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TRANSCRIPT
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Design flow of amplifiers for maximum output power
• Active devices (FET or BJT) are selected for design requirements
(Output power, efficiency, gain, IMD, bandwidth)
• Topology and Class are chosen in accordance with power, efficiency,
linearity
• Evaluation of optimum output termination (with respect to output
power, PAE, IMD)
• Design of output matching network
• Evaluation of optimum termination (with respect to gain and gain
ripple)
• Synthesis of input matching network to obtain:
– Gain and gain ripple;
– Stability
2
Power amplifiers design requirements
• In a power amplifier, the output has to be matched to the optimum load
for power delivery
• Output matching network cannot be used to compensate gain roll-off
in the bandwidth and to achieve stability
• Input matching network is used to fulfill at the same time requirements
on: VSWR, GT, Gain Flatness, stability
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3
Power amplifier design methodology
• In the design of a power amplifier the main challenge is to
determine the optimum load termination
• It can be done by:
– Load Pull measurements (either passive or active)
– Cripps method (analytic formula and Smith chart)
– Non-linear simulation in frequency domain (Harmonic
Balance)
4
Setup for Load pull measurement
DUT
Controller
Steppingmotors
DC BiasInput Tuner
Output Tuner
Power Source
Power meter
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5
Passive Load pull: main steps for measurements
• The input tuner has small effect on delivered power POUT : it is changed in
order to guarantee good input matching at specified frequencies
• The output tuner is varied in order to deliver the required POUT
• The DUT is disconnected, and input impedance of the tuner is measured
• The same operations are performed in order to draw the -X dB locus on
Smith chart
6
Passive Load pull measurements
• Operations described in previous slide are performed at
different power levels
• Typically, the designer is interested to locus at POPT-{1,2,3} dB
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7
Typical “Load Pull contours” (L88016 device from Polyfet )
8
Limitations of passive Load pull measurements
• Due to loss, passive Load-Pull is not able to fully cover
the unitary circle on the Smith chart
• Mechanical tuners are not able to guarantee measurement
accuracy
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9
Active Load pull set-up: operation principle
• A controlled power source is added at the output
• Active tuning is obtained by varying amplitude and phase of the output power
source
• Better Smith chart coverage is obtained: a variable complex termination (Z1) is
seen if the ratio I2/I1 is complex (amplitude and phase variation)
• No mechanical tuners are needed
Gen1 R Gen2
I2I1
1
21 1
I
IRZ
10
Active Load pull set-up: RF implementation
VDD
RFC
VGS
RF Load
(Power meter)
F
PowerAmp
Phaseshifter Attenuator
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11
Load Pull methodology features
• It is possible to enter instability regions of the DUT
• Measurement procedure is long and tedious if manual, expensive if automated
• It is required to de-embed parasitic elements (wire bond inductance, electrical
length of test fixture, …)
• When a new device is developed, load-pull measurements are the most
powerful tool to forecast non-linear device performance
12
Computer-assisted design of power amplifiers
• If an accurate non-linear model of the device is available, it is possible to
exploit for design:
– “Harmonic Balance” simulators, to achieve steady-state solution in
frequency domain
– “Time-domain” simulators, that allow to derive also transient response till
steady-state condition
• Even if non-linear simulators are a powerful design tool, a fully automated
design flow is still not possible
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13
Load-line Cripps theory
• Load-line Cripps theory can be used with linear models and simulators
• It is able to take into account of main limitations to output power delivery
(limitations on voltage and current dynamic)
• Output power prediction is often very accurate
14
Comparison of compression curves for maximum GT and for maximum POUT
A
A’B
B’ C’
C
PIN (dB)
PO
UT
(dB)
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15
“Linear” approximation for load-pull locus drawing
• Points A and A’ of previous slide are the maximum power in linear condition
• Points B and B’ of previous slide are the -1 dB compression point
• Matching for maximum GT produces a -1 dB compression point 2-3 dB lower
than matching for maximum POUT
16
“Linear” approximation for load-pull locus drawing
• Matching for maximum power POUT shows different benefits:
– increase of maximum power in linear condition (points A e A’);
– increase of -1 dB compression point (points B e B’);
– increase of saturation power (points C e C’).
