iew-15.pdf

Demystifying Measurements

from HBM and CDM ESD Testers

(How to Solve ESD Problems by

Thinking in Pictures)

Tim Maloney

Intel Corp.

IEW Seminar

Outline

Tim Maloney, Intel Corp. 2

•Quick review of math background

•Complex impedances, Laplace Transforms, s-domain •HBM as a 2-pole circuit

•Extracting RC and network from waveform •Measurement aside: Transformer effects •CDM as a 2-pole circuit

•Network from I_pk, Q_fp, Q_total and V₀

•Spark rise time and scope response may add poles •Oscilloscope bandwidth limits and effects on waveforms

•Measurement aside: 2-pole oscilloscope model •Other ESD examples: CCDM, TLP

•Antenna response to CDM events •ESD chair antenna

•Appendices:

Flow Chart of Outline

Tim Maloney, Intel Corp. 3

Math review

HBM 2009-11

Antennas 2011-13

Thermal 2013-15

IEC, Chairs

2015 TLP, CCDM

2012

CDM 2013-14

Math Review

Tim Maloney, Intel Corp. 4

Double Exponential (e.g., HBM) F(s) f(t) Differentiate: s d/dt

Integrate: 1/s ∫ Unit step: 1/s

Impulse: 1 ↑

Laplace Transform and inverse:

Gaussian Convolution

Tim Maloney, Intel Corp. 7

⇔

“flip and slide”*

) exp(

)

( ₂

2

f

t t

f

τ

− =

2 2

2

*g f g

f τ τ

τ = +

Source: www.wolfram.com

Convolution in Excel

Tim Maloney, Intel Corp. 8

Suppose you have two functions with identical time steps and want to convolve them—these could be, for example, a digital record of a scope trace and a calculated filter impulse response in the time domain.

Once you have the filter impulse response, there’s no reason to sweat FFTs, number of points in powers of 2 (don’t you hate that?), etc., you can install free “Excellaneous” VB macros, at

http://www.bowdoin.edu/~rdelevie/excellaneous/#downloads , into Excel and use one of several convolution macros.

Macrobundle12 is the latest, appearing as a Word file.

This is what was done for modeling oscilloscope response in a 2011 EOS/ESD paper,

Flow Chart of Outline

Tim Maloney, Intel Corp. 9

Math review

HBM 2009-11

Antennas 2011-13

Thermal 2013-15

IEC, Chairs

2015 TLP, CCDM

2012

CDM 2013-14

Two-pole HBM (or CDM) Network

,

1

)

(

+

+

=

RCs

LCs

Cs

s

Y

)

(

)

(

),

1

(

)

(

a

s

s

aV

s

V

e

V

t

V

+

=

−

=

C

L

R

So

)

)(

)(

(

)

(

c

s

b

s

a

s

abc

CV

s

I

+

+

+

=

_and

c

b

RC

=

1

+

1

(Vieta, 1579)

Slide 11

Elmore Theorem (1948)

∫

=

dt

e

t

h

s

H

(

)

(

)

∫

⋅

⋅

⋅

+

−

+

−

=

6

2

1

st

s

t

s

t

dt

t

h

(

)[

]

∑

_∫

−

=

1

dt

t

h

t

s

k

!

(

)

.

)

(

A waveform

h(t)

transforms and expands in powers of t:

0

_{moment (k=0) is integral (total charge Q}

t

for HBM)

Waveform Moments

∫

∞

=

0

0

I

t

dt

m

(

)

= charge Q

= centroid (time)

= Elmore Delay

= 2

_moment

For a current function,

I(s)=

m

(

1+m

s+m

s

_+…+m

n

s

)

∫

∫

=

−

dt

t

I

dt

t

tI

m

)

(

)

(

∫

∫

=

2

I

t

dt

dt

t

I

t

m

)

(

!

)

(

Slide 13

Simple extraction of any moment

Start with a network function (impulse response),

N(s)=

a

+a

s+a

s

_+…+a

n

s

Step response (HBM, CDM, TLP) is

M(s)=

a

/s+a

+a

s+…+a

s

DC offset (

a

/s)

gives

a

,

remove to

get

L(s)=

a

+a

s+…+a

s

Now integrate to get

K(s)=

a

/s+a

+…+a

s

DC offset gives a

.

Successive

Integration*

Tim Maloney, Intel Corp. 14

) ) 2 ) (( ) ( 1 ( ) 2 1 )( 1 ( ) ( 2 2 2 0 2 2 2

0 ≈ − + + + − − − +_

+ + +

+

= CV RC s RC LC RC s

s s LCs RCs CV s

I _{e e e} e

e e τ τ τ τ τ τ

0.5 1.0 1.5 2.0

2 2 4 6 8

Expand I(s) around s=0, using

Now integrate, i.e., multiply by 1/s, to get charge Q(s):

) ) 2 ) (( ) ( 1 ( ) ( ) ( 2 2

0 − + + + − − − +

≈

= RC RC LC RC s

s CV s s I s Q e e e e τ τ τ τ

This is primarily a step of height CV₀ , the total charge Q

0.5 1.0 1.5 2.0

2.5 Now suppose we normalize, remove the

step and integrate again… current charge  + + + + = − 3 2 1 1 1 x x x x

Time Constant from Integration

Tim Maloney, Intel Corp. ₁₅

 + − − − + + + − ≈       − ) 2 ) (( ) ( 1 ) (

1 2 2

0 e e e e RC LC RC s RC s CV s Q s τ τ τ τ

Again, suppose we normalize, remove the step and integrate again…

This is a (negative) step that measures RC+τ_e (=A₁-A₂+A₃ areas):

0.5 1.0 1.5 2.0

0.2 0.4 0.6 0.8 1.0 1.2 A₁ A₂ A₃ Q(t)/CV₀ nsec

We now have C from V₀ and integrating I(t) to get Q. We get R from the 2nd

integration if we can estimate

About Successive Integration

Tim Maloney, Intel Corp. 16

0.5 1.0 1.5 2.0

2 2 4 6 8 nsec amps

Centroid, 267 psec

RC+τ_e is the centroid (“balance point”) of the original current waveform

Remove the time constant, continue the integration and you get an expression including inductance L. This can be repeated to the limits of measurement accuracy.

