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POWER AMPLIFIER DESIGN
forHigh-Frequency Circuit Design Elective
byMichael Tse
September 2003
M. Tse: Power Amplifier Design 2
ContentsScattering Parameters
Relationship with voltage and currentMeanings of s-parametersIntuitive Stability Issues
Gain Definitions of Power AmplifiersTransducer Power Gain of Two-Port Circuits
Using s-parametersSignal Flow Graph ReductionMatching for Maximum Transducer Power Gain
Stability of AmplifierConditions for StabilityConditional and Unconditional StabilityDetermining Stability RegionsRollett CriteriaStabilizing Amplifiers by Neutralization via y-parameters
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M. Tse: Power Amplifier Design 3
Scattering ParametersWhen a wave arrives at a circuit, its energy is being “scattered” and partitioned intomany possible outgoing waves. Scattering parameters of a circuit describe how a setof incoming waves is scattered.
a8
a7
a3
a1 a2
a6 a5
a4
b3
b4
b5b6
b7
b8
b2b1
an: incident wave at port nbn: reflected wave at port n
B = SA
[S]
†
b1
b2
M
bn
È
Î
Í Í Í Í
˘
˚
˙ ˙ ˙ ˙
a1
a2
M
an
È
Î
Í Í Í Í
˘
˚
˙ ˙ ˙ ˙
S is called scattering matrix
M. Tse: Power Amplifier Design 4
Two-port Scattering Parameters
[S]a1 a2
b1 b2
†
b1
b2
È
Î Í
˘
˚ ˙ =
s11 s12
s21 s22
È
Î Í
˘
˚ ˙
a1
a2
È
Î Í
˘
˚ ˙
†
b1 = s11a1 + s12a2
b2 = s21a1 + s22a2or
s11, s12, s21 and s22 are called scattering parameters. They completely characterizesthe two-port circuit.
Suppose a1 is the incident wave at port 1, b1 is the reflected wave at port 1; b2 is the incident wave at port 2, b2 is the reflected wave at port 2.
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M. Tse: Power Amplifier Design 5
Relationship with voltage and current
†
a1 =V1 + ZoI1
2 2Zo
b1 =V1 - ZoI1
2 2Zo
a2 =V2 + ZoI2
2 2Zo
b2 =V2 - ZoI2
2 2Zo
Note: √Zo is the normalizationfactor such that aa* and bb* givethe incident and reflected powers.
Special case: a one-port scattering parameter is simply the reflection coefficient!
†
s = G =z -1z +1
where z =ZL
Zo
Similarly, for n-port circuits,
[S] = [Z – 1][Z + 1]–1
unit matrix
M. Tse: Power Amplifier Design 6
Meanings of s-parameters
Suppose port 2 is matched with a resistor. So, there is no reflection back into thecircuit, i.e., a2 = 0.
[S]a1
b1
b2
In this case, the reflected wave at port 1 is s11a1, and the wave transmitted tothe load resistor is s21a1.
So, basically, s11 and s21 tell how power is split between the possible outputports (here port 1 and port 2 are possible output ports).
If this circuit is a “good” amplifier, we want a small s11 and a large s21.
Also, if the circuit is lossless and gainless, input power must equal outputpower, so |s11|2 + |s21|2 = 1.
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M. Tse: Power Amplifier Design 7
Intuitive Stability IssuesLet’s look at a one-port circuit. There is only one s-parameter whichis s11.
If this s11 is large (say >1), then it actually reflects more power or ithas a reflection gain. Stability can be a problem.
If this one-port is connected to a load (say a transmission line) ofreflection coefficient g1 and if |s11g1| > 1, the circuit will oscillate.This is just the same Barkhausen criterion for oscillation, which saysthat if the round trip gain is one and phase shift is 2π, oscillationoccurs.
Let’s look at a two-port circuit. A large s11 or s22 can causeinstability for some source impedance or load impedance.
