ndf stability analysis of linear networks from return ratiosndf polar plots • regardless of...

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


• 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


• 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


• 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


• 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


• 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


+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)


Scale = 10

44

NDF Plot (scale = 2)


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


• 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


• 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


• 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


• 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


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


DC_FeedDC_Feed8

DC_BlockDC_Block7

DC_FeedDC_Feed7

Vd


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

.

.




Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n

Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n

.

.




Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n

Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n

.

.




Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n

Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n

.

.




Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n

Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n

.

.




Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n

Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n

.

.




Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n

Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n

.

.




Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n

Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n

.

.




Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n

Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n

.

.




Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n

Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n

.

.




Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n

Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n

.

.




Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n

Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n

.

.




Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n

Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n

.

.




Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n

Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n

.

.




Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n

Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n

.

.




Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n

Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n

.

.




Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n

Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n

.

.




Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n

Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n

.

.




Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n

Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n

.

.




Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n

Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n

.

.




Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n

Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n

.

.




Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n

Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n

.

.




Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n

Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n

.

.




Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n

Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n

.

.




Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n

Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n

.

.




Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n

Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n

.

.




Y11 Y12 ………… …. Y1i Y1j………Y1k Y1l ……..Y1n

Y21 Y22 ………… Y2i Y2j ……..... Y2k Y2l………Y2n

.

.




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


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


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:


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


• 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


• 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


• 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


• 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


• 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


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


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


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


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


i1=0i2=0

in-3

0

.

.

.

v1

vn-3

v2

vn-2

short @ ACopen at DC

M-nodelinear

Networkwith


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.

ndf stability analysis of linear networks from return ratiosndf polar plots • regardless of...

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