1 ee 616 computer aided analysis of electronic networks lecture 9 instructor: dr. j. a. starzyk,...

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EE 616 Computer Aided Analysis of Electronic Networks

Lecture 9

Instructor: Dr. J. A. Starzyk, ProfessorSchool of EECSOhio UniversityAthens, OH, 45701

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Outline

Sensitivities

-- Network function sensitivity

-- Zero and pole sensitivity

-- Q and sensitivity Multiparameter Sensitivity Sensitivities to Parasitics and Operational Amplifiers

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3

Sensitivities

Normalized sensitivity of a function F w.r.t parameter

h

F

F

h

hln

FlnS F

h

Two semi-normalized sensitivities are discussed when either For h is zero.

h

Fh

h

FS Fh

ln

and

h

F

Fh

FS Fh

1ln

F can be a network function, its pole or zero, Q etc., while h can be component value, frequency s, operating temperature or humidity, etc.

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SENSITIVITIES - Example

Resonant circuit2o

o22

Qss

s

C

1

LC

1

C

Gss

s

C

1Z

where

LC

10

L

C

G

1

GL

1

G

CQ

o

o

We haveCln

2

1Lln

2

1ln o

Lln2

1Cln

2

1GlnQln

so

2

1S o

L &

2

1SSS Q

CQL

oC

also 1S0S QG

oG

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SENSITIVITIES

The use of sensitivities can be demonstrated when we replace differentials by increments. Using the above example we have

Q

L

L

QSQ

L

and since 2

1SQ

L => L

L

2

1

Q

Q

Assume that there is a 1% increase of 01.0L

LL

then %5.001.02

1

Q

Q

so we can expect Q to decrease by 0.5%.

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Network function sensitivity

If network function is D

NT

then DlnNlnTln

so Dh

Nn

Tn SS

hln

TlnS

if )(exp jTT then jTlnTln so

h

Th

Tn SjS

hj

h

TS

lnln

ln

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Example:

we haveKCL at node v1 :

(v1 - E)G1 + (v1 - vout)G2 + (v1 - v2)sC2 + v1sC1 = 0

KCL at node v2 : (v2 - v1) sC2 + v2G3 = 0

or

322

222121

GsCsC

sCAG)CC(sGG

0

EG

v

v 1

2

1

and from here the transfer function

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Example (cont’d)

222322121

21out

sCAGsCGsC)CC(sGG

sCAG

E

vT

2133123212212

21

GGGGCAGGGGCsCCs

sCAG

For C1 = C2 = 1, G1 = G2 = G3 = 1, and A = 2 we have

2s2s

s2)S(T

2

22

21

2222121

ss

ssCG

D

AsCG

sCGA

ASSS DA

NA

TA

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Zero and pole sensitivity

Zeros and poles give good characterization of network response for different frequencies.

The sensitivity of the zero of a polynomial is obtained through expressing zero as function of parameter h.

0)(, zshshP

Since zero of the polynomial is not known analytically (it can be obtained by nonlinear iterations), the problem which must be solved is how to find derivative for evaluation of its sensitivity without explicit knowledge of the zero or its functioan dependence on the parameter h.

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Zero and pole sensitivity (cont’d)

Differentiating P w.r.t. h gives

0

zsdh

ds

s

P

h

P => zszs sP

hP

dh

dz

dh

ds

/

/

This expression is valid for simple zeros and can be used to get

dh

dz

z

hSz

h

if z = a + jb we obtain

dh

dzIm

b

hS&

dh

dzRe

a

hS b

hah

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Zero and pole sensitivity - example

Suppose a transfer function of the network is (compare with the previous example)

2133123212212

21)(GGGGCAGGGGCsCCs

sCAGsT

j1sj12

s2

D

N

Find the sensitivity of a pole sp = -1+j w.r.t. A

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Zero and pole sensitivity - example

