Berkeley Power Gain and Stability Prof. Ali M. Niknejad U.C. Berkeley Copyright c 2014 by Ali M. Niknejad September 17, 2014 Niknejad Power Gain
Berkeley
Power Gain and Stability
Prof. Ali M. Niknejad
U.C. BerkeleyCopyright c© 2014 by Ali M. Niknejad
September 17, 2014
Niknejad Power Gain
Power Gain
Niknejad Power Gain
Power Gain
+vs
!
YS
YL
!y11 y12
y21 y22
"
Pin PL
Pav,s Pav,l
We can define power gain in many different ways. The powergain Gp is defined as follows
Gp =PL
Pin= f (YL,Yij) 6= f (YS)
We note that this power gain is a function of the loadadmittance YL and the two-port parameters Yij .
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Power Gain (cont)
The available power gain is defined as follows
Ga =Pav ,L
Pav ,S= f (YS ,Yij) 6= f (YL)
The available power from the two-port is denoted Pav ,L
whereas the power available from the source is Pav ,S .
Finally, the transducer gain is defined by
GT =PL
Pav ,S= f (YL,YS ,Yij)
This is a measure of the efficacy of the two-port as itcompares the power at the load to a simple conjugate match.
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Bi-Conjugate Match
When the input and output are simultaneously conjugatelymatched, or a bi-conjugate match has been established, wefind that the transducer gain is maximized with respect to thesource and load impedance
GT ,max = Gp,max = Ga,max
This is thus the recipe for calculating the optimal source andload impedance in to maximize gain
Yin = Y11 −Y12Y21
YL + Y22= Y ∗
S
Yout = Y22 −Y12Y21
YS + Y11= Y ∗
L
Solution of the above four equations (real/imag) results in theoptimal YS ,opt and YL,opt .
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Calculation of Optimal Source/Load
Another approach is to simply equate the partial derivatives ofGT with respect to the source/load admittance to find themaximum point
∂GT
∂GS= 0;
∂GT
∂BS= 0
∂GT
∂GL= 0;
∂GT
∂BL= 0
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Optimal Power Gain Derivation (cont)
Again we have four equations. But we should be smarterabout this and recall that the maximum gains are all equal.Since Ga and Gp are only a function of the source or load, wecan get away with only solving two equations. For instance
∂Ga
∂GS= 0;
∂Ga
∂BS= 0
This yields YS,opt and by setting YL = Y ∗out we can find the
YL,opt .
Likewise we can also solve
∂Gp
∂GL= 0;
∂Gp
∂BL= 0
And now use YS ,opt = Y ∗in.
Niknejad Power Gain
Optimal Power Gain Derivation
Let’s outline the procedure for the optimal power gain. We’lluse the power gain Gp and take partials with respect to theload. Let
Yjk = mjk + jnjk
YL = GL + jXL
Y12Y21 = P + jQ = Le jφ
Gp =|Y21|2
|YL + Y22|2<(YL)
<(Yin)=|Y21|2
DGL
<(
Y11 −Y12Y21
YL + Y22
)= m11 −
<(Y12Y21(YL + Y22)∗)
|YL + Y22|2
D = m11|YL + Y22|2 − P(GL + m22)− Q(BL + n22)
∂Gp
∂BL= 0 = −|Y21|2GL
D2
∂D
∂BL
Niknejad Power Gain
Optimal Load (cont)
Solving the above equation we arrive at the following solution
BL,opt =Q
2m11− n22
In a similar fashion, solving for the optimal load conductance
GL,opt =1
2m11
√(2m11m22 − P)2 − L2
If we substitute these values into the equation for Gp (lot’s ofalgebra ...), we arrive at
Gp,max =|Y21|2
2m11m22 − P +√
(2m11m22 − P)2 − L2
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Final Solution
Notice that for the solution to exists, GL must be a realnumber. In other words
(2m11m22 − P)2 > L2
(2m11m22 − P) > L
K =2m11m22 − P
L> 1
This factor K plays an important role as we shall show that italso corresponds to an unconditionally stable two-port. Wecan recast all of the work up to here in terms of K
YS ,opt =Y12Y21 + |Y12Y21|(K +
√K 2 − 1)
2<(Y22)
YL,opt =Y12Y21 + |Y12Y21|(K +
√K 2 − 1)
2<(Y11)
Gp,max = GT ,max = Ga,max =Y21
Y12
1
K +√
K 2 − 1
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Maximum Gain
The maximum gain is usually written in the followinginsightful form
Gmax =Y21
Y12(K −
√K 2 − 1)
For a reciprocal network, such as a passive element, Y12 = Y21
and thus the maximum gain is given by the second factor
Gr ,max = K −√
K 2 − 1
Since K > 1, |Gr ,max | < 1. The reciprocal gain factor isknown as the efficiency of the reciprocal network.