• The hypothesis is made that output power increase obtained by switching from
gain matching condition to power matching condition (A-A’, B-B’, C-C’) is
the same
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17
“Linear” approximation for load-pull locus drawing
• Cripps theory of load-line is based on linear approximation, i.e. the maximum
power in linear condition (point A’) is evaluated
• It is assumed that the termination that allows maximization of power in linear
condition, is the same for maximum P-1dB and PSAT
• Very good agreement is found between load-pull measurements and Cripps
theory
18
Load-line Cripps theory
VDD
RFC
ZL
VGS
VDS
ZL=ROPT
Device parasitics at output side are included into the load
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Load-line Cripps theory: ideal models for FET devices
IDVGS
VK VDSmax
VGS
(linear
steps)
ID
Transconductance is linear, except for interdiction and saturation
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Load-line Cripps theory
VK
2VDD-VK
VDSpeak
IDpeak
0
IF
t
tROPT
F
Kdd
Dpeak
DSpeakOPT
I
)VV(2
I
VR
2
FI
VDD
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Load-line Cripps theory
• If we assume that RF trajectories in output plane of voltage and current
waveforms are the same of DC ones (quasi-static approximation);
• If we assume perfectly symmetric waveforms (harmonics are neglected);
• for maximum power in linear condition we get:
• and:
F
KddOPT
I
)VV(R
2
2
I)VV(
2
1P F
KddOPT
22
Load-line Cripps theory
• In order to draw the “Load Pull” locus, let’s see what happens when real and
imaginary parts of ZL are varied with respect to the optimum value
• The goal is to derive the locus of constant output power as a function of ZL
• For each value of p coefficient (p > 1) we get a “Load pull” locus where POUT
is constant: this locus is not a circle!
p
PP OPTOUT
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Load-line Cripps theory
• For a given p value, 2 resistance values exist that produce an output power
equal to:
• Such resistance values are one greater and one lower than ROPT:
p
PP OPTOUT
OPTHi RpR p
RR OPT
Lo
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Load-line Cripps theory
• In case of lower resistance:
– the FET device shows full current dynamic;
– voltage dynamic is reduced for a factor p
• In case of higher resistance :
– the FET device shows full voltage dynamic;
– current dynamic is reduced for a factor p
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Load-line Cripps theory
ROPT
RHIRLOExample case p=2 (-3 dB)
Two points of the -3dB locus are drawn
In order to complete the locus drawing, we need to consider also complex terminations that allows to obtain an output power equal to POPT/2 (i.e. to add the proper reactive parts to RLO and to RHI)
26
Load-line Cripps theory: series reactance added to RLO
• Let’s consider also reactive part of the load:
– if |ZL|<ROPT, linear power is reduced due to a lower voltage dynamic
– the maximum current dynamic can be considered, and voltage dynamic
can be increased by adding a series reactance: however, output power is
not affected, as it depends only on resistive part
22LL
OPT
KDDLLL XR
R
VVZIV
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Load-line Cripps theory: series reactance added to RLO
• The maximum value of added reactance is the one that allows maximum
voltage dynamic:
KDDLL
OPT
KDDL VVXR
R
VVV
22
222LOPTL
RRX p
RR OPT
L
mLm XXX 2
11
pRX OPTm
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Load-line Cripps theory: series reactance added to RLO
• The first half of load pull locus is obtained: it is a part of R=RLO circle,
comprised between -Xm and Xm values
ROPT
RHIRLO
Xm
-Xm
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Load-line Cripps theory: parallel susceptance added to RHI
