Lect 14 2014
Lect 14 2014
Solution
1. The normalized source and load impedances are:
zS Z L / Z 0 0.5 j1 yS 3 j 0.8
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Example – 1 (contd.)
2. Mark zS and then plot (zL)*=0.5+j1
circles of constant
resistance and
conductance that passes
through zS. A D
zS=1+j0.5
B
3. Mark (zL)* and then plot C
circles of constant
resistance and
conductance that passes
through (zL)*.
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Example – 1 (contd.)
4. The intersection points of these circles are A, B, C and D with the
normalized impedances and inductances as:
z A 0.5 j 0.6 y A 0.8 j1 z B 0.5 j 0.6 yB 0.8 j1
5. There are four intersection points and therefore four L-type matching
circuit configurations are possible.
zS → zA → (zL)* Shunt L, Series L
zS → zB → (zL)* Shunt C, Series L
zS → zA → (zL)*
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Example – 1 (contd.)
Similarly:
zS → zB → (zL)* zS → zC → (zL)*
zS → zD → (zL)*
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Forbidden Region, Frequency Response, and Quality Factor
Self Study - Section 8.1.2 in the Text Book
• The “nodal quality factor” and loaded quality factor are related as:
True for any L-type matching network
Q
QL n
2 For more complicated networks, QL = Qn
2
1 1
i2 r 1
𝑸𝒏 circles! Qn Qn2
Example – 2
• Design a T-type matching network that transforms a load impedance ZL =
(60 – j30)Ω into a Zin = (10 + j20)Ω input impedance and that has a
maximum Qn of 3. Compute the values for the matching network
components, assuming that matching is required at f = 1GHz.
Solution
• Several possible configurations! Let us focus on just one!
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Example – 2 (contd.)
General Topology of a T-matching Network Gives the
name T
Example – 3
• For a broadband amplifier, it is required to develop a Pi-type matching
network that transforms a load impedance ZL = (10 – j10)Ω into an input
impedance of Zin = (20 + j40)Ω. The design should involve the lowest
possible Qn. Compute the values for the matching network components,
assuming that matching is required at f = 2.4GHz.
Solution
• Several Configurations possible (including the forbidden!). One such is below:
Z2
Z3 Z1
• Since the load and source impedances are fixed, we can’t develop a
matching network that has Qn lower than the values at locations ZL and Zin
• Therefore in this example, the minimum value of Qn is determined at the
input impedance location as Qn = |Xin|/Rin = 40/20 = 2
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Example – 3 (contd.)
• In the design, we first plot
constant conductance circle g = zin
gin and find its intersection with
Qn=2 circle (point B) →
determines the value of Z3
• Next find the intersection point
(labeled as A) of the g=gL circle
and constant-resistance circle zL
that passes through B →
determines value of Z2 and Z1
Final solution at 2.4 GHz
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Example – 3 (contd.)
• It is important to note that the relative positions of Zin and ZL allows only
one optimal Pi-type network for a given specification.
• All other realizations will result in higher Qn → essentially smaller BW!
• Furthermore, for smaller ZL the Pi-matching isn’t possible!
• Inductors are avoided in these designs as they tend to have higher resistive
losses as compared to capacitors.
• In principle, only one shunt capacitor with two TL segments connected in
series on both sides is sufficient to transform any given load impedance to
any input impedance.
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Microstrip Line Matching Networks (contd.)
• Similar to the L-type matching network, these configurations may also
involve the additional requirement of a fixed Qn, necessitating additional
components to control the bandwidth of the circuit.
• In practice, these configurations are extremely useful as they permit
tuning of the circuits even after manufacturing → changing the values of
capacitors as well as placing them at different locations along the TL
offers a wide range of flexibility → In general, all the TL segments have
the same width to simplify the actual tuning →the tuning ability makes
these circuits very appropriate for prototyping.
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Example – 4
Design a hybrid matching network that transforms the load ZL = (30 + j10) Ω to
an input impedance Zin = (60 + j80) Ω. The matching network should contain
only two series TL segments and one shunt capacitor. Both TLs have a 50Ω
characteristic impedance, and the frequency at which the matching is required
is f = 1.5 GHz
Solution
• Mark the normalized load impedance (0.6 + j0.2) on the Smith chart.
• Draw the corresponding SWR circle.
• Mark the normalized input impedance (1.2 + j1.6) on the Smith chart.
• Draw the corresponding SWR circle.
• The choice of the point from which we transition from the load SWR circle
to the input SWR circle can be made arbitrarily.
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Example – 4 (contd.)
Normalized load
to the point A
gives length of
point B to
the first segment
normalized input
of TL
impedance gives
length of the
second segment
of TL
A to B provides
the necessary
susceptance
value for the
shunt capacitor