The first factor, on the other hand, is a measure of thenon-reciprocity.
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Unilateral Maximum Gain
For a unilateral network, the design for maximum gain istrivial. For a bi-conjugate match
YS = Y ∗11
YL = Y ∗22
GT ,max =|Y21|2
4m11m22
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Ideal MOSFET
roCgs gmvin
+vin
−Cds
The AC equivalent circuit for a MOSFET at low to moderatefrequencies is shown above. Since |S11| = 1, this circuit hasinfinite power gain. This is a trivial fact since the gatecapacitance cannot dissipate power whereas the output candeliver real power to the load.
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Real MOSFET
Cgs gmvin
+vin
−
RiRi
CdsRds Rds
+vs
−
A more realistic equivalent circuit is shown above. If we makethe unilateral assumption, then the input and output powercan be easily calculated. Assume we conjugate match theinput/output
Pavs =|VS |28Ri
PL = <( 12 ILV ∗
L ) = 12
∣∣∣∣gmV1
2
∣∣∣∣2
Rds
GTU,max = g 2mRdsRi
∣∣∣∣V1
VS
∣∣∣∣2
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Real MOSFET (cont)
At the center resonant frequency, the voltage at the input ofthe FET is given by
V1 =1
jωCgs
VS
2Ri
GTU,max =Rds
Ri
(gm/Cgs)2
4ω2
This can be written in terms of the device unity gainfrequency fT
GTU,max =1
4
Rds
Ri
(fTf
)2
The above expression is very insightful. To maximum powergain we should maximize the device fT and minimize theinput resistance while maximizing the output resistance.
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Stability of a Two-Port
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Stability of a Two-Port
A two-port is unstable if the admittance of either port has anegative conductance for a passive termination on the secondport. Under such a condtion, the two-port can oscillate.Consider the input admittance
Yin = Gin + jBin = Y11 −Y12Y21
Y22 + YL
Using the following definitions
Y11 = g11 + jb11
Y22 = g22 + jb22
Y12Y21 = P + jQ = L∠φ
YL = GL + jBLNow substitute real/imag parts of the above quantities intoYin
Yin = g11 + jb11 −P + jQ
g22 + jb22 + GL + jBL
= g11 + jb11 −(P + jQ)(g22 + GL − j(b22 + BL))
(g22 + GL)2 + (b22 + BL)2
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Input Conductance
Taking the real part, we have the input conductance
<(Yin) = Gin = g11 −P(g22 + GL) + Q(b22 + BL)
(g22 + GL)2 + (b22 + BL)2
=(g22 + GL)2 + (b22 + BL)2 − P
g11(g22 + GL)− Q
g11(b22 + BL)
D
Since D > 0 if g11 > 0, we can focus on the numerator. Notethat g11 > 0 is a requirement since otherwise oscillationswould occur for a short circuit at port 2.
The numerator can be factored into several positive terms
N = (g22 +GL)2 +(b22 +BL)2− P
g11(g22 +GL)− Q
g11(b22 +BL)
=
(GL +
(g22 −
P
2g11
))2
+
(BL +
(b22 −
Q
2g11
))2
−P2 + Q2
4g 211
Niknejad Power Gain
Input Conductance (cont)
Now note that the numerator can go negative only if the firsttwo terms are smaller than the last term. To minimize thefirst two terms, choose GL = 0 and BL = −
(b22 − Q
2g11
)
(reactive load)
Nmin =
(g22 −
P
2g11
)2
− P2 + Q2
4g 211
And thus the above must remain positive, Nmin > 0, so
(g22 +
P
2g11
)2
− P2 + Q2
4g 211
> 0
g11g22 >P + L
2=
L
2(1 + cosφ)
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Linvill/Llewellyn Stability Factors
Using the above equation, we define the Linvill stability factor
L < 2g11g22 − P
C =L
2g11g22 − P< 1
The two-port is stable if 0 < C < 1.
