Mode CircuitZ044 Vs 2V3oZ0The Univ. of KansasDept. of EECS

4/14/2009Wilkinson Divider Even and Odd Mode Analysis.doc3/14Note the circuit has odd symmetry, and thus the plane ofsymmetry becomes a virtual short, and in this case, a virtualVsground!o2V2λo2Z0 V1Z04Z02 Z0V 02Z0 V1oZ02 Z0λ Vs 2V3o4Z0Dividing the circuit into two half-circuits, we get:λZ04 2Z 0V1 o 2 Z0Z0 2Z04 V1 oVs λ2Z 0 -V2o 2 Z0 Z0V3o - Vs 2 Note we have again drawn the bottom conductor of thetransmission line (a ground plane) to enhance clarity (I hope!).Jim StilesThe Univ. of KansasDept. of EECS

4/14/2009Wilkinson Divider Even and Odd Mode Analysis.doc4/14Analyzing the top circuit, we find that the transmission line isterminated in a short circuit in parallel with a resistor of value2Z0. Thus, the transmission line is terminated in a short circuit!λZ04 V1 o 0Z02 Z0V2o -Vs2 This of course makes determining V1 o trivial (hint: V1 o 0 ).Now, since the transmission line is a quarter wavelength, thisshort circuit at the end of the transmission line transforms toan open circuit at the beginning!Z0 Z0V2o -Vs2As a result, determining voltage V2ois nearly as trivial as determiningvoltage V1 o . Hint: V2o VsZ0V s2 Z0 Z0 4And from the odd symmetry of the circuit, we likewise know:VV3o V2o s4Now, let’s turn off the odd mode sources, and turn back on theeven mode sources.Jim StilesThe Univ. of KansasDept. of EECS

4/14/2009Wilkinson Divider Even and Odd Mode Analysis.doc5/14VsV2eEven Mode CircuitλV1 eZ042 Z0Z02 Z02 Z0λ2VsV3e42Z0Note the circuit has even symmetry, and thus the plane ofsymmetry becomes a virtual open.VsV2eλ2Z0 V1e2Z042 Z0Z0I 02Z0 V1e2 Z0λZ0VsV3e42Z0Dividing the circuit into two half-circuits, we get:λ 2Z 0V1 e2 Z0 Jim StilesZ04Z0 V2e -Vs2 The Univ. of KansasDept. of EECS

4/14/2009Wilkinson Divider Even and Odd Mode Analysis.docλZ04 2Z 0V1 e6/14Z02 Z0 -V3eVs2 Note we have again drawn the bottom conductor of thetransmission line (a ground plane).Analyzing the top circuit, we find that the transmission line isterminated in a open circuit in parallel with a resistor of value2Z0. Thus, the transmission line is terminated in a resistorvalued 2Z0.λZ04 2Z 0V1 eV2e2 Z0 -Vs2 Now, since the transmission line is a quarter wavelength, the2Z0 resistor at the end of the transmission line transforms tothis value at the beginning:Zin Z0 eZ0V2 -Vs22Z 0)2 Z0Voltage V2 e can again be determined byvoltage division:V2e Jim Stiles(2Z 0The Univ. of KansasVsZ0V s2 Z0 Z0 4Dept. of EECS

4/14/2009Wilkinson Divider Even and Odd Mode Analysis.doc7/14And then due to the even symmetry of the circuit, we know:V3e V2e Vs4Q: What about voltage V1 e ? What is its value?A: Well, there’s no direct or easy way to find this value. Wemust apply our transmission line theory (i.e., the solution to thetelegrapher’s equations boundary conditions) to find thisvalue. This means applying the knowledge and skills acquiredduring our scholarly examination of Chapter 2!λ 2Z 0V1 eZ042 Z0V2e -Vs2 If we carefully and patiently analyze the above transmissionline circuit, we find that (see if you can verify this!):V1 e jVs2 2And thus, completing our superposition analysis, the voltagesand currents within the circuit is simply found from the sum ofthe solutions of each mode:Jim StilesThe Univ. of KansasDept. of EECS

4/14/2009Wilkinson Divider Even and Odd Mode Analysis.docV1 V1 o V1 o 0 V2 V2o V2o jVs2 2Vs4 Vs4 8/14jVs2 2Vs2V VV3 V3o V3o s s 04λV1 jVsZ0V2 Vs 2Z042 Z02 24 Vs-2 Z02 Z0λ4V3 0Z0Note that the voltages we calculated are total voltages—thesum of the incident and exiting waves at each port:V1 V1 ( z1 z1P ) V1 ( z1 z1P ) V1 ( z1 z1P )V2 V2 ( z2 z2P ) V2 ( z2 z2P ) V2 ( z2 z2P )V3 V3 ( z3 z3P ) V3 ( z3 z3P ) V3 ( z3 z3P )Since ports 1 and 3 are terminated in matched loads, we knowthat the incident wave on those ports are zero. As a result, thetotal voltage is equal to the value of the exiting waves at thoseports:Jim StilesThe Univ. of KansasDept. of EECS

