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1
NDF Stability Analysisof Linear Networksfrom Return Ratios
Wayne Strubleand
Aryeh Platzker
Appendix 1
2
Abstract
• The purpose of this presentation is to show a method of determining the stability (or instability) of linear, passively terminated, multiport networks from the normalized network determinant function (NDF).
• This method is rigorous and can be calculated within commercially available steady-state circuit simulators.
3
Outline
• Background on stability concepts• Normalized Determinant Function (NDF)• Use and calculation of the NDF• Calculating the NDF from return ratios• A ring oscillator example• Single active element example• NDF guidelines
4
BACKGROUND
5
Background
• Most GaAs MMIC designers use the well known Linvill or Rollett stability criteria (C or K) to determine the stability of linear two-port networks. This procedure is not rigorous!
• Platzker et al. have shown that a separate test is required to determine the stability of the network; i.e. test for the existence of any zeroes of the network determinant in the right half plane (RHP), before the Linvill or Rollett stability criteria can be applied.
6
Background
• The network is stable if and only if no zeroes of the full network determinant exist in the RHP. If any zeroes do exist in the RHP, the network is unstable. This is RIGOROUS!
• Platzker’s test applies the Principle of the Argument theorem to a normalized network determinant function (NDF) to determine the number of zeroes in the RHP of the full network determinant (and thus poles of the network).
7
WHAT IS THE NDF?
8
What is the NDF
• Platzker’s normalized network determinant function NDF is simply the full network determinant, including all port terminations, divided by the resulting passive network determinant when all dependent sources (i.e. N voltage controlled or current controlled sources) contained within the network are set equal to zero.
N0NDF
9
What is the NDF
• Note that any linear network parameters such as Y, Z, H etc. can be used to calculate the above determinants.
• 0N represents the determinant of a passive network and cannot contain any zeroes in the RHP. Therefore, zeroes in the RHP of the NDF correspond to zeroes in the RHP of the full network determinant.
10
HOW IS THE NDFUSED TO DETERMINE
STABILITY?
11
Using the NDF
• To determine stability, the complex quantity NDF is calculated for a given network along the frequency axis from - to + and its locus is plotted in the complex plane.
• If the locus of the NDF encircles the origin (0,0) in a clockwise direction, the network determinant contains zeroes in the RHP. The number of encirclements is equal to the number of zeroes in the RHP.
12
Using the NDF
• A counterclockwise encirclement of the origin (from - to + ) is not possible by construction of the NDF.
• This NDF test can be used on any physically realizable, passively terminated, linear multi-port network.
13
HOW IS THE NDFCALCULATED?
14
Calculating the NDF
• Using the brute force approach, the NDF can be found by calculating the full network determinants and 0N at each frequency .
• A faster approach to calculating the NDF is to first reduce the network to a parallel connection of two networks, one totally passive and one totally active.
15
Calculating the NDF
PassiveComponents
ActiveComponents
16
Calculating the NDF
• This reduces the number of nodes in the network, and thus the sizes of the relevant matrices, to between two and four times the number of dependent sources contained in the network (2N-4N) depending on connectivity.
• The network cannot be reduced any further than this due to the possibility of introducing poles in . This will result in pole-zero cancellation in the NDF.
17
Calculating the NDF
• In networks containing more than five dependent sources, direct calculation of the NDF becomes difficult using commercially available circuit simulators.
• This is due to the limitation in the sizes of matrices that can be saved by the simulators for external calculations.
• Also, calculation of the NDF inside a circuit simulator is desirable since it allows for sensitivity analysis or optimization.
18
Calculating the NDF
• To this end, we have derived an alternative way of easily calculating the NDF within commercial circuit simulators using the concept of Return Ratios.
19
A METHOD OFCALCULATING THE NDFUSING RETURN RATIOS
20
NDF from Return Ratios
• The Return Ratio (RR) of a dependent source was first presented by Bode and is defined as the Return Difference minus one or,
• Where is the full network determinant and 0represents the full network determinant where the dependent source is set equal to zero.
RR 0
1
21
NDF from Return Ratios
• For networks containing a single dependent source, the Return Ratio is equivalent to the NDF and can be used as a rigorous check for stability.
• If the network contains more than one dependent source, 01 may contain zeroes in the RHP due to other dependent sources, and a single Return Ratio calculation is not sufficient for a rigorous assessment of stability.
NDF RR 01
1 1
22
( )RR1 011
NDF from Return Ratios
• However, the concept can be extended to networks with N dependent sources by rearranging the previous equation into the form,
• and realizing that,
• Where RR2 is the Return Ratio of a second dependent source in the network with the first dependent source set to zero.
01 2 021 ( )RR
23
NDF from Return Ratios
• By substituting 01 into the previous equation,
• By continuous substitution,
• or,
( )( )( ). . .( )RR RR RR RR N N1 2 3 01 1 1 1
01 2 021 ( )RR 0221 )1RR)(1RR(
NDF RR RR RR RR N ( )( )( ). . .( )1 2 31 1 1 1
24
NDF from Return Ratios
• For each successive Return Ratio calculation RRi(i = 2-N), the network is physically changed by setting all previous dependent sources to zero.
25
HOW IS THERETURN RATIO OF ADEPENDENT SOURCE
CALCULATED?
26
Return Ratio of a Dependent Source
• The Return Ratio of a dependent source embedded within a network is calculated by replacing the dependent source, which is controlled by an internal voltage or current, with an identical source that is controlled by an external voltage or current.
• The stimulus from this new source will result in some amount of feedback to the controlling voltage or current of the original dependent source.