Normally, an amplifier has large s21 (of course). If it has small s12and a mismatched input or output, the circuit may oscillate if theround trip gain exceeds one. Typically, if the input reflectioncoefficient is g1 and output reflection coefficient is g2, theninstability occurs if |s21s12g1g2| > 1.
s11G = g1
round-trip gain = |s11g1|
[S]G = g1 G = g2
round-trip gain = | s21s12g1g2 |
MORE DETAILS LATER
M. Tse: Power Amplifier Design 8
Gain Definitions in Power Amplifiers
Power Gain =
Available Gain =
Exchangeable Gain =
Insertion Gain =
Transducer Power Gain =
†
power dissipated in load ZL
power delivered to amplifier
†
amplifier output poweravailable power from generator source
†
output exchangeable powerinput exchangeable power
†
exchangeable power of source = V 2
4Re ZG[ ]where
†
output powerpower dissipated in load if the amplifier were absent
†
power delivered to loadavailable power from source
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M. Tse: Power Amplifier Design 9
Transducer Power Gain of Two-Port Circuit
Using z-parameters
[Z] V2
+
–V1
+
–
I1 I2
ZL
†
Zin = z11 -z12z21
z22 + ZL
ZG
Power delivered to load =
†
12
I22Re ZL[ ]Power available from source =
†
VG2
8Re ZG[ ]
VG
†
GT =4Re[ZL ]Re[ZG ]z21
2
ZG + z11( ) ZL + z22( ) - z21z122Transducer Power Gain
M. Tse: Power Amplifier Design 10
Using s-parameters
[S] ZL
ZG
VGa1
b1
a2
b2
Suppose the circuit is matched, i.e., GG = Gi* The available power Pa is
GG Gi
†
Pa =
12
bG2
1- GG2
bG
Power delivered to load PL is
†
PL =12
b22 1- GL
2( )
Transducer power gain is
†
GT =b2
2
bG2 1- GL
2( ) 1- GG2( )
Problem: can we express |b2/bG| in terms of s-parameters?
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M. Tse: Power Amplifier Design 11
Signal Flow Graph (SFG) Reduction
• •• X Y XY=Rule 1:
Rule 2:
Rule 3:
Rule 4:
• •X
Y
••
••
=X + Y
• •• XYZ • •• Y
†
X1-Y
=
• •• X Z
•
•
Y
W
• •• X Z
•
•
Y
W
X •
=We can use these rulesto find |b2/bG|.
M. Tse: Power Amplifier Design 12
SFG for the s-parameter circuit
• •
• •
•bG= a1
GG
s12
s22s11
s211
GL
a1 b2
a2b1
b2 • •
•• s12
†
bG
1- GGs11
†
s21
1- GLs22
†
s21
1- GLs22
†
GG
1- GGs11
b2
=
From the reduced SFG, we get
†
b2 =
bG
1- GGs11
1-s21s12GLGG
1- GLs22( ) 1- GGs11( )
s21
1- GLs22
b2
bG
=s21
1- GLs22( ) 1- GGs11( ) - s21s12GLGG
†
fi
Transducer power gain is
†
GT =s21
2 1- GG2( ) 1- GL
2( )1- GLs22( ) 1- GGs11( ) - s21s12GLGG
2
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M. Tse: Power Amplifier Design 13
Extensions:
Unilateral power gain
†
Gu = GT s12 = 0
†
=s21
2 1- GG2( ) 1- GL
2( )1- GLs22( ) 1- GGs11( )
2
NOTE:s12 = 0 means that there is no internal feedback within the two-port. This isimpossible to achieve at Ghz range. For practical transistors, s12 is verysmall but never 0. The above does not really exist in practice.
Maximum unilateral power gain occurs when s12 = 0, GG = s11* and GL = s22
*.
†
Gu,max =s21
2
1- s112( ) 1- s22
2( )
M. Tse: Power Amplifier Design 14
Matching for Maximum Power Gain
2-port
GG Gi Go GL
ZLZG
Maximum power gain occurs when both terminals are matched.
