ECE 451 Circuit Synthesisemlab.uiuc.edu/ece451/notes/synth.pdfECE 451 –Jose Schutt‐Aine 18 In...

ECE 451 – Jose Schutt‐Aine 1

ECE 451Circuit Synthesis

Jose E. Schutt-AineElectrical & Computer Engineering

University of [email protected]


MOR via Vector Fitting

• Rational function approximation:

• Introduce an unknown function σ(s) that satisfies:

• Poles of f(s) = zeros of σ(s):

• Flip unstable poles into the left half plane.

1

( )N

n

n n

cf s d sh

s a

1

1

( ) ( )

( )1

Nn

n n

Nn

n n

cd sh

s as f ss c

s a

1

1 1

1 1

( )1

N Nn

nn n n

N Nn

nn nn

cd sh s z

s af s

cs z

s a

s-domain

Re

Im


Passivity Enforcement

• State‐space form:

• Hamiltonian matrix:

• Passive if M has no imaginary eigenvalues.

• Sweep:

• Quadratic programming:– Minimize (change in response) subject to (passivity compensation).

x Ax Buy Cx Du

T T

T T T T

A BKD C BKBM

C LC A C DKB

1 1T TK I D D L I DD

Δ Δ Δmin subject to ( ).vec vec G vec T( C) H ( C) = C

( ) ( )Heig I S j S j

3/28


• Time-Domain simulation using recursive convolution

• Frequency-domain circuit synthesis for SPICE netlist

Use of Macromodel

Macromodel Circuit Synthesis


• Circuit can be used in SPICE

Objective: Determine equivalent circuit from macromodel representation*

Motivation

Macromodel Circuit Synthesis

*Giulio Antonini "SPICE Equivalent Circuits of Frequency-Domain Responses", IEEE Transactions on Electromagnetic Compatibility, pp 502-512, Vol. 45, No. 3, August 2003.

• Generate a netlist of circuit elements

Goal


Circuit RealizationCircuit realization consists of interfacing the reduced model with a general circuit simulator such as SPICE

11 1

1

N

N NN

s s s sS s

s s s s

Model order reduction gives a transfer function that can be presented in matrix form as

or

11 1

1

N

N NN

y s y sY s

y s y s


Y-Parameter - Circuit Realization

Each of the Y-parameters can be represented asL

kij

k 1 k

ay ( s ) ds p

where the ak’s are the residues and the pk’s are the poles. d is a constant



We need to determine the circuit elements within yijk

The realized circuit will have the following topology:



We try to find the circuit associated with each term:

Lk

ijk 1 k

ay ( s ) ds p

1. Constant term d

ijdy ( s ) d

2. Each pole-residue pair

kijk

k

ay ( s )s p



In the pole-residue case, we must distinguish two cases

k k k kijk

k k k k

j jy ( s )s j s j

(a) Pole is real kijk

k

ay ( s )s p

(b) Complex conjugate pair of poles

In all cases, we must find an equivalent circuit consisting of lumped elements that will exhibit the same behavior


Circuit Realization – Constant Term

1Rd


Z sL R

Consider the circuit shown above. The input impedance Z as a function of the complex frequency s can be expressed as:

Circuit Realization - Real Pole

1//

LY ss R L

kijk

k

ay ( s )s p

kL 1 / a k kR p / a


21 1

2 2

11/ 1

RZ sL R sL RR sC sCR

1 2 2

2

11

R sL sCR RZ

sCR

Consider the circuit shown above. The input impedance Z as a function of the complex frequency s can be expressed as:

Circuit Realization - Complex Poles


2

1 22 1 2

2 2

1/1 s CRY

L R RL CR Rs sLR C LR C

2 2

1 22 1 22

2 2

1/CR s CRY

R RL CR RLCR s sLR C LR C



1 2

1 2

ˆ r rYs p s p

Circuit Realization - Complex Poles Each term associated with a complex pole pair in the expansion gives:

Where r1, r2, p1 and p2 are the complex residues and poles. They satisfy: r1=r2* and p1=p2*

It can be re-arranged as:

1 2 2 1 1 2

1 2 21 2 1 2

/ˆ s r p r p r rY r r

s s p p p p


2

1 22 1 2

2 2

1/1 s CRY

L R RL CR Rs sLR C LR C

1 2 2 1 1 2

1 2 21 2 1 2

/ˆ s r p r p r rY r r

s s p p p p

1 2p p p

1 2g p p

1 2a r r

1 2 2 1x r p r p

product of poles sum of residues

sum of poles cross product

DEFINE

Circuit Realization - Complex Poles We next compare

and


1/L a1 2

x gRa a

2 2

p x gRx a a

2

1pa gCx a x


We can identify the circuit elements


In the circuit synthesis process, it is possible that some circuit elements come as negative. To prevent this situation, we add a contribution to the real parts of the residues of the system. In the case of a complex residue, for instance, assume that

