polynomial coefficient based multi-tone testing of analog ...vagrawal/talks/natw09/pan_natw.pdf ·...

Motivation Our Idea Generalization Results Fault Diagnosis Conclusion and Future Work

Polynomial Coefficient Based Multi-ToneTesting of Analog Circuits

Suraj Sindia Virendra Singh Vishwani D. AgrawalIndian Institute of Science Auburn University

Bangalore, India Auburn, AL, USA

18th North Atlantic Test WorkshopNew York, NY

May 14, 2009

Suraj Sindia @ NATW 2009 Non-Linear Analog Circuit Testing


Analog Circuit Testing

To determine catastrophic (open or short) faults and fractionaldeviations in circuit components from their nominal values.



Analog Circuit Testing

To determine catastrophic (open or short) faults and fractionaldeviations in circuit components from their nominal values.

In this talk

To propose a method to detect fractional deviations of circuitcomponents from their nominal values in a large class ofcircuits.



Outline

1 Motivation

2 Our Idea

3 Generalization

4 Results

5 Fault Diagnosis

6 Conclusion and Future Work



Motivation

To Develop an Analog Circuit test scheme

Suitable for large class of circuits



Motivation



Detects sufficiently small parametric faults



Motivation




Small area overhead – requires little circuit augmentation



Motivation




Small area overhead – requires little circuit augmentation

Large number observables – handy in diagnosis



Previous Approaches

Important previous techniques

IDDQ based test – Intrusive, Area overhead is high[Chakravarty ’97]

Signal flow graph – Complexity order is high[Bushnell et al. ’97]

Transfer function based test – Valid only for LTI systems[Savir and Guo ’03]

Digital assisted analog test – Intrusive[Tim Cheng et al. ’06]

Polynomial coefficient based test – DC test[Sindia et al. ’09]



Transfer Function Coefficient Based Test

Vin Vout

R1 R2

C1 C2

Second order low pass filter

H(s) =1

(R1R2C1C2) s2 + (R1C1 + (R1 + R2)C2) s + 1



Our Idea

Taylor series expansion of circuit function about vin = 0 at Multitones

vout = f (vin)

vout = f (0) + f ′(0)1! vin + f ′′(0)

2! v2in + f ′′′(0)

3! v3in + · · · + f (n)(0)

n! vnin + · · ·



Our Idea

Taylor series expansion of circuit function about vin = 0 at Multitones

vout = f (vin)

vout = f (0) + f ′(0)1! vin + f ′′(0)

2! v2in + f ′′′(0)

3! v3in + · · · + f (n)(0)

n! vnin + · · ·

Ignoring the higher order terms we have

vout ≈ a0 + a1vin + a2v2in + · · · + anvn

in

where every ai ∈ < and is bounded between its extreme valuesfor

ai ,min < ai < ai ,max ∀i 0 ≤ i ≤ n



Our Idea (Contd..)

In a nutshell

Find the Vout v/s V in relationship at DC and “relevant”frequencies.



Our Idea (Contd..)

In a nutshell


Compute the coefficients of fault-free circuit



Our Idea (Contd..)

In a nutshell



Repeat the same for CUT by curve fitting the I/O response



Our Idea (Contd..)

In a nutshell




Compare each of the obtained coefficients with fault-freecircuit range



Our Idea (Contd..)

In a nutshell




Compare each of the obtained coefficients with fault-freecircuit range

Classify CUT as Good or Bad



Cascaded Amplifiers

Vdd

R2R1 IM1 IM2

M1 M2

Vin

Vout

Two stage amplifier with 4th degree non-linearity in V in

vout = a0 + a1vin + a2v2in + a3v3

in + a4v4in



Polynomial Coefficients

a0 = VDD − R2K(

WL

)

2

[

(VDD − VT )2+ R2

1K 2(

WL

)2

1 V 4T

−2(VDD − VT )R1(

WL

)

1 V 2T

]

a1 = R2K(

WL

)

2

[

4R21K 2

(

WL

)2

1V 3

T + 2(VDD − VT )R1K(

WL

)

1VT

]

a2 = R2K(

WL

)

2

[

2(VDD − VT )R1K(

WL

)

1− 6R2

1K 2(

WL

)2

1V 2

T

]

a3 = 4VT K 3(

WL

)2

1

(

WL

)2

2R2

1R2

a4 = −K 3(

WL

)2

1

(

WL

)2

2R2

1R2



Nature of Polynomial Coefficients

x0 x1 x2 x3x4 x5

x

f(x) Decreasing

Increasing

Non-linear, Non-monotonic function is decomposed intopiecewise monotonic functions



MSDF Calculation

Definition

Minimum Size Detectable Fault(ρ) of a circuit parameter isdefined as its minimum fractional deviation to force atleast oneof the polynomial coefficients out of its fault free range



MSDF Calculation

Definition


Overview of MSDF calculation of R1 with VDD=1.2V, VT=400mV,

(

WL

)

1= 1

2

(

WL

)

2= 20, and K = 100µA/V2



MSDF Calculation

Definition


Overview of MSDF calculation of R1 with VDD=1.2V, VT=400mV,

(

WL

)

1= 1

2

(

WL

)

2= 20, and K = 100µA/V2

Maximize a0{

1.2 − R2,nom(1 + y)

(

2.56x10−3 + R21,nom(1 + x)21.024x10−7

−5.12x10−4R1,nom(1 + x)

)}

subject to a1, a2, a3, a4 being in their fault free ranges and

−α ≤ x , y ≤ α



MSDF Calculation (contd..)