Using the derived formula we have

jsss

p

GCAGGGGCCsC

GsC

s

D

A

D

dA

ds

p

1312321221

22

2/

For C1 = C2 = 1, G1 = G2 = G3 = 1, and A = 2 we have

jj

j

dA

ds p2

1

2

1

222

1

so the zero sensitivity w.r.t. A is

jjj

jdA

ds

s

AS p

p

sAp

22

1

2

1

1

2

and for sp=a+jb=-1+j

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1*

1

2Re

dA

ds

a

AS paA , and 1

2

1*

1

2Im dA

ds

b

AS pbA

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Q and sensitivity0

In filter design Q and o are easier to work with. For a pair of

complex zeros _

zandz

2o

o2__

2_

sQ

szzs)zz(s)zs()zs(

where

22o_

o zw&)zz(

Q

or for 222o

o baa2

Qjbaz

using zero's sensitivity we obtain

ah

oh

Qh SSS b

hbh

2ah

22

o

oh SSbSa

1S

22 bafor (high Q circuits)

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Q and sensitivity (cont’d)0

Derivation

h

ba

ba

h

h

baS oh

22

22

22

2

1

ln

ln

h

b

b

hb

h

a

a

ha

h

bb

h

aah

oo

2222

12

2

2

2

1

bhah

o

SbSa 222

1

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Example

In the case of transfer function from previous example

22

22

ss

sT

we have z = a+jb = -1+jso

2

1

)1(2

22222

Qandbao

Using 1aAS and 1bAS we have

0)11(4

11 222

bA

aA

oA SbSaS o

In this case bAA SS o but Q was low so approximation did not hold.

110SSS ah

oA

QA

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Example 2

Derive the transfer function of the network shown in figure. Find the transfer function sensitivity T

hS w.r.t. the capacitors and the amplifier

KCL at v1:

EGvsCvGvsCGG out 11221121

KCL at v2: 022212 vsCGvG

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Example 2 (cont’d)

so

A

V

G

sC1v

G

sC1v out

2

22

2

21

AsCG

G

sC1sCGG

AG

E

vT

122

2121

1out

2112122121

21

AGsC)sCGG(sCGsCGG

GAG

Using the formula for transfer function sensitivity

D

AGsCCCsGsC

C

D

D

CSSS DC

NC

TC

21212

21

1

1111

D

sCGGsC

C

D

D

CSSS DC

NC

TC

)( 1212

2

2222

D

AsC1

A

D

D

A

A

N

N

AS 1T

A

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Multiparameter Sensitivity

The function F generally depends on several parameters

hFhhhFF m ,...,, 21

The change in F due to infinitesimally small changes in parameters is expressed by the total differential

m

1i

hidhi

FdF

or

m

i

Fhi hi

hidS

hi

hid

F

hi

hi

F

F

dF

1

To compare different designs we introduce multiparameter sensitivity measures.

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Multiparameter Sensitivity (cont’d)

The worst case multiparameter sensitivity

FhiSWCMS

For incremental changes of parameters within their tolerance

ii

i th

h

we have i

Fhi tS

F

F

or in case all ti are equal to t

WCMS*tF

F

This is a very pessimistic estimate of the function deviation from its nominal value.

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In IC fabrication design parameters like resistor or capacitor values track each other – i.e. change in their values are strongly correlated. So, to design these circuits we use the multiparameter tracking sensitivity

ktypeofelementsall

i

Fhik SMTS

1

Since all elements of the same kind (e.g. capacitors) have similar values of hi/hi

and for such elements (only)

FhiS

h

h

F

F

then, for all types of elements, the worst case variation with tracking is given by

kkk MTS*t

F

F

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hi/hiHowever, worst case situation is very unlikely to happen in practice. Fabricated device parameter deviations follow a statistical distribution. Two commonly used distributions to model parameter deviations are uniform and normal distributions

h

hprob

-t 0 t h

h

For uniform distribution:

oth

hprob

*2

1

22


For normal distribution:

2

2

1

2

1)(

h

h

eh

hprob

The function deviation F/F becomes a random variable with its own distribution. For large circuits F/F has approximately

normal distribution with zero mean and variance h

hFhi

F

F S 22

provided that the component variations are statistically independent, where

ondistributinormalfort

ondistributiuniformfort

i

i

hi

hi

9

32

2

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If all the tolerances are equal, and hi/hi have the same distribution then the standard deviation can be calculated from

MSSh

h

F

F

where MSS is the multiparameter statistical sensitivity

i

2FhiSMSS

Actual variation will lie in the interval F

F 68% of the time,

F

F2 95%, and F

F2 99.7%.