 + − − − + + + − ≈       − ) 2 ) (( ) ( 1 ) (

1 2 2

Slide 17

Obtaining true RC time constant

RC=A/Q

, the Elmore Delay of the integrated HBM current

t t

Q

Q

RC

∫

∫

∞













−

=

dt

d

τ

τ

h

t

)

(

Current Transformers for HBM Waveforms

)

)(

(

)

(

b

s

a

s

s

s

T

+

+

=

Transfer function:

where

a

= 1/

τ

(lo-freq cutoff)

b

= 1/

τ

(hi-freq cutoff )

s

=

σ

+ j

ω

(complex frequency)

Zero at

s

= 0 guarantees

eventual zero integral

Slide 19

Tektronix CT1, CT2 Current Probes*

*figures used with permission from Tektronix, Inc.

1 MHz

_{10 KHz}

Current Probe Step Response

•

For CT1,

τ

≈

6.35

µ

sec

•

For CT2,

τ

> 100

µ

sec

•

This has considerable effect as “real” waveform transforms

to measured waveform

– Impulse response is derivative of step response

– Convolution means eventual negative values measured

V

t

τ

V

t

τ

step

impulse

Slide 21

HBM 0

Ω

Waveform, CT1 Convolution

∫

∞

−

=

0

1

(

t

)

HBM

(

τ

)

CT

1

(

t

τ

)

d

τ

HBM

HBM waveform, calculated

0 1 2 3 4 5 6 7

0 100 200 300 400 500 600 700 800 900

time, ns c u rr e n t, a rb u n it s

HBM and CT1 convolution

-0.02 -0.01 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07

660 680 700 720 740 760 780

time, ns c u rr e n t, a rb u n it s CT1 HBM pure HBM

Calculation shows zero crossing of measured waveform

Slide 22

4-pole HBM model, with socket cap C

₂

C

C

R

L

C

R

When R

= 500

Ω

,

,

1

)

1

(

)

(

s

b

s

b

s

b

s

b

s

C

R

C

V

s

I

− − − −

+

+

+

+

+

=

b

=R

(C

+C

)+R

(C

+C

),

Slide 23

0 0.5 1 1.5 2 2.5 3 3.5

0 100 200 300 400 500

time, nsec

Cu

rre

n

t,

A

4 kV 0

Ω

HBM Waveform with CT1

t

= 126 nsec

decay constant t

is out of spec

CT2 removes droop and t

is in spec

36.8% I

Cu

rre

n

t,

Effects of Low-Frequency Cutoff

CT2 and CT1, 500 ohm

-0.01 -0.005 0 0.005 0.01 0.015 0.02 0.025 0.03

760 780 800 820 840 860 880 900 920

time, nsec

cu

rr

en

t,

A

CT1

_CT2

CT2 provides a more accurate waveform tail

Slide 25

0

Ω

and 500

Ω

HBM waveforms

)

C

(C

R

C

R

C

R

b

C

R

2 hb

1 hb

hb 1

hb 500

1 500

hb hb

+

+

=

−

=

=

−

τ

τ

Socket Capacitance C

₂

, from Waveforms

hb

C

R

C

=

−

−

500 0 500

2

τ

τ

Tester

C

, pF

R

, k

Ω

τ

, nsec

τ

, nsec

C

, pF

MK-4

114

1.412

161

225

13.9

Zapmaster

512

101

1.445

146

222

51

Zapmaster

Slide 27

Designing for Reduced Socket Capacitance

C₁

C_hb

R_hb

Z₀

R₁

V₀

Z_in

Most tester “socket” cap is distributed, can benefit

from high Z

Could raise Z

with ferrites…

2 1

2 1

2 0 2

1

1

)

1

(

1

1

C

s

R

R

Z

sC

R

Z

Y

in

in

=

≈

+

−

=

+

α

α

, reduces effective cap

2 1 0

C

L

Slide 28

Using Distributed Resistance in Tester

α

C

Z₀’_{, R}

a

Y

R

R

Take some of

R

and

distribute in-line

























+

′

−

+

+

+

≈

)

(

1

1

1

R

R

R

Z

R

R

R

sC

R

R

Y

C

L

Z

=

α

, reduces effective cap

For

α

<0.5, need

2 1 2

0 2

2

Z

R

R

+

′

>

If R

= 500

Ω

,

Slide 29

Additional Discussions in the Paper

•

Expression for full loop HBM current I

(s) for 4-pole

model

•

Expansion of H(s) transfer functions

•

Pole-zero sum rules

– From Vieta, 1579

– Use sum rule to calculate effect of inductance L

on 0

Ω

HBM waveform rise time

•

These and other concepts adopted from the signal

integrity field

HBM Paper Conclusions

•

Current probe transformers examined

– Tek CT2 is more agreeable for HBM time

decay spec, little droop due to low-f cutoff

• CT1 still ok for most spec-related measurements

• CT2 is now called out in JS-001 HBM spec (2012)

– Tek CT2 also works better for integrals,

centroids needed for HBM circuit model

elements (accurate asymptotic properties)

•

4-pole HBM model with socket cap re-examined

– Analysis: Time constants (centroids) of 0

Ω

and 500

Ω

waveforms readily yield socket cap

value C

Slide 31

HBM Paper Conclusions, cont’d

•

Synthesis: Design strategies for reducing tester C

– “Socket” cap is mostly distributed along line

– Analytical model points to lower effective C

:

• Effective line Z

can be raised

– Ferrite loading

• Part of R

(e.g., 300 out of 1500

Ω

) can be distributed

along that line

• Effective cap can thus be reduced by half or more

More HBM Trickery (and pictures)

See Appendix II, IEW11 presentation

•

HBM double exp: What are the poles?