Assuming single parametric faults, ρ for R1

ρ = (1 − α)1.5 − 1 ≈ 1.5α − 0.375α2



MSDF Calculation (contd..)

Assuming single parametric faults, ρ for R1

ρ = (1 − α)1.5 − 1 ≈ 1.5α − 0.375α2

MSDF for Cascaded Amplifier with α = 0.05

Circuit parameter %upside MSDF %downside MSDFResistor R1 10.3 7.4Resistor R2 12.3 8.5



Let us Generalize



Generalization – Fault Simulation

1 Start

2 Choose frequency for fault simulation

3 Apply sweep to input and note corresponding outputvoltage levels

4 Polynomial Curve fit the obtained I/O data – find thecoefficient values of fault free circuit

5 Simulate for all parametric faults at the simplex ofhypercube

6 Find min-max values of each coefficient (Ci ) fromi = 1 · · ·N across all simulations

7 Repeat process at all chosen frequencies

8 Stop



Generalization – Test Procedure

1 Start

2 Choose a frequency

3 Sweep input and note corresponding output voltage levels

4 Polynomial Curve fit the obtained I/O data. Obtain coefficientsCi ∀ i = 1 · · ·N

5 Start with first coefficient

6 Consider next coefficient Ci+1

7 |Ci | > |Ci,max | or |Ci | < |Ci,min|?If True go to step 11

8 i < N? If True go to step 6

9 Repeat steps 2–8 at all desired frequencies

10 Subject CUT to further tests. Stop

11 CUT is faulty. Stop



Results – Elliptic filter

−

+−

+−

+Vout

VinR1

R2

R4

R5

R3 R7

R6

R8

R9

R10

R11 R12

R13

R14

R15

C1

C3

C4

C5

C6 C7

C2



Results - Curve fitting at DC

0 1 2 3 4 5−3

−2

−1

0

1

2

3

4

5

Input DC voltage (Vin)

Out

put V

olta

ge(V

out)

Simulated

5th degree polynomial

a5 = 0.039463a4 = −0.50514a3 = 2.1309a2 = −2.5487a1 = −3.498a0 = 4.5341

vout=4.5341−3.498vin−2.5487v2in+2.1309v3

in−0.50514v4in+0.039463v5

in



Results - Curve fitting at 100Hz

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5−1

0

1

2

3

4

5

Input Voltage, Vin(v)

Out

put V

olta

ge, V

out(

v)a5 = 0.049 a4 = − 0.78 a3 = 4.4 a2 = − 11 a1 = 7.9 a0 = 3

Simulated

5th degree Polynomial

vout=3 − 7.9vin − 11v2in + 4.4v3

in − 0.78v4in + 0.049v5

in




0 1 2 3 4 5−2

−1

0

1

2

3

4

5


Out

put V

olta

ge, V

out(

v)

a5 = 0.054 a4 = − 0.77 a3 = 4 a2 = − 8.6 a1 = 5.4 a0 = 2.5

Simulated 5th degree Polynomial

vout=2.5 + 5.4vin − 8.6v2in + 4v3

in − 0.77v4in + 0.054v5

in




0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5−0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8


Out

put V

olag

e, V

out(

v)

Simulated


a5 = 0.024a4 = −0.35a3 = 1.8a2 = −3.9a1 = 2.4a0 = 1.2

vout=1.2 + 2.4vin − 3.9v2in + 1.8v3

in − 0.35v4in + 0.024v5

in




0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5−0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4


Out

put V

olta

ge, V

out(

v)

a5 = 0.0043 a4 = − 0.063 a3 = 0.34 a2 = − 0.74 a1 = 0.48 a0 = 0.23

Simulated


vout=0.23 − 0.48vin − 0.74v2in + 0.34v3

in − 0.063v4in + 0.0043v5

in



Results at 1000Hz – Elliptic filter

Parameter Combinations Leading to Max Values of Coefficientswith α = 0.05 at 1000Hz

Circuit Parameters (Resistance in ohm,Capacitance in farad)Nominal Values a0 a1 a2 a3 a4 a5

R1 = 19.6k 18.6k 18.6k 20.5k 20.5k 20.5k 18.6kR2 = 196k 205k 205k 205k 205k 186k 186kR3 = 147k 139k 139k 154k 139k 139k 139kR4 = 1k 950 950 1.05k 1.05k 1.05k 1.05k

C4 = 2.67n 2.5n 2.8n 2.5n 2.5n 2.5n 2.5nC5 = 2.67n 2.5n 2.5n 2.5n 2.5n 2.5n 2.8nC6 = 2.67n 2.5n 2.8n 2.5n 2.8n 2.5n 2.8nC7 = 2.67n 2.5n 2.8n 2.8n 2.8n 2.8n 2.5n