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Stability (cont)
It’s more common to use the inverse of C as the stabilitymeasure
2g11g22 − P
L> 1
The above definition of stability is perhaps the most common
K =2<(Y11)<(Y22)−<(Y12Y21)
|Y12Y21|> 1
The above expression is identical if we interchange ports 1/2.Thus it’s the general condition for stability.
Note that K > 1 is the same condition for the maximumstable gain derived earlier. The connection is now moreobvious. If K < 1, then the maximum gain is infinity!
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Stability From Another Perspective
We can also derive stability in terms of the input reflectioncoefficient. For a general two-port with load ΓL we have
v−2 = Γ−1
L v +2 = S21v +
1 + S22v +2
v +2 =
S21
Γ−1L − S22
v−1
v−1 =
(S11 +
S12S21ΓL
1− ΓLS22
)v +
1
Γ = S11 +S12S21ΓL
1− ΓLS22
If |Γ| < 1 for all ΓL, then the two-port is stable
Γ =S11(1− S22ΓL) + S12S21ΓL
1− S22ΓL=
S11 + ΓL(S21S12 − S11S22)
1− S22ΓL
=S11 −∆ΓL
1− S22ΓL
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Stability Circle
To find the boundary between stability/instability, let’s set|Γ| = 1 ∣∣∣∣
S11 −∆ΓL
1− S22ΓL
∣∣∣∣ = 1
|S11 −∆ΓL| = |1− S22ΓL|After some algebraic manipulations, we arrive at the followingequation
∣∣∣∣ΓL −S∗
22 −∆∗S11
|S22|2 − |∆|2∣∣∣∣ =
|S12S21||S22|2 − |∆|2
This is of course an equation of a circle, |ΓL − C | = R, in thecomplex plane with center at C and radius R
Thus a circle on the Smith Chart divides the region ofinstability from stability.
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Example: Stability Circle
CS
RS
|S11| < 1
stable reg
ion
unstable reg
ion
In this example, the originof the circle lies outsidethe stability circle but aportion of the circle fallsinside the unit circle. Isthe region of stabilityinside the circle oroutside?
This is easily determined ifwe note that if ΓL = 0,then Γ = S11. So ifS11 < 1, the origin shouldbe in the stable region.Otherwise, if S11 > 1, theorigin should be in theunstable region.
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Stability: Unilateral Case
Consider the stability circle for a unilateral two-port
CS =S∗
11 − (S∗11S∗
22)S22
|S11|2 − |S11S22|2=
S∗11
|S11|2
RS = 0 |CS | =1
|S11|The cetner of the circle lies outside of the unit circle if|S11| < 1. The same is true of the load stability circle. Sincethe radius is zero, stability is only determined by the locationof the center.
If S12 = 0, then the two-port is unconditionally stable ifS11 < 1 and S22 < 1.
This result is trivial since
ΓS |S12=0 = S11
The stability of the source depends only on the device and noton the load.
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Mu Stability Test
If we want to determine if a two-port is unconditionally stable,then we should use the µ test
µ =1− |S11|2
|S22 −∆S∗11|+ |S12S21|
> 1
The µ test not only is a test for unconditional stability, butthe magnitude of µ is a measure of the stability. In otherwords, if one two port has a larger µ, it is more stable.
The advantage of the µ test is that only a single parameterneeds to be evaluated. There are no auxiliary conditions likethe K test derivation earlier.
The derivation of the µ test proceeds as follows. First letΓS = |ρs |e jφ and evaluate Γout
Γout =S22 −∆|ρs |e jφ1− S11|ρs |e jφ
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Mu Test (cont)
Next we can manipulate this equation into the following circle|Γout − C | = R
∣∣∣∣Γout +|ρs |S∗
11∆− S22
1− |ρs ||S11|2∣∣∣∣ =
√|ρs ||S12S21|
(1− |ρs ||S11|2)
For a two-port to be unconditionally stable, we’d like Γout tofall within the unit circle
||C |+ R| < 1
||ρs |S∗11∆− S22|+
√|ρs ||S21S12| < 1− |ρs ||S11|2
||ρs |S∗11∆− S22|+
√|ρs ||S21S12|+ |ρs ||S11|2 < 1
The worse case stability occurs when |ρs | = 1 since itmaximizes the left-hand side of the equation. Therefore wehave
µ =1− |S11|2
|S∗11∆− S22|+ |S12S21|
> 1
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K-∆ Test
The K stability test has already been derived using Yparameters. We can also do a derivation based on Sparameters. This form of the equation has been attributed toRollett and Kurokawa.