27
Return Ratio of a Dependent Source
• The negative ratio of the voltage or current returned, to the external voltage or current stimulus, is the Return Ratio of the dependent source.
28
Return Ratios of Dependent Sources
+Vin-
•Vin => •Vext VVRRext
in
Iin •Iin => •Iext IIRRext
in
VCCS
CCCS
29
Return Ratios of Dependent Sources
+Vin-
•Vin => •Vext VVRRext
in
Iin •Iin => •Iext IIRRext
in
+-
+-
VCVS
CCVS
30
...
V13 gm.V13
+
-
1 2
3 4
Port 1 Port 2 Port N
ORIGINAL NETWORK
...Port 1 Port 2 Port N
RETURN RATIONETWORK
~
RR VVext
13
Return Ratio of a Dependent Source
+V13-
gm•Vext
1
3 4
2
+Vext-
31
RING OSCILLATOREXAMPLE
32
Ring Oscillator Example
• The following example is a ring oscillator containing two dependent sources.
• The NDF of the oscillator is calculated from two Return Ratios (one Return Ratio for each dependent source in the network).
• The order of the Return Ratio calculations is unimportant as long as after calculating the Return Ratio of any given dependent source, it is set to zero in the Return Ratio calculations of all following dependent sources.
33
Cf
Cf
L1
L1
R1
C1V2
+
-
gm2*V2
R1
C1
gm1*V1
V1
+
-
R2 R2
R1 = 10R2 = 50L1 = 560pHC1 = 16pFCf = 0.1pFgm1 = 500mSgm2 = 400mS
Ring Oscillator Schematic
34
Ring Oscillator k-factor and |S|
0
50
100
150
200
0.000
0.005
0.010
0.015
0.020
0 50 100 150 200 250
k-fa
ctor
Mag
nitu
de o
f the
det
erm
inan
t of
[S]
Frequency (GHz)
35
Ring Oscillator k-factor and |S|
0
50
100
150
200
0.000
0.005
0.010
0.015
0.020
0 2 4 6 8 10
k-fa
ctor
Mag
nitu
de o
f the
det
erm
inan
t of
[S]
Frequency (GHz)
36
Cf
Cf
L1
L1
R1
C1V2
+
-
gm2*V2
R1
C1
gm1*Vext
V1
+
-
R2 R2
50 50
~ Vext
+
-
RR VVext
11
Return Ratio RR1 Schematic
37
Cf
Cf
L1
L1
R1
C1V2
+
-
gm2*Vext
R1
C1
R2 R2
50 50
~ Vext
+
-
RR VVext
22
NDF RR RR ( )( )1 21 1
Return Ratio RR2 Schematic
gm1=0
38
-1.5 -1.0 -0.5 0.0 0.5 1.0 1.5-2.0 2.0
freq (10.00kHz to 100.0GHz)
ND
FN
DF1
Positive FrequenciesNegative Frequencies
Scale = 2.0Increasing Frequency
+100 GHz
NDF Plot of Ring Oscillator
Increasing Frequency
-100 GHz
39
NDF Encirclement Plot ( = 0 to +)
10 20 30 40 50 60 70 80 900 100
0.2
0.4
0.6
0.8
1.0
0.0
1.2
freq, GHz
unw
rap(
phas
e(N
DF)
)/-36
0
40
SINGLE ACTIVEELEMENT EXAMPLE
41
R1 R1
R2 C1 C2
+V1
-
gmV1
R1 = 100
R2 = 10
C1 = 0.1pF
C2 = 10pF
gm = 1.0S
Z0 = 50 = 180o @ 1GHz
Single Active Element Example
42
k-factor and |S|
0
50
100
150
200
0.000
0.050
0.100
0.150
0.200
0 2 4 6 8 10
k-fa
ctor
Mag
nitu
de o
f the
det
erm
inan
t of
[S]
Frequency (GHz)
43
NDF Plot (scale = 10)
Positive FrequenciesNegative Frequencies
Scale = 10
44
NDF Plot (scale = 2)
Positive FrequenciesNegative Frequencies
Scale = 2
45
NDF GUIDELINES
46
NDF Polar Plots
• Regardless of stability, the polar plot for physically realizable networks will always begin and end at = +/- on the real axis at (1,0) by construction.
• The NDF does not specifically determine the frequency of oscillation (if an encirclement occurs). However, the frequency where the NDF plot crosses 180 degrees is generally close to the frequency of oscillation.
47
NDF Polar Plots
• One exception to this rule is if the NDF plot touches the origin (0,0) at some frequency. If this happens, then this frequency is at least one frequency of oscillation.
• Encirclements of the origin (from = - to +) in a counterclockwise direction are not possible by construction of the NDF.
48
NDF Encirclement Plots
• A stable network will always have a cumulative NDF phase/-360 of zero. That is, sweeping from - to + the plot will begin (by definition) and end at zero encirclements.
• It is unimportant if the encirclement plot rises above +1 or below -1 so long as it returns to zero at = +.
49
NDF Encirclement Plots
• A unstable network will always have a cumulative NDF phase/-360 of some multiple of 2. That is, sweeping from = - to + the plot will begin at zero and end at +2,4,6... encirclements.
• Since the NDF over negative frequencies is the complex conjugate of positive frequencies, one can look for encirclements from = 0 to + only. This will always show half the encirclements (as above, but capture the same stability information).
• Negative numbers of encirclements at = + are not possible by construction of the NDF.
