Transcript ppt

ESE370:
Circuit-Level
Modeling, Design, and Optimization
for Digital Systems
Day 31: November 22, 2010
Inductive Noise
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Today
•
•
•
•
•
Inductive Responses
Calculating L
Where do inductances show up
Impact of inductance on digital circuits
How address
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Response
• What happens here?
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V2
dI
L
LC Response
V

V
dt 2
dV2 
I  C dt 


d 2V2 
CL

V

V

2
dt 

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
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LC Response
V2  A  Be
wt
d 2V2 
CL

V

V

2
dt


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LC Response
V2  A  Be
dV2
2
wt

w
Be
dt
wt
d 2V2 
CL

V

V

2
dt


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LC Response
V2  A  Be
wt
2
d V2

w
Be
dt
2
wt
d 2V2 
CL

V

V

2
dt
  
CLw Be
2
wt
 A  Be
wt
V
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LC Response
CLw Be
2
V A
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wt
 A  Be
wt
V
CLw 1  0
2
1
w i
CL
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LC Response
V A
V2  A  Be
V2  V Be
wt
1
w i
CL
 1 
i
t
 CL 
d 2V2 
CL

V

V

2
dt


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LC Response
i
e  cos( )  isin(  )
V2  V  Be
 1 
i
t
 CL 
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
LC Response
i

V2  V  Be
1 
t
CL 

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Response?
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V2
RLC Response
L dI dt  IR  V2  V
dV2 
I  C dt 


d 2V2 
dV2 

 V2  V
CL

RC

dt 
 dt 


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RLC Response
V2  A  Be
wt
dV2
2

wBe
dt
d V2
wt

w
Be
dt
2
wt
d 2V2  
dV2 

 V2  V
CL

RC

dt 
 dt 


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RLC Response
dV2
V2  A  Be
wt
2
d V2

wBe
dt
wt

w
Be
dt
2
wt
d 2V2 


dV
RC  2  V  V
CL


2
dt
dt




CLw Be
2
wt

wt
wt
 RC wBe  A  Be  V



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Solving for w
CLw Be
2
 RC wBe  A  Be
CLw  RC w 1  0
wt
wt
wt
V
2
R
1
w  w  
0
L
LC
2

Penn ESE370 Fall2010 -- DeHon
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RLC
R 
R
4
    
L  LC
L
w
2
2
R
1
w  w  
0
L
LC
2
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RLC
R 
R
4
    
L  LC
L
w
2
2
V2  A  Be
wt
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RLC
• For
R 
R
4
    
L  LC
L
w
2
2
V2  A  Be
4L
R
C
• Oscillation
• Decay
wt 
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RLC Response (R=100)
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When Oscillate
4L
R
C

4L
 200
C
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RLC Response
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Inductance of Wire
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Lwire
CL  
C and L per unit length
L
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
C
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Chip Inductance
• Cwire = 0.16 pF for the 1mm)
• Cwire = 0.16nF/m
• Permeability 0≈ Si02=12.6×10-7H/m
• Permitivity ox=3.5×10-11F/m
L
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
C
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On Chip
• Cwire = 0.16 pF for the 1mm)
• Cwire = 0.16nF/m
• Permeability 0≈ Si02=12.6×10-7H/m
• Permitivity ox=3.5×10-11F/m
 276pH (for 1 mm)
L
Penn ESE370 Fall2010 -- DeHon

C
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Comparisons
• 5mil trace on PCB
– About 2.7nH/cm
– http://www.pcb123.com/help/calculators/microstrip.html
• Protoboard wires (0.6mm diameter)
– About 7nH/cm
– http://www.consultrsr.com/resources/eis/induct5.htm
• On chip wire
– 0.28nH/mm = 2.8nH/cm
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Inductors
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Bond pads
Chip leads
Long wire runs
Cables
Src: http://en.wikipedia.org/wiki/File:Wirebonding2.svg
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Where Arise
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Signal Path
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Power Ground
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Estimate
• Req, Ceq for gates in parallel
– R0 = 25K W
– C0 = 0.01 fF
• say 10C0=0.1fF for typical load
• 250 gates switching at clock
• Req = 100WCeq=25fF
• Assume L=1nH
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Power Ground
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RLC Response
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Today’s Chips
• How many gates?
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Multiple Power/Ground Pins
• Use many power/ground pins
• How many pins on a package?
• Divide switching gates by pins
– To get effective load on each pin
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How Improve
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Minimize the L
• Make wires short
• Use power and ground planes
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Add Good C’s
• Bypass Capacitors
– On board
– On chip
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With Bypass
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Minimize Current Draw
• More Power/Ground Pins
• Slower rise/fall times
• Spread out switching
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Admin
• Wednesday
– Homework due
– Lecture
• Friday holiday
• Lecture on Monday
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Idea
• Long wires are inductive
– Avoid them
– Especially on power supplies
• Bypass capacitors help
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