• In this case:
– if |ZL|>ROPT, linear power is reduced due to a lower current dynamic
– the maximum voltage dynamic can be considered, and current dynamic
can be increased by adding a parallel susceptance: however, output power
is not affected, as it depends only on resistive part
22LLKDDLLL BGVVYVI
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Load-line Cripps theory: parallel susceptance added to RHI
• The maximum value of added susceptance is the one that allows maximum
current dynamic:
222LOPTL
GGB
OPTOPT
RG
1
mLm BBB 2
11
pGB OPTm
22
22 FLLOPT
FL
IBGR
II
p
GG OPT
L
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Load-line Cripps theory: parallel susceptance added to RHI
• The second half of load pull locus is obtained: it is a part of G=GHI circle,
comprised between -Bm and Bm values
It can be demonstrated that the impedance RLO+jXm
corresponds just to the admittance GHI+jBm
ROPT
RHIRLO
RLO+jXm
RLO-jXm
32
Drawing of “load Pull contours” on Smith chart
• In order to draw “Load Pull” contours we need to:
– Evaluate ROPT and POPT according to expressions presented above;
– Choose p coefficient according to design requirements;
– Evaluate RLO and RHI (locus points on real axis);
– Starting from RLO, consider constant-R circle (R= RLO) up to +/- Xm
values;
– Starting from RHI , consider constant-G circle (G= GHI) up to +/- Bm
values;
– Repeat the same procedure for different p values
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Comparison between “load Pull contours” and mismatch circles on
Smith chart
Load pull locus does not show circular behavior; moreover, -X dB region is smaller than the corresponding region for mismatch circles
34
Package effects: Load-line Cripps theory improvements
• Load-Pull loci of previous slide are not dependent on frequency
• Indeed, measured loci are strongly dependent on frequency
• Dependence on frequency is related to the choice of reference plane for
impedance measurement
• In order to have a meaningful comparison between theory and measurements,
effects of device output impedance and package effects have to be considered
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Package effects: Load-line Cripps theory improvements
• The proposed approach permit evaluation of loci at reference plane A
• In order to allow evaluation at plane B, de-embedding of parasitics (XC) has to
be made
• De-embedding is made by considering a transformation in the admittance
plane:
ID
“A” “B”
XC ZL
CAL
BL
YYY
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Package effects: Load-line Cripps theory improvements
De-embedding of device output
capacitance is obtained by means of a
rotation around the parallel
conductance circle
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Package effects: Load-line Cripps theory improvements
“A”
COUT
ZL
“B”LOUT
Package
- Firstly, rotation due to COUT (see previous slide)
- Then, de-embedding of wire bond inductance is
obtained by means of a rotation around the series
resistance circle
Also packageeffects
38
Package effects: Load-line Cripps theory improvements
• Good agreement between theory
and measurements
• Theory is more pessimistic with
respect to measurements
• Theory is a good initial point also
for broad-band design
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Accuracy and limitations of the proposed approach
• For methodology practical implementation, we need:
– Static curves of active device
– Knowledge of device output impedance and package parasitics (a
model of package can be extracted from measured S parameters)
• Even if different approximations have been made, the proposed
methodology is a good starting point for power amplifier design
• Availability of both a non-linear model and a non-linear simulator