Results at 1000Hz – Elliptic filter

Parameter Combinations Leading to Min Values of Coefficientswith α = 0.05 at 1000Hz

Circuit Parameters (Resistance in ohm,Capacitance in farad)Nominal Values a0 a1 a2 a3 a4 a5

R1 = 19.6k 18.6k 18.6k 18.6k 18.6k 20.5k 20.5kR2 = 196k 205k 186k 186k 205k 205k 205kR3 = 147k 139k 139k 154k 139k 139k 139kR4 = 1k 1.05k 950 1.05k 950 950 1.05k

C4 = 2.67n 2.5n 2.5n 2.8n 2.5n 2.5n 2.8nC5 = 2.67n 2.8n 2.8n 2.8n 2.8n 2.8n 2.8nC6 = 2.67n 2.5n 2.5n 2.8n 2.8n 2.8n 2.5nC7 = 2.67n 2.8n 2.5n 2.5n 2.5n 2.5n 2.8n



Results at DC – Elliptic filter

Parameter Combinations Leading to Max Values of Coefficientswith α = 0.05 at DC

Resistance a0 a1 a2 a3 a4 a5

(ohm)R1 = 19.6k 18.6k 20.5k 20.5k 20.5k 18.6k 18.6kR2 = 196k 186k 205k 186k 186k 186k 205kR3 = 147k 139k 154k 154k 154k 139k 154kR4 = 1k 950 1010 1010 1010 1010 1010

R5 = 71.5 70 80 80 70 80 70R6 = 37.4k 37.4k 37.4k 37.4k 37.4k 37.4k 37.4kR7 = 154k 161k 161k 146k 161k 146k 146kR11 = 110k 115k 115k 104k 115k 104k 104kR12 = 110k 104k 115k 104k 104k 104k 104k



Fault Detection at Multi-Tones

Injected fault Coefficients out of Bounds at DetectDC f1=100Hz f2=900Hz f3=1000Hz

R1 down 15% a0 − a4 a1 − a4 a3, a5 a2, a4 YesR2 down 5% a2, a5 a1, a3 a1, a5 a1, a2, a5 YesR3 up 10% a1, a2, a3 a3, a5 a0, a3, a4 a1, a3, a4 Yes

R4 down 20% a0 − a3 a1 − a2 a2, a3 a1, a2, a3 YesC5 up 5% − a0, a1 a1, a5 a1, a2 Yes

C6 up 15% − a3, a4 a1, a2, a4 a3, a4, a5 YesC7 up 15% − a1, a4 a1, a3, a4 a1, a3, a5 Yes



Fault Diagnosis

Definition

To determine the circuit parameters responsible for deviation ofcircuit from its desired behavior.



Fault Diagnosis

Definition

To determine the circuit parameters responsible for deviation ofcircuit from its desired behavior.

Sensitivity based diagnosis

SCipk

=pk

Ci

∂Ci

∂pk

P(δpk|δCi) = φ

(

SCipk

δpk

δCi

)

φ is any probability measure. We chose negative exponentialfunction



Fault Diagnosis

p1

p2

p3

...

pk

C1

C2

C3

.

.

Ci

.

Cn

1

1

CpS

1

k

CpS

n

k

CpS

2

2

CpS

3

k

CpS

Parameter space

Coefficient space



Fault Diagnosis at Multi-Tones

Parametric Fault Diagnosis with Confidence Levels ≈ 98.9%

Injected fault Diagnosed fault sites at DeductionDC 100Hz 900Hz 1000Hz

R1 down 15% R1, R4 R1 R1, R2 R1, R2, C1 R1

R2 down 5% R2 R2, C1 R2, R3, C1 R2, R3 R2

R3 up 10% R1, R3 R3, C3 R3, R4, C3 R3 R3

R4 down 20% R1, R4 R1, R4 R2, R4, C1 R1, R2, R4 R4

C5 up 5% − C5 R12, C5 C5 C5

C6 up 15% − R10, C6 C6, C7 C6, C7 C6

C7 up 15% − C6, C7 C7 C6, C7 C7



Conclusions

Technique for parametric fault detection in analog circuits –faults as small as 10% were uncovered



Conclusions


Could uncover parametric deviations in reactive elementsas input is swept at DC and selected frequencies



Conclusions


Could uncover parametric deviations in reactive elementsas input is swept at DC and selected frequencies

Technique to diagnose faults through multi-frequencyexcitation using sensitivity



Future Work

“Monotonizing” the coefficient variation with circuitparameter



Future Work


Increasing the sensitivity of coefficients to parameters (toenhance fault coverage), using “smart” transforms



Future Work


Increasing the sensitivity of coefficients to parameters (toenhance fault coverage), using “smart” transforms

Techniques for optimal choice of frequencies at which CUTcould be excited



Acknowledgments

Department of Science and Technology, Govt. of India

Pawan Kumar, SERC, IISc

Pramod Subramanyan, SERC, IISc

Thanks for your Attention!


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