Single – Stage Transistor Amplifier Design
Ro
Input
Matching
Circuit
Gs
Transistor
[S]
Go
S
in
Output
matching
circuit
GL
out
Zo
L
Maximum Gain Consideration
A maximum power transfer from the input matching network to the
transistor will occur when
in  S *
A maximum power transfer from the transistor to the output matching
network will occur when
out  L *
If we assume that the matching sections at the input are lossless , we get
1
2 1  L

S 21
1  S
1  S 22 L
2
GT max
S  in  S11 
*
S12 S 21L
1  S 22 L
L  out  S 22 
*
2
S12 S 21s
1  S11s
S 1*2 S *2 1
S  S 1 1 
1 *  S*
22
L
*
L 
*
S 22  ( S11S 22  S 21S12 ) s S 22  s

1  S11s
1  S11s
S 12* S *21
S  S 11 
(1  S11s )
 S *22
S 22  s
*
 S 11 
S 1*2 S *2 1 ( S 22  s )
*
1  S11s  S *2 2 ( S 22  s )
S (1  S 22 )  S2 ( S *22  S11 )  S ( S 11* S *22  S11  S 12* S *21 )
2
2
 S 11* (1  S 22 )  S 12* S *21 S 22
2
( S11  S *2 2 ) S2  (   S11  S 22  1) S  ( S 1*1  * S 22 )  0
2
2
2
The relation for S is
S 
B1  B12  4 C1
2
2C1
Where B1  1  S11  S 22  
2
2
2
C1  S11  S *22
Similarly , we obtain L as
L 
B2  B22  4 C2
2
2C2
Where B2  1  S 22  S11  
2
2
2
C2  S22  S 11*
In the unilateral case , we set S12  0 , we then have
*
*
S  S11
, L  S 22
As a result , we get
GTU max 
1
1  S11
2
S 21
2
1
1  S 22
2
Example :
Design a amplifier for maximum gain at 4.0GHz using
single – stub – matching sections . The AaAs FET has the following S
parameter(50  )
f(GHz)
S11
S 21
S12
S22
3.0
0.80  89 0
2.86990
0.03560
0.76  410
4.0
0.72  1160 2.60760
0.03570
0.73  540
5.0
0.66  1420 2.39540
0.03620
0.72  680
In the maximum gain , we have
S 
B1  B12  4 C1
2
2C1
B1  1  S11  S 22    1  (0.72) 2  (0.73) 2  0.525  170 0  0.0781330
2
2
2
=1+0.518-0.533-0.297=0.748
C1  S11  S *22  0.72  1160  (0.465  j0.145)0.73540
 0.72( 0.441  j 0.897)  ( 0.465  j 0.145)0.73(0.586  j 0.81)
 (0.317  j 0.646)  ( 0.113  j 0.337)
 ( 0.204  j0.309)  0.37  1230
0.748  0.7482  0.37 2
0.748  0.11
S 

 0.861230
0
0
2  0.37  123
2  0.37  123
L 
B2  B22  4 C2
2C2
2
 0.87610
2
GS 
1
1  S
2

1
 3.84  5.84dB
1  0.86 2
GO  S 21  6.76  8.3dB
2
GL 
1  L
2
1  S 22 L
2
 1.67  2.22dB
The maximum transducer gain is
GT max  5.84  8.3  2.22  16.36dB
0.206
50 ohm
0.120
50 ohm
50 ohm
50 ohm
50 ohm
50 ohm
0.206
0.206
Stability of a Two – Port Network
The two – port network is said to be unconditionally stable at a given
frequency if the real ports of Z in and Z out are greater than zero for all
passive load and source impedances .
Potentially unstable : some passive load and source terminations can
produce input and output impedances having a negative real part .
Conditions for unconditional stability at a given frequency
S  1
L  1
S S 
12 21 L
where in  S11  1  S 
22
in  1
out  1
out  S 22 
L
S12 S 21s
1  S11s
Graphical Analysis Of Stability Problems
(1)
To determine the regions where values of L and S produce
in  1 and out  1
Setting in  1 and solving for L
* *
( S 22  DS11
)
L 
S 22  D
2
2

