Transcript Impedance Matching

TELECOMMUNICATIONS
Dr. Hugh Blanton
ENTC 4307/ENTC 5307
Impedance Matching
• The power delivered to the variable load, RL
is calculated by defining the output voltage
first.
VOUT
RL
 Vs
Rs  RL
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• Then, the power delivered into RL is:
2
POUT

VOUT 2

RL

 RL 


RL  Rs 

Vs 2

RL
2


V
s
2
RL  Rs 
RL
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• Plotting Pout versus RL shows that maximum power
is dissipated in RL when it is equal to Rs.
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• RL = Rs is proved by differentiating P0UT
with respect to RL and setting it equal to
zero.
POUT Rs  RL   2 RL Rs  RL 

0
4
RL
Rs  RL 
2
 Rs  RL   2 RL Rs  RL   0
2
 Rs  RL   2 RL  Rs  RL  0
Rs  RL
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• Power transfer is maximized when the
source is “conjugate matched” to the load.
• In case of resistive terminations, the source
resistance must be equal to the load resistance
for maximum power transfer.
• High resistance load lead to high voltage but low
current across the load.
• Low resistance load result in high current but low
voltage.
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• Real life terminations generally represent
complex impedances, and their real parts
may not be equal.
• In such case, an impedance matching circuit is
required to eliminate the mismatch.
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• For example, if Rs = RL = 50W, and the load reactance (XL)
is presented by a 1.59 pF series capacitors, the matching
reactance must “negate” the load reactance.
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• At 100MHz the necessary reactance is,
  j159 
 159X L 
   j1000W  L  
  1590nH
jX L  
 f
 f GHz C pF 
 GHz 


 1.59H
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• The bandwidth is determined by the Q of the
circuit.
QRESONANT
BW3dB 
XC
XL
1000



 10
Rs  RL Rs  RL
10
Fo
QRESONANT
100MHz

 10MHz
10
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• Perfect match (zero reflection coefficient) can
only be achieved at selected single frequencies.
• Matching a source to a complex load require two tasks:
1. The imaginary part of the load must be negated, or
“tuned out.”
2. The real parts must have equal values.
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• Small series parasitic capacitance or large series
inductance leads to high-Q condition, leading to
narrow-bandwidth frequency response.
• If the reactive parts of the terminations are in different
configuration (i.e. one parallel, one series) a series-to parallel conversion must be used before choosing the
matching element.
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Matching Network Frequency Response
• Analyzing the previously shown circuit
verifies the computed 10MHz 3dB
bandwidth at the 100MHz center frequency.
• At the band edges the reflected energy |s11|, and
the transmitted energy, |s21|, are exactly the
same.
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Impedance Matching Procedure Development
• Recalling the series-parallel 1 GHz circuit
equivalence, we now develop a generalized
procedure to match two resistors at a given
frequency.


RP  Q 2  1 Rs
Q
RP
1
Rs
Note: RP>Rs
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• Using the above, match 50 W to 100 W.
1.
2.
4.
3.
5.
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• If the conversion is done by the above
procedure, the series and parallel Q’s are
equal.
• If two uneven resistive terminations are to be
matched, one can always be transformed by the
above concept - to become identical to the other at a specified frequency.
• By replacing one of the two reactive elements with its
opposite type, conjugate match is established between
the two circuits.
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•
If the two terminations are real but have different
values, they can be matched at any single
frequency by an appropriate “L-Network,”
observing the following procedure:
1.
Add a shunt reactance (capacitor or inductor) to the larger
termination, such that
XP 
2.
RLarger
QP
, where QP 
RLarger
RSmaller
1
Add a series reactance. (opposite kind of what selected in step 1),
to the smaller termination, such that
X s  RsQs
Qs  QP
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• Compute the matching element values:
C pF 
159
f GHz X C
LnH 
0.159X L
f GHz
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• Using ideal matching elements, the insertion loss
is zero at the center frequency.
• Bandwidth depends on Q:
• low Q results in wide bandwidth,
• increasing Q decreases the bandwidth.
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• Design a circuit to match a 10W source to a 50W
load at 400MHz. Assume that the source and the
load need to be DC-coupled, therefore use a
lowpass circuit.
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Solution:
• The need for a DC path between the source and
the load dictates the need for an inductor in the
series leg.
• The Q is computed as:
Qs  QP 