50
References[1] J.G. Linville and L.G. Schimpf, “The Design of Tetrode Transistor Amplifiers”,
Bell System Technical Journal, Vol 35, July 1956, pp. 813-840.[2] J.M. Rollett, “Stability and Power-Gain Invariants of Linear Two-Ports”, IRE
Transactions on Circuit Theory, Vol CT-9, March 1962, pp. 29-32.[3] H. Nyquist, “Regeneration Theory”, Bell System Technical Journal, Vol 11,
January1932, pp. 126-147.[4] H.W. Bode, “Network Analysis and Feedback Amplifier Design”, D. Van Nostrand
Co. Inc., New York, 1945.[5] E.J. Routh, “Dynamics of a System of Rigid Bodies”, 3rd Edition, Macmillan,
London, 1877.[6] A. Platzker, W. Struble, and K. Hetzler, “Instabilities Diagnosis and the Role of K
in Microwave Circuits”, IEEE MTT-S International Microwave Symposium Digest, 1993, pp. 1185-1188.
[7] W. Struble and A. Platzker, “A Rigorous Yet Simple Method for Determining Stability of Linear N-Port Networks”, GaAs IC Symposium Digest, 1993, pp. 251-254.
[8] A. Platzker and W. Struble, “Rigorous Determination of the Stability of Linear N-Node Circuits From Network Determinants and the Appropriate Role of the Stability Factor K of Their Reduced Two-Ports”, 3rd International Workshop on Integrated Nonlinear Microwave and Millimeter-wave Circuits, University of Duisberg, Germany, Oct. 1994, pp. 93-107.
51
Dead (passive) “black box”Transistor Models
Wayne Strubleand
Aryeh Platzker
Appendix 2
52
Dead “black box” Transistor Models
• How do we make a “black box” transistor in a linear network passive (no RHP zeroes).
• Consider the simplified network representation of an active transistor (at any single given frequency) (D.J.H. Maclean [1]). This is valid for any transistor type (FET or Bipolar). (can also be a [S]-parameter file)
Ya
Yb
Ycgmv1
+v1-
+v2-
i1 i2
2
1
2
1
ii
vv
YYYgYYY
cbbm
bba
221
121
ivYYvYgivYvYY
cbbm
bba
53
Dead “black box” Transistor Models
• We want to set gm = 0 at all frequencies (other than DC) to make the transistor passive.
Ya
Yb
Yc
+v1-
+v2-
i1 i2
2
1
2
1
ii
vv
YYYYYY
cbb
bba
Ya
Yb
Ycgmv1
+v1-
+v2-
i1 i2
2
1
2
1
ii
vv
YYYgYYY
cbbm
bba
54
Dead “black box” Transistor Models
• We need to subtract gmv1 from i2 in the simplified model to achieve this. This is done using two additional (identical) transistors that use the following AC terminal conditions (DC are the same as original):
• First, in identical transistor 1, set the input voltage equal to v1 (sampled from the original transistor terminals), short v2and find the short circuit current i2.
• The short circuit current is (gm-Yb)v1.• Next we need to get Ybv1.
12' vYgi bm Ya
Yb
Ycgmv1
+v’1=v1
-
+v’2=0
-
i’1 i’2
55
Dead “black box” Transistor Models
• Next, in identical transistor 2, set the output voltage equal to v1 (sampled from the original transistor terminals) and find the short circuit input current i1.
• The short circuit current is -Ybv1.• Next we simply subtract i2’ and add i1’’ back to original
transistor model as shown on the next slide.
Ya
Yb
Ycgmv1
+v’’1=0
-
+v’’2=v1
-
i’’1 i’’2
11'' vYi b
56
Dead “black box” Transistor Models
• Subtract i’2 and add i’’1 to the original model like so…
Ya
Yb
Ycgmv1
+v1-
+v2-
i1 i2
i’2 i’’1
21221
121
''' iiivYYvYgivYvYY
cbbm
bba
221
121
ivYYvYivYvYY
cbb
bba
12' vYgi bm
11'' vYi b
Ya
Yb
Yc
+v1-
+v2-
i1 i2
Same asgm=0
This results in
57
Vg
Vs
Vd
VARVAR1
G1=if (Dead > 0.5) then 1.0 else 0.0 endifW1=if (W < 5e-6) then 5e-6 else W endif
EqnVar
PortP3Num=3
PortP1Num=1
PortP2Num=2
tqped_ehssQ1
Ng=NgW=W1
Vd
VsNonlinVCVSCSRC2Coeff=list(0,1)
Vg
VsNonlinVCVSCSRC1Coeff=list(0,1)
DC_BlockDC_Block8
Vs
Vd
NonlinCCCSCSRC4Coeff=list(0,-G1)
Vs
Vd
NonlinCCCSCSRC3Coeff=list(0,G1)
DC_FeedDC_Feed6
DC_BlockDC_Block6
tqped_ehssQ2
Ng=NgW=W1
tqped_ehssQ3
Ng=NgW=W1
Vg
VsNonlinVCVSCSRC6Coeff=list(0,1)
DC_FeedDC_Feed8
DC_BlockDC_Block7
DC_FeedDC_Feed7
Vd
VsNonlinVCVSCSRC5Coeff=list(0,1)
ADS Implementation of “dead” Transistor Models(uses 3 identical transistor models)
• Model gives same DC currents as active model, but AC gm = 0 (no gain at AC frequencies). Same can be done for BJTs, [S]-parameter files.