4/14/2009Wilkinson Divider Even and Odd Mode Analysis.doc9/14 jVsV1 ( z1 z1P ) 0V1 ( z1 z1P ) V3 ( z 3 z 3P ) 0V3 ( z3 z 3P ) 02 2The problem now is to determine the values of the incident andexiting waves at port 2 (i.e., V2 ( z 2 z 2P ) and V2 ( z 2 z 2P ) ).Recall however, the specific case where the source impedanceis matched to transmission line characteristic impedance (i.e.,Z s Z 0 ). We found for this specific case, the incident wave“launched” by the source always has the value Vs 2 at thesource:Z0 Vs -V Z0( z z s ) Vs 2 zz zsNow, if the length of the transmission line connecting a sourceto a port (or load) is electrically very small (i.e., β A 1 ), thenthe source is effectively connected directly to the source (i.e,βz s βz P ):Z0And thus the total voltage is: Vs -V z zs zPJim StilesZinV V ( z z P ) V ( z z P ) V Vs (z zS ) V (z z P )2 VThe Univ. of Kansas (z zP )Dept. of EECS

4/14/2009Wilkinson Divider Even and Odd Mode Analysis.doc10/14For the case where a matched source (i.e. Z s Z 0 ) is connecteddirectly to a port, we can thus conclude:V ( z z P ) Vs 2V ( z z P ) V Vs 2Thus, for port 2 we find:V2 ( z 2 z2P ) Vs 2V2 ( z2 z2P ) V2 Vs 2 Vs 2 Vs 2 0Now, we can finally determine the scattering parametersS12 , S22 , S32 :V1 ( z1 z1P ) jVs 2 jS12 V2 ( z2 z2P ) 2 2 Vs2V2 ( z2 z2P )2S22 (0) 0V2 ( z2 z2P )VsV3 ( z 3 z3P )2S32 (0) 0V2 ( z2 z2P )VsQ: Wow! That seemed like a lot of hard work, and we’re only13of the way done. Do we have to move the source to port 1 andthen port 3 and perform similar analyses?Jim StilesThe Univ. of KansasDept. of EECS

4/14/2009Wilkinson Divider Even and Odd Mode Analysis.doc11/14A: Nope! Using the bilateral symmetry of the circuit( 1 1, 2 3, 3 2 ), we can conclude:S13 S12 j2S33 S22 0S23 S32 0and from reciprocity:S21 S12 jS31 S13 2 j2We thus have determined 8 of the 9 scattering parametersneeded to characterize this 3-port device. The remainingholdout is the scattering parameter S11. To find this value, wemust move the source to port 1 and analyze.Port 2λ-Vs Z042 Z0Port 1Z02 Z02 Z0λ4Port 3Z0Note this source does not alter the bilateral symmetry of thecircuit. We can thus use this symmetry to help analyze thecircuit, without having to specifically define odd and even modesources.Jim StilesThe Univ. of KansasDept. of EECS

4/14/2009Wilkinson Divider Even and Odd Mode Analysis.doc12/14Since the circuit has (even) bilateral symmetry, we know thatthe symmetry plane forms a virtual open.V2λ Vs V1Z04Z02 Z02Z0I 02Z0Z02 Z0V1 Vs λV34Z0Note the value of the voltage sources. They have a value of Vs(as opposed to, say, 2Vs or Vs/2) because two equal voltagesources in parallel is equivalent to one voltage source of thesame value. E.G.: 5V - -5V5V -5V Now splitting the circuit into two half-circuits, we find the tophalf-circuit to be:λVs42Z 0 -V12 Z0Z0Z0 Jim StilesThe Univ. of KansasDept. of EECS

4/14/2009Wilkinson Divider Even and Odd Mode Analysis.doc13/14Which simplifies to:λ42Z 0 VsZ02 Z0V1 - And transforming the load resistor at the end of theline back to its beginning:Vs2Z 0 - 2Z 0V1λ4waveFinally, we use voltage division todetermine that: Vs2Z 0 Z Z2220 0 V1 Vs Port 2Thus,λ-Vs V1 VsZ042 Z02Z02 Z02 Z0λ4Port 3And since the source is matched:Z0V1 ( z1 z1P ) Vs 2V1 ( z1 z1P ) V1 Vs 2 Vs 2 Vs 2 0Jim StilesThe Univ. of KansasDept. of EECS

4/14/2009Wilkinson Divider Even and Odd Mode Analysis.doc14/14So our final scattering element is revealed!V1 ( z1 z1P )2S11 (0) 0V1 ( z1 z1P )VsSo the scattering matrix of a Wilkinson power divider has beenconfirmed: 0 S j 2 j 2 j002 j 0 0 2His worsthandout ever!Oh no, I’ve seenmuch worse.So, what’dya think?Jim StilesThe Univ. of KansasDept. of EECS

4/14/2009 7_3 The Wilkinson Power Divider.doc 1/1 Jim Stiles The Univ. of Kansas Dept. of EECS 7.3 - The Wilkinson Power Divider Reading Assignment: pp. 318-323 The Wilkinson power divider is the most popular power divider designs. It is very similar to a lossless 3dB divider, but has one additional component! HO: THE WILKINSON POWER DIVIDER