•

Short

τ

(a few nsec) from I(t) rise time

•

Long

τ

(about 150 nsec) from

integrated current Q(t) rise time

•

Adjustments for 3-pole HBM

Flow Chart of Outline

Tim Maloney, Intel Corp. 33

Math review

HBM 2009-11

Antennas 2011-13

Thermal 2013-15

IEC, Chairs

2015 TLP, CCDM

2012

CDM 2013-14

Slide 34

CDM Tester simulates event

Vf

Cg 300 MΩ

Cf Cfrg

1 ohm disk resistor here dielectric to

field plate

top gnd plane

. *

   

 

+ +

   

 

+ =

Cfrg Cf

Cfrg Cf

Cg Cf

Cg Cf Vf

Q_imm

Immediate charge packet is

3 capacitors collapse to one equivalent “fast” cap

Simplified CDM Network

Slide 35

2

1

)

(

LCs

RCs

Cs

s

Y

+

+

=

2 0

1 ) ( ) ( )

(

LCs RCs

CV s

Y s V s

I

+ +

= =

R

L

C

V(s)

≈

V

/s

•

Examine step response of this network

•

R is spark resistance, ~25-30 ohms

2-pole First Peak charge, Q

_fp

First half-cycle is what causes damage

2 4 6 8 10

0.1 0.1 0.2 0.3 0.4 0.5

2 4 6 8 10

0.2 0.4 0.6 0.8 1.0 1.2

Q_fp (area)

Q_fp (16% overshoot)

amps

nC

t, nsec

t, nsec Example for D=0.5, RC₀=1 nsec, C₀V₀=1 nC:

1

),

1

exp(

1

<

−

−

+

=

D

D

D

V

C

Q

π

( 0 0 ₂

s s s V C s Q + + = 2 0 0 1 ) ( s s V C s I + + =

See Appendix A, esd13

0 0

2

LC

RC

D

=

2-pole I

_max

= f(R

_eq

,D)

Slide 37

•

For D>1

•

For D<1

))

1

(

tanh

1

exp(

2

2 0 max

D

D

D

D

D

R

V

I

−

−

−

=

))

1

(

tan

1

exp(

2

2 0 max

D

D

D

D

D

R

V

I

−

−

−

=

Also maps to

package size

I

trend

Lower C



lower

D, lower I

0 1 2 3 4 5

g

(D)

D -- Damping factor

I_max = g(D) *V₀/R_eq

See Appendix B, esd13

0 0

2-pole RLC calculation

Q_a

Q_fp V₀ I_max

C₀

L_eq D

C_max

R_eq

derived

quantity

measured

quantity

0 0

2

LC

RC

D

=

0 0

V

Empirical Fit for C

₀

vs. Target Size

•

Metal targets (JEDEC, P4, P6) represent a worst

case for equivalent capacitance of package area

•

Tester inductance and spark resistance give the

rest of the circuit model

Slide 39 y = 0.6696x

0 5 10 15 20 25 30

0 10 20 30 40 50

C

o

, p

F

Sqrt(Area), mm

Co vs. Target Size

250V 500V

Linear (500V)

τ

= R

_eq

*C

₀

is linear vs. C

₀

small target

large target

Leq=2.39±0.6nH

Leq=4.26±0.4nH

•

Dielectric thickness variation, 14-59 mils, several

companies

•

Linear Req*C

happens repeatedly

•

JEDEC targets, no ferrites; note low inductance Leq

•

20.6

Ω

slope, 26 psec offset

More

τ

= R

_eq

*C

₀

linearity

Slide 41 y = 20.562x + 68.704

0 100 200 300 400 500 600

0 5 10 15 20 25

ta u = R e q *C o , p s e c Co, pF Orion2, 250V

Leq=10.33±1.49 nH

y = 27.798x + 8.8219

0 100 200 300 400 500 600 700 800

0 5 10 15 20 25 30

ta u = R e q *C o , p s e c Co, pF Orion2, 500V

Leq=12.03±2 nH

•JEDEC tester with ferrite (higher Leq)

•JEDEC coins plus P4, P6 targets

20.6 Ω, 68.7 psec offset

τ

= R

_eq

*C

₀

linearity for 7 metal targets

y = 27.482x + 56.849

0 100 200 300 400 500 600 700 800

0 5 10 15 20 25 30

ta

u

=

R

e

q

*C

o

, p

s

e

c

Co, pF

JEDEC, 500V

•JEDEC tester with ferrite, 500V

•27.5

Ω

, 56.8 psec offset

Air Spark + 25 ohms

Slide 43

Leq=11.62±1.16 nH

•

Terminate CDM2 test head with SMA tee, 50

Ω

in each branch

•

50

Ω

termination and 50

Ω

scope channel; 25

Ω

instead of 1

Ω

•

Test like regular CDM, with air spark in series

•

No ferrites but extra Leq seems due to 50 ohm mismatch at top

of test head

•

Req = 48.9

Ω

; 25

Ω

plus air spark

Example of 2-pole RLC fit

•

500V, 25

Ω

+ air spark, 4.57pF small target

•

I

and Q

match are guaranteed by method

2-pole RLC fit for large JEDEC target

•

JEDEC tester with ferrite, 500V

•

RLC fit is 28.6

Ω

, 11.7 nH, 16.3 pF

Slide 45

-4 -2 0 2 4 6 8 10 12

0 1 2 3 4

cu

rr

en

t, A

time, nsec

Large JEDEC target, 500V

measured 2-pole fit

What can we do so far?

