EE316 Microwave Engineering
Lab 4: Design of impedance transformers
记录微波作业，从我做起????
这次作业需要查阅一篇论文、具体需要的部分在theory部分已经给出。
Problem.
Design impedance transformers to match a source of 50 Ω 50\Omega 50Ω to a load of 110 Ω 110\Omega 110Ω at a frequency of 2 G H z 2GHz 2GHz (See the attached paper). Plot the simulated S-parameter.
Solution.
1. Theory.
In order to design an impedance transformer with wide-band, maximally flat real-to-real impedance
matching, we consider the following two-section quarter-wave transformer:
[
1
]
{}^{[1]}
[1]
In Fig.1, we have following parameters:
1️⃣
f
0
f_0
f0 is the center frequency at which electrical length of two transmission lines
Z
1
Z_1
Z1 and
Z
2
Z_2
Z2 is equal to
9
0
∘
90^{\circ}
90∘ ;
2️⃣
R
L
R_L
RL is the load impedance,
Z
L
Z_L
ZL is is the impedance seen at the input of the transmission line;
3️⃣
Z
i
n
Z_{in}
Zin is the input impedance, and
Z
i
n
Z_{in}
Zin equals to
Z
0
Z_0
Z0 at the center frequency
f
0
f_0
f0;
There is a formula to calculate
Z
1
Z_1
Z1 and
Z
2
Z_2
Z2:
[
1
]
{}^{[1]}
[1]
Z
0
=
Z
1
2
R
L
Z
2
2
(1)
Z_0=\frac{Z_1^2R_L}{Z_2^2}\tag{1}
Z0=Z22Z12RL(1)
Z 1 = Z 0 2 1 + 1 k 1 + k (2) Z_{1} = {Z_{0}\over\sqrt{2}}{1 + {1\over\sqrt{k}}\over\sqrt{1 + \sqrt{k}}}\tag{2} Z1=2 Z01+k 1+k 1(2)
Z 2 = Z 1 k (3) Z_{2} = {Z_{1}\over\sqrt{k}}\tag{3} Z2=k Z1(3)
where k is the ratio of
Z
0
Z_0
Z0 and
R
L
R_L
RL such that :
k
=
Z
0
/
R
L
(4)
k=Z_0/R_L\tag{4}
k=Z0/RL(4)
2. Simulation.
A. Simulation settings:
-
Set f 0 = 2 G H z f_0=2GHz f0=2GHz, Z 0 = 50 Ω Z_0=50\Omega Z0=50Ω, R L = 110 Ω R_L=110\Omega RL=110Ω.
-
According to equation ( 1 ) (1) (1) to ( 4 ) (4) (4), we have：
k = Z 0 / R L = 5 11 Z 1 = Z 0 2 1 + 1 k 1 + k = 50 2 1 + 1 5 / 11 1 + 5 / 11 = 67.85314747 Ω Z 2 = Z 1 k = 100.642819 Ω \begin{aligned} k&=Z_0/R_L=\frac{5}{11}\\ Z_1&={Z_{0}\over\sqrt{2}}{1 + {1\over\sqrt{k}}\over\sqrt{1 + \sqrt{k}}}={50\over\sqrt{2}}{1 + {1\over\sqrt{5/11}}\over\sqrt{1 + \sqrt{5/11}}}=67.85314747\Omega\\ Z_2&={Z_{1}\over\sqrt{k}}=100.642819\Omega \end{aligned} kZ1Z2=Z0/RL=115=2 Z01+k 1+k 1=2 501+5/11 1+5/11 1=67.85314747Ω=k Z1=100.642819Ω -
Substrate Properties:
Dielectric constant Substrate thickness Loss tangent Copper thickness 2.2 2.2 2.2 0.508 m m 0.508 mm 0.508mm 0.0009 0.0009 0.0009 0.018 m m 0.018mm 0.018mm
B. ADS Design:
- Set substrate and conductor parameters in MSUB according to table 1.
- Use Line Calculator to synthesize the width and Length of transmission lines
Z
1
Z_1
Z1 and
Z
2
Z_2
Z2:
- Width of Z 1 Z_1 Z1: 0.936261 m m 0.936261 mm 0.936261mm; Length of Z 1 Z_1 Z1: 27.805300 m m 27.805300 mm 27.805300mm;
- Width of Z 2 Z_2 Z2: 0.421578 m m 0.421578 mm 0.421578mm; Length of Z 2 Z_2 Z2: 28.395000 m m 28.395000 mm 28.395000mm;
- Schematic considered in this lab:
- Plot of S parameters:
- Analysis: When frequency equals 2.0 G H z 2.0GHz 2.0GHz, it could be seen that S ( 1 , 1 ) ≪ − 30 d B S(1,1)\ll -30dB S(1,1)≪−30dB , and S ( 1 , 2 ) ≈ 0 d B S(1,2)\approx0dB S(1,2)≈0dB. In other word, the design of impedance transformer is quite successful.
Reference
[1] R. Darraji, M. M. Honari, R. Mirzavand, F. M. Ghannouchi and P. Mousavi, “Wideband Two-Section Impedance Transformer With Flat Real-to-Real Impedance Matching,” in IEEE Microwave and Wireless Components Letters, vol. 26, no. 5, pp. 313-315, May 2016, doi: 10.1109/LMWC.2016.2548997.