S12 S 21
S 22  D
2
2
where D  S11S22  S12 S21
Setting out  1 and solving for S
* *
( S11  DS 22
)
S 
S11  D
2
2

S12 S 21
S11  D
2
2
L value for in  1 (output stability circle) :
S12 S 21
rL 
cL 
S 22  D
2
2
* *
( S 22  DS 11
)
S 22  D
2
2
S valued for out  1 (input stability circle)
rS 
cS 
S11 
S12 S 21
S11  D
2
2
* *
( S11  DS 22
)
S11  D
2
2
S12 S 21L
1
1  S 22 L
or S11 (1  S22 L )  S12 S21L  1  S22 L
Now we define
  S11S22  S12 S21
We then obtain
S11  L  1  S22 L
Square both sides and simplify to obtain
*
*
S11   L  ( L S11
 * L* S11 )  1  S 22 L  ( S 22
L*  S 22 L )
2
2
2
2
2
*
*
( S 22   ) L L*  ( S 22  S11
)L  ( S 22
 * S11 ) L*  S11  1
2
2
L L 
*
2
*
*
( S 22  S11
)L  ( S 22
 * S11 )L*
S 22  
2
2
S11  1
2

S 22  
2
Now complete the square by adding
L 
( S 22  S )
2
* *
11
2
S 22  
2
or L 
2

* *
( S 22  S11
)
S 22  
2
S11  1
2
S 22  
2

2

*
S 22  S11
2
22

2
( S 22   ) 2
2
*
S 22  S11
S
2
2
to both side .
2

2 2
S12 S 21
S 22  
2
2
Figure : Smith Chart illustration stable and unstable regions in the
In the
L
in  1
L
plane
plane , on one side of the stability circle boundary , we will have
and on the other side in  1
S S 
12 21 L
Consider the origin point ( L  0 ) , corresponding in  S11  1  S   S11 .
22
L
Therefore , if S11  1 then in  1 i.e. the origin represents a stable
operating point . On the other hank , if S11  1 , then in  1 , the origin
represents an unstable operating point .
The figure illustrates the two cases discussed .
Similar argument can be extended to S plane to find the stable and
unstable region for out .
They are shown in the next figure .
Figure : Smith Chart illustration stable and unstable regions in the S
plane .
For unconditional stability any passive load or source in the network must
produce a stable condition .
From a graphical point of view , for S11  1 and S 22  1 , we want the
stability circles to fall completely outside the Simth Chart (as shown in the
next figure ) or to completely enclose the Smith Chart (as shown in the next
2nd figure)
Therefore , the conditions for unconditional stability for all passive sources
and loads can be expressed in the form
cL  rL  1
or rL  cL  1
for S11  1
and
cS  rS  1
or rS  cS  1
for S 22  1
Figure : Conditions for unconditional stability : (a)
L
plane ; (b)
S
plane
Figure : Conditions for unconditional stability : (a)
L
plane ; (b)
S
plane
If either S11  1 or S 22  1 , the network cannot be unconditional stable
because the termination
L  0 or S  0 will produce in  1 or out  1 .
Conditions for Unconditional Stability
The conditions for unconditional stability , in terms of reflection coefficient ,
are
S  1
L  1
in  1
out  1
This can be rephrased in terms of S parameters of the two – port as
(1) S11  1
S 22  1
* *
S12 S 21  ( S11  DS 22
)
S11  D
2
2
* *
S12 S 21  ( S 22  DS11
)
S 22  D
2
2
1
1
( rS  c S  1 )
rL  c L  1
reference : R.W. Anderson . “ S parameter techniques for faster and more
accurate network design , “ HP Application Note 95
, 1967 .
This can be concluded from a graphical point of view .
(2)
k>1
1  S11  S12 S 21
2
1  S 22  S12 S 21
2
Reference : K . Kurokawa , “ Power waves and the scattering matrix ,
“IEEE Transactions on Microwave Theory and Techniques , vol . MTT-13 ,
pp . 194-202 , March 1965 .
D . Woods . “ Reappraisal of the unconditional stability criteria for active
2 – port networks in terms of S parameters , “IEEE Transactions on circuits
and systems , vol . CAS -23 , no .2 , pp. 73-81 , Feb. 1976 .
(3) k > 1 B1  0
B2  0
or k >1
Reference : G. E Bodway , “Two port power flow analysis using generalized
scattering parameters , “Microwave Journal , vol . 10 , pp . 61-69 , May
1967 .
(3)
k>1
D 1
Reference : D . Woods , “ Reappraisal of the unconditional stability
criteria for active 2- port networks in terms of S parameters , “ IEEE
Transactions on Circuits and Systems , vol . CAS -23 , no . 2 . pp . 73-81 ,
Feb . 1976
Example : Stability Circles of a Microwave Amplifier
A certain GaAs MESFET has the following S parameters measured at
9GHz with a 50-  reference :
S11  0.64  1700
S12  0.05150
S21  2.10300
S 22  0.57  950
Compute (a) the delta factor  , and (b) the stability factor K. (c) Find
the center and radius of the input stability circle and plot the circle . (d)
Determine the center and radius of the output stability circle and plot the
circle .
Solution : (a) The delta factor is
  S11S22  S12S21
 0.64  1700  0.57  950  0.05150  2.10300
0.30 111.450
 =0.30<1
 =0.09
2
(b) The stability factor K is
1    S11  S 22
2
K
2
2 S12 S 21
2
1  0.30  0.64  0.57
2