•

With these methods, we can readily:

•

Estimate equivalent cap C

from package size

•

Estimate R

from slope-intercept of

τ

= R

C

as

measured by metal targets

•

Estimate L

for ferrite/no ferrite situation

•

Therefore, generate a complete set of worst-case

CDM waveforms for a given product situation

– Focus on I

and Q

at test voltage V

I

_max

vs. Q

_fp

, 25 ohm + air spark

Slide 47

sm

lg

P4 P6

•Come close to green (JEDEC) point at intermediate voltage for

each target, closest approach being a least squares fit

•JEDEC legacy result expected when I

, Q

match

•Data and analysis for 50

Ω

and 25

Ω

CCDM/CDM2 could be even

better. See III.5 of text, esd13 paper.

RLC element trend: Compression of D

•

Why is the damping factor D compressed into a

mid-range for so many measurements?

Slide 48

(a)

0 0.2 0.4 0.6 0.8 1 1.2

0 2 4 6 8 10 12 14

da

m

pi

ng

fa

ct

or

, D

Co, pF

D vs. Co, dielectric thickness variation

(b)

0 0

2

LC

RC

D

=

Damping factor and EM field dissipation

•

D=1/

√

2 maximizes field energy burn rate of spark,

thus minimizing the time integral, A

(=

√

2)

•

With optimized D(t) it is possible to beat

√

2…

Slide 49 0 0

2

LC

RC

D

=

0 1 2 3 4 5 6

Po

w

er

time units

I²R vs. time

D=0.5 D=0.707 D=1

A

0 1 2 3 4 5

fr ac tio n o f E time units

Remaining Field Energy vs. t

D=0.707

(b)

A

See Appendix C, esd13

0

LC

Why you should read the Appendices

•

Appendix A:

First Peak Charge (Q

)

– Take generalized i(t) expression for RLC,

integrate first cycle, find that Q

/C

V

= f(D)

•

Appendix B:

Peak Current (I

)

– Differentiate i(t) to find that I

is V

/R

times a

unique monotonic function g(D)

•

Appendix C:

Maximum Burn Rate of Field Energy

– Find i

_(t)R

eq

=P(t), convert to P(s), find that field

collapse time is

τ

=D+1/2D, min at D=1/

√

2

Conclusions

•

RLC calculation scheme for CDM waveforms

– Used 4 measured quantities, V

, Q

, Q

, I

– Spreadsheet extraction of RLC



D to fit all 4

quantities precisely

•

Linear trends allow RLC prediction:

– C

goes as

√

Area of target/package

–

τ

= R

*C

vs. C

is linear (slope-intercept)

• Fits JEDEC, air spark + 25

Ω

, ferrite-free tests

•

JEDEC legacy conditions (I

,Q

) approximated

– With air spark + 25

Ω

, possibly better with CCDM

•

Damping factor D stays near maximum burn rate

Tim Maloney, Intel Corp. 52

Simplified CDM Network (again)

2

1

)

(

LCs

RCs

Cs

s

Y

+

+

=

Examine step response of this network

R is spark resistance, ~25-30 ohms

V(s)=Vo/s

2 0

1 ) ( ) ( )

(

LCs RCs

CV s

Y s V s

I

+ +

Tim Maloney, Intel Corp. 53

2-pole CDM solutions

•

The poles are

•

where the damping factor, and

•

the characteristic frequency

•

Note only two independent variables for an

RLC waveform profile

•

Thus the same waveform results with

–

L1=

α

L, C1=C/

α

, R1=

α

R,

α

>0

–

many ways to get the same profile

[

1

]

2 ,

1

=

−

D

±

D

−

P

ω

LC

RC

D

2

=

LC

1

=

ω

54

Definitions of CDM Tester Parameters

Tim Maloney, Intel Corp. d2-fplt

(effective)

±V

d1-fplt (probe)

upper gnd plate (Area=Afplt)

field plate

t-fplt

(dielectric thickness)

target dia

targ-th

probe radius = a

L and C Model Parameters

Tim Maloney, Intel Corp. 55

Textbook references for C (e.g., Kaiser):

“Our” Bob Renninger for L, 1991 EOS/ESD:

d=d1, probe length

2a

Measurements of Voltages on IC Leads

Slide 56

• Device held by vacuum in a Discharge Test Fixture simulating

an automated handler pick and place mechanism

• Device held at a distance of 2.5 mm from ground

• Device leads charged to a known voltage

Discharge

Test Fixture

Discharge Target

•

Made with FR-4 double-sided PC board material

•

SMA connector on bottom

•

50-ohm cable to LeCroy oscilloscope with 50-ohm input.

