ECE 5317-6351
This presentation is the property of its rightful owner.
Sponsored Links
1 / 37

Notes 19 PowerPoint PPT Presentation


  • 177 Views
  • Uploaded on
  • Presentation posted in: General

ECE 5317-6351 Microwave Engineering. Fall 2011. Prof. David R. Jackson Dept. of ECE. Notes 19. Power Dividers and Couplers Part 1. Three-port networks. Power Dividers and Directional Couplers. General 3-port:. Power Dividers and Directional Couplers (cont.).

Download Presentation

Notes 19

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Notes 19

ECE 5317-6351

Microwave Engineering

Fall 2011

Prof. David R. Jackson

Dept. of ECE

Notes 19

Power Dividers and Couplers

Part 1


Power dividers and directional couplers

Three-port networks

Power Dividers and Directional Couplers

General 3-port:


For all ports matched reciprocal and lossless

Power Dividers and Directional Couplers (cont.)

For all ports matched, reciprocal, and lossless:

Not physically possible

(There are three distinct values.)

Lossless  [S] is unitary

These cannot all be satisfied.

(If only one is nonzero, we cannot satisfy all three.)

 At least 2 of S13, S12, S23must be zero.

(If only one is zero (or none is zero), we cannot satisfy all three.)


Now consider a 3 port network that is non reciprocal all ports matched and lossless

Power Dividers and Directional Couplers (cont.)

Now consider a 3-port network that is non-reciprocal, all ports matched, and lossless:

“Circulator”

These equations will be satisfied if:

(There are six distinct values.)

1

2

Lossless

or

Note that Sij Sji.


Notes 19

Circulators

Clockwise (LH) circulator

1

2

Note: We have assumed here that the phases of all the S parameters are zero.

Circulators can be made using biased ferrite materials.

Counter clockwise (RH) circulator


Power dividers

Power Dividers

T-Junction: lossless divider

If we match at port 1, we cannot match at the other ports!


Power dividers cont

Power Dividers (cont.)

Assuming port 1 matched:

We can design the splitter to control the powers into the two output lines.


Power dividers cont1

Power Dividers (cont.)

For each port we have:


Power dividers cont2

Power Dividers (cont.)

Also, we have


Power dividers cont3

Power Dividers (cont.)

If port 1 is matched:

The output ports are not isolated.


Power dividers cont4

Power Dividers (cont.)

Powers:

Hence


Power dividers cont5

Power Dividers (cont.)

Summary

  • The input port is matched, but not the output ports.

  • The output ports are not isolated.


Power dividers cont6

Power Dividers (cont.)

Example: Microstrip T-junction power divider


Resistive power divider

Resistive Power Divider

Same for Zin1 and Zin2

 All ports are matched.


Resistive power divider cont

Resistive Power Divider (cont.)

By reciprocity and symmetry


Resistive power divider cont1

Resistive Power Divider (cont.)

Hence we have

All ports are matched, but 1/2 Pinis dissipated by resistors, and the output ports are not isolated.


Even odd mode analysis

Even-Odd Mode Analysis

(This is needed for analyzing the Wilkenson.)

Obviously,

Example: We want to solve for V.

using even/odd mode analysis

“even” problem

“odd” problem


Even problem

Even-Odd Mode Analysis (cont.)

“Even” problem


Odd problem

Even-Odd Mode Analysis (cont.)

“Odd” problem

short circuit (SC) plane of symmetry


Notes 19

Even-Odd Mode Analysis (cont.)

“odd” problem

“even” problem

By superposition:


Notes 19

Wilkenson Power Divider

Equal-split (3 dB) power divider

  • All ports matched (S11 = S22 = S33 = 0)

  • Output ports are isolated (S23 = S32 = 0)

Note: No power is lost in going from port 1 to ports 2 and 3.

Obviously not unitary


Notes 19

Wilkenson Power Divider (cont.)

Example: Microstrip Wilkenson power divider


Notes 19

Wilkenson Power Divider (cont.)

  • Even and odd analysis is required to analyze structure when port 2 is excited.

 To determine

  • Only even analysis is needed to analyze structure when port 1 is excited.

 To determine

The other components can be found by using symmetry and reciprocity.


Top view

Wilkenson Power Divider (cont.)

Top view

A microstrip realization is shown.

Split structure along plane of symmetry (POS)

Even  voltage even about POS  place OC along POS

Odd  voltage odd about POS  place SC along POS


Notes 19

Wilkenson Power Divider (cont.)

How do you split a transmission line? (This is needed for the even case.)

top view

Voltage is the same for each half of line (V)

Current is halved for each half of line (I/2)

(magnetic wall)

Z0microstrip line

For each half


Even problem1

Wilkenson Power Divider (cont.)

“Even” Problem

Note: The 2Z0 resistor has been split into two Z0 resistors in series.

Ports 2 and 3 are excited in phase.


Odd problem1

Wilkenson Power Divider (cont.)

“Odd” problem

Note: The 2Z0 resistor has been split into two Z0 resistors in series.

Ports 2 and 3 are excited 180o out of phase.


Even problem2

Wilkenson Power Divider (cont.)

Even Problem

Port 2 excitation

Port 2


Odd problem2

Wilkenson Power Divider (cont.)

Odd Problem

Port 2 excitation

Port 2


Notes 19

Wilkenson Power Divider (cont.)

We add the results from the even and odd cases together:

Note: Since all ports have the same Z0, we ignore the normalizing factor Z0in the S parameter definition.

In summary,


Notes 19

Wilkenson Power Divider (cont.)

Port 1 excitation

Port 1

When port 1 is excited, the response, by symmetry, is even. (Hence, the total fields are the same as the even fields.)

Even Problem


Even problem3

Wilkenson Power Divider (cont.)

Even Problem

Port 1 excitation

Hence


Even problem4

Wilkenson Power Divider (cont.)

Even Problem

Port 1 excitation

Along g/4 wave transformer:

(reciprocal)


Notes 19

Wilkenson Power Divider (cont.)

For the other components:

By symmetry:

By reciprocity:


Notes 19

Wilkenson Power Divider (cont.)

All three ports are matched, and the output ports are isolated.


Notes 19

Wilkenson Power Divider (cont.)

  • When a wave is incident from port 1, half of the total incident power gets transmitted to each output port(no loss of power).

  • When a wave is incident from port 2 or port 3, half of the power gets transmitted to port 1 and half gets absorbed by the resistor, but nothing gets through to the other output port.


Notes 19

Wilkenson Power Divider (cont.)

Figure 7.15 of PozarPhotograph of a four-way corporate power divider network using three microstrip Wilkinson power dividers. Note the isolation chip resistors.Courtesy of M.D. Abouzahra, MIT Lincoln Laboratory.


  • Login