2
2
2 0.05  2.10
=1.71 > 1
(c) The input stability circle is determined as follows :
*
C1*  S11
 * S 22  0.64170 0  0.30  111.450  0.57  950
 0.48176.52 0
The center of the input stability circle is
cS 
C1*
S11  
2
2

0.48176.42 0
0.64  0.30
2
2
 1.50176.42 0
The radius is
S12 S 21
rS 
S11  
2
2

0.05  2.10
0.64  0.30
2
2
 0.33
(d) The output stability circle is found as follows :
*
C2*  S 22
 * S11  0.57950  0.30  111.450  0.64  1700
 0.39103.32 0
The center of the output stability circle is
cL 
C2*
S 22  
2
2

0.39103.32 0
0.57  0.30
2
2
 1.70103.32 0
2
 0.48
The radius is
rL 
S12 S 21
S 22  
2
2

0.05  2.10
0.57  0.30
2
Both the input and output stability circles are completely outside the
Smith Chart , as shown in Fig S3 , sot the amplifier is unconditionally
stable .
Figure . S3 Stability circles for Example
Note : 1. Even when the selection of L and S produces in  1 or
out  1 , the circuit can be made stable if
Re( Z S  Z in )  0 and Re( Z L  Z out )  0
2. A potentially unstable transistor can be made unconditionally stable y
either resistively loading the transistor or by adding negative feedback.
Example :
The S parameters of a transistor at f=800MHz are
S11  0.65  950
S12  0.035400
S21  51150
S 22  0.8  350
Determine the stability and show how resistive loading can stabilize the
transistor .
Solution :
Since K<1 , the transistor is potentially unstable at f=800MHz .
The input and output stability circles are calculate
C S  1.79122 0
rS  1.04
CL  1.3480
rL  0.45
Figure S1 shows the plot of the stability circles , together with the stable
region . It can be seen that a series resistor of approximately 29  , or a
shunt resistor of approximately 500  at the output , produces stability at
the output . The three choices of resistive loading are shown in Fig . S2 .
The most popular is the shunt resistor configuration in Fig . S2 (c)
For the stabilized shunt resistor configuration in Fig . S2 (c) (i.e. , with a
500  shunt resistor ) , the resulting S parameters are
S11  0.65  940
S12  0.03241.20
S21  4.62116.20
S22  0.66  360
and from (3.3.13) and (3.3.17) K=1.04 and   0.409250.130 , which show
that the stabilized network in Fig . S2 (c) is unconditionally stable at
f=800MHz .
Figure S1 Input and output stability circles .
Figure S2 Three types of resistive loading to improve stability .
Simultaneous Conjugate Match – Bilateral Case
in  S11 
S12 S 21L
1  S 22 L
out  S 22 
S12 S 21s
1  S11s
for simultaneous Conjugate match
*
, in  S* , out  L , then
( S11  S* )(1  S22 L )  S12 S21L  0 …
(3)
( S 22  L* )(1  S11S )  S12 S 21S  0 … (4)
from (3) (4) , we have
L 
S*  S11
S* S 22  D
S 
… (5)
L*  S 22
….(6)
L* S11  D
Substitute (5) into (4) , then
( S11  S11S 22 S *22  S12 S 21S *22 )S2  ( S 22  S11  1  S11S 22 D *  S12 S 21D * )S
2
 S 11*  D * S 22  0
2
*
or C1 S  B1 S  C1  0
The solution for (7) is
sm 
B1
1