EFET_NDFX252
Dead=DeadNg=1W=46 um
-i’2
i’’1
v’2 shorted
v’’1 shortedv’’2 = v1
Dead = 0 ActiveDead = 1 Passive (no RHP zeroes)
Original modelv’1 = v1
58
References
[1] D.J.H. Maclean, “Broadband Feedback Amplifiers”, Research Studies Press, 1982
59
NDF Stability Analysisof Linear Networks fromCurrent/Voltage Ratios
Wayne Strubleand
Aryeh Platzker
Appendix 3
NDF calculation using currents ratioWith a dependent current source gm(vi-vj) connected between nodes k and l, and an independent source Is connected in parallel with it, the voltage nodes can be calculated from the matrix equation using the Y matrix below
Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n
Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n
.
.
Yk1 Yk2 ….... -gm+Yki Ykj+gm…..Ykk Ykl ………….Ykn
Yl1 Yl2……….. gm+Yli Ylj-gm……Ylk Yll ……. .Yln
Yn1 Yn2 …………. Yni Ynj ………..Ynk Ynl Ynn
Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n
Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n
.
.
Yk1 Yk2 ….... -gm+Yki Ykj+gm…..Ykk Ykl ………….Ykn
Yl1 Yl2……….. gm+Yli Ylj-gm……Ylk Yll ……. .Yln
Yn1 Yn2 …………. Yni Ynj ………..Ynk Ynl Ynn
Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n
Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n
.
.
Yk1 Yk2 ….... -gm+Yki Ykj+gm…..Ykk Ykl ………….Ykn
Yl1 Yl2……….. gm+Yli Ylj-gm……Ylk Yll ……. .Yln
Yn1 Yn2 …………. Yni Ynj ………..Ynk Ynl Ynn
Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n
Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n
.
.
Yk1 Yk2 ….... -gm+Yki Ykj+gm…..Ykk Ykl ………….Ykn
Yl1 Yl2……….. gm+Yli Ylj-gm……Ylk Yll ……. .Yln
Yn1 Yn2 …………. Yni Ynj ………..Ynk Ynl Ynn
Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n
Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n
.
.
Yk1 Yk2 ….... -gm+Yki Ykj+gm…..Ykk Ykl ………….Ykn
Yl1 Yl2……….. gm+Yli Ylj-gm……Ylk Yll ……. .Yln
Yn1 Yn2 …………. Yni Ynj ………..Ynk Ynl Ynn
Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n
Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n
.
.
Yk1 Yk2 ….... -gm+Yki Ykj+gm…..Ykk Ykl ………….Ykn
Yl1 Yl2……….. gm+Yli Ylj-gm……Ylk Yll ……. .Yln
Yn1 Yn2 …………. Yni Ynj ………..Ynk Ynl Ynn
Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n
Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n
.
.
Yk1 Yk2 ….... -gm+Yki Ykj+gm…..Ykk Ykl ………….Ykn
Yl1 Yl2……….. gm+Yli Ylj-gm……Ylk Yll ……. .Yln
Yn1 Yn2 …………. Yni Ynj ………..Ynk Ynl Ynn
Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n
Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n
.
.
Yk1 Yk2 ….... -gm+Yki Ykj+gm…..Ykk Ykl ………….Ykn
Yl1 Yl2……….. gm+Yli Ylj-gm……Ylk Yll ……. .Yln
Yn1 Yn2 …………. Yni Ynj ………..Ynk Ynl Ynn
Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n
Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n
.
.
Yk1 Yk2 ….... -gm+Yki Ykj+gm…..Ykk Ykl ………….Ykn
Yl1 Yl2……….. gm+Yli Ylj-gm……Ylk Yll ……. .Yln
Yn1 Yn2 …………. Yni Ynj ………..Ynk Ynl Ynn
Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n
Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n
.
.
Yk1 Yk2 ….... -gm+Yki Ykj+gm…..Ykk Ykl ………….Ykn
Yl1 Yl2……….. gm+Yli Ylj-gm……Ylk Yll ……. .Yln
Yn1 Yn2 …………. Yni Ynj ………..Ynk Ynl Ynn
Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n
Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n
.
.
Yk1 Yk2 ….... -gm+Yki Ykj+gm…..Ykk Ykl ………….Ykn
Yl1 Yl2……….. gm+Yli Ylj-gm……Ylk Yll ……. .Yln
Yn1 Yn2 …………. Yni Ynj ………..Ynk Ynl Ynn
Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n
Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n
.
.
Yk1 Yk2 ….... -gm+Yki Ykj+gm…..Ykk Ykl ………….Ykn
Yl1 Yl2……….. gm+Yli Ylj-gm……Ylk Yll ……. .Yln
Yn1 Yn2 …………. Yni Ynj ………..Ynk Ynl Ynn
Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n
Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n
.
.
Yk1 Yk2 ….... -gm+Yki Ykj+gm…..Ykk Ykl ………….Ykn
Yl1 Yl2……….. gm+Yli Ylj-gm……Ylk Yll ……. .Yln
Yn1 Yn2 …………. Yni Ynj ………..Ynk Ynl Ynn
Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n
Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n
.
.
Yk1 Yk2 ….... -gm+Yki Ykj+gm…..Ykk Ykl ………….Ykn
Yl1 Yl2……….. gm+Yli Ylj-gm……Ylk Yll ……. .Yln
Yn1 Yn2 …………. Yni Ynj ………..Ynk Ynl Ynn
Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n
Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n
.
.
Yk1 Yk2 ….... -gm+Yki Ykj+gm…..Ykk Ykl ………….Ykn
Yl1 Yl2……….. gm+Yli Ylj-gm……Ylk Yll ……. .Yln
Yn1 Yn2 …………. Yni Ynj ………..Ynk Ynl Ynn
Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n
Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n
.
.
Yk1 Yk2 ….... -gm+Yki Ykj+gm…..Ykk Ykl ………….Ykn
Yl1 Yl2……….. gm+Yli Ylj-gm……Ylk Yll ……. .Yln
Yn1 Yn2 …………. Yni Ynj ………..Ynk Ynl Ynn
Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n
Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n
.
