Transcript ppt

ESE370:
Circuit-Level
Modeling, Design, and Optimization
for Digital Systems
Day 39: December 5, 2014
Repeaters in Wiring
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Penn ESE370 Fall2014 -- DeHon
Previously
• Transmission line (LC wire) wire delay
scales as Length
• Unbuffered RC wire delay scales as
Length2
– 0.5 Rwire Cwire
– 0.5 L2 Ru Cu
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Today
• RC (on-chip) Interconnect Buffering
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Back to RC Wire
(on-chip, no inductance, L)
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Delay of Wire
•
•
•
•
Long Wire: 1mm
Ru = 60K W per 1mm of wire
Cu = 0.16 pF per 1mm of wire
Driven by inverter
– R0 = 25K W
– C0 = 0.01 fF
– Assume velocity saturated, sized Wp=Wn=1
• Loaded by identical inverter
Penn ESE370 Fall2014 -- DeHon
5
Should be able to do these calculations on final.
Formulate Delay
Delay of inverter driving wire?


Rbuf  Cself  Cwire  Cload  0.5Rwire  Cwire  Rwire  Cload
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Penn ESE370 Fall2014 -- DeHon
Calculate Delay
• Cload = 2 C0
• Rbuf = R0
• Cself = g 2 C0 = 2 C0


Rbuf  Cself  Cwire  Cload  0.5Rwire  Cwire  Rwire  Cload
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Buffering Wire
• Complete Preclass Table
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N Buffers
• Delay Equation for N buffers?




 Rwire Cwire  Rwire
Cwire
N Rbuf  Cself 
 Cload   0.5

 Cload 



 N
N
N  N




N  Rbuf  Cself  Cload  Rbuf
 Rwire  Cwire 
 Cwire  0.5
  Rwire  Cload


N
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Minimize Delay


N  Rbuf  Cself  Cload  Rbuf
 Rwire  Cwire 
 Cwire  0.5
  Rwire  Cload


N
• How determine N to minimize delay?
• Derivative with respect to N

Rbuf  Cself  Cload
 Rwire  Cwire 
 0.5
0
2


N

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Penn ESE370 Fall2014 -- DeHon
Solve for N

Rbuf  Cself  Cload
 Rwire  Cwire 
 0.5
0
2


N


Rwire  Cwire

N  0.5
R  C C
self
load
 buf






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Minimize Delay

Rwire  Cwire

N  0.5
R  C C
self
load
 buf



N  Rbuf  Cself  Cload  Rbuf






 Rwire  Cwire 
 Cwire  0.5
  Rwire  Cload


N

  Rbuf  Cself  Cload  Rbuf  Cwire  Rwire  Cload
2 0.5Rwire  Cwire
Penn ESE370 Fall2014 -- DeHon
Equalizes delay in buffer and wire
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Calculate: Delay at Optimum
Stages for Example
•
•
•
•
Ru = 60K W per 1mm of wire
Cu = 0.16 pF per 1mm of wire
Rbuf=R0 = 25K W
Cself=Cload=2(C0 = 0.01 fF)=0.02fF


2 0.5Rwire  Cwire  Rbuf  Cself  Cload  Rbuf  Cwire  Rwire  Cload
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Segment Length
• Rwire = L×Runit
• Cwire = L×Cunit

Rwire  Cwire

N  0.5
R  C C
self
load
 buf



Ru  Cu

N  L 0.5
R  C C
self
load
 buf






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




Optimal Segment Length
• Delay scales linearly with distance once
optimally buffered
*
seg
L


R  C C
L
buf
self
load

  2

N
Ru  Cu


Ru  Cu

N  L 0.5
R  C C
self
load
 buf
Penn ESE370 Fall2014 -- DeHon











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Buffer Size?
• How big should buffer be?
– Rbuf = R0/W
– Cload = 2 W C0 (assuming velocity saturation)
– Cself = g 2 W C0


2 0.5Rwire  Cwire  Rbuf  Cself  Cload  Rbuf  Cwire  Rwire  Cload
R0
R0
2 0.5Rwire  Cwire 
 1 g 2WC0 
 Cwire  Rwire  2WC0
W
W
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Penn ESE370 Fall2014 -- DeHon
Implication W
• Rwire = L×Runit
• Cwire = L×Cunit
•  W independent of Length
– Depends on technology
R0  Cwire
W 
Rwire  2C0
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Penn ESE370 Fall2014 -- DeHon
Delay at Optimum W
2 0.5Rwire  Cwire  R0  1 g 2C0  2 R0  Cwire  Rwire  2C0
• With g=1, 1+g=2
• Same size as first term
4 R0  Cwire  Rwire  2C0
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Ideas
• Wire delay linear once buffered
• Optimal buffering matches
– Buffer delay
– Delay on wire between buffers
– Delay of wire driving buffer
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Penn ESE370 Fall2014 -- DeHon
Final
• Everything
– Including today
• Focus on wiring,
memory
– Crosstalk
– Transmission lines
• Delay
• Energy
•
•
•
•
•
•
•
Static CMOS
Precharge
Pass Transistors
Ratio
Clocking
Restoration
Buffering
2010, 2011, 2012, 2013 finals all good content
--2011 many “small” problems – good coverage
Penn ESE370 Fall2014 -- DeHon
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Admin
• Ron Review Monday (12/8)
– Talk with him about final Q&A session
• Final (12/18) noon Towne 303
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Penn ESE370 Fall2014 -- DeHon