B12  4C1C1*
2C1 2C1
2
B1
B12
1


2 C1
1
2
2C1 2C1
4 C1
C
B
 1  1
2C1 C1
B12
4 C1
2
1

 B
C1*  B1

  1
C1  2 C1
 2 C1

2


  1




Similarly , if we substitute (6) into (3)
, we have
C2 L2  B2 L  C2*  0
The solution for (8) is

 B
C 2*  B2
sm 
  2
C2  2 C2
 2 C2

2


  1




B1
From (8) , if B1 >0 and 2 C  1
1
then the minus sign  sm  1
the plus sign  sm  1
if B1 <0 and
B1
1
2 C1
then the minus sign  sm  1
the plus sign  sm  1
if
B1
 1 , then  sm  1 , no matter what B1 positive or negative
2 C1
Similar considerations apply to (10)
Bi
Since
2Ci
therefore
2
 1 (i=1,2)
 k2 1 ,
Bi
 1 (i = 1,2)  k  1 ,
2Ci
TABLE : Four Case of (8)
B1  0 Normal Condition
B1  0
Case 1
Case 2
Case 3
Case 4
B1
1
2 C1
B1
1
2 C1
B1
1
2 C1
B1
1
2 C1
k>1
k<1
k>1
k<1
Useful solution
Not useful
Potentially
sm  1 not
given in
sm  1
unstable even
useful
though k>1
A simultaneous conjugate match having unconditional stability is
possible if k>1 and D  1 , i.e. , k>1 and B1  0 or k>1
and B2  0 .
The minus sign must be chosen in (8) and (10) . In what follows any
reference to a simultaneous conjugate match assumes that the two – port
network is unconditionally stable . In a potentially unstable situation the
design procedure is bet done in terms of G P and G A
The maximum transducer power gain , under simultaneous conjugate
match conditions ( s  Sm , L  Lm ) is
(1  Sm ) S 21 (1  Lm )
2
GT ,max 
S 21 (1  Lm )
2
(1  Sm ) 1  S 22 Lm
2