.
Yk1 Yk2 ….... -gm+Yki Ykj+gm…..Ykk Ykl ………….Ykn
Yl1 Yl2……….. gm+Yli Ylj-gm……Ylk Yll ……. .Yln
Yn1 Yn2 …………. Yni Ynj ………..Ynk Ynl Ynn
Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n
Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n
.
.
Yk1 Yk2 ….... -gm+Yki Ykj+gm…..Ykk Ykl ………….Ykn
Yl1 Yl2……….. gm+Yli Ylj-gm……Ylk Yll ……. .Yln
Yn1 Yn2 …………. Yni Ynj ………..Ynk Ynl Ynn
Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n
Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n
.
.
Yk1 Yk2 ….... -gm+Yki Ykj+gm…..Ykk Ykl ………….Ykn
Yl1 Yl2……….. gm+Yli Ylj-gm……Ylk Yll ……. .Yln
Yn1 Yn2 …………. Yni Ynj ………..Ynk Ynl Ynn
Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n
Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n
.
.
Yk1 Yk2 ….... -gm+Yki Ykj+gm…..Ykk Ykl ………….Ykn
Yl1 Yl2……….. gm+Yli Ylj-gm……Ylk Yll ……. .Yln
Yn1 Yn2 …………. Yni Ynj ………..Ynk Ynl Ynn
Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n
Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n
.
.
Yk1 Yk2 ….... -gm+Yki Ykj+gm…..Ykk Ykl ………….Ykn
Yl1 Yl2……….. gm+Yli Ylj-gm……Ylk Yll ……. .Yln
Yn1 Yn2 …………. Yni Ynj ………..Ynk Ynl Ynn
Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n
Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n
.
.
Yk1 Yk2 ….... -gm+Yki Ykj+gm…..Ykk Ykl ………….Ykn
Yl1 Yl2……….. gm+Yli Ylj-gm……Ylk Yll ……. .Yln
Yn1 Yn2 …………. Yni Ynj ………..Ynk Ynl Ynn
Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n
Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n
.
.
Yk1 Yk2 ….... -gm+Yki Ykj+gm…..Ykk Ykl ………….Ykn
Yl1 Yl2……….. gm+Yli Ylj-gm……Ylk Yll ……. .Yln
Yn1 Yn2 …………. Yni Ynj ………..Ynk Ynl Ynn
Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n
Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n
.
.
Yk1 Yk2 ….... -gm+Yki Ykj+gm…..Ykk Ykl ………….Ykn
Yl1 Yl2……….. gm+Yli Ylj-gm……Ylk Yll ……. .Yln
Yn1 Yn2 …………. Yni Ynj ………..Ynk Ynl Ynn
Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n
Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n
.
.
Yk1 Yk2 ….... -gm+Yki Ykj+gm…..Ykk Ykl ………….Ykn
Yl1 Yl2……….. gm+Yli Ylj-gm……Ylk Yll ……. .Yln
Yn1 Yn2 …………. Yni Ynj ………..Ynk Ynl Ynn
Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n
Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n
.
.
Yk1 Yk2 ….... -gm+Yki Ykj+gm…..Ykk Ykl ………….Ykn
Yl1 Yl2……….. gm+Yli Ylj-gm……Ylk Yll ……. .Yln
Yn1 Yn2 …………. Yni Ynj ………..Ynk Ynl Ynn
Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n
Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n
.
.
Yk1 Yk2 ….... -gm+Yki Ykj+gm…..Ykk Ykl ………….Ykn
Yl1 Yl2……….. gm+Yli Ylj-gm……Ylk Yll ……. .Yln
Yn1 Yn2 …………. Yni Ynj ………..Ynk Ynl Ynn
Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n
Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n
.
.
Yk1 Yk2 ….... -gm+Yki Ykj+gm…..Ykk Ykl ………….Ykn
Yl1 Yl2……….. gm+Yli Ylj-gm……Ylk Yll ……. .Yln
Yn1 Yn2 …………. Yni Ynj ………..Ynk Ynl Ynn
Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n v1 0
Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n v2 0
. . . . . . . 0
. . . . . . . = 0
Yk1 Yk2 ….... -gm+Yki Ykj+gm…..Ykk Ykl ………….Ykn vk -Is
Yl1 Yl2……….. gm+Yli Ylj-gm……Ylk Yll ……. .Yln vl Is
. . . . . . . . 0
Yn1 Yn2 …………. Yni Ynj ……… vn 0
NDF calculation using currents ratio
Y11 Y12 ………… Y1i Y1j………Y1k Y1l ……..Y1n 0 v1 0Y21 Y22 ………… Y2i Y2j …….. Y2k Y2l……...Y2n 0 v2 0. . . . . . . . .. . . . . . . =Yk1 Yk2 ….... Yki Ykj……….Ykk Ykl … Ykn 1 vk 0Yl1 Yl2……….. Yli Ylj… …….Ylk Yll ………Yln -1 vl 0. . . . . . . . . . Yn1 Yn2 ………….Yni Ynj ………Ynk Ynl ….. Ynn 0 vn 00 0 gm -gm 0 0 0 1 I Is
Introduce the sum of the independent and the dependent currents I as a new variable I = Is – gm (vi-vj) and obtain the circuit description with the augmented Y matrix, Yaug
NDF calculation using currents ratio
Y11 Y12 ………… Y1i Y1j………Y1k Y1l ……..Y1n 0Y21 Y22 ………… Y2i Y2j …….. Y2k Y2l……...Y2n 0. . . . . . . . .. . . . . . . Yk1 Yk2 ….... Yki Ykj……….Ykk Ykl … Ykn 0Yl1 Yl2……….. Yli Ylj… …….Ylk Yll …… Yln 0. . . . . . . . . . Yn1 Yn2 ……… ….Yni Ynj ………Ynk Ynl ….. Ynn 00 0 ………….. gm -gm 0 0 ……..0 Is
Yaug I =
By Cramer’s rule the variable I is given by
From which Is / I = Yaug / Yaug(gm=0)
i.e. Is / I is the Normalized Determinant Function (NDF) of the circuit
NDF calculation using currents ratio
Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n 0
Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n 0
. . . . . . .