permits refinements and optimization of the design
40
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Matching networks - 1
In RF amplifier design, matching networks are used to provide
fulfillment of gain/power requirements. Design methodology is
quite different from the one used or low-frequency amplifiers
(for instance OPAMP and audio amplifiers)
Review: at low frequency, matching is obtained by:
•Zg<<Zl Voltage matching
•Zg>>Zl Current matching
42
Matching networks - 2
Example: ideal transformer with given coil number ratio n
g2
2LOAD
Rn'Rg
nVg'Vg
n
RZin
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Matching networks - 3
Ideal transformer and, in general, loss-less networks are able
to preserve the available power:
g2
22
Rn'Rg
Rg8
Vg
'Rg8
'VgPavnVg'Vg
The same available power is referred to a different internal
source impedance
44
Narrow-band Matching networks
• Requirements: f0, Q, transform ratio RS / RL
• L-shape (2 elements)
• Tapped resonant, T-shape and -shape (3 elements)
• Transformer-based networks
– Single-tuned transformers
– Tapped coil transformers
– Double-tuned transformers
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L-shape networks /1
• The matching network is composed of 2 elements:
the first one allows impedance transformation, the
second one provides tuning to f0
• WARNING: It is not possible independent choice
of Q and f0
46
L-shape networks /2
Configurations for downward resistance transformation
RL RLR’ R’
LR'R
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LR'R
RL RLR’ R’
L-shape networks /3
Configurations for upward resistance transformation
48
L-shape networks /4
• Design example:
– Transformation from a 50 load to 250 at 50 MHz
50
S->P transformation
250
4Q)Q1(R'R 22
H318.0LR/LQ
pF5.25)'L/(1C 2
0
H3975.0)Q/11(L'L 2
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• The matching network is composed of 3 elements, the first
2 ones allow load transformation, the third one tuning to f0
• It is possible independent choice of Q and f0
• The resistance is transformed two times (both downward
and upward
49
Tapped resonant, T-shape and -shape
R” R”R R
T-shape -shape
B2 B1B3
X1 X2X3
50
-shape networks /1
R” RB1B3
X2
R’ - j / B1’
R’ - j / B3’
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-shape networks /2
• A first P->S transformation allows R decrease, the second
one (S->P) resistance increase
)Q1/(R'R 2
1
)Q1('R"R 2
2
)Q1(
)Q1(R"R
2
1
2
2
• In case of downward transformation: Q1 > Q2
• In case of upward transformation: Q2 > Q1
52
-shape networks /3
• Q1 is related to R and to the first element B1 of the network:
• The impedance seen at the right side is:
• where:
• and therefore the reactance seen at the right side is:
RBQ 11
)Q/11(B'B 2111
11
1 Q'R'B
1'X
R’ + j X1’= R’ - j / B1’
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-shape networks /4
• Q2 is related to R” and to the third element B3 of the network:
• and therefore the reactance seen at the left side is:
• For matching at the tuning frequency f0 we get:
"RBQ 32
)Q/11(B'B 2233
23
3 Q'R'B
1'X
0'XX'X 321 )QQ('RX212
54
-shape networks /5
• In case of upward transformation, maximum Q is equal to Q2. In case of downward transformation, maximum Q is equal to Q1
• In both cases we have: Qmax = 2Q (loaded Q is given) where approximation is made that loaded Q is equal to Qmax / 2
DESIGN PROCEDURE
R"R 2max QQ
R"R 1max QQ
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-shape networks /6
• The intermediate resistance R’ is chosen:
• The other Q value is then
chosen from transform ratio:
• B1 and B3 are determined: B1 = Q1 / R B3 = Q2 / R”
• X2 is found: X2 = R’·(Q1 + Q2)
DESIGN PROCEDURE
)Q1(
)Q1(R"R
2
1
2
2
R"R
R"R )Q1/(R'R 2
1
)Q1/("R'R 22 The maximum
Q is chosen
56
-shape networks /7
DESIGN EXAMPLE