( G p ,max )
2
S 21 (1  Sm )
2

2
(1  S11Sm )(1  S 22 Lm )  S12 S 21Sm Lm
2

2
2
(1  Lm ) 1  S11Sm
S 21
S12
2
( k  k 2  1)
2
( G A,max )
2
Transducer Power Gain Circles – Bilateral Case
(1) Unconditionally stable bilateral case , k>1 and D  1
In this situation , the terminations Sm and Lm will produce a
simultaneous conjugate match which results in the maximum value of the
transducer power gain .
If the design not for a maximum , a constant gain circle procedure can
be used . Let
GT  GS GO GL
where GS 
GO  S 21
and GL 
1  S
2
1  in S
2
2
1  L
2
1  S 22 L
2
The procedure :
(1)
From G L , the constant gain circles for GL can be drawn using
g i , Ri , d i , and  i (i=L) . Select the desired L for a given
G L gain.
(2)
S S 
12 21 L
Calculate in  S11  1  S 
22
in depends on L ; therefore ,
L
this value is the maximum available power gain , and the
maximum operation power gain .
(3)
S
21
When k=1 , we have the maximum stable gain as G MSG  S
12
GMSG is a figure of merit that represents the maximum value that GT ,nax
can have . It can be achieved by resistively loading the two – port (i.e. , the
transistor) to make k=1 , or by using feedback.
G S depends on
GL
(3) From G S , the constant gain circles for G S can be drawn using g i ,
Ri , d i , and  i (i=L) . Select the desired S for a given G S gain . The
value of G S might not be satisfactory and will require the selection of
another
L
and the procedure repeated .
(4) Design the matching networks .
The outlined procedure is not recommended for a practical design , since
in is a function of L , making the G S function dependent of the G L
function . Furthermore , the centers of the gain circles do not give GT ,nax .
In fact , the graphical approach becomes tedious because of the iterative
process required for obtaining the desired gain .
It is recommended to use operation power gain for the design of a
microwave transistor amplifier in the unconditional stable bilateral case .
Conclusion : When transistor S12 cannot be neglected (bilateral case)
unconditionally stable :
(1) GT ,nax : Simultaneous conjugate match
Sm and Lm
(2) other from GT ,nax : (i) output constant gain circles
(ii) given a L calculate in
(iii) input constant gain circles
(iv) matching networks .
or alternatively , input part first , then output port later .
Potentially unstable :
operating power gain or available power gain
Example :
Design a microwave amplifier using a GaAs FET to operate at f = 6Hz with
maximum transducer power gain . The transistor S parameters at the linear
bias point , VDS  4V and I DS  0.5I DSS , are
S11  0.641  171.30
S12  0.057116.30
S 21  2.05828.50
S 22  0.572  95.70
Solution :
We obtain k=1.504 and   0.3014109.880
Since k>1 and   1 , the GaAs FET is unconditionally stable .
Next , we must decide if the amplifier can be considered unilateral
U=0.1085 , and
-0.89dB<
GT
GTU
<1dB
The inequality above shows that S12 cannot be neglected .
The reflection coefficients for a simultaneous conjugate match are
calculated as follows :
B1 =0.9928
B2 =0.8255
C1  0.4786182.70
C2  0.3911256.10
MS  0.762177.30
and ML  0.718103.90
The maximum transducer power gain ,
GT ,max 
2.058
(1.504  (1.504) 2  1)  13.74
0.057
or 11.38 dB
The design of the matching networks using microstrip lines is illustrated in
Fig. p1 , where the admittances associated with MS and ML are
YMS 
7.2  j1.23
 (144  j 24.6)  10 3 S
50
and YML 
0.414  j1.19
 (8.28  j 23.8)  10 3 S
50
The input matching network can be designed with an open shunt stub of
length 0.185  and a series transmission line of length 0.0615  . The
output matching network is designed with an open shunt stub of length
0.176  and a series transmission line of length 0.169  .
The ac amplifier schematic is shown in Jig. p2. Using Duroid(  r =2.23 , h=
0.7874 mm )for the board material , we find that W=2.41mm for a
characteristic impedance of 50  ,  ff =1.9052 , and  =0.7245 0 where
0 =5cm at f=6GHz.The microstrip lengths at f=6GHz are
0185  =6.70mm
0.0615  =2.23mm
Figure p1 (a) Design of the input matching network ; (b) design of the
output matching network .
0.169  =6.12mm
0.176  =6.38mm
The design for GT ,max will MS and ML , at 6GHz , assures that the input
and output VSWR are 1 .
Finally , we should point out that the stability must be checked at all
frequencies , so that the reflection coefficients
stable operation .
MS and ML provide
Figure p1 (continued)
Figure p2 The ac schematic of a GaAs FET microwave amplifier . All
microstrip lines have a characteristic impedance of 50  .
Operating Power Gain Circles
The operating power gain circle procedure for both unconditionally stable
and potentially unstable transistors is simple and practical .
(1) unconditionally stable bilateral case
S 21 (1  L )
2
GP 
2
where
gp 
 S 21 g p
2
(1  in ) 1  S 22 L
2
2
1  L
2
2
S  DL
(1  11
) 1  S 22 L
1  S 22 L