The conditions are
†
Gi = GG* = s11 +
s21s12GL
1- GLs22
Go = GL* = s22 +
s21s12GG
1- GGs11and
We want to find GG and GL that satisfy the above conditions.
(a)
(b)
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M. Tse: Power Amplifier Design 15
Putting (b) in (a), we get
†
GG* =
s11(1- GG*s11
* ) - D(s22* - GG
*D*)1- GG
*s11* - s22
*+ s22D
*GG*
†
GG2(s22
* D - s11) + GG (1- s222
+ s112
- D2) - s11
* + D*s22 = 0
†
fi
where
and D is the determinant of [S].
To find GG, we need to solve the above quadratic equation. The answer is
†
GG =C1
*
2C12 B1 ± B1
2 - 4 C12È
Î Í ˘ ˚ ˙
†
-GG2C1 + GGB1 - C1
* = 0
†
fi
†
C1 = s11 - Ds22*
B1 =1+ s112
- s222
- D2
M. Tse: Power Amplifier Design 16
Similarly, we get GL as
†
GL =C2
*
2C22 B2 ± B2
2 - 4 C22È
Î Í ˘ ˚ ˙
where
†
C2 = s22 - DS11*
B2 =1+ s222
- s112
- D2
To choose the appropriate sign in the above formulae, we simply checkwhether B1 and B2 are positive or negative. If Bi > 0, we take the – sign.Otherwise, we take the + sign.
Finally, the matching impedances can be found using
†
ZG = Z01+ GG
1- GG
ZL = Z01+ GL
1- GL
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M. Tse: Power Amplifier Design 17
Stability of Amplifier
Problem: unwanted oscillation may occur if there is a feedback loop.
Amplifier
feedback
• External circuit• Feedback parasitic elements• Internal feedback path, e.g., Cµ
M. Tse: Power Amplifier Design 18
Conditions for Stability
Requirement:
†
Gi £1 and Go £1
Gi Go
2-portHence, the stability boundary isrepresented by circles correspondingto
†
Gi =1 and Go =1
Smith chart
†
Gi =1
†
Go =1
†
GL = 0
(un)stable(un)stable
(un)stable (un)stable
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M. Tse: Power Amplifier Design 19
The condition is equivalent to
†
Gi =1
†
s11 - DGL
1- GLs22
=1
†
(1- GLs22)(1- GL*s22
* ) = (s11 - DGL )(s11* - D*GL
*)
Mathematical manipulation gives
†
GL +D*s11 - s22
*
s222
- D2
Ê
Ë Á Á
ˆ
¯ ˜ ˜ GL
* +Ds11
* - s22
s222
- D2
Ê
Ë Á Á
ˆ
¯ ˜ ˜ =
s12s122
s222
- D2( )
2
which is a circle equation with
†
cL =s11D
* - s22*
D2
- s222centre and radius
†
rL =s12s12
s222
- D2( )
M. Tse: Power Amplifier Design 20
Similar, for the input side, we have the stability circle’s centre and radiusgiven by
†
cG =s22D
* - s11*
D2
- s112
†
rG =s21s12
D2
- s112and
Smith chart
†
Gi =1
†
Go =1
†
GL = 0
(un)stable(un)stable
(un)stable (un)stable
Question: Which side is stable?
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M. Tse: Power Amplifier Design 21
Determining Stability Regions
Simple trick:
We know that when the load (or input)side is matched, we have GL = 0 (or GG= 0) which is the centre of the Smithchart.
But, when the load is matched, we haveGi = s11. So, if | s11 | < 1, then we knowthat the centre of the Smith chart isstable, and so is the whole region thatcovers the centre.
Similar reasoning applies for the inputside.