• Matching of a 50 load to a 2.5 source termination at a 10.0 MHz
frequency, with loaded Q equal to 2.5
• Transformation produces resistance decrease, and therefore:
Qmax = Q1 = 2·2.5
• A network with a capacitor as a first element is chosen:
0C1 = Q1/R => C1 = 1.59 nF
• Q2 is chosen starting from Q1 and from transform ratio: Q2 = [12.5 / 50
·(1 + 5 2)] 0.5 = 2.35
• Reactance of the 2nd element is chosen:
X2 = 50 (5 + 2.35) / (1 + 52) = 14.13 = 0L2 => L2 = 225 nH
• Tuning capacitance is chosen: 0C3 = Q2/R” => C3 = 3 nF
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-shape networks /8
DESIGN EXAMPLE
58
-shape networks /9
DESIGN EXAMPLE
Q = 3.1
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T-shape networks /1
R” R
X1X3
B2G’ - j / X1’G’ - j / X3’
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T-shape networks /2
• A first S->P transformation allows R increase, the second
one (P->S) resistance decrease
• In case of downward transformation: Q1 < Q2
• In case of upward transformation: Q2 < Q1
)Q1(R'R 2
1
)Q1/('R"R 2
2
)Q1(
)Q1(R"R
2
2
2
1
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T-shape networks /3
• Q1 is related to R and to the first element X1 of the network:
• The susceptance seen at the right side is:
R/XQ11
)Q/11(X'X 2
111 'R/Q
'X
1'B 1
11
62
T-shape networks /4
• Q2 is related to R” and to the third element X3 of the network:
• and therefore the susceptance seen at the left side is:
• For matching at the tuning frequency f0 we get:
"R/XQ32
)Q/11(X'X 2
233 'R/Q
'X
1'B 2
33
0'BB'B 321 'R/)QQ(Y212
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T-shape networks /5
• In case of upward transformation, maximum Q is equal to Q1. In case of downward transformation, maximum Q is equal to Q2
• In both cases we have: Qmax = 2Q (loaded Q is given) where approximation is made that loaded Q is equal to Qmax / 2
DESIGN PROCEDURE
R"R 1max QQ
R"R 2max QQ
64
T-shape networks /6
• The intermediate resistance R’ is chosen:
• The other Q value is then
chosen from transform ratio:
• X1 and X3 are determined: X1 = Q1·R X3 = Q2·R”
• B2 is found: B2 = (Q1 + Q2) / R’
DESIGN PROCEDURE
The maximum
Q is chosen
)Q1(
)Q1(R"R
22
21
R"R
R"R )Q1("R'R 22
)Q1(R'R 21
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Tapped resonant circuits /1
b) Tapped-capacitor circuita) Tapped-inductor circuit
C1
L1RSRL
C2
L2
RL
C L
RS
66
Tapped resonant circuits /2
• A first Parallel->Series transformation (resistance is
lowered):
22C
L2
2C
LLS
222C
22C
2S2
Q
R
Q1
1RR
CQ
Q1CC
C1
RLS
C2SL
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Tapped resonant circuits /3
• A second Series -> Parallel transformation
(resistance is enhanced):
RTOTCL
2
2C
2C
L22C
2C
L2CLSTOT
S21
S21
2C
2C
S21
S21
Q
QR
Q1
Q1RQ1RR
CC
CC
Q1
Q
CC
CCC
68
Tapped resonant circuits /4
• The 2nd transformation produces a parallel resonant circuit, where
input resistance is just the matching resistance at resonance
• If we consider Q expressions we find;
• The matching network produces multiplication of load resistance for a
factor depending on the ratio of capacitances
2
1
2L
2
1
21LTOT
C
C1R
C
CCRR
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Tapped resonant circuits /5
• Qtot = 2Q = 2f0 / BW is chosen (in order to account for partition due to
matched load).
• C value is calculated:
• L value is calculated :
• QC2 value is calculated :
Design procedure
S0
tot
R
QC
C
1L
20
1R/R
Q1Q
LS
2tot
2C
70
Tapped resonant circuits /6
• C2 value is calculated :
• C2S value is calculated :
• C1 value is calculated :
Design procedure
CC
CCC
S2
S21
L0
2C2
R
QC
22C
22C
2S2Q
Q1CC
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Tapped resonant circuits /7
• We need to match a 50 load to a 4K source resistance at 3.0 MHz
resonance frequency, with a loaded Q = 7.5.
• Unloaded Q value is: Qtot = 2·7.5 = 15.