•

Current calculation based on 50 ohms

Slide 57

Top

Bottom

Waveform Analysis

•

I

, Q

, and Q

are calculated from

oscilloscope data, while V

is the initial

charging voltage

•

C

V

= fC

* V

/f where f = 1.5 for

charging at 2.5 mm from ground

•

Previously described computation

methods solve for R,L, and C

Slide 58

Test fixture discharge for 500V, 64-pin device.

Qa

Qfp

Vo

Imax

Co

Leq D

Cmax

Req

derived quantity measured quantity

LC

RC

D

2

=

Time (500 psec/div) Pulse Height

2.27 Amps

Blind Alleys and Dubious Ventures

in CDM Modeling

Tim Maloney, Intel Corp. 59

•Method of Moments for fixture capacitance

•Parallel plate and fringing field estimates work well enough (but we’re glad MoM was done at least once) •Plasma physics modeling of spark

•Too many unknown quantities; may as well use linearized models with R_spark and τ_spark

•1990s publications can help with the linearization •Field and ground plates are an open-circuited radial

transmission line but does it really have any ¼-wave effects? •Modeling hasn’t shown that the device feels much

•Maybe try a shorted package trace stub in conjunction… •Device could feel resonances that may not be

CDM Summary

Tim Maloney, Intel Corp. 60

• CDM tester configuration and RLC equivalent • RLC circuit extraction from waveforms

• Exact matching of charge and peak current • Remarkable result: RC = R₀C + τ₀

• Describes how R varies with capacitance • Trends in damping factor D

• Estimating L and C from hardware properties

Flow Chart of Outline

Tim Maloney, Intel Corp. 61

Math review

HBM 2009-11

Antennas 2011-13

Thermal 2013-15

IEC, Chairs

2015 TLP, CCDM

2012

CDM 2013-14

The Problem

Same CDM pulse on fast and slow oscilloscopes

Filter models needed for measurement channel

-4 -2 0 2 4 6 8 10

0 500 1000 1500 2000 2500

c

u

rr

e

n

t,

A

m

p

s

time, psec

500V small JEDEC target

8 GHz Tek scope

1 GHz Tek scope

Use LTI* System Theory

Slide 63

*LTI=Linear, Time-Independent

Generalized transfer function:

Expansion in powers of s:

Use to model voltage step and scope

m>n

Tim Maloney, Intel Corp. 64

Treats Gaussian, 1-pole, and 2-pole response functions:

2 2 D for 2 is ; 2 2 1 2 2 1 2 0 3 4 2 2 0

3 = − + ⋅ − + = π =

ω π

ω

dB

dB D D D f

f 2-pole bandwidth: Gaussian bandwidth: πτ 2 177 . 1 3dB =

f _{rise time} τ_10-90%=1.812τ

Bandwidth-Rise Time Product

Tim Maloney, Intel Corp. 65

for Gaussian and for 2-pole with 0.6<D<1. For pseudo-Gaussian, D=0.707.

3396 .

0 % 90 10

3dB ⋅

τ

f

Convolved 10-90% rise times of Gaussian,

pseudo-Gaussian add in quadrature:

2 2 2

1 2

12

τ

τ

τ

0 0

% 90 10

2 2D

RC ; 509 . 1 2

2 3396 . 0

ω ω

π

τ ₋ = ⋅ RC = RC = =

Waveform Analysis

Tim Maloney, Intel Corp. 66

C

L

R

τ_10-90%=1.509τ_e for a pseudo-Gaussian ) 2 1 )( 1 ( ) ( ) ( ) ( 2 2 2 0 s s LCs RCs CV s Y s V s I e e τ τ + + + + = = Thus

Now has spark rise time. Add a scope filter (2 more poles, pseudo-Gaussian) if scope speed is an issue

Waveform and Peak Current

Tim Maloney, Intel Corp. 67

Total charge Q(nC) for voltage V

84 psec rise time 3 GHz scope RLC series circuit

In Mathematica*, Ipeak can be extracted with a 1-line command:

For quadratic terms, s (σ+jω) in GHz and coefficients in nsec and nsec2

In this case, 60 mil diel thickness gives Ipk=5.495 amps and Q=0.8509 nC for 500V

*Mathematica is licensed software; some progress is possible with free I.L.T. app as cited in references.

      + + + + + + = ) 1 )( 0028 . 075 . 1 )( 0016 . 056 . 1 ( )

( ₂ 0 _{2 2}

LCs RCs s s s s CV t

1-Whoa! What did he just do with

that I(t) function?

Tim Maloney, Intel Corp. 68

Introduce 84 psec spark rise time:

3 GHz scope response:

0.2 0.4 0.6 0.8 1.0 0.2 0.4 0.6 0.8 1.0 1.0 nsec

0.2 0.4 0.6 0.8 1.0 2 4 6 8 1.0 nsec Scope modeling: https://sites.google.com/site/esdpubs/ documents/esd11.pdf         + + ⇒         + + ⇒ 2 0 2 0 0 2 0 2 0 0 0 2 1 2 1 ω ω ω ω s s s V s s D s V s V 2 0 2 0 2 1 1 ω ω s s+ +

Gaussian Convolution

Slide 69

Resulting Width:

Bessel-Thomson filtering CDM RLC

Tim Maloney, Intel 70 Using FFPA book values L=6 nH, C_sm=5 pF, C_lg=15.76 pF,

Calculated ratio R_lg=0.9, R_sm=0.75. Curve is nearly insensitive to D when D<1

Measured ratio R_lg=0.899, R_sm=0.74 (5 or 6 samples) J-lg

J-sm

Ratio=Ipk_1GHz/Ipk_8GHz

JEDEC CDM targets

noted

Presented at Sept 2013 CDM

TDMR version of esd13 (CDM models)