1 2

1 2

ˆ r rYs p s p

1 2

1 2 1 2

ˆ

Augmented Circuit Compensation Circuit

r rYs p s p s p s p


Can show that both augmented and compensation circuits will have positive elements


S-Parameter - Circuit Realization

Each of the S-parameters can be represented asL

kij

k 1 k

as ( s ) ds p

where the ak’s are the residues and the pk’s are the poles. d is a constant


Realization from S-Parameters

We need to determine the circuit elements within sijk

The realized circuit will have the following topology:



We try to find the circuit associated with each term:

Lk

ijk 1 k

as ( s ) ds p

1. Constant term d

ijds ( s ) d

2. Each pole and residue pair

kijk

k

as ( s )s p



In the pole-residue case, we must distinguish two cases

k k k kijk

k k k k

j js ( s )s j s j

(a) Pole is real kijk

k

as ( s )s p

(b) Complex conjugate pair of poles

In all cases, we must find an equivalent circuit consisting of lumped elements that will exhibit the same behavior


S- Circuit Realization – Constant Term

o1 dR Y1 d


S-Realization – Real Poles


1

1 2 1 21

1 22

sR R R R C

YR R s

R C


Admittance of proposed model is given by:



ˆo

s aY Ys b

, andk k k ka p r b p r

1 2

1 2o

R RYR R

2o

b aC

b Z

21R

bC

kijk

k

rs ( s )s p

1 21R R

aC

From S-parameter expansion we have:

which corresponds to:

where

from which


Realization – Complex Poles

2

22 1 2 1 2

11 1a

o o

sCRY YR s LCR s L R R C R R R

2 1 2 2 1 2

2 2

2 1 2 1 2

2 2

1o o

o

o

L R R C R R C R R Rs sLCR LCR

YR L R R C R Rs s

LCR LCR

Proposed model

which can be re-arranged as:


1 2 1 2 2 11 22

1 2 1 2 1 2

ˆ s r r r p r pr rSs p s p s s p p p p

2

2

1ˆ1ˆˆ1 1

o o

sa xS s sg pY Y Ysa xS

s sg p

which corresponds to an admittance of:

From the S-parameter expansion, the complex pole pair gives:



22

2 2ˆ

o o

s s g a p xs sg p sa xY Y Ys sg p sa x s s g a p x

1 2p p p

1 2g p p

1 2a r r

1 2 2 1x r p r p

product of poles sum of residues

sum of poles cross product

WE HAD DEFINED


The admittance expression can be re-arranged as


1o

o

RY

1 2

2

oR R Rp xLCR

1 2

2

R Rp xLCR

1 2

2

2 22 oR R RpLCR

2

2 oRxLCR


Matching the terms with like coefficients gives



2oRLa

11 22

oo

gR LxR Ra a

2 1

1

2

opRxR R

2

aCR x

Solving gives


Typical SPICE Netlist* 32 -pole approximation *This subcircuit has 16 pairs of complex poles and 0 real poles .subckt sample 8000 9000 vsens8001 8000 8001 0.0 vsens9001 9000 9001 0.0 *subcircuit for s[1][1] *complex residue-pole pairs for k= 1 residue: -6.4662e-002 8.1147e-002 pole: -4.4593e-001 -2.4048e+001 elc1 1 0 8001 0 1.0 hc2 2 1 vsens8001 50.0 rtersc3 2 3 50.0 vp4 3 4 0.0 l1cd5 4 5 1.933e-007 rocd5 4 0 5.000e+001 r1cd6 5 6 5.895e+003 c1cd6 6 0 3.474e-015 r2cd6 6 0 -9.682e+003 :*constant term 2 2 -6.192e-003 edee397 397 0 9001 0 1.0e+000 hdee398 398 397 vsens9001 50.0 rterdee399 398 399 50.0 vp400 399 400 0.0 rdee400 400 0 49.4 *current sources fs4 0 8001 vp4 -1.0 gs4 0 8001 4 0 0.020 fs10 0 8001 vp10 -1.0 gs10 0 8001 10 0 0.020 fs16 0 8001 vp16 -1.0 gs16 0 8001 16 0 0.020 fs22 0 8001 vp22 -1.0 gs22 0 8001 22 0 0.020 fs28 0 8001 vp28 -1.0 gs28 0 8001 28 0 0.020


Realization from Y-Parameters

Recursive convolution SPICE realization



Recursive convolution SPICE realization

ECE 451 Circuit Synthesisemlab.uiuc.edu/ece451/notes/synth.pdfECE 451 –Jose Schutt‐Aine 18 In...

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