irps13-poster2.pdf

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Timothy J. Maloney

Intel Corporation, Santa Clara, CA

Lei Jiang, Steven S. Poon, Krishna B.

Kolluru, and AKM Ahsan

Intel Corporation, Hillsboro, OR

Achieving Electrothermal Stability in

Interconnect Metal During ESD

(2)

Outline

•Partitioned wires as a heating slab •More surface/volume than ever •Neighboring metal is a heat sink

•1-dimensional R-C transmission line models for thermal behavior •Feedback model for metal heating

•Self-consistent T(t) expression

•From metal tempco and thermal Ohm’s Law; T(s)=P(s)Z(s)

•Positive (tempco) and negative (electrical circuit, e.g., TLP) feedback •Thermal impulse response Z(s)⇔Z(t)

•Pre-silicon: finite element modeling

•Also thermal circuit models and known limiting conditions

•Post-silicon: Transmission Line Pulse (TLP) measurements

•Derive Z(s) from T(s)=P(s)Z(s), having measured T(t) and P(t) •Pole-zero expression for complex thermal impedance

•Corresponds to 5-element R-C model •ESD predictions in Excel using Z(t) and convolution software

•Convolve Z(t)*P(t) for HBM and CDM ESD conditions

•Obtain complete T(t) waveform from feedback equation and

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Metal Self-Heating Test Pattern

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

(4)

Wires Embedded in ILD Oxide

M5

w_m+g

…… ……

w_m

t_ox hm

C_metal

Z₀₁, γ_oxtox

ZL Z02, γoxg/2

P(t)

Pattern C as shown

Thermal circuit model:

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

(5)

Thermal Feedback Model

+

Z(s)

α

T(t)P

₀

(t)

P

₀

(t)

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

α

_{= metal tempco}

T = temp (“voltage”)

P

₀

= I

2

_R

0

in (“current”)

))

(

1 (

)

(

t

R

₀

T

t

R

=

+

α

α = metal tempco=0.0025/°C for Cu

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

(t P₀ t Z t T t P₀ t Z t

T = +

α

[

]

) ( ) ( * ) ( ) ( 1 ) ( * ) ( ) ( 0 0 t T t Z t P t T t Z t P 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

(6)

General Feedback Network

+

Z(s)

FB(t)

P

₀

(t)

Convolve

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

T(t)

Z(s)⇔Z(t)

For TLP:

( )2

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

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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)

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

(8)

0 o_C

430 o_C

Close-up of metal after 200 nsec

FEM results

M4 M5 M6

(9)

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

(10)

TLP data

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

P

+ +

=

) 58 1 (

440 )

(

s s

C s

T

+

= 

normalized A_c, A_pwr give time constants

)

(

)

(

)

(

s

P

s

T

s

(11)

Thermal Impedance

2 0 0 2 0 0 ) 50 ( R R V P +

= _{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

₀

= 1.1 nJ/

°

C (metal + oxide)

R

₂

= 32.7

°

C/W (oxide)

C

₁

= 20 nJ/

°

C

R

₁

= 2.52

°

C/W

(12)

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 = − + −

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

(13)

TLP on Patterns C and D, 60V, 1

µ

_m

2

_{cross-section}

Pattern C width = X

T

_final

= 440

°

C (“volts”)

Z

_0C

= 35.26

°

C/W (“ohms”)

P

₀

= 8.24W, P

_final

=12.16W

A

_C

(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

_D

=2g

_C

)

T

_final

= 225

°

C (“volts”)

Z

_0D

= 23.2

°

C/W (“ohms”)

P

₀

= 8.24W, P

_final

=10.99W

A

_D

(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 ….

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Thermal Circuit Elements

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

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Stability Criterion, Step Current

th th th final final Z P Z P Z P T 0 0 0 0 0

1−α

= =

Eq. (10), steady state:

W i J  = W z R_el  0

ρ

= W z J W W z J R i

P _el 

  0 2 2 2 0 2 2 0 ρ ρ = = = electrical thermal zW K Z ox th δ 1 0 =

T_final is finite unless

1

0

Z

th

>

P

α

1 0 2 > ox K J ρ δ

α

or

D.G. Pierce (1982) derived the same

condition for “unbounded” solutions:  δ αρ₀

2 Kox

J >

Pierce, EOS/ESD 1982

heated metal

oxide

heat sink l

δ

W depth=z

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0 20 40 60 80 100 120

0 100 200 300 400 500

de lta -T, d eg C nsec

HBM 1kV, Pattern C

FEM

TLP

HBM Temperature Waveforms

0 100 200 300 400 500 600 700 800 900

0 100 200 300 400 500

de lta -T, d eg C nsec

HBM 2kV, Pattern C

FEM

TLP

Convolve Z(t) with Human Body Model P₀(t) function and

solve feedback equation for self-consistent T(t) in Excel

        + + − = + −

−t t t

e e e t P 60 15 201 60 15 201 4 / 0 2 201 1040 ) (

For 1kV HBM,

P_0-2kV(t)=4P_0-1kV(t) but T_max is higher due to positive

feedback effect

(17)