Tim Maloney, Intel Corp. 71 convolved

(a)

next slide

CDM I

_max

multiplier for D

≈

0.707

Tim Maloney, Intel Corp. 72 (b)

Good estimate of peak current change

given scope bandwidth and FWHM, using 2-pole

pseudo-Gaussian scope model

Further notes on CDM waveform

xforms

FWHM adds in quadrature, as with rise time formula

from Gaussian math in 1999 M-S paper:

•

FWHM of f

=1 GHz is 507 psec; scale from there

•

Q

(first peak charge) not affected much; often <1%

•

If FWHM > 1/f nsec, Imax multiplier

≈

1, no correction

•

Our CDM committee’s Orion2 data fits pretty well

•

Large target correction for 1 GHz is 10-13%

•

Small target should use 2-3 GHz scope & correct

•

Small target is at least 40% low for 1 GHz

4-pole Bessel-Thomson scope filter

Tim Maloney, Intel Corp. 74

)

1

)(

2

1

)(

2

1

(

)

(

LCs

RCs

s

s

D

s

s

D

CV

s

I

+

+

+

+

+

+

=

ω

ω

ω

ω

For f_3dB=1 GHz,

ω₀₁=10.07439 GHz D₁=0.620703

ω₀₂=8.9862 GHz D₂=0.957974

Scalable to any f_3dB!

Flow Chart of Outline

Tim Maloney, Intel Corp. 75

Math review

HBM 2009-11

Antennas 2011-13

Thermal 2013-15

IEC, Chairs

2015 TLP, CCDM

2012

CDM 2013-14

TLP s-domain Model

Tim Maloney, Intel Corp.

R_d

Z₀, τ

V₀/s Z_in of an open-circuited

transmission line of

impedance Z₀, transit time

τ, is Z₀ coth(τs)

Wave series with time step 2τ, round-trip transit time Simple square pulse if reflection coefficient = 0:

[

]

2 0 0 0 0 _, ) 1 ( 1 ) ( ) coth( ) ( Z R Z R e e R Z s V s Z R s V s I d d s s d d + − = − − + = +

= −₋

ρ

ρ

τ

τ

[

]

k d k t u k t u Z R V t

I

∑

CCDM in the s-domain

Tim Maloney, Intel Corp. 77

Z₀

Z₀, τ

V₀/s

Z_L CCDM Z_L is (nearly)

sL +1/sC but could be anything for this general expression. ) sinh( ) cosh( ) sinh( ) cosh( ) ( ) sinh( ) cosh( ) sinh( ) cosh( 1 1 ) ( 0 0 0 0 0 0 0 s s s Z Z s Z Z s V s Z s Z s Z s Z Z s V s I L L L

L τ τ

τ τ τ τ τ τ + + ⋅ + =       + + + ⋅ = s L L e Z Z s V s sZ V s s s s Z Z s V s sZ V _τ τ τ τ τ τ τ 2 0 0 0 0 0 0 0 0 ) ( 1 ) coth( 1 ) sinh( ) cosh( ) sinh( ) cosh( ) ( 1 ) coth(

1 _⋅ −

+ + + ⋅ = + − ⋅ + + + ⋅ =

CCDM 2-pole Model

Tim Maloney, Intel Corp. 78

C

V(s)

≈

Vo/s

L

Z

•The transmission line impedance Z₀ replaces the spark resistance, aside from relay spark

•Z₀=50 ohms is higher than usual air spark R, resulting in lower Ipeak for given V₀, but stability is good

Flow Chart of Outline

Tim Maloney, Intel Corp. 79

Math review

HBM 2009-11

Antennas 2011-13

Thermal 2013-15

IEC, Chairs

2015 TLP, CCDM

2012

CDM 2013-14

Objectives (from esd12 & related pubs)

•

Assembly factory

in situ

static event monitoring

• Need for 3-D chip assembly with hi-speed I/O

•

Antenna and detector arrangement in factory

• Create CDM events at will using CDMES (event

simulator) with antenna and detector

•

Detector calibrated with a “standard” pulse

• Artificial, reproducible antenna-like pulse

•

Present theory of

Outline

•

CDM as a 2-pole circuit

• Add spark rise time, map to time domain

•

CDM as a source of dipole radiation

• Detect with monopole antenna

• s-domain expressions for everything

•

Measured antenna signal, agrees with theory

•

Artificial antenna signals with “monocycle” pulser

• Theory and experiment, compared favorably

• Calibration of MiniPulse detector with monocycle

pulser

Simplified CDM Network

•

Examine step response of this network

•

R is spark resistance, ~25-60 ohms

V(s)=Vo/s

C

L

CDM dipole radiation, monopole

antenna

Slide 83

Experimental arrangement of CDM electric dipole

initial source

p

and 6 mm coaxial antenna.

to 50

Ω

scope

p

15 cm

150 cm monopole antenna

•- - - - - -

Charge plate

Coax to 50 ohm scope

Ground plate

CDM Event Simulator (CDMES)

CDM pulse generator. Charge plate probe hits

pedestal and dipole collapses, with current pulse and

dipole radiation.

10 Meg

+V

+++++ +++++ - - - - - -

Simco-Ion CDMES Model

7” long. Voltage cable and signal coax cable shown.

10 Meg

Charge plate

Coax to 50 ohm scope +V

Ground plate

-Slide 87

Electric Dipole “equatorial” field

sinθ=1 at equator

In the s-domain,

s=σ+jω

But we know that

So

Solve in time domain with inverse Laplace Transform;

pole-zero expansion in natural frequencies

s=σ+jω

Calculated current and field

Slide 89

Calculated CDM current pulse,

1 nsec full scale.