Zo
GL = 0
Smith chart
load stability circle
unstable if |s11| < 1
stable if |s11| < 1
M. Tse: Power Amplifier Design 22
Unconditional Stability
Smith chart
load stability circle
unstable if |s11| < 1
stable if |s11| < 1
input stability circle
unstable if |s11| < 1 |Gi| = 1
|Go| = 1
The amplifier is unconditionally stable if the stability circles are outside theunit circle (Smith chart boundary) and |s11|<1 and |s22|<1.
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M. Tse: Power Amplifier Design 23
Conditional Stability
Smith chart
load stability circlestable if |s11| < 1
input stability circle
unstable if |s11| < 1|Gi| = 1
|Go| = 1
The amplifier is conditionally stable if the stability circles overlap with theunit circle (Smith chart boundary) and |s11|<1 and |s22|<1. We should onlyoperate the amplifier with terminal impedances located further away from thestability circles.
unstable if |s11| < 1
M. Tse: Power Amplifier Design 24
Stability Criteria in Algebraic Form
It can be shown mathematically that the unconditional stability criteriacan be translated to the following equivalent form:
†
k =1- s11
2- s22
2+ D
2
2 s12s21
≥1
D £1and
The above is called the Rollett criteria, and k is called the stability factor,sometimes called the k-factor.
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M. Tse: Power Amplifier Design 25
Stabilizing a Transistor by Neutralization
A transmitter can be inherently unstable due to internal feedback of Cµ.Suppose a transistor has the following scattering parameters:
†
[S] =0.73 -102˚ 0.10 48˚2.21 104˚ 0.47 – 48˚
È
Î Í
˘
˚ ˙
Hence, we can find the stability factor and determinant as
k = 0.752 and |D| = 0.294.
So, Rollett criteria are not satisfied, and the amplifier is NOTunconditionally stable! That means, the amplifier can be unstable forcertain load and input impedances.
How can we make it unconditionally stable?
M. Tse: Power Amplifier Design 26
One way to solve this problem is to use a shunt feedback to neutralize theinternal feedback. This method is okay up to VHF range and is valid fornarrowband only.
Amplifier[Ya]
external feedback circuit [Yf]
YG YL
Since the feedback is shunt-shunt type, y-parameters should be moreconveniently used in analysis.
The overall [YT] is [YT] = [Ya] + [Yf]
[YT]YG YL=
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M. Tse: Power Amplifier Design 27
yfb
Suppose a simple feedback circuit is used:
†
[Yf ] =yfb -yfb
-yfb yfb
È
Î Í
˘
˚ ˙
†
fi
Also, from the s-parameters, we can find the y-parameters of the amplifier:
†
[Ya ] =5.5307 ¥10-3 + j1.9049 ¥10-2S 3.9086 ¥10-4 - j2.3092 ¥10-3S4.7114 ¥10-2 - j2.1376 ¥10-2S 5.4445 ¥10-3 + j5.1841¥10-3S
È
Î Í
˘
˚ ˙
Thus, the overall y-parameters are
†
[YT ] =5.5307 ¥10-3 + j1.9049 ¥10-2 + yfb S 3.9086 ¥10-4 - j2.3092 ¥10-3 - yfb S4.7114 ¥10-2 - j2.1376 ¥10-2 - yfb S 5.4445 ¥10-3 + j5.1841¥10-3 + yfb S
È
Î Í
˘
˚ ˙
M. Tse: Power Amplifier Design 28
Our aim is to neutralize the internal feedback using the external feedback yfb.Obviously, if we set yfb = –j2.3092 S, then the internal feedback is cancelled.This gives
†
[YT ] =5.5307 ¥10-3 + j1.6739 ¥10-2 S 3.9086 ¥10-4 - j0 S4.7114 ¥10-2 - j1.9067 ¥10-2 S 5.4445 ¥10-3 + j2.8750 ¥10-3 S
È
Î Í
˘
˚ ˙
We can now convert it back to s-parameters, and find the stability factorand determinant again. We get
k = 2.067 and |D| = 0.4037.
These satisfy the Rollett criteria and the amplifier becomes unconditionallystable.