1  L
2
2
1  S 22 L  S11  DL
2
1  L
2
1  S11  L ( S 22  D )  2 Re( L C 2 )
2
2
2
2
where C2  S 22  DS11*
Let L  u L  jvL
Re( LC2 )  uL Re[C2 ]  vL Im[ C2 ]
2




g P Re[ C 2* ]
g P Im[ C 2* ]
  v L 

then u L 
2
2
2
2
1  g P ( S 22  D ) 
1  g P ( S 22  D ) 



2

2
1

2
k
S
S
g

S
S
gP
12
21
P
12
21


2
2

1  g P ( S 22  D )


1
2




2
2
This equation is a family of circles in the uL  vL plane with g P as a
parameter . The centers of the circles are located at
P  uP  jvP 
g P Re[ C2* ]
1  g P ( S22  D )
2
2
+j
g P Im[ C2* ]
1  g P ( S22  D )
2
2
The distance from the origin of the Smith Chart to the centers of the circles
is given by d P  P  u  v 
2
2
P
P
g P C 2*
1  g P ( S 22  D )
2
2
The radii of the circles are given by
RP
1  2k S

12 S 21 g P  S12 S 21 g P
2
2

1
2
1  g P ( S 22  D )
2
2
The centers of the power gain circles are located along a line with an angle
  tan
1
Im[ C 2* ]
Re[ C 2* ]
We can also expect that the maximum operation power gain occurs when
RP =0 . Therefore ,
g P2 ,max S12 S 21  2k S12 S 21 g P ,max  1  0
2
The solution for this equation for unconditional stability is
g P ,max 
1
( k  k 2  1)
S12 S 21
G P ,max 
S 21
S12
( k  k 2  1)
For a given G P , L is selected from the constant operating power gain
circles . G P ,max results when L is selected at the distance where
g P ,max =
GP ,max
S 21
2
.The maximum output power results when a conjugate
*
match is selected at the input (i.e. , S  in )and the input power is equal
to the maximum available input power . Therefore , under these
circumstances the maximum transducer power gain ( GT ,max ) and the
maximum operating power gain are equal , and the values of S and L
that result in G P ,max are identical to Sm and Lm , respectively .
The procedure for drawing a constant operating power gain circles in the
Smith Chart is as follows :
(1) For a given G P , the radius and center of the constant operating power
gain circles are given by P and R P .
(2) Select the desired L .
(3) For the given L , maximum output power is obtained with a conjugate
match at the input . ( S  in* ) . This value of S produces the transducer
power gain GT  GP .
Example :
Design the amplifier in last example to have an operating power gain of
9bB instead of GT ,max  GP ,max  11.38dB.
Solution : Since
S 21  ( 2.058) 2  4.235 or 6.27dB
2
then g P 
GP
S 21
2

7.94
 1.875
4,235
From the results in Example P1 , k=1.504 ,   0.3014 , and
C2  0.3911256.10 . Therefore , the radius and center of the 9-dB operating
power gain circle ,
RP  0.431 and C P  0.508103.9 0 .
Figure P3 Operating power gain circle for GP  9dB . The load reflection
coefficient can be selected at point A , namely L  0.3647.50 . Then the
required S for maximum output power is
*

S S  
S  in*   S11  12 21 L   0.629175.510
1  S 22 L 

The location of GP ,max  11.38dB can be found as follows :
g P ,max = GP,max
S 21
2

13.74
 3.24
(2.058) 2
RP ,max =0
and g P ,max 
g P ,max C 2*
1  g P ,max ( S 22   )
2
2
 0.718103.9 0
At the location of G P ,max , we obtain from Fig. P3 L,max  0.718103.90 .
This value of L ,max is identical to the value ML found in last example .
The associated S for maximum output power is
*
S ,max


S S 
  S11  12 21 L,max   0.762177.30
1  S 22 L,max 

which is identical to the value of MS in last example .
Example : Constant Operating Power – Gain Circles
A certain AaAs MESFET has the following S parameters measured at 8
GHz with a 50-  resistance :
S11  0.26  550
S12  0.0880 0
S 21  2.14650
S 22  0.82  30 0
Plot the power – gain circles .
Solution :
(1) Compute the delta factor  and the stability factor k :
 = 0.26  550  0.82  30 0 - 0.0880 0  2.14650  0.36  62.7 0
  0.36  1
  0.13
2
1  0.36  0.26  0.82
2
k
2
2
 1.15  1
2  0.08  2.14
(2) The maximum operating power gain is
GP ,max 