The idea is very simple and similar to the µ test. We simplyrequire that all points in the instability region fall outside ofthe unit circle.
The stability circle will intersect with the unit circle if
|CL| − RL > 1
or|S∗
22 −∆∗S11| − |S12S21||S22|2 − |∆|2
> 1
This can be recast into the following form (assuming |∆| < 1)
K =1− |S11|2 − |S22|2 + |∆|2
2|S12||S21|> 1
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Two-Port Power and Scattering Parameters
The power flowing into a two-port can be represented by
Pin =|V +
1 |22Z0
(1− |Γin|2)
The power flowing to the load is likewise given by
PL =|V −
2 |22Z0
(1− |ΓL|2)
We can solve for V +1 using circuit theory
V +1 + V −
1 = V +1 (1 + Γin) =
Zin
Zin + ZSVS
In terms of the input and source reflection coefficient
Zin =1 + Γin
1− ΓinZ0 ZS =
1 + ΓS
1− ΓSZ0
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Two-Port Incident Wave
Solve for V +1
V +1 (1 + Γin) =
VS(1 + Γin)(1− ΓS)
(1 + Γin)(1− ΓS) + (1 + ΓS)(1− Γin)
V +1 =
VS
2
1− ΓS
1− ΓinΓS
The voltage incident on the load is given by
V −2 = S21V +
1 + S22V +2 = S21V +
1 + S22ΓLV −2
V −2 =
S21V +1
1− S22ΓL
PL =|S21|2
∣∣V +1
∣∣2
|1− S22ΓL|21− |ΓL|2
2Z0
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Operating Gain and Available Power
The operating power gain can be written in terms of thetwo-port s-parameters and the load reflection coefficient
Gp =PL
Pin=
|S21|2 (1− |ΓL|2)
|1− S22ΓL|2 (1− |Γin|2)
The available power can be similarly derived from V +1
Pavs = Pin|Γin=Γ∗S
=
∣∣V +1a
∣∣2
2Z0(1− |Γ∗
S |2)
V +1a = V +
1
∣∣Γin=Γ∗
S=
VS
2
1− Γ∗S
1− |ΓS |2
Pavs =|VS |28Z0
|1− ΓS |2
1− |ΓS |2
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Transducer Gain
The transducer gain can be easily derived
GT =PL
Pavs=|S21|2 (1− |ΓL|2)(1− |ΓS |2)
|1− ΓinΓS |2 |1− S22ΓL|2
Note that as expected, GT is a function of the two-ports-parameters and the load and source impedance.
If the two port is connected to a source and load withimpedance Z0, then we have ΓL = ΓS = 0 and
GT = |S21|2
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Unilateral Gain
+vs
−
Z0
M1 M2
GS GL|S21|2
!S11 0S21 S22
"Z0
If S12 ≈ 0, we can simplify the expression by just assumingS12 = 0. This is the unilateral assumption
GTU =1− |ΓS |2
|1− S11ΓS |2|S21|2
1− |ΓL|2
|1− S22ΓL|2= GS |S21|2 GL
The gain partitions into three terms, which can be interpretedas the gain from the source matching network, the gain of thetwo port, and the gain of the load.
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Maximum Unilateral Gain
We know that the maximum gain occurs for the biconjugatematch
ΓS = S∗11
ΓL = S∗22
GS ,max =1
1− |S11|2
GL,max =1
1− |S22|2
GTU,max =|S21|2
(1− |S11|2)(1− |S22|2)
Note that if |S11| = 1 of |S22| = 1, the maximum gain isinfinity. This is the unstable case since |Sii | > 1 is potentiallyunstable.
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Design for Gain
So far we have only discussed power gain using bi-conjugatematching. This is possible when the device is unconditionallystable. In many case, though, we’d like to design with apotentially unstable device.
Moreover, we would like to introduce more flexibility in thedesign. We can trade off gain for
bandwidthnoisegain flatnesslinearityetc.
We can make this tradeoff by identifying a range ofsource/load impedances that can realize a given value ofpower gain. While maximum gain is acheived for a singlepoint on the Smith Chart, we will find that a lot moreflexibility if we back-off from the peak gain.