• Overall capacitance is: C = 15 / (0RS) = 200 pF.
• Tuning inductance is: L = 1 / (02 C) = 14 uH.
• If the other expressions previously presented are applied, we find:
– QC2 = 1.34
– C2 = 1.4 nF
– C2S = 2.2 nF
– C1 = 0.22 nF
DESIGN EXAMPLE
72
Tapped resonant circuits /8
DESIGN EXAMPLE
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Tapped resonant circuits /9
DESIGN EXAMPLE
Q = 7.14
74
Tapped resonant circuits /10
A first Parallel->Series transformation is done (resistance is
lowered):
and we get:
22L
L2
2L
LLS
222L
22L
2S2
Q
R
Q1
1RR
LQ1
QLL
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75
Tapped resonant circuits /11
A second Series -> Parallel transformation is done: resistance
is enhanced, and Q value is unchanged
.
2
2L
2L
L22L
2L
L2LLSTOT
S21TOT
Q
QR
Q1
Q1RQ1RR
LLL
Under high-Q (Q<10)
hypothesis we get:
2
2
1L
2
2
21LTOT
L
L1R
L
LLRR
76
Design with complex termination
• At a first step, reactive parts of both load and source
terminations are neglected: the network is designed for
pure resistive transformation
• Then, the 1st and the 3rd element are modified to account
for reactive parts
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T-shape networks
DESIGN WITH COMPLEX
TERMINATIONS
RL
X1 XL
X3
Xin X3’
R” + j XinB2 ZL
"RQ'X23
LL11XRQX
78
T-shape networks
DESIGN EXAMPLE WITH COMPLEX TERMINATIONS
• A matching network is designed at 5 MHz in order to match a load ZL =
10 + j 10 to a source ZS = 50 - j 40 , with a 2.5 loaded Q
• Q of the circuit is: Qmax = 2·2.5 = 5.
• Resistance increase is produced: Qmax = Q1.
• Overall reactance at load side is X1 + XL = 5·10 = 50 , and we get:
X1 = 40 L1 = 1.27 uH.
• Q2 is determined: 1 + Q22 = (1 + 52) ·10 / 50 Q2 = 2.05.
• Overall reactance at source side (that sets resonance) is X3 = 2.05·50 =
103 , from which: L3 = 3.3 uH. A reactance equal to -XS = 40 is
added
• Finally, tuning parallel capacitor is evaluated:
C2 = (5 + 2.05) / [(2·5 MHz)·10 ·(1 + 52)] = 0.86 nF
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T-shape networks
DESIGN EXAMPLE WITH COMPLEX TERMINATIONS
80
T-shape networks
DESIGN EXAMPLE WITH COMPLEX TERMINATIONS
Q = 2.63
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81
-shape networks
DESIGN WITH COMPLEX
TERMINATIONS
"R/Q'B23
LL11
BR/QB
RLB1 BL
B3
Bin B3’G” + j Bin
X2
YL
82
Transformer-based matching networks
:
BW
Matched transformer at f0
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83
Tapped coil with mutual coupling /1
Matching to the amplifier input or output impedance
84
Tapped coil with mutual coupling /2
1) V1 = j (L1 + L2 + 2 M) I1 + j (M + L2) I2
2) V2 = j (L2 + M) I1 + j L2 I2 = - I2 R0 (only R0 at a first step)
Matching to the amplifier output impedance
I1
I2
M
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85
Tapped coil with mutual coupling /3
By substituting 2 into1, the input impedance Zin() is obtained:
3) Zin() = jL + (R0 - j L2) (L2 + M)2 2 / |R0 + j L2|2=
jL + (R0 - j L2) (L2 + M)2 / [L22 (1 + QL2
2)]
where:
L = L1 + L2 + 2M
QL2 = R0 / L2
Matching to the amplifier output impedance
86
Tapped coil with mutual coupling /4
Transformer expressions:
1] L1 = N12 / (N1 + N2)
2 L 2] L2 = N22 / (N1 + N2)
2 L
where N = (N1 + N2) / N2 is the transform ratio
3] Mutual inductance M = k (L1 L2)1/2, k is the coupling factor
By using 1]-2]-3] with k = 1 (approximation) we get:
4) (L2 + M)2 / L22 = N2
That can be substituted into 3) in order to evaluate the input impedance:
5) Zin() = jL + (R0 - j L2) N2 / (1 + QL2
2)
Matching to the amplifier output impedance
