Chapter 4. Impedance Matching Networks
4.1 L Matching Networks
4.2 The π Matching Network
4.3 The T Matching Network
4.4 Wideband Matching Networks
4.5 Matching Network Using Smith Chart
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Reasons for Impedance Matching
1. Maximum power transfer
To ensure minimum losses due to impedance mismatch.
2. Impedance transformation
Usually for Rx LNA, we need a source resistance to obtain the best NF. (This
source resistance is different from that to give maximum power transfer).
3. As an aid in filtering off unwanted signals
Due to the frequency response of the matching networks.
4.1 L Matching Networks
• A 2 elements L network may be configured as LP or HP depending on design
requirements:
If a DC path is required between source and load, the LP version must be used.
If a DC block is required, HP version must be used.
Fig.4-1: The L network
• The shunt arm or reactance must be connected to the higher resistance of Rs or
RL. The series element and the shunt element must be of opposite polarity.
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• The Q of the matching network is:
Q=
RL
−1
Rs
Once the source and load resistance is fixed, Q is fixed. This means that the
bandwidth of the network can’t be controlled by design. (This limitation can be
overcome by using 3 elements matching networks).
• L matching networks are always wider band compared to 3 elements matching
networks like T or π.
Note: 1. LP matching networks can be a source of help in harmonic filtering.
2. So far the matching is done only for resistive source and load. For
complex source and load, the absorption and resonant methods are
used.
Derivation of Q formula for L matching network
Fig.4-2: L matching network
1000 − j 333
Z1 =
= 315∠ − 71.58° = 100 − j 300 Ω
1000 − j 300
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Fig.4-3: Equivalent circuit
Rp
Qp =
Xp
'
X
X
Qs = s' = s = Q s
Rs
Rs
'
i.e. Qp = Qs = Q
Since Rp =Rs (Q2 + 1), we obtain
Q=
Rp
−1
Rs
Note: matching using LC is true for one frequency only.
4.2 The π Matching Network
• It is two “back to back” L networks that are both configured to match load and
source to an “invisible” or “virtual” resistance.
The virtual resistance R must be smaller than either Rs or RL as it is “connected”
to the series arm of each section i.e. R ≤ Rs, R ≤ RL
Fig.4-4: π matching network
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• In practice, the rule of thumb for loaded Q for the π network is:
Q=
RH
−1
R
Where
RH = largest terminating impedance of Rs or RL
R = virtual resistance
The Q of the network is therefore controllable by design. The lowest Q is equal
to that of the L network.
• Suitable for cases where Rs and RL are large values.
4.3 The T Matching Network
• Similar to the π network except that the virtual resistance is larger than either
the load or source resistance.
Fig.4-5: The T matching network
• As a rule of thumb, the loaded Q is:
Q=
R
Rsmall
−1
Where
R = virtual resistance
Rsmall = smallest terminating resistance of Rs or RL
The Q of the network is therefore controllable by design. The lowest Q is equal
to that of the L network.
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• Suitable for cases where Rs and RL are small values.
4.4 Wideband Matching Networks
• Cascade L sections to reduce the effective Q of the matching networks.
Fig.4-6: Cascaded L networks for wideband matching
• As in the previous cases, R must be larger than the smallest terminating
resistance and smaller than the largest terminating resistance.
• Maximum bandwidth (minimum Q) is obtained when:
Loaded Q then becomes:
Q=
R
Rsmall
−1 =
Rl arg e
−1
R
Where
R = virtual resistance
Rsmall = smallest terminating resistance
Rlarge = largest terminating resistance
• If even wider bandwidth were required, more L networks can be cascaded with
virtual resistances between each networks.
For optimum BW
Rl arg e
R
R1
R
= 2 = 3 ==
Rsmall R1 R2
Rn
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Where
= smallest terminating resistance
Rsmall
= largest terminating resistance
Rlarge
R1, R2, …, Rn = virtual resistance of each L network
Example 4-1:
Design a LP L network to match a source resistance of 100Ω to a load resistance of
1000Ω. Assume that frequency of operation is 100MHz.
