Download When tow unshielded transmission lines are close together, power

1. Introduction
1.1 Edge-coupled directional coupler
Microstrip directional couplers are extensively used in microwave engineering to monitor
forward and backward traveling waves on a microstrip line.
For instance a directional coupler placed on the output of a power amplifier can detect if
the load is disconnected, which can be further used to shut down the amplifier to prevent
and damge.
Below is a typical microstrip directional coupler implemented using coupled microstrip
lines. when two unshielded transmission lines are close together, power can be coupled
between the lines due to the interaction of the electromagnetic fields of each line. Such
lines are referred to as coupled transmission lines.
The microstrip lines are parallel to each other for this reason this circuit is also described
as an edge-coupled circuit. As shown in the graph the separation between the lines is s
and the width of the lines in the coupling region is w.
We can apply an even-odd mode analysis to a length of coupled line to arrive at the
design equations for a single-section coupled line coupler. The four-port network is
terminated in the impedance Zo at three of its ports and driven with a voltage generator of
2V and internal impedance Zo at port 1.
The maximum coupling from port 1 to port 3 occurs when the coupling length is one
quarter-wavelength, i.e. θ= π/2.
It should be noted that port 4 always has zero output, irrespective of the electrical length
of the coupling region. In practical circuits, a major cause of the poor isolation may be
unequal even and odd mode phase velocities.
The maximum coupling to port 2 occurs at the frequency that gives quarter wave coupling
length. This will be the mid-band frequency. Because of this property, these couplers are
also known as quarter wavelength couples.
At this maximum coupling frequency, the through-line voltage V2 is 90 degree out of
phase with the coupled line voltage V3, i.e. this coupler may be described as a quadrature
coupler. The coupled voltage V3 is in phase with V1 and thus V 2 lags V1 by 90 degree.
1.2 broadband directional coupler
As the coupling of a single-section coupled line coupler is limited in bandwidth due to the
quarter wavelength requirement, to increase the bandwidth, multiple sections are used in
couplers.
Because the phase characteristics are usually better, multisection coupled line couplers
are generally made with an odd number of sections. We assume that N is odd and each
section is quarter wavelength at the center frequency, where N is the total number of
sections in the coupler.
Multisection couplers of this form can achieve decade bandwidths. But coupling levels
must be low. Because of the longer electrical length, it is more critical to have equal even
and odd mode phase velocities than it is for the single section coupler. Mismatched phase
velocities will degrade the coupler directivity, as will junction discontinuities, load
mismatches and fabrication tolerances.
2. Design method
2.1 Design of coupled-line 10 dB directional coupler, 11.5 -12.5 GHz
To calculate the values of S and W, first we need to know the even and odd mode
impedance of the coupled lines ( Z 0e and Z 0O ). While by leaving complicated
mathematic aside, the two can be obtained from the following equations.
Z 0  Z 0 e Z 0O
C
Z 0e  Z 0o
Z 0e  Z 0o
so
Z 0e  Z 0
1 C
1 C
Z 0O  Z 0
1 C
1 C
Z 0 is the characteristic impedance of our design, or 50 Ω, specifically. C is the
coupling coefficient, whose unit is usually in dB
c  xdB  10  X / 20
Based on the calculated even and odd mode impedance Z 0e and Z 0O , the ratios of
and
S
h
W
can be read from figure 1. Since h, height of the substrate is already known,
h
S W can be found.
Figure 1
To find the coupling length L, we start by finding the ¼ wavelength of central frequency
and then divide it by the root square of effective dielectric constant εeff, namely
l
1
1 0

……………………(2.X)
4
4 e f f
o 
c
f
( c is the velocity of light in free space )
To find εeff
 eff 
 eff ( e )   eff ( o )
2
εeff
for the even and odd mode, namely  eff
(e )
&  eff
(o )
, can be read from graph below
Coupling coefficient c
2.2 Broadband 10 dB directional coupler, 6- 18GHz
Multisection coupled line couplers are generally made with an odd number of coupled
line sections.
If the multisection coupler is symmetrical, then C1=CN, C2=CN-1 and so on. We have the
Fourier series form
1
V3  jV1 sin e1 jN [C1 cos( N  1)  C 2 cos( N  3)  ...  C M ………(b)
2
At the center frequency, the voltage coupling factor C0 
V3
V1
  / 2
In our design, we use three sections, with section 1 and 3 being symmetrical. For a
three-section (N=3) coupler, we require
dn
C ( )
d n
  / 2
 0 , for n=1, 2.
From (b),
C
V3
1 