50
1
10
42
10W
50W
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• Calculating the series and parallel reactances
XP 
RP
50

 25W (capacitive)
QP
2
X s  Qs Rs  (2)(10)  20W (inductive)
• Generally two component combinations exist:
• lowpass or
• highpass topologies.
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• The component values at 400MHz are:
0.159X L 0.15920
L

 7.95nH
FGHz
C
0.4
159
159

 15.9 pF
FGHz X C 0.425
• The final circuit with ideal components is:
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• Parasitic source inductance or load capacitance
may be “absorbed” into the matching network:
• Case I: Source with series inductance
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• Case II: Load with shunt capacitance:
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• Impedance matching takes place at 400MHz
only, and mismatch occurs at all other
frequencies.
• Of course, real physical circuits have frequency
dependent dissipative losses that also affect the
frequency response.
• Absorbing the source or load parasitics into
the matching network does not change the
bandwidth of the frequency response.
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Impedance Matching - Complex Loads
•
There are two basic approaches in
handling complex impedances
1. Absorption - Stray reactances are absorbed
into the impedance-matching network, up to
the maximums, that are equal to the matching
component values.
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2. Resonance - Beyond the limits of
maximum absorption, the excessive
parasitics may be resonated with an
equal and opposite reactance at the
frequency of interest.
•
Once this is done, the matching network
design can proceed for two pure resistances.
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• In the Resonance technique, LR resonates CM at
the frequency of interest, leaving a resistive load.
• For parasitic inductance, resonance is achieved by
using a capacitor.
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• The resonating inductance (or capacitance
when applicable):
LRnH 
25.33
2
FRGHz
CPpF
or
CRpF 
25.33
2
FRGHz
LRnH
• Resonating a parasitic inductance or
capacitance of a complex termination always
leads to reduced bandwidth.
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Resonance Matching
1. Another approach is to fully resonate the
parasitic portion first.
2. Then a suitable matching topology is selected
with one component identical to the resonating
element.
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• Finally, the matching and resonating elements are
combined to save a component.
• BUT, LR and 50 W define a Q equal to 20 pF and 50 W.
• Paralleling LR with CM will result in a higher loaded Q for
certain.
• Although this approach saves a component, the
bandwidth is not quite as wide as it was in the
previous case.
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Absorption Matching Network Example
• A complex source of 50W parallel with 5.9pF
capacitance is to be matched to a load of 10W
resistance in series with 3.98nH inductance.
• Design two matching networks (one lowpass and one
highpass) on the Smith Chart, and compute the
component values at 400MHz (at that frequency the
inductor represents XL = +j10 W, xL = j0.2 W.)
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Absorption Matching Network Example
In order to match
impedances, we
must create an
reactance that
negates the
reactance of the
load.
zL
z L*
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• Lowpass solution: Selecting the shunt C-series L
network:
7.96xLs 7.960.2
L 

 3.98nH
S
CP 
FGHz
3.13bCp
FGHz
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0.4
3.13(1.26)

 10 pF
0.4
36
Z L  10  j10  zL  0.2  j0.2 
*
zL
 0.2  j0.2
ys  Gs  jBs  1  j0.74
y  Gs  jBs  1  j 0.74  j1.26  1  2 j
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y
ys
zL
z L*
z
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• The highpass option may also be viewed as
resonance matching since the 3.98nH load
inductance is resonated by part of the series
capacitor Cs.
• The 5.9pF source capacitance is resonated by part of
the parallel inductor LP.
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• Highpass solution: Selecting the shunt L-series C
network:
7.96
7.96
LP 

 7.26nH
bLp FGHz 2.740.4
Cs 
3.183
3.183

 13.26 pF
X Cs FGHz 0.60.4
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Absorption Matching Network Example
In order to match
impedances, we
must create an
reactance that
negates the
reactance of the
load.
zL
z L*
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ys  Gs  jBs  1  j0.74
y  Gs  jBs  jBL  1  j0.74  j 2.74  1  2 j
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