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Example

We have KCL1:

021121 outvsCEGvsCG

KCL2:

0GvvGG 3out143

so

EGvsCGG

GsCGv

GG

Gv outout 12

43

321

43

31 &

4231

431out

GsCGG

GGG

E

vT

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Example (cont’d)

Let us assume that all elements have tolerances t = 1% and s = 1. Let’s calculate various multiparameter sensitivities and use them to predict deviations of the transfer function T from its nominal value.

3

4

3

4

GsCGG

GsCG

D

G1SSS

4231

423

1D

1GN

1GT

1G

3

4

3

4sG

D

CSSS 4

2D

2CN

2CT

2C

15

8

3

1

5

1G

D

G

GGG

GGSSS 1

3

431

13D

3GN

3GT

3G

15

8

3

4

5

4sC

D

G

GGG

GGSSS 2

4

431

14D

4GN

4GT

4G

15

8

3

4

5

4sC

D

G

GGG

GGSSS 2

4

431

14D

4GN

4GT

4G

15

56

15

282

15

8

15

8

3

4

3

4SWCMS i

3

4SSSMTS T

4GT

3GT

1Gi 03.22252259

16

9

16 44 GG

MSS

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Example (cont’d)

For the nominal values the transfer function can be evaluated as

667.13

5T

Let us discuss the effect of 1% changes assumingG1 = .99 C2 = 1.01 G3 = 1.01 G4 = 3.96

The actual transfer function value can be calculated as

640.13

92.4

96.301.101.199.

)96.301.1(99.'T

so

%6.1016.0667.1

0267.0

T

T

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Example (cont’d)

while the estimate for such a change using different multiparameter sensitivities is as follows.Worst case analysis

%73.315

56*%1*

WCMSt

T

T

Worst case analysis with tracking

%67.23

42%1** 21

MTStMTSt

T

T (still too big)

(too pessimistic)

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Statistical analysis

If tolerances are distributed uniformly then the standard deviation

0117.003.23

tMSS

h

h

T

T

and if the tolerances are distributed normally then thestandard deviation

0068.003.23

tMSS

h

h

T

T

indicating that 95% of the time T

T will be less than %34.22 T

T

in the uniform case and less than %35.12 T

T in the normal case

Since the true deviation that was 1.6% exceeded the 95% limits for the standard deviation of the normal distribution so our case was rather uniform than normal.

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Sensitivities to Parasitics and Operational Amplifiers

Since parasitics have nominal values equal zero we cannot calculate sensitivities to these elements in the regular way. Denote parasitics by i . We have

parasiticsall

ii

i

elementsnozeroall

i F

F

hi

hi

F

hi

hi

F

F

F

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1

or equivalent

hi

hiS

F

F Fhi i

FiS

semi-normalized sensitivity

hi

hi as well as i are fixed for a specific technology, so the

only way to reduce functional variation of F is to have design with small F

hiS and FiS

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Sensitivities to Parasitics and Operational Amplifiers (cont’d)

To evaluate FiS we analyze the network in the regular way, calculate

i

F

and finally substitute i = 0 at the final result.

In the case of Op Amp we may consider the inverse of its amplification as a parasitic

we have

Avvv jik

or 0Bvvv kji

where

A

1B

B is parasitic. If B 0 then we obtain ideal Op Amp.

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Example:

Find the sensitivity TBS for the transfer function T of the

network shown where A

1B

from the previous example

2112122121

21out

GsCsCGGsCGsCGGB

GG

E

vT

TD

sCGGsCGsCGGGG

TB

TS BTB 2

12122121210

1

21

121221210B GsC

sCGGsCGsCGG

D

1

B

D

1 ee 616 computer aided analysis of electronic networks lecture 9 instructor: dr. j. a. starzyk,...

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