. . . . . . . Yk1 Yk2 ….... -gm+Yki Ykj+gm…..Ykk Ykl ………….Ykn 0
Yl1 Yl2……….. gm+Yli Ylj-gm……Ylk Yll ……. .Yln 0
. . . . . . . . Yn1 Yn2 ………….Yni Ynj ………..Ynk Ynl ……... Ynn 0 0 0 gm -gm 0 0 0 1
It is easy to show that Y = Yaug by observing that subtracting in Yaug its k’th row from its (n+1)’th row and adding its l’th row to its (n+1)’th row leads Yaug to become:
Since the row adding operations we performed do not change the value of the determinant, we readily see by expanding the determinant along the (n+1)’th column that Y = Yaug
Y1
Y2 Y6
Y4
Y3
Y5
Is
I
gm(V1-V2)
1
2 4
3
By Cramer’s Rule we get:
NDF calculation using currents ratio
sIIVVgm
IVYVVYVVYIVVYVVY
VVYVYVVYVVYVVY
21
46345244
435133
42422121
313211
0000
IsIVVVV
gmgmYYYYY
YYYYYYYYY
YYYY
0000
10010
100000
4
3
2
1
65454
5533
44211
3131
)0(
gm
s
YY
IINDF
Y1
Y2 Y6
Y4
Y3
Y5
Vs -Vb+vm(V1-V2)
1
2 4
3
+-
-+
Ib
By Cramer’s Rule we get:
NDF calculation using voltages ratio
sb
b
b
b
VVVVvmVVVIVYVVYVVYIVVYVVY
VVYVYVVYVVYVVY
21
34
46345244
435133
42422121
313211
00000
sb
b
VVIVVVV
vmvm
YYYYYYYYYYYYYY
YYYY
00000
1000101100
010010000000
4
3
2
1
65454
5533
44211
3131
)0(
vmb
s
YY
VVNDF
(from Wolfram Math World)
69
NDF Stability Analysisof Linear Networks from
Network Admittances
Wayne Strubleand
Aryeh Platzker
Appendix 4
70
Y1
Y2 Y6
Y4
Y3
Y5
Is
+
Vs
-
gm(V1-V2)
1
2 4
3
Y1
Y2 Y6
Y4
Y3
Y5
Is
+
Vs
-
gm(V1-V2)
1
2 4
3Add current source Is
in parallel with dependentSource and measure Vs
Admittance = Is/Vs
Relationship Between Network Determinants and Admittance
[Y][V] = [I]
s
s
II
VVVV
YYYYYgmgmYYYgmYgmYYYYY
YYYY00
00
4
3
2
1
65454
5533
44211
3131
71
Relationship Between Network Determinants and Admittance
• First redefine the node voltages in terms of Vs
where V4 = V3-Vs
• So, column 3 in the matrix becomes the sum of columns 3 and 4, and column 4 changes sign as shown below.
s
s
s II
VVVV
YYYYYYgmgmYYgmYgmYYYYYY
YYYY000
3
2
1
654644
533
444211
3131
72
Relationship Between Network Determinants and Admittance
• Next, we replace row 3 in the matrix with the sum of rows 3 and 4 to eliminate Is on the right side.
• And multiplying row 4 by -1,
ss IVVVV
YYYYYYgmgmYYYYYYY
YYYYYYYYYY
0000
3
2
1
654644
6464343
444211
3131
ss IVVVV
YYYYYYgmgmYYYYYYY
YYYYYYYYYY
0000
3
2
1
654644
6464343
444211
3131
73
Relationship Between Network Determinants and Admittance
• Now using Cramer’s rule.
• Next, we recognize that the determinant in the numerator is equal to the determinant of the original network with nodes 3 and 4 shorted (i.e. dependent source is shorted)…
YYYYYY
YYYYYYYYY
I
V
s
s64343
44211
3131
74
Relationship Between Network Determinants and Admittance
• Original network with nodes 3 & 4 shorted
• Therefore the admittance,
• Note that this shorted network is passive (no RHP zeroes), but is not of the same order as the original network.
Y1
Y2 Y6
Y4
Y3
1
2
3
Y1
Y2 Y6
Y4
Y3
1
2
3
3
2
1
3
2
1
64343
44211
3131
III
VVV
YYYYYYYYYYYYYY
shortedVVs
s
YY
VIA
43&
75
Relationship Between Network Determinants and Admittance
• Next, we calculate admittance (at the same nodes) from a known stable operating point (gm=0 for example).
• Since,
• Because the dependent source has been shorted.
shortedVVstable
stable
s
s
YY
VIA
43&0
shortedVVs
s
YY
VIA
43&
shortedVVshortedVVstable YY
4343 &&
stableYY
AANDF
0
76
Rigorous LinearStability Analysis
Wayne Struble
Appendix 5
77
Rigorous Linear Stability Analysis(networks with many dependent sources)
The perturbation nodes are chosenat the outputs of all active devices
M-nodelinear
Networkwith
N/2 activedevices
i1
.