Solution:
Since RL > Rs, shunt arm is at RL. We have
Q=
RL
− 1 = 1000 − 1 = 3
100
Rs
Qs = Xs / Rs, Xs = Qs x Rs
Therefore
Xs = 3 x 100 =300 Ω
L = Xs / ω = 300 / (2π x 100 x 106) = 477 nH
Qp = RL / Xp , Xp = RL / Qp
Xp = 1000 / 3 = 333 Ω
C = 1 / ωXp = 1 / [(2π x 100 x 106) x 333] = 4.8 pF
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Example 4-2:
Design a π network to match a source resistance of 100 Ω to a load resistance of
1000 Ω. Assume a loaded Q of 15.
Solution:
We have
Q=
RH
−1
R
Therefore
R=
RH
= 1000
= 4.42 Ω
2
Q + 1 152 + 1
Load Side:
X p2 =
R p RL 1000
=
=
= 66.6 Ω
15
Q
Q
Xs2 = Q x R = 15 x 4.42 = 66.3 Ω
Source side:
Q=
X p1 =
Rs
− 1 = 100 − 1 = 4.6
R
4.42
R p 100
=
= 21.7 Ω
Qs 4.6
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Xs1 =Qs x R = 4.6 x 4.42 = 20.51 Ω
Example 4-3:
Use the absorption method to match the complex source and load at 100MHz.
Solution:
1. Ignore the source and load reactances and design a L network to match Rs and
RL.
2. Refer to earlier worked example where we have done the L network matching
from the source resistance of 100 Ω to load resistance of 1000 Ω.
Series L = 477 nH
Shunt C = 4.8 pF
j126 Ω @100 MHZ = 200 nH
Therefore the elements of this L network are
series L = 477 – 200 = 277 nH
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shunt C = 4.8 –2 = 2.8 pF
Example 4-4:
Design an impedance matching network that will block the flow of DC from the
source to the load. The frequency of operation is 75 MHz. Use the resonant
approach.
Solution:
1. The need to block DC dictates a series C and shunt L matching network.
2. Use the resonant method to remove the stray capacitance of 40 pF
L=
1
1
=
= 112.6 nH
6
ω Cstray  2π × 75 × 10  2 × 40 × 10−12


2
3. Proceed to match 50 Ω source to 600 Ω load.
Qs = Q p =
Rp
− 1 = 600 − 1 = 3.32
50
Rs
Xs = Qs x Rs = 3.32 x 50 = 166 Ω
Xp = Rp / Qp = 600 / 3.32 = 181 Ω
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Cs = 1 / (ωXs) = 12.78 pF
Lp = Xp / ω = 384 nH
Leqv = L1L2 /(L1 + L2) = 87 nH
4.5 Matching Network Using Smith Chart
• The Smith Chart is the most convenient method for design of all matching
networks.
The procedure for matching network design is as follows:
1. Plot the normalized load impedance and complex conjugate of the source.
2. Determine if the L network is sufficient or if Q need to be controlled.
L network
Join the load and conjugate source points together by 2 arcs.
3 elements network
a)
b)
Plot the constant Q circle for the specified Q.
Ensure that one out of the 3 network arcs touches the constant Q circle
when the graphical manipulation is completed.
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Fig.4-7: Summary of component addition on a Smith Chart
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Fig.4-8: VSWR circles and constant Q circles
Example 4-5:
Given that the source impedance is Zsn = 1.5 + j1 (normalized) and the load
impedance ZLn = 0.2 – j0.2 (normalized). Design a 2 elements matching network
for maximum power transfer.
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Solution:
Z1 = 0.2 – j0.2 - j0.43
= 0.2 – j0.63
= 0.661∠ -72.38°
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Z2 = Z1 // j0.88
= (0.661∠-72.38°) x (0.88∠ 90°) / (0.2 – j0.63 + j0.88)
= (0.582∠ 17.62°) / (0.32∠ 51.34°)
= 1.818∠ -33.72°
= 1.5 –j 1
= Zs*
There we have maximum power transfer.
Example 4-6:
Investigate the performance of various matching networks in matching a load of
388 + j245.5Ω at 500MHz to a 50Ω source.
Consider the following matching networks:
a. 2 elements
b. 3 elements
c. multi elements wideband
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Solution:
2 elements matching network
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3 elements matching network (a)
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3 elements matching network (b)
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Multi elements wideband matching network (a)
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Multi elements wideband matching network (b)
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Multi elements wideband matching network (c)
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