INDIAN INSTITUTE OF TECHNOLOGY ROORKEE
ECN 342 RFM
Impedance Matching
Prof. Darshak Bhatt
Stub Matching Example
• A 50 Ω transmission line is connected to a cellular phone
antenna with load impedance ZL =25-j50 Ω. Find the
position and the length of a shunt short-circuit stub required
to match the 50 Ω line.
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0.115λ
d
B
C
D
A
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Solution for point C:
~
0.115λ
d
y d = 1 + j 1.6 point C located at 0.178λ
0.178λ
d(B,C)=(0.178 - 0.115)λ= 0.063λ
B
~
C
In order to match: yin = 1
~
But
=>
~
~
~
yin = y d + y s
~
y s = - j 1.6
= 1 + j1.6 + y s = 1
E
So we need a stub with admittance –j1.6.
The normalized admittance of a short is -j
D
A
located at point E
Starting from E, we move twd’s generator
F
~
until y = -j1.6 → point F (0.34λ).
Distance E-F gives stub length:
l = 0.34λ - 0.25 λ =0.09 λ
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Solution for point D:
~
0.115λ
d
y d = 1 - j 1.6 point D located at 0.321λ
G
0.178λ
d(B,D)=(0.321 - 0.115)λ= 0.206λ
~
B
C
In order to match: yin = 1
The needed normalized input admittance
E
~
of the stub is y s = + j 1.6,
located at point G (0.16λ on WTG scale)
The normalized admittance of a short is -j
located at point E
Starting from E, we move twd’s generator
~
D
A
F
until y = +j1.6 → point G (0.25+0.16 = 0.41λ)
Distance E-G gives stub length:
l = 0.25λ + 0.16 λ =0.41λ
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Parallel RLC Tank
• Admittance of RLC tank:
• At resonance:
• 1-nH inductor and a 1-pF capacitor resonate at 5 GHz to an
excellent (better than 1%) approximation.
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Quality Factor Q
• Q is dimensionless, and that it is proportional to the ratio of
energy stored to the energy lost, per unit time.
• the voltage across the network is simply 𝐼𝑖𝑛 𝑅.
• The total network energy:
• The average power dissipated in the resistor at resonance is
therefore simply :
• The Q of the network at resonance is then
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• The quantity
𝐿
has the dimensions of resistance, and is
𝐶
sometimes called the characteristic impedance of the
network
• additional expressions for the Q of our parallel RLC network
at resonance:
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BRANCH CURRENTS AT RESONANCE
• Since the inductive and capacitive reactances are equal at
resonance, the inductive and capacitive branch currents will
be equal in magnitude:
• That is, the current flowing in the inductive and capacitive
branches is Q times as large as the net current.
• if Q = 100 and we drive the network at resonance with a oneampere current source, that one ampere will flow through the
resistor, but one thousand amperes will flow through the
inductor and capacitor
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BANDWIDTH AND Q
• let 𝜔 = 𝜔0 + Δ𝜔, Then we may rewrite our expression for the
admittance as
• This admittance behavior is exactly the same as that of a
resistor of value R in parallel with a capacitor of value 2C,
except with Δ𝜔 replacing 𝜔0 .
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Series RLC network
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OTHER RESONANT RLC NETWORKS
• equate the impedances of the series and parallel LR
sections:
•
Equating
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• The Generalised form of Transformation:
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THE MAXIMUM POWER TRANSFER
THEOREM
• The power delivered to the load impedance is entirely due to
RL.
• To maximize the power delivered to RL, it's clear that XL and
Xs should be inverses so that they sum to zero.
• RL should equal Rs.
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Impedance matching (L Match)
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Example
• center frequency of 1GHz, R1= 50 ohm, R2 = 5 ohm,
• If BW is narrow ?
• Example: If bandwidth of 25 MHz, Q = 40.
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𝝅 − 𝒎𝒂𝒕𝒄𝒉
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Example
• The required Q is still 40.
• A Q of 40 is large enough to calculate the image resistance
with the approximate formula
• The capacitors are calculated as follows: 𝑅𝑝𝑙𝑒𝑓𝑡 =
5, 𝑅𝑃𝑟𝑖𝑔ℎ𝑡 = 50, 𝑄𝑙𝑒𝑓𝑡 = 9.57, 𝑄𝑟𝑖𝑔ℎ𝑡 = 30.4
• Finally, the required inductance is
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T match
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TAPPED CAPACITOR RESONATOR AS AN
IMPEDANCE MATCHING NETWORK
• that the network transforms a resistance Rin downward to a
value R2, or a resistance R2 upward to a value Rin,
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• Converting Parallel RC to series:
Q2 is the Q of the parallel RC section.
• The series resistor may also be viewed as the result of
transforming Rin:
• Equating the two expressions for Rs and solving for Q2
yields
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• The only remaining undetermined element is C1. To derive
an equation for its value, first express the series combination
of C1 with C2s as a single capacitance:
• The network Q then can be expressed as
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Example
• 𝑅𝑖𝑛 = 50, 𝑅2 = 5, 𝑄 = 40
• Next, we find the bottom capacitor, C2.
• finding the other capacitor's value:
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TAPPED INDUCTOR MATCH
• R2 must be less than Rin because, once again, we have a
voltage divider.
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DOUBLE-TAPPED RESONATOR
• This circuit boosts R2 to a larger effective parallel resistance
across the whole tank than in a standard tapped capacitor
network, and then reduces this parallel resistance by the
tapped inductors to the desired value Rin.
• This technique therefore increases the required inductance
and simultaneously reduces the required capacitance,
potentially bringing both closer to comfortably realizable
values.
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Representing Q on smith-chart
For series
For Parallel
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• For x>0, the center in the Γ plane is at (0,−1/𝑄𝑛 )
• For x<0, the center in the Γ plane is at (0,+1/𝑄𝑛 )
• The radius of the circle is
• For 𝑄𝑛 = 5 contour, the upper and lower part of the contour
are simply one half of a circle centered at (0,±0.2), with
radius of 1.02
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