E-field pulse E

at 15 cm from CDM

current source; 1 nsec full scale.

Antenna Transfer Function

From Ref. 10, esd12

(2007 Trans. EMC)

Transfer function in terms of initial

dipole source p

Slide 91

Calculated antenna response to E-field,15

cm from CDM source, 1.5 nsec full scale

Adjust the current function slightly…

Slide 93

Measured current (top) and antenna response (bottom) to

E-field at 15 cm, using artificial CDM source; 2 nsec/division

-100V pulse -0.496 A

max

125 mV =Vp-p -166.4 pC

antenna, 15 cm current

1.664 pF

Antenna problem—ESD effects in chairs

Tim Maloney, Intel Corp. 94

Math review

HBM 2009-11

Antennas 2011-13

Thermal 2013-15

IEC, Chairs

2015 TLP, CCDM

2012

CDM 2013-14

Recent Vermillion-Smith paper*

95

500mV/div, 20 ns/div, 50 Ω antenna termination Antenna is 15 cm long. Suppose it is #18AWG wire.

With concepts in this presentation, you can estimate the initial static field (in kV/cm) at the antenna. The

field then collapses over the time scale shown. You can also prove that the “radiation” is quasi-static, with

very little signal due to 𝑝̇ and 𝑝̈.

ESD Publications

Tim Maloney, Intel Corp. 96

Download most from TJM ESD publications page:

https://sites.google.com/site/esdpubs/documents

HBM:

https://sites.google.com/site/esdpubs/documents/esd09.pdf https://sites.google.com/site/esdpubs/documents/esd10.pdf

https://sites.google.com/site/esdpubs/documents/MR13.pdf (nearly the same as esd10)

plus TJM, IEW 2011 (unpublished)

CDM:

https://sites.google.com/site/esdpubs/documents/esd13.pdf

https://sites.google.com/site/esdpubs/documents/tdmr14.pdf (augmented esd13) https://sites.google.com/site/esdpubs/documents/esd14.pdf (with Arnie Steinman)

https://sites.google.com/site/esdpubs/documents/Ipk-engnrg-2.xlsx (CDM tester modeling program, Excel file)

B. Atwood, et al., “Effect of Large Device Capacitance on FICDM Peak Current”, 5A.1, 2007 EOS/ESD Symposium, pp. 273-282.

Oscilloscope filter models:

https://sites.google.com/site/esdpubs/documents/esd11.pdf

Antennas:

https://sites.google.com/site/esdpubs/documents/pulsdipole11-emc.pdf https://sites.google.com/site/esdpubs/documents/esd12.pdf

https://sites.google.com/site/esdpubs/documents/emc13-proof.pdf (augmented esd12)

Publications, cont’d

Tim Maloney, Intel Corp. 97

Thermal, s-domain:

https://sites.google.com/site/esdpubs/documents/irps13.pdf

plus Industry Council EOS Whitepaper section (not yet published)

TLP, CCDM, s-domain:

2012 TJM Seattle ESDA Tutorial (unpublished)

Pre-work for 2012 Seattle Tutorial (with RLC net practice):

https://sites.google.com/site/esdpubs/documents/prework-tjm1.pdf

IEC, s-domain:

TJM Abstract, 2015 EOS/ESD Symposium:

Other References

Tim Maloney, Intel Corp. 98

• Free online inverse Laplace Transform applet:

http://www.eecircle.com/applets/007/ILaplace.html • Free “Excellaneous” VB macros for Excel, at

http://www.bowdoin.edu/~rdelevie/excellaneous/#downloads

• Rise time-bandwidth, quad addition, pseudo-Gaussians, etc.: C. Mittermayer and A. Steininger, “…Dynamic Errors for Rise Time Measurement with an Oscilloscope”, IEEE Trans. Instrumentation and Measurement", Vol. 48, pp. 1103-1107, December 1999.

• 1998 Lucent paper: R.E. Carey and L.F. DeChiaro, “…Physical Design Parameters in FICDM ESD Simulators …", 1998 EOS/ESD Symposium Proceedings, pp. 40-53.

•Y. Ismail, et al., “Equivalent Elmore Delay for RLC Trees”, ACM DAC, 1999.

•K. Verhaege, et al., “Analysis of HBM ESD Testers…”, EOS/ESD Symposium Proc., 1993, pp. 129-137.

Appendix I—Metal heating, thermal effects,

temp waveforms—from IRPS13

Tim Maloney, Intel Corp. 99

Math review

HBM 2009-11

Antennas 2011-13

Thermal 2013-15

IEC, Chairs

2015 TLP, CCDM

2012

CDM 2013-14

Metal Self-Heating Test Pattern

100

M5

t V RC x

V

∂ ∂ =

∂ ∂

2 2

heat flow for 1-D heat slab:

Cp sK Z_th

ρ

1

=

s=σ+jω

K Cp s

th

ρ

γ = _{Electrical: Thermal:}

Volts ⇒ °C, temperature (usually a ∆T from room T) Amps ⇒ Watts

Coulombs ⇒ Joules

Ohms ⇒ _{thermal impedance} °_C/W

Farads ⇒ Joules/°C

1 µm2 _{metal cross section; electrical current through M5}

K=thermal conductivity Cp=heat capacity

ρ=mass density

Wires Embedded in ILD Oxide

101

M5

w_m+g

…… ……

w_m

t_ox hm

C_metal

Z₀₁, γ_oxtox

ZL Z02, γoxg/2

P(t)