2.14
1.15  (1.15) 2  1
0.08

=15.52=11.91dB
(3) Compute g P ,max :
15.52
g P ,max 
2.14
2
 3.39
(4) The distance of the circle center is computed as
C2*  S *22  S11  0.82300  0.3662.7 0  0.26  550
= 0.7332.830
1  2  1.15 0.08  2.14  3.39  0.08  2.14 (3.39) 2
2
d pt 

1  3.39 0.82  0.36
2
2

 0.00
(6) Compute the distances and the radii of the circles for the power gains of
10, 8, and 6 dB
(7) The computed values are tabulated in Table
TABLE
COMPUTED VALUES
GP ,max
(dB )
(Numerical)
gP
dP
rP
11.91
15.52
3.39
0.87
0
10
10
2.18
0.73
0.25
8
6.31
1.38
0.58
0.41
6
3.98
0.87
0.43
0.56
(8) Plot the power - gain circles on the Smith Chart as shown in Fig. P4
(9) The normalized load impedance for 11.91-dB power gain is read from
the plot at l as
z l  0.80  j3.20
The load impedance is Z l  40  j160
(10) For each value of the distance d p used for the power gain desired ,
maximum output power occurs with a conjugate match at the input for
*
which in  s .
We have
0.0880 0  2.14650  0.8732.830 *
s  (0.26  55 
)
0
0
1  0.82  30  0.8732.83
0
 0.31  76.870
Then read
zs
at s as zs  0.95  j0.65
and the input impedance is zs  47.5  j32.5
Figure P4 Constant operating power – gain circles for Example
(2) Potentially Unstable Bilateral Case
Design Procedures
1. Draw the constant operating power – gain circle for a given power gain
GP
in decibels .
2. Draw the output stability circles .
3. Choose l in the stable region .
4. Compute in and determine if a conjugate match at the input is
possible .
*
5. Draw the input stability circle and determine if s  in is within the
input stable region .
6. If s  in* is not in the stable region (or in the stable region but very
close to the input stability circle) , a new value of s must be selected
arbitrarily or a value of G P is chosen again . Be careful to ensure that the
values of l and s should not be too close to their respective stability
circles , because oscillation may occur when the input and output circles are
not matched .
Design Example : Operating Power – Gain Circles for Potentially Unstable
Bilateral Case
A certain GaAs MESFET has the following S parameter measured at 9GHz
with a 50-  resistance reference :
S11  0.45  600
S12  0.0970 0
S 21  2.5074 0
S 22  0.80  50 0
Design an amplifier for an operating power gain of 10 dB
Solution :
1. Compute the delta factor  and the stability factor k
 = 0.45  60 0  0.80  50 0 - 0.0970 0  2.5074 0
 0.20  70.7 0
  0.20  1
  0.04
2
1  0.04  0.45  0.80
2
k
2  0.09  2.5
2
 0.44  1
The device is potentially unstable because k<1
2. Compute the maximum stable power gain at k=1 .
Gmsp 
S 21
S12

2.5
 27.78  14.44dB
0.09
A 10- dB power gain is possible .
3. Compute d p and rp for the 10- dB operating power – gain circle .
gp 
GP
S 21
2

