POWER AMPLIFIER DESIGN - Michael Tsecktse.eie.polyu.edu.hk/eie403/PowerAmplifier.pdf · Gain...

1

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

2


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


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.

3


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


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.

4


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


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

5


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


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?

6


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|.


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

7


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( )


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)

8


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


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

9


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


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

10


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( )


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?

11


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


Unconditional Stability

Smith chart

load stability circle


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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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.



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.

13


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?


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=

14


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

˘

˚ ˙


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.

POWER AMPLIFIER DESIGN - Michael Tsecktse.eie.polyu.edu.hk/eie403/PowerAmplifier.pdf · Gain...

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