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Unilateral Design
No real transistor is unilateral. But most are predominantlyunilateral, or else we use cascades of devices (such as thecascode) to realize such a device.
The unilateral figure of merit can be used to test the validityof the unilateral assumption
Um =|S12|2 |S21|2 |S11|2 |S22|2
(1− |S11|2)(1− |S22|2)
It can be shown that the transducer gain satisfies thefollowing inequality
1
(1 + U)2<
GT
GTU<
1
(1− U)2
Where the actual power gain GT is compared to the powergain under the unilateral assumption GTU . If the inequality istight, say on the order of 0.1 dB, then the amplifier can beassumed to be unilateral with negligible error.
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Gain Circles
We now can plot gain circles for the source and load. Let
gS =GS
GS ,max
gL =GL
GL,max
By definition, 0 ≤ gS ≤ 1 and 0 ≤ gL ≤ 1. One can show thata fixed value of gS represents a circle on the ΓS plane
∣∣∣∣ΓS −S∗
11gS
|S11|2 (gS − 1) + 1
∣∣∣∣ =
∣∣∣∣∣
√1− gS(1− |S11|2)
|S11|2 (gS − 1) + 1
∣∣∣∣∣
More simply, |ΓS − CS | = RS . A similar equation can bederived for the load. Note that for gS = 1, RS = 0, andCS = S∗
11 corresponding to the maximum gain.
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Gain Circles (cont)
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AN
GLE O
F TRAN
SMISSIO
N C
OEFFIC
IENT IN
DEG
REES
AN
GLE O
F REF LECTIO
N C
OEFFIC
IENT IN
DEG
REES
—>
WA
VELE
NG
THS
TOW
ARD
GEN
ERA
TOR
—>
<— W
AVE
LEN
GTH
S TO
WA
RD L
OA
D <
—
IND
UCT
IVE
REAC
TANCE
COM
PONEN
T (+jX
/Zo), O
R CAPACITIVE SUSCEPTANCE (+jB/Yo)
CAPACITIVE REACTANCE COMPONENT (-jX/Z
o), OR IN
DUCTIVE
SUSC
EPTA
NCE
(-jB
/Yo)
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
S∗11
gS = −1 dB
gS = 0dB
gS = −2 dB
gS = −3 dB
All gain circles lie on the line given by the angle of S∗ii . We
can select any desired value of source/load reflectioncoefficient to acheive the desired gain. To minimize theimpedance mismatch, and thus maximize the bandwidth, weshould select a point closest to the origin.
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Extended Smith Chart
For |Γ| > 1, we can still employ the Smith Chart if we makethe following mapping. The reflection coefficient for anegative resistance is given by
Γ(−R + jX ) =−R + jX − Z0
−R + jX + Z0=
(R + Z0)− jX
(R − Z0)− jX
1
Γ∗ =(R − Z0) + jX
(R + Z0) + jX
We see that Γ can be mapped to the unit circle by taking1/Γ∗ and reading the resistance value (and noting that it’sactually negative).
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Potentially Unstable Unilateral Amplifier
For a unilateral two-port with |S11| > 1, we note that theinput impedance has a negative real part. Thus we can stilldesign a stable amplifier as long as the source resistance islarger than <(Zin)
<(ZS) > |<(Zin)|
The same is true of the load impedance if |S22| > 1. Thus thedesign procedure is identical to before as long as we avoidsource or load reflection coefficients with real part less thanthe critical value.