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Tapped coil with mutual coupling /5
Matching to the amplifier output impedance
By using 2] we get:
6) Zin() = jL QL22 / (1 + QL2
2)+ R0 N2 / (1 + QL22)
For usual Q values (QL2 >> 1) the input impedance is:
7) Zin() = jL + R0 N2 / QL22
and therefore on the primary coil we see the series connection of L (the total
primary coil inductance) and a resistance R’0 equal to:
8) R’0 = R0 N2 / QL22
The related Q factor is just:
9) Q = L / R’0 = L QL22 / (R0 N2) = QL2
where expression 2] has been used
88
Tapped coil with mutual coupling /6
Matching to the amplifier output impedance
A Series -> Parallel transformation allows to get at the primary coil:
10) L0 = L (1 + Q2) / Q2 L
11) Rin0 = R’0 (1 + Q2) R’0 Q2 = R0 N2
where 8 and 9 have been used.
Also C is transformed at the same way:
12) Cin0 = C0 / N2
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89
Tapped coil with mutual coupling /7
Matching to the amplifier output impedance
The equivalent circuit at the primary side is:
- We need Cin0 << Cp so that the tuning is mainly due to Cp: high transform
ratio N helps us to reduce effects of parasitic capacitance
- High Rino means high Q
Rin0 = R0 N2
Cin0 = C0 / N2
90
Single-tuned transformer /1
Matching to the amplifier input or output impedance
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91
Single-tuned transformer /1
Matching to the amplifier input impedance
13) V1 = j L I1 + j M I2
14) V2 = j L3 I2 + j M I1 = - I2 Rin2 (only Rin2 at a first step)
92
Single-tuned transformer /2
Matching to the amplifier input impedance
By substituting 14 into 13 the input impedance Zin() is obtained :
15) Zin() = jL + (Rin2 - j L3) ( M)2 / |Rin2 + j L3|2 =
jL + (Rin2 - j L3) M2 / [L32 (1 + QL3
2)]
where QL3 = Rin2 / L3
Again M2 / L32 is just the transform ratio N’2 and therefore we get:
16) N’ = N1 / N3
and:
17) Rin0 = Rin2 N’2 Cin0 = Cin2 / N’2
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93
Single-tuned transformer /3
Matching to the amplifier input impedance
The equivalent circuit at the primary side is:
- We need Cin0 << Cp so that the tuning is mainly due to Cp: high transform
ratio N helps us to reduce effects of parasitic capacitance
- High Rin0 means high Q
Rin0 = Rin2 N’2
Cin0 = Cin2 / N’2
94
Interstage matching network /1
Tapped coil with mutual coupling and Single-tuned transformer
are used together to match the output of the first stage and the
input of the 2nd stage to a common tuning capacitance at the
primary side
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95
Interstage matching network /2
- Results for Tapped coil with mutual coupling can be used
- Results for Single-tuned transformer can be used with:
L = L1 + L2 + 2M
N’ = (N1 + N2) / N3
96
Interstage matching network /3
N and N’ have to be chosen suitably high, so that:
- C1 and C2 are small with respect to Cp, and resonance frequency is
mainly determined by Cp value
- R1 and R2 don’t lower the coil Q factor
Equivalent circuit at the primary side
Coil
Rin2 N’2
Cin2 / N’2
R0 N2
C0 / N2
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97
Double-tuned transformer
98
Equivalent scheme
2secsec1L
2
sec
2
2
21prpr1
I)Cj/1Lj(MIj)RC/(I
MIjI)Cj/1Lj(V
At resonance we have:
1L
2
sec
3
2
21
IRCjMI
MIjV
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99
Maximum power transfer
secpr
crS
2
pr
2
0in
1k)RC/(1)(Z
Input impedance at resonance:
Maximum power transfer condition:
L
2
sec
24
0in RCM)(Z
100
Insertion loss at resonance
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101
Transmission line trasformers /1