10
2.5
2
 1.60
*
C2*  S 22
 * S11  0.8050 0  0.2020 0  0.45  60 0  0.7354.550
d pt 
1.60  0.7354.550
1  1.6  ( 0.8  0.20 )
2
2
 0.6054.550
1  2  0.44  0.09  2.5  1.6  0.09  2.5  (1.6) 2
2
rp 
1  1.6( 0.80  0.20 )
2
2
 0.46
The 10- dB power gain circle is plotted on the Smith Chart shown in Fig. p5
4. Compute d l and rl for the output stability circle .
C2*  0.7354.550
Cl 
rl 
0.7354.550
0.80  0.20
2
2
0.09  2.50
0.80  0.20
2
2
 1.2254.550
 0.38
The output stability circle is plotted on the same Smith Chart.
5. Since S11  1 , the stable region is the region outside the output stability
circle .
The load reflection coefficient l is chosen on the 10- dB power – gain
circle at the location A .Then
l  0.381 0 04
z l  0.70  j0.50
Z l  35  j 25
Figure P5 Potential stability for Example
6. Compute s for a possible conjugate match.
0.0970 0  2.5074 0
s    (0.45  60 
) *  0.30  266.2 0
0
0
1  0.80  50  0.38104
*
in
0
7. Compute d s and rs for the input stability circle .
C1*  0.4560 0  0.2070.7 0  0.80  50 0  0.3576.37 0
cs 
0.3576.37 0
0.45  0.20
2
2
 2.1976.37 0
rs 
0.09  2.50
0.45  0.20
2
2
 1.41
The input stability circle is plotted on the same Smith Chart .
8. From Fig P5 s is a stable source reflection coefficient because it is
located in the stable region outside the input stability circle .
Example :
The S parameters of a GaAs FET at I D  50% I DSS , I DSS  10mA ,
VDS  5V and f=8GHz are
S11  0.5  180 0
S12  0.0830 0
S 21  2.570 0
S 22  0.8  100 0
Design an amplifier with GP  10dB .
Solution :(1) First check the stability of the transistor . We obtain k=0.4 and
D=0.223 62.12 0 . This GaAs FET is potentially unstable .
G MSG 
S 21
S12
The
 31.25 or 14.9 dB .
(2) In order to design for GP  10dB (4.9dB less than the GMSG ) , the
10  dB operating power gain circle and the output stability circle must be
calculated . The radius and center of the 10  dB power gain circle are
RP  0.473 and C P  0.57297.2 0 . The radius and center of the output
stability circle are rL  0.34 and C L  1.1897.2 0 .
(3) The Smith Chart in Fig. P6 shows the construction of the 10  dB
operating power gain circle and the output stability circle . Since S11  1 ,
the stable region is the region outside the output stability circle . L is
selected on the 10  dB power gain circle at location A , namely
L  0.197 0 or Z L  50(0.96  j0.19) .
(4) For a conjugate match at the input , S is given by
S  in*  0.52179.32 0 and we must determine if the value of S is in
the stable region . The radius and center of the input stability circle are
rS  1.0 and C S  1.671710 , where the stable region is the region outside
the input stability circle . Therefore , S is a stable source reflection
coefficient .
Figure: p6 Power and stability circles construction for Example
Available Power Gain Circles
The available power gain is independent of the load impedance . The
derivation of the constant available power gain circles is similar to that of
the operating power gain circles .
(1) unconditionally stable bilateral case
S 21 (1  S )
2
GA 
2
(1  out ) 1  S11S
where
2
ga 
 S 21 g a
2
2
1  S
2
2
S  DS
(1  22
) 1  S11S
1  S11S


1  S
2
2
1  S11S  S 22  DS
2
2
1  S
2
1  S 22  S ( S11  D )  2 Re( S C1 )
2
2
2
2
Let S  u S  jvS
Re( S C1 )  u S Re[C1 ]  v S Im[ C1 ]
2




g a Re[ C1* ]
g a Im[ C1* ]
  v S 

then u S 
2
2
2
2
1  g a ( S11  D ) 
1  g a ( S11  D ) 



 1  2k S S g  S S g 2
12 21 a
12 21 a

2
2

1  g a ( S11  D )


1
2




2
2
This equation is a family of circles in the u S , v S plane with g a as a
parameter . The centers of the circles are located at
a  u a  jva 
g a C1*
1  g a ( S11  D )
2
2
The distance from the center of the Smith Chart to the centers of the circles
is given by
d a  a  u  v 
2
a
2
a
g a C1*
1  g a ( S11  D )
2
2
The radii of the circles are given by
Ra
1  2k S

S 21 g a  S12 S 21 g
2
12
2
a

1
2
1  g a ( S11  D )
2
2
The centers of the available power gain circles are located along a line with
an angle
  tan 1
Im[ C1* ]
Re[ C1* ]