Pattern C as shown

Pattern D is wider, g_D=2g_C

Thermal circuit model:

Note open circuit b.c. Thermal Ohm’s Law:

Thermal Feedback Model

102

+

Z(s)

α

T(t)P

(t)

P

(t)

P(s)Z(s)=T(s)

α

_{= metal tempco}

T = temp (“voltage”)

P

= I

_R

0

in (“current”)

))

(

1

(

)

(

t

R

T

t

R

=

+

α

metal resistance ) ( * )] ( ) ( [ ) ( * ) ( )

(t P₀ t Z t T t P₀ t Z t

T = +

α

[

]

or =

∫

t s Z s P d Z t P t Z t P 0 0 0

0( )* ( ) ( τ) (τ) τ ( ) ( )

For “current"

source P₀(t):

mixer

thermal Ohm’s Law

General Feedback Network

103

+

Z(s)

FB(t)

P

(t)

Convolve

P(t)*Z(t)=T(t)

T(t)

Z(s)⇔Z(t)

For TLP:

(

)

0 0 2 0 0 0 50 ) ( + = = R R V P t P               −                     + + + = 1 50 ) ( 1 ) ( 1 ) ( ₂ 0 0 0 R t T R t T P t FB α α

•Because of source resistance Z_s=50Ω, TLP introduces negative

feedback, and when R(t)>50 Ω, it becomes net negative

•Current source produces

positive feedback

•Voltage source produces

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

0 50 100 150 200 250 300

Amp lit ud e nsec exp fit FEM data 118 . 0 432 . 0 142 . 0 )

(t = e−t/2.88 + e−t/71.4 + Z

Z(t) from Finite Element Modeling

differentiate and normalize

Step response gives

thermal

impulse response

Z(t)

104 0 100 200 300 400 500 600 700 800

0 200 400 600 800 1000

de lta -T, d eg C nsec

FEM raw data

8.44W step, M6 8.44W, M6-M7-M8

105

0 o_C

430 o_C

Close-up of metal after 200 nsec

FEM results

M4 M5 M6

Transmission Line Pulsing (TLP)

•

Transmission line pulsing generates brief, high current (several ampere)

pulses; same current/time scale as ESD

Setup:

•

Equivalent circuit

50 ohms Device

Scope

L

V 10 Meg

(R_device<50 Ω):

50 ohms

Device

Scope

I_device=(V-V_device)/50

t_pulse = 2L/c, c=20 cm/nsec

Apply to metal lines and use Cu tempco (α=0.0025) for in-situ T measurement

TLP data

107 A_C

calculated from FB(t)

C e

t

T( ) ≈ 440(1− −t/58)

measured 0 2 4 6 8 10 12 14

0 50 100 150 200 250

W

atts

nsec

TLP Power, M5-C-60V

Apwr

measured

Pattern C, 60V

) 47 1 ( 09 . 4 39 . 8 ) ( s s W s W s P + + = ) 58 1 ( 440 ) ( s s C s T + = 

Thermal Impedance

= _{for TLP}

2 0 1 2 1 0 2 1 1 1 1 2 1 2 1 ) ) ( ( 1 ) 1 )( ( ) ( s C C R R s C R R C R s C R R R R s Z + + + + + + =

•Z(s) from slide 10 has 2 poles and 1 zero; 5-element RC network •For TLP, temp should flatten out at P_final(R₁+R₂) = P_finalZ₀ = T_final

P(t) ⇔ P(s) C₀

C₁

R₁ R₂

C

= 1.1 nJ/

°

C (metal + oxide)

R

= 32.7

°

C/W (oxide)

C

= 20 nJ/

°

C

R

= 2.52

°

C/W

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.91

0 50 100 150 200

Am

pl

itu

de

nsec

Z(t) from TLP data (°C/nJ)

) 00718 . 0 0185 . 0 ( 26 . 35 )

(t e t/31.6 e t/58

Z = − + −

109 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6

0 5 10 15 20

Am

pl

itu

de

nsec

Z(t) at short times

Dt model TLP data

Dt model _adiabatic

Thermal Impulse Response

) 58 1 )( 6 . 31 1 ( 47 1 26 . 35 ) ( s s s s Z + + + =

from °C/W

thermal sheath for short time: Dt …… …… ) ( ) ( s P s T =

TLP on Patterns C and D, 60V, 1

µ

_m

_{cross-section}

110

Pattern C width = X

T

= 440

°

C (“volts”)

Z

= 35.26

°

C/W (“ohms”)

P

= 8.24W, P

=12.16W

A

(normalized) = 58 nsec

-100 0 100 200 300 400 500

0 50 100 150 200

de lta -T, d eg C nsec M5-C-60V A_C

Pattern D width = 1.48X (g

=2g

)

T

= 225

°

C (“volts”)

Z

= 23.2

°

C/W (“ohms”)

P

= 8.24W, P

=10.99W

A

(normalized) = 46 nsec

-50 0 50 100 150 200 250 300 350

0 50 100 150 200

de lta -T, d eg C nsec M5-D-60V A_D

α= 0.0025

g_C ….

Thermal Circuit Elements

111

P(t) ⇔ P(s) C₀

C₁

R₁ R₂

Pattern TLP Volts C₀ R₂ R₁ C₁ T_final

C 60 1.1 nF 32.7Ω 2.52Ω 20 nF 440 °C C 70 1.09 nF 33.9Ω 2.84Ω 21.7 nF 645 °C D 60 1.53 nF 21.9Ω 1.33Ω 29.5 nF 225 °C

from TLP data

Ω

⇒

°

C/W

nF

⇒

nJ/

°

C