lcosIlsinZ
jVI
lsinIZjlcosVV
20
21
2021
Equations of a loss-less transmission line:
where Z0 is the characteristic impedance and is the (real) propagation
constant
If the output termination is just Z0, the input impedance is equal to Z0 for
any value of length l
102
Transmission line trasformers /2
A Balanced-to-Unbalanced (Balun) converter can be obtained: it works as a
broadband 1:1 transformer
:
This kind of transormers is widely used in broadband RF power amplifiers in
Class A and Class B and AB push-pull
+
V1
-
+
V2
-
I1 I2
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103
Transmission line trasformers /3
Another widely used topology in power amplifiers is the 1:4 impedance
transformer
I the two transmission lines show characteristic impedance Z0, and the
termination load is ZL =2 Z0, at the input the parallel connection of two Z0
impedances is seen and therefore ZIN =1/2 Z0
Moreover, if the input termination is ZS =1/2 Z0, at the input the series
connection of two Z0 impedances is seen, and therefore ZOUT = 2 Z0
Both the balanced and the unbalanced version of 1:4 impedance transformer
can be found: the former is used in push-pull amplifiers
104
Transmission line trasformers /4
Single-ended
transformer
Balanced
transformer
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105
Transmission line trasformers /5
RF transmission line trasformers are usually made with handformable or
semi-rigid coaxial cables, available with different characteristic impedance
values (www.micro-coax.com)
Vbias
106
Transmission line trasformers /6
The external conductor shows magnetic loss, modelled by means of a parallel
inductor
Such inductance produces a low-frequency pole-zero couple: the bandwidth is
reduced, as a low-frequency cut-off frequency is introduced
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107
Transmission line trasformers /7
Inductance is found by integrating the magnetic density flux B created by a
perfect cylindrical conductor (with length Len and radius c) in which current I
flows, in the region surrounding the coaxial cable itself
The expression of flux B(y0,R0) in a point y0 in the interval [-Len/2,Len/2] on
y axis, and in a point R0 on an axis orthogonal to y, is:
20
20
0
20
20
0
0
00
)y2Len(R4
y2Len
)y2Len(R4
y2Len
R4
I)R,y(B
108
Transmission line trasformers /8
If c << Len and ferrite is not used, L is obtained by integration of B (and by
considering that the integral is equal to LI) in an unlimited volume with
length Len and radius comprised in the interval [c , ):
If length Len is partially surrounded by ferrite, the integral in this region has
to consider the complex permeability (’ + j ”) of the ferrite itself:
1
c
Len2lnLen
2L
ferriteferrite L'j'
''LLjZ
Integral in ferriteIntegral in air
The real part
causes resistive
loss: heat
dissipation issue
Also ferrites
are placed on
dissipator
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109
Transmission line trasformers /9
Circuital implementation in CAD Keysight ADS: Parasitic
impedance due
to magnetic loss
Parasitic inductance
of internal conductor
portion uncovered
for soldering
110
Transmission line trasformers /10
Push-pull amplifier based on 1:4 transmission line transformers:
VbiasIN VbiasOUT
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111
Transmission line trasformers /11
Photo of a 20-500MHz 100W Push-pull amplifier:
A lossy input matching network (not shown) is put at the input side in broadband
amplifiers for gain slope adjust
A local parallel-parallel
feedback on each transistor
is often used