.
.
i3
in-1
v1
v3
vn-1
Ground(datum node)
v2
v4
vn
i2
i4
in
nnnnn
n
i
i
v
v
yy
yy
11
1
111
• Consider the network with many dependent sources (i.e. transistors) and matrix equation shown below.
78
Rigorous Linear Stability Analysis
M-nodelinear
Networkwith
N/2 activedevices
i1
i3
in-1
Ground(datum node)
i2
i4
in
-(i1+i2+… in)
M-nodelinear
Networkwith
N/2 activedevices
i1
i3
in-1
v1-vn
v3-vn
vn-1-vn
Ground(datum node)
v2-vn
v4-vn
-vn
i2
i4
in
-(i1+i2+… in)
EquivalentNetwork.
.
.
v1
v3
vn-1
v2
v4
vn
.
.
.
• From network theory, any node contained within a network can be designated as the datum node (ground node) without affecting the network (and thus the network determinant).
79
Rigorous Linear Stability Analysis• The indefinite [Y] matrix is defined as:
• Note that the datum node (ground) contribution to the indefinite network matrix is a linear summation of rows and/or columns of the definite [Y] matrix.
n
n
kk
nnnn
n
jnj
n
n
jj
n
iin
n
ii
n
jiij
i
i
i
v
v
yyy
yyy
yyy
1
11
11
1111
1
111
11 0
80
From The Properties of Determinants• 1. Switching two rows or columns changes the sign.
• 2. Scalars can be factored out from rows and columns.
• 3. Multiples of rows and columns can be added together without changing the determinant's value.
• 4. Scalar multiplication of a row by a constant C multiplies the determinant by C.
• 5. A determinant with a row or column of zeros has value 0.
• 6. Any determinant with two rows or columns equal has value 0.
81
Rigorous Linear Stability Analysis
Ground(datum node)
-in-1
0
EquivalentNetwork
M-nodelinear
Networkwith
N/2 activedevices
i1=0
i3=0
Ground(datum node)
i2=0
i4=0
in-1
-(i1+i2+… in)
.
.
.
v1
v3
vn-1
v2
v4
vn
M-nodelinear
Networkwith
N/2 activedevicesin-1
v1-vn
v3-vn
vn-1-vn
v2-vn
v4-vn
-vn
.
.
.
i1=0
i3=0
i2=0
i4=0
• So, if one stimulates nodes n and n-1 with external perturbation current source in-1 (and all other node current stimulations at are zero), we get…
82
Rigorous Linear Stability Analysis
1
11
1
1,11
,11,11,11
,1
,11,1111
1
1,
11,
11
11
00
0)(
)(
n
nnn
n
n
nnnnn
n
jnj
nnnnn
n
jjn
nn
n
jj
n
ini
n
ini
n
ii
n
jiij
iivv
vvv
yyyy
yyyy
yyyy
yyyy
11
1
1,11,11
,1
1,1111
1
11,
11
11
00
)(
)(
nnn
n
n
nnn
n
jjn
n
n
jj
n
ini
n
ii
n
jiij
ivv
vvv
yyy
yyy
yyy
Indefinite [Y] matrix
definite [Y] matrix
Ground(datum node)
-in-1
0
M-nodelinear
Networkwith
N/2 activedevicesin-1
v1-vn
v3-vn
vn-1-vn
v2-vn
v4-vn
-vn
.
.
.
i1=0
i3=0
i2=0
i4=0
• The network [Y] matrix equations are:
83
Rigorous Linear Stability Analysis• Partition the definite [Y] matrix as follows:
• Where:
NN
NN
nnn
n
jjn
n
n
jj
n
ini
n
ii
n
jiij
DCBA
yyy
yyy
yyy
1,11,11
,1
1,1111
1
11,
11
11
1,2
1,1
11,
nn
n
n
ini
N
y
y
y
B
2,21,21
,2
2,1111
1
12,
11
11
nnn
n
jjn
n
n
jj
n
ini
n
ii
n
jiij
N
yyy
yyy
yyy
A
2,11,1
1,1 nnn
n
jjnN yyyC 1,1 nnN yD
84
Rigorous Linear Stability Analysis• Now, solving for the admittance in-1/(vn-1-vn):
• Expanding gives:
• Solve for [V1] and substitute:
22
1 0IV
VDCBA
NN
NN
)(
)(
2
11
nn
n
n
vv
vvv
V
)( 12 nn vvV
12 niI
Where:
021 VBVA NN
221 IVDVC NN
21
1 VBAV NN
NNNN
nn
n BACDvv
iVI 1
1
1
2
2
Ground
(datum node) 0
M-nodelinear
Networkwith
N/2 activedevicesin-1
v1-vn
v3-vn
vn-1-vn
v2-vn
v4-vn
-vn
.
.
.
i1=0
i3=0
i2=0
i4=0
85
Rigorous Linear Stability Analysis• The result, [DN]-[CN][AN]-1[BN] is a scalar (1x1 matrix) so the determinant
is equal to the value of the scalar.
• Therefore:
• From the theory of determinants of a subdivided matrix ref [10]:
• The admittance below is related to the full network determinant:
NNNNnn
n BACDvv
i 1
1
1
ABCADDCBDADCBA
Y 11
NNNNnn
n BACDvv
i 1
1
1
Where |AN| may have RHP zeroes due to other activetransistors resulting in pole-zero cancellation in |Y|/|AN|.