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Pot. Unstable Unilateral Amp Example
Consider a transistor with the following S-Parameters
S11 = 2.02∠− 130.4◦
S22 = 0.50∠− 70◦
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1.0
1.0
1.0
1.2
1.2
1.2
1.4
1.4
1.4
1.6
1.6
1.6
1.8
1.8
1.8
2.0
2.0
2.0
3.0
3.0
3.0
4.0
4.0
4.0
5.0
5.0
5.0
10
10
10
20
20
20
50
50
50
0.2
0.2
0.2
0.2
0.4
0.4
0.4
0.4
0.6
0.6
0.6
0.6
0.8
0.8
0.8
0.8
1.0
1.0
1.0
1.0
90-9
085
-85
80-8
0
75-7
5
70-7
0
65-6
5
60-6
0
55-5
5
50-5
0
45
-45
40
-40
35
-35
30
-30
25
-25
20
-20
15
-15
10
-10
0.04
0.04
0.05
0.05
0.06
0.06
0.07
0.07
0.08
0.08
0.09
0.09
0.1
0.1
0.11
0.11
0.12
0.12
0.13
0.13
0.14
0.14
0.15
0.15
0.16
0.16
0.17
0.17
0.18
0.18
0.190.19
0.20.2
0.21
0.210.22
0.220.23
0.230.24
0.24
0.25
0.25
0.26
0.26
0.27
0.27
0.28
0.28
0.29
0.29
0.3
0.3
0.31
0.31
0.32
0.32
0.33
0.33
0.34
0.34
0.35
0.35
0.36
0.36
0.37
0.37
0.38
0.38
0.39
0.39
0.4
0.4
0.41
0.41
0.42
0.42
0.43
0.43
0.440.44
0.45
0.45
0.46
0.46
0.47
0.47
0.48
0.48
0.49
0.49
0.0
0.0
AN
GLE O
F TRAN
SMISSIO
N C
OEFFIC
IENT IN
DEG
REES
AN
GLE O
F REFLECTIO
N C
OEFFIC
IENT IN
DEG
REES
—>
WA
VELE
NG
THS
TOW
ARD
GEN
ERAT
OR
—>
<— W
AVE
LEN
GTH
S TO
WA
RD L
OA
D <
—
IND
UCT
IVE
REAC
TANCE
COM
PONEN
T (+jX
/Zo), O
R CAPACITIVE SUSCEPTANCE (+jB/Yo)
CAPACITIVE REACTANCE COMPONENT (-jX/Z
o), OR IN
DUCTIVE
SUSC
EPTA
NCE
(-jB
/Yo)
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
CS
RS
stable region
1
S∗11
GS = 5dB
ΓS
S12 = 0
S21 = 5.00∠60◦
Since |S11| > 1, theamplifier is potentiallyunstable. We begin byplotting 1/S∗
11 to find thenegative real inputresistance.
Now any source inside thiscircle is stable, since<(ZS) > <(Zin).
We also draw the sourcegain circle for GS = 5dB.
Niknejad Power Gain
Amp Example (cont)
The input impedance is read off the Smith Chart from 1/S∗11.
Note the real part is interpreted as negative
Zin = 50(−0.4− 0.4j)
The GS = 5dB gain circle is calculated as follows
gS = 3.15(1− |S11|2)
RS =
√1− gS(1− |S11|2)
1− |S11|2 (1− gS)= 0.236
CS =gSS∗
11
1− |S11|2 (1− gS)= −.3 + 0.35j
We can select any point on this circle and obtain a stable gainof 5 dB. In particular, we can pick a point near the origin (tomaximize the BW) but with as large of a real impedance aspossible:
ZS = 50(0.75 + 0.4j)
Niknejad Power Gain
Bilateral Amp Design
In the bilateral case, we will work with the power gain Gp.The transducer gain is not used since the source impedance isa function of the load impedaance. Gp, on the other hand, isonly a function of the load.
Gp =|S21|2 (1− |ΓL|2)(
1−∣∣∣S11−∆ΓL
1−S22ΓL
∣∣∣2)|1− S22ΓL|2
= |S21|2 gp
It can be shown that gp is a circle on the ΓL plane. The radiusand center are given by
RL =
√1− 2K |S12S21|gp + |S12S21|2g 2
p∣∣∣−1− |S22|2 gp + |∆|2 gp
∣∣∣2
CL =gp(S∗
22 −∆∗S11)
1 + gp(|S22|2 − |∆|2)
Niknejad Power Gain
Bilateral Amp (cont)
We can also use this formula to find the maximum gain. Weknow that this occurs when RL = 0, or
1− 2K |S12S21|gp,max + |S12S21|2g 2p,max = 0
gp,max =1
|S12S21|(
K −√
K 2 − 1)
Gp,max =
∣∣∣∣S21
S12
∣∣∣∣(
K −√
K 2 − 1)
The design procedure is as follows1 Specify gp
2 Draw operating gain circle.3 Draw load stability circle. Select ΓL that is in the stable region
and not too close to the stability circle.4 Draw source stability circle.5 To maximize gain, calculate Γin and check to see if ΓS = Γ∗
in isin the stable region. If not, iterate on ΓL or compromise.
Niknejad Power Gain