 2 sin  C1 cos 2  C 2 
V1
2 

 C1 sin 3  (C 2  C1 ) sin 
………………………… (3)
To find second derivative of (b)
dC
 3C1 cos 3  (C 2  C1 ) cos    / 2  0
d
d 2C
  9C1 s i n3  (C 2  C1 ) s i n   / 2  0 ……………………(4)
d 2
Based on equation (3) and (4), with  = π/2, relation between C1 and C2can be found
C  C2  2C1
10C1  C2  0
Because C1 =C3, coupling factor of each section is known. Then dimension of S W & L
for each section is calculated the same with narrow band design.
3. Calculation
3.1 Coupled-line 10 dB directional coupler, 11.5 -12.5 GHz, on a substrate with a 50 Ω
characteristic impedance system. The substrate permittivity is 2.5. The substrate
height is 0.75mm.
Solution:
Start out by finding the coupling coefficient
C of 10 dB gives
c  10dB  1010 / 20  0.316
The even and odd mode characteristic impedances are
1 C
 69.4
1 C
Z 0e  Z 0
Z 0O  Z 0
1 C
 36
1 C
From figure 1, the dimension of microstrip coupled lines can be obtained
S
=0.12
h
S  0.12h  0.09
W
= 2.1
h
W  2.1h  1.575
From figure 2,
 eff (e)  0.86 r  2.15
 eff (o)  0.78 r  1.95
 eff 
 eff ( e )   eff ( o )
2
 1.431
o 


 eff

c
3  108

 0.025mm
f 12  10 9
0.025
 0.01747 m  17.47 mm
1.431
l

4
 4.37 mm
3.2 Broadband Coupled-line 10 dB directional coupler, 6 - 18 GHz, on a substrate with a
50 Ω characteristic impedance system. The substrate permittivity is 10 and the
substrate height is 0.75 mm.
Solution:
Start by finding C1 C2 C3 since C = 0.316
10C1  C2  0
C2  2C1  0.316
It can be found
C1  C3  0.0395
C2= 0.395
The even and odd mode characteristic impedances are
Z 0e  Z 0e  52
1
3
Z 0o  Z 0o  48
1
3
Z 0e  76
2
Z 0o  32.9
2
From figure 1
S1 S 3

 2.796 S1 = S3 = 2.097
h
h
W1
W3
 2.8 W1 = 2.1
h

h
S2
 0.07
h
S2 = 0.0525
W2
 2.1 W2 = 1.575
h
For C1 and C3 =0.0395, to find coupling length
 eff (e)  0.87 r  2.175
 eff ( o )  0.82 r  2.05
 eff ( e )   eff ( o )
 eff 
2
 1.45
o 
c
3  108

 0.025mm
f 12  10 9

0
0.025

 0.0172m  17.2mm
1.45
 eff
l

4
 4.3mm
For C2 = 0.395, to find coupling length
 eff ( e)  0.7 r  1.75
 eff (o)  0.58 r  1.45
 e f (fe )   e f (fo )
e f f 
2
 1.2 6 5
c
3  108
o  
 0.025mm
f 12  10 9

o
0.025

 0.01976m  19.76mm
 eff 1.265
l

4
 4.94mm
3.3 Broadband Coupled-line 10 dB directional coupler, 6 - 18 GHz, on a substrate with a
50 Ω characteristic impedance system. The substrate permittivity is 10. The substrate
height is 25 mil (0.635 mm).
Solution:
Start by finding C1 C2 C3 since C = 0.316
10C1  C2  0
C2  2C1  0.316
It can be found
C1  C3  0.0395
C2= 0.395
The even and odd mode characteristic impedances are
Z 0e  Z 0e  52
1
3
Z 0o  Z 0o  48
1
3
Z 0e  76
2
Z 0o  32.9
2
From figure 1
S1 S 3

 2.11 S1 = S3 = 1.34
h
h
W1
h

W3
 0.99
h
W1 = W3 = 0.63
S2
 0.13 S2 = 0.083
h
W2
 0.75 W2 = 0.473
h
For C1 and C3 = 0.0395, to find the coupling length
 eff (e)  0.7 r  7
 eff ( o )  0.61 r  6.1
 eff ( e )   eff ( o )
 eff 
2
 2.56
c
3  108
o  
 0.025mm
f 12  10 9
0
0.025

 0.00976m  9.76mm
2.56
 eff

l

4
 2.44mm
For C2 = 0.395, to find the coupling length
 eff ( e)  0.66 r  6.6
 eff ( o )  0.54 r  5.4
 eff 
o 

