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Lect 23

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Lecture 23: The (3-port) T-Junction Power Divider

Lossless 3-port netowork


The discontinuity at the
junction create fringing
fields and higher order
modes, and leads to stored
energy which can be
accounted for by a lumped E plane waveguide T H plane waveguide T
susceptance.

Microstrip T-junction

Transmission line model of a


lossless T-junction.
ELEC4630, Shu Yang, HKUST 1
If we relax the requirement for matched ports, we can realize
tee networks that divide power and provide a match at one
port (typically chosen as the input port).

1 1 1
Yin jB
Z1 Z 2 Z 0
For a lossless network, all the characteristic impedances are
real, and we can assume B = 0. Then Yo = Y1 + Y2 .
The voltage V0 at the junction is the same for line 1 and line
2, and the power in the lines is
P1 = V02Y1/2 and P2 = V02Y2/2
P2 Y2 Z1
so the power division ratio r is r
P1 Y1 Z 2
ELEC4630, Shu Yang, HKUST 2
Substituting Y2 = rY1 we have Yo = Y1(1+r), so Z1 = Zo(1+r)
1+r
and Z2 = Zo r

The output line impedances Z1 and Z2 can be selected to provide


various power division ratio. If we let r = 1, we get even power
division (power split, -3 dB), and Z1 = 2 Z0, Z2 = 2 Z0 . If needed,
the quarter-wave transformers can be used to bring the output line
impedances back to the desired level.
If B is not negligible, some type of tuning element can usually be
added to the divider to cancel this susceptance.
Output ports are not isolated.
Example 7.1 of Pozar!

ELEC4630, Shu Yang, HKUST 3


Resistive Divider
If we relax the requirement for zero loss,
we can realize a matched, reciprocal
power divider.

Such a divider
can be realized
by connecting
three Zo An equal-split three-port
transmission resistive power divider.
lines to a star
circuit consisting
The impedance looking into the resistor
of three resistors
followed by the output line, is
R = Zo/3.
Z0 4Z0
Z Z0
3 3
ELEC4630, Shu Yang, HKUST 4
The input impedance of the divider is
Z0 2Z0
Z in Z0 matched
3 3
Since the network is symmetric from all three ports, the
output ports are also matched.

Output Power
V1
The output voltage from port 2 and 3 are, V2 V3
2
Thus, S21 = S31 = S23 = 1/2, which indicates 6 dB below the
input power level. We have
0 1 1 1
[S ] 1 0 1 Port 2 and 3 are
1 P2 P3 Pin
2 not isolated. 4
1 1 0
ELEC4630, Shu Yang, HKUST 5
The Wilkinson Power Divider
Motivation: to solve the problem of output isolation
The Wilkinson power divider is a three-port that has all ports
matched with isolation between the two output ports.

An equal-split Microstrip Wilkinson Equivalent circuit


power divider
The Wilkinson power divider can be made to give arbitrary
power division. But an equal-split one is used here for
analysis. ELEC4630, Shu Yang, HKUST 6
Features of the Wilkinson dividers
1. By choosing the impedance of the /4 lines to be 2 Zo, the
matched output loads ZL = Zo are transformed to 2Zo so they can
be placed in parallel to equal 2Zo/2 = Zo creating a matched
condition at the input port.
2. Any mismatched power returned from a load at either output
port is divided equally between the load on the input port (the
generator source impedance, presumed to be Zg = Zo) and the
resistor R.
3. None of the reflected power from a mismatched load is
dissipated in the other load, so the output ports are isolated (S23 =
S32 = 0).

ELEC4630, Shu Yang, HKUST 7


Even-Odd Mode Analysis
Preparing procedures:
Step 1: Draw a
symmetric equivalent
circuit across the mid-
plane.

Step 2: Normalize all


the impedances and
resistance.
Step 3: Define even
Even mode: Vg 2 Vg 3 2V
and odd modes
Odd mode: Vg 2 Vg 3 2V

ELEC4630, Shu Yang, HKUST 8


Even mode: Vg 2 Vg 3 2V
No current (o.c.) flows through the resistors between port 2 and 3
or the short circuit at port 1.

The
impedance
looking into
port 2 is
2
Z
Z in
e x 0
Bisection of the equivalent circuit for
2 even mode excitation.
Quarter-wave
transformer For a matched port 2, we have Z 2
so that V2e V and Z ine 1
ELEC4630, Shu Yang, HKUST 9
Setting up x coordinate as shown in the figure. Looking
toward the left from port 1, the transmission line voltage can
be written as
V ( x ) V ( e j x e j x )
Then, V2e V ( / 4) jV (1 ) V
1
V1 V (0) V (1 ) jV
e

1
2 2
We have therefore, V1 jV 2
e
2 2

ELEC4630, Shu Yang, HKUST 10


Odd mode: Vg 2 Vg 3 2V
From symmetry point of view, the middle point
of the equivalent circuit has zero potential.

The
impedance
looking into
port 2 is

Z ino r / 2 Bisection of the equivalent circuit for


A short- odd mode excitation.
circuited For a matched port 2, we have r2
quarter-wave
line is open. so that V V
2
o

ELEC4630, Shu Yang, HKUST 11


Then port 2 and 3 are terminated in matched
loads, so that we can find S11.

1
Z in ( 2 ) 2 1
2
Summary:
S11 0
symmetric
S 22 S33 0
V1e V1o
S12 S 21 e
V2 V2o
j/ 2
S13 S31 j / 2
S 23 S32 0 ELEC4630, Shu Yang, HKUST 12
When the Wilkinson power divider is driven at port 1 and the
outputs are matched, no power is dissipated in the resistor.
Thus the divider is lossless when the outputs are matched;
only reflected power from ports 2 or 3 is dissipated in the
resistor. Since S23 = S32 = 0, ports 2 and 3 are isolated.

Example 7.2 of Pozar.


Unequal Power Division and N-Way Wilkinson Dividers
Design equations: K 2 P3 / P2
1 K 2
Z 03 Z 0
K3
Z 02 K 2 Z 03 Z 0 K (1 K 2 )
1 Unequal division
R Z0 K
K ELEC4630, Shu Yang, HKUST 13
An N-way, equal-split Wilkinson Power Divider
All ports are matched,
with isolation between all
ports.

Disadvantage: crossovers among the


resistors, result in difficult fabrication
process.
Solution: using stepped multiple
sections.

ELEC4630, Shu Yang, HKUST 14

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