CDM Temperature Waveforms

17

0 20 40 60 80 100 120 140 160 180

0 10 20 30 40 50

de

lta

-T,

d

eg

C

nsec

CDM 250V, 15 pF

TLP FEM

FEM

TLP

0 200 400 600 800 1000 1200 1400

0 10 20 30 40 50

de

lta

-T,

d

eg

C

nsec

CDM 500V, 15 pF

TLP FEM

FEM

TLP

Using Z(t) for Pattern C, 1 µm2_Cu

Excel calculations

amps

0.5 1.0 1.5 2.0 2.5 3.0 1

2 3 4

15 pF, 250V3.75 nC, ~4.5A Ipeak

CDM current waveform:

At 500V (I_peak=9A), feedback pushes Cu metal temperature beyond the melting

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1D representation of heat flow

P₀(t)

αP₀(t)T(t)

T(t)

Cap_metal

Res from M6 to Top metal

Distributed RC of ILD Oxide to M4

open

Distributed RC of ILD Oxide to M6

Cap_M7-M8

Metal under test 1D SPICE-like circuit model captures cumulative temperature gradient.

Effects of vias and hot/cold spots can be captured using 2D model.

feedback due to metal tempco α

Res_M4-heatsink

Cap_M4&below

Distributed RC of interstitial oxide

Cap_M6

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Results : FEM vs 1D model, step response

(pattern C)

Close-up of metal after 200 nsec

M4 M5 M6

M3 M2 M1 Contact

The temperature profile of M6 & above and M4 & below of the 1D model matches closely with FEM (see slide 7).

0 100 200 300 400 500 600

del

ta

-T

, d

eg

C

nsec

M5-C-60V

M5

heatsink(M4 & below) M6 & above

g_C ….

….

(20)

Impulse Response

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6

0 5 10 15 20

Amp

lit

ud

e

nsec

Z(t) at short times : Pattern C

Model Data

adiabatic

g_C ….

….

g_D ….

….

Pattern D width = 1.48X (g_D=2g_C) Pattern C width = X

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6

Amp

lit

ud

e

Z(t) at short times : Pattern D

Model data

adiabatic

From SPICE-like circuit model, no feedback

(21)

CDM & HBM response: Pattern C

0 10 20 30 40 50 60 70 80 90 100

0 100 200 300 400 500

Te mp era tu re (d eg C ) Time(nsec) HBM 1kV M5

heatsink (M4 & below) M6 & above

0 100 200 300 400 500 600 700 800

0 100 200 300 400 500

Te mp era tu re (d eg C ) Time(nsec) HBM 2kV M5

0 20 40 60 80 100 120 140 160

0 20 40 60 80 100

Te mp era tu re (d eg C ) CDM 250V,15pF M5

From SPICE-like circuit model, with feedback 0 200 400 600 800 1000 1200

0 10 20 30 40 50 60 70 80 90 100

Te mp era tu re (d eg C ) CDM 500V,15pF M5

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Conclusions

•Feedback model for metal heating presented

•Self-consistent T(t) expression from thermal Ohm’s Law and α •Positive feedback from pure current source

•Negative feedback from pure voltage source

•Electrical source/load impedance determines net feedback •Thermal impulse response Z(s)⇔Z(t) is central to solving for T(t)

•Pre-silicon: finite element modeling (FEM) or SPICE-like circuit model

•Differentiate response to heat step to get Z(t)

•Post-silicon: Transmission Line Pulse (TLP) measurements

•Measure T(t) and input power P(t)

•Maps to 5-element R-C model for each waveform •Add simple “heat sheath” model for 0<t<10 nsec

•Meshes nicely with TLP data

•ESD predictions in Excel using Z(t) and convolution software

•Convolve Z(t)*P(t), solve for T(t) for HBM and CDM ESD conditions •1 µm2 Cu x-section gives T_max<800°C for 2kV HBM

•But 500V CDM (7.5nC) melts Cu with T_max≈1200-1300°C