 eff ( e )   eff ( o )
2
 2.45
c
3  108

 0.025m
f 12  10 9
o
0.025m

 10.2mm
2.45
 eff
l

4
 2.55mm
4. Evaluation by simulations
4.1 Coupled-line 10 dB directional coupler, 11.5 -12.5 GHz
a. Layout for single section directional coupler
2-D layout
3-D layout
b. narrow band directional coupler performance
Comments: the above graph of 10 dB directional coupler is based on minor tuning.
Calculated value
Actual value
Discrepancy
S
0.09 mm
0.09 mm
No tuning
W
1.575 mm
1.67 mm
0.095 mm
L
4.37 mm
4.35 mm
0.02 mm
4.2 Broadband 10 dB directional coupler, 6- 18GHz, on a 2.5 permittivity and 0.75 mm
height substrate
a. Layout for 3-section directional couple
3-D layout
2-D layout
b. Broadband directional coupler performance
Comments: to obtain requested bandwidth, proper tuning is needed for this design.
Calculated value
Actual value
Discrepancy
S1, S3
2.097 mm
2.097 mm
No tuning
W1, W3
2.1 mm
1.46 mm
0.64 mm
L1, L3
4.3 mm
1.29 mm
3.01 mm
S2
0.0535 mm
0.0535 mm
No tuning
W2
1.575 mm
1.37 mm
0.25 mm
L2
4.94 mm
4.94 mm
No tuning
It’s found that for the central section, S W &L is around theoretical value, however
the length of each end section differs a lot from calculated value. The reason will be
elaborated in the discussion part.
4.3 Broadband 10 dB directional coupler, 6- 18GHz, on a 10 permittivity and 0.635 mm
height substrate
a. Layout for 3-section directional couple
2-D layout
2-D layout
b. Broadband directional coupler performance
Comments: to obtain requested bandwidth, proper tuning is needed for this design.
Calculated value
Actual value
Discrepancy
S1, S3
1.34 mm
1.25 mm
0.09 mm
W1, W3
0.63 mm
0.689 mm
0.059 mm
L1, L3
2.44 mm
1.31 mm
1.13 mm
S2
0.083 mm
0.082 mm
0.001 mm
W2
0.473 mm
0.504 mm
0.031 mm
L2
2.55 mm
2.87 mm
0.32 mm
Same as previous design, L1, L3 differs greatly from calculated values.
5. Discussion and recommendation
1. How the coupling coefficient is being affected by the width, spacing and length?
Try to have an intuitive understanding while does not rely heavily on mathematics,
and then verify it by software simulation.
The 3-section coupler has been chosen instead of the single section model as the
latter’s change is not as obvious as the former.
The spacing will shift the whole curve up and down, or in other words, it change
the value of C at center frequency, while it doesn’t change the shape of curve too
much.
Larger S means less coupling effect, or larger C. While in the other hand, if the
spacing decreases, the coupling effect increases, or a smaller C.
The width changes the curve a lot. when the width is increased, the flat
broadband curve will concave a bit downward while if the width is decreased, the
flat curve convex a bit upward around the center frequency.
The length is a critical value related to frequency as it equals quarter wavelength.
for smaller L, the curve will has a positive slope instead of being flat. While for a
larger L, the slope is a little negative rather than being flat. As a result, large
bandwidth could only be obtained once proper length has been chosen.
2. when designing the multisection coupler, it has been noted that for the left and
right section, the length usually differs a lot from the theoretical value, try to
explain this discrepancy and then come out with one possible solution.
For an ideal multi-section directional coupler, the calculation is based on the
assumption that the connections between each section do not have coupling effect,
or at least be negligible.
Original design with large bends
Revised design with smaller bends
However, in our design, the center spacing is less than one tenth of the two end
spacing, which as a result, results in extended junction length between different
sections. And in such a case, the coupling effect is no longer negligible. What’s
more, as the spacing between two junctions are usually smaller, as shown from
the graph above, the microstrip length is being decreased even further.
One possible solution is to have smaller and thinner bends, as what’s shown in the
graph. As all the bends provided in AWR software package are either bulky or
EM based, so this solution is only theoretically true. However, as with smaller
bends, this is no reason to doubt that the coupling effect due to the bends will be
much smaller, which means the end section’s length is much more near what’s
calculated.
3. When the directional coupler is being manufactured, some specifications for the
coupler must be noted, such as the dimension of the coupler as it needs to be fit into a
box and so on. Take the 3rd design as an example and then check whether your design
can meet such specifications. If not, try to propose some possible solutions.
The substrate chosen is 0.5 × 0.25 inch, as the coupler is around 0.5 inch long and 0.2
inch wide. It’s being noted that the distance between two adjacent ports is less than
half a inch, which do not accord with the manufacturing requirement. However this
problem can be overcome by adding more bends.
Verified by simulation software, the coupling coefficient graph changes less than 3 %,
which is, by and large, negligible. As a result, we conclude this design can meet the
manufacturing specification and is a feasible design.
4. In the broadband directional coupler design, how can we achieve the required
performance by implementing a Schiffman sawtooth pattern?
A Schiffman sawtooth directional coupler uses multistage coupled lines, with a set of
small zig-zag patterns joining together. Using this method, the mechanical length
required for a given electrical length can be reduced.
The Schiffman sawtooth pattern was designed using an empirical method. “It was not
possible to analyze this structure using the available software tools”. [1] In microwave
office software package, the zig-zag pattern isn’t provided as it’s too complicated,
which was also verified by the microwave technician.
So we didn’t include the realization of schiffman coupler in our design. However in
the empirical design, the total length along the zig-zag path is made equal to the
straight line length of this section when implementing an ordinary coupled line. This
can reduce the total length by half. The spacing between straight lines is maintained
between the interlocking teeth.
6. Reference
1.Liao, Samuel Y. “Microwave devices and circuits”, Prentice Hall, 1990
2.Trinogga, L. A. (Lothar Alfred), “Practical microstrip circuit design”, E. Horwood,
1991
3.Howe, Harlan H. “Stripline circuit design”, Artech House,1974
4.Pozar, David M. “Microwave engineering”, Wiley, 1998
5. Dana Brady, “The Design, Fabrication and Measurement of Microstrip Filter and
Coupler Circuits”, CAP Wireless, Inc. 2002