Nnn
n
AY
avv
i
1
1
1
86
Rigorous Linear Stability Analysis• However, if we now short circuit the first perturbed nodes (at AC frequencies
only, so the transistor biases are unaffected) we get…
• This isolates, but does not change, the sub-network [AN] from the full network [Y], in affect, creating a new network [AN] that is of order 1 smaller than [Y].
Partition the Network Matrix as:
2
11
0 IIV
DCBA
NN
NN
11 0 IBVA NN
21 0 IDVC NN
0
0
IVAN
-(i1+i2+…in-2)Ground
(datum node)
M-nodelinear
Networkwith
(N/2-1) activedevices
v1-vn
v3-vn
v2-vn
v4-vn
-vn
.
.
.
i1
i3
i2
i4
87
Rigorous Linear Stability Analysis• Remember the simplified network representation of an active transistor
(ala D.J.H. Maclean [4]).
Ya
Yb
Ycgmv1
+v1-
+v2-
i1 i2
2
1
2
1
ii
vv
YYYgYYY
cbbm
bba
221
121
ivYYvYgivYvYY
cbbm
bba
88
Rigorous Linear Stability Analysis• By shorting the dependent source of the active transistor model at AC
frequencies only (v2=0), we have made the transistor passive so it can no longer contribute any RHP zeroes to the remaining network.
Ya
Yb
Ycgmv1
+V1
-
+V2=0
-
i1 i2
Ya
Yb
+V1
-
+V2=0
-
i1 i2
Equivalent network @AC frequencies
89
Rigorous Linear Stability Analysis• Next, we simply repeat the procedure (perturb at a second active
transistor contained in sub-network [AN] with the first active transistor shorted at AC frequencies)…
EquivalentNetwork
short @ ACopen at DC
M-nodelinear
Networkwith
(N/2-1) activedevices
i1=0i2=0
in-3
0
.
.
.
v1
vn-3
v2
vn-2
short @ ACopen at DC
M-nodelinear
Networkwith
(N/2-1) activedevices
i1=0i2=0
0
.
.
.
v1 –vn-2
vn-3-vn-2
v2 –vn-2
in-3
in-3
-vn-2
90
)(
)(
24
21
2
1
nn
n
n
vv
vvv
V
)( 232 nn vvV
32 niI
Where:
Rigorous Linear Stability Analysis• Partition the remaining [AN] network matrix as:
• Solving for the admittance [I2]/[V2]:
• Multiplying this admittance result by the previous one we get:
22
1
11
11 0IV
VDCBA
NN
NN
02111 VBVA NN
22111 IVDVC NN
IVAN
Results in:
12
23
3
N
N
nn
n
AA
avv
i
1121
NN
N
N AY
AA
AY
aa
91
Rigorous Linear Stability Analysis• If we continue to repeat the procedure until we have perturbed all active
transistors contained in network [Y] (and short circuited all subsequent perturbed nodes at AC frequencies)…
• |A1| contains no RHP zeroes (all active transistors have been shorted), so |Y|/|A1| cannot have any pole-zero cancellations.
• Now, repeat the process for the same linear network from a known stable operating point (for normalization) and we get…
121 A
Yaaa N
100201 A
Yaaa stable
N
92
Rigorous Linear Stability Analysis• Use this result to normalize and we get the Normalized Determinant
Function.
• Where N is the number of active transistors in the linear network, ai is the measured admittance of the ith active transistor due to a perturbing current with all previous transistors shorted at AC frequencies, and a0iis the measured admittance of the same transistor due to the same perturbing current with all previous transistors shorted at AC frequencies, where the network is operating from a known stable state.
N
i i
i
stable aa
YY
NDF1 0
93
References[1] E. Routh, “Dynamics of a System of Rigid Bodies”, 3rd Ed., Macmillan, London, 1877[2] H. Nyquist, “Regeneration Theory”, Bell System Technical Journal, Vol. 11,
pp. 126-147, Jan. 1932[3] H. Bode, “Network Analysis and Feedback Amplifier Design”, D. Van Nostrand
Co. Inc., New York, 1945[4] D.J.H. Maclean, “Broadband Feedback Amplifiers”, Research Studies Press, 1982[5] A. Platzker, W. Struble, and K. Hetzler, “Instabilities Diagnosis and the Role of
K in Microwave Circuits”, IEEE MTT-S Digest, vol. 3, pp. 1185-1188, Jun. 1993[6] W. Struble and A. Platzker, “A Rigorous Yet Simple Method For Determining
Stability of Linear N-port Networks”, 15th Annual GaAs IC Symposium Digest,pp. 251-254, Oct. 1993
[7] A. Platzker and W. Struble, “Rigorous Determination of The Stability of Linear N-nodeCircuits From Network Determinants and The Appropriate Role of The StabilityFactor K of Their Reduced Two-Ports”, 3rd International Workshop on IntegratedNonlinear Microwave and Millimeterwave Circuits, pp. 93-107, Oct. 1994
[8] J. Jugo, J. Portilla, A. Anakabe, A. Suarez and J.M. Collantes, “Closed-LoopStability Analysis of Microwave Amplifiers”, Electronics Letters, vol.37 No. 4,pp. 226-228, Feb. 2001
[9] C. Barquinero, A. Suarez, A. Herrera and J.L. Garcia, “Complete Stability Analysisof Multifunction MMIC Circuits”, IEEE Tran. Microwave Theory and Techniques,vol.55 No. 10, pp. 2024-2033, Oct. 2007
[10] I. Schur, “Ueber Potenzreihen die im Innern des Einheitskreises beschraenkt sind”,Journalfuer Reine und Angewandte Mathematik, 1917, 147, 205-232.
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