ESE370:
Circuit-Level
Modeling, Design, and Optimization
for Digital Systems
Day 21: October 28, 2011
Distributed RC Delay
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Penn ESE370 Fall2011 -- DeHon
Today
• Estimate delay in RC Network
– Elmore delay calculation
• Wire Delay
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Penn ESE370 Fall2011 -- DeHon
What is response?
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Penn ESE370 Fall2011 -- DeHon
What is response?
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Penn ESE370 Fall2011 -- DeHon
What is response?
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Penn ESE370 Fall2011 -- DeHon
SPICE Response
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Penn ESE370 Fall2011 -- DeHon
What is response?
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Penn ESE370 Fall2011 -- DeHon
SPICE Response
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Penn ESE370 Fall2011 -- DeHon
Intuition
• Look at series of R’s on path
– Must move Q=V(SC) across each R
• Not as much as if both R’s precede C’s
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Penn ESE370 Fall2011 -- DeHon
Elmore Delay
• For each resistor Ri in path
– Compute CRi = sum of all C’s downstream of Ri
– Delay through Ri is Ri×CRi



Delay  
R

C

j 
 i
path
path
i 
 j 
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Penn ESE370 Fall2011 -- DeHon
Superposition
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Penn ESE370 Fall2011 -- DeHon
Superposition
R1
R1
C1
R2
C2
R2
R1
C1
C2
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Penn ESE370 Fall2011 -- DeHon
Superposition
R1*C1
(R1+R2)*C2
R1
R1
C1
C2
R2
R1
C1
Penn ESE370 Fall2011 -- DeHon
R2
C2
R1*(C1+C2)+R2*C2
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Superposition not concurrent
• Don’t happen concurrently since must
divide current
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Penn ESE370 Fall2011 -- DeHon
Superposition
• For R1=R2=R, C1=C2=C
– Delay = 3RC
R2
R1
C1
Penn ESE370 Fall2011 -- DeHon
C2
R1*(C1+C2)+R2*C2
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SPICE Response
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Penn ESE370 Fall2011 -- DeHon
Apply to Y
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Penn ESE370 Fall2011 -- DeHon
Apply Y
• 1000W×3pF
• +1000W×1pF
• =4ns
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Penn ESE370 Fall2011 -- DeHon
SPICE Response
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Penn ESE370 Fall2011 -- DeHon
Elmore Delay
• For each resistor Ri in path
– Compute CRi = sum of all C’s downstream of Ri
– Delay through Ri is Ri×CRi



Delay  
R

C

j 
 i
path
path
i 
 j 
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Penn ESE370 Fall2011 -- DeHon
Wire
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Penn ESE370 Fall2011 -- DeHon
Wire Capacitance
Penn ESE370 Fall2011 -- DeHon
A
Cr0
d
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Wire Resistance
R
Penn ESE370 Fall2011 -- DeHon
L
A
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Wire as RC Ladder
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Penn ESE370 Fall2011 -- DeHon
Wire Delay as f(L)
• Measure wire length in units
– Say l
– Each lambda have Cunit, Runit
• Capacitance and resistance of wire of length l
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Penn ESE370 Fall2011 -- DeHon
Wire Delay
• Wire N units long:
Runit*(N*Cunit)
+Runit((N-1)*Cunit
+Runit*(N-2)*Cunit+…
+Runit*Cunit
=(Runit*Cunit)*(N+N-1+N-2+….1)
Penn ESE370 Fall2011 -- DeHon
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Sum of integers
• What’s the sum of the integer 1 to N?
N+N-1+N-2+…1
N
k
k 0
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Penn ESE370 Fall2011 -- DeHon
Sum of integers
• What’s the sum of the integer 1 to N?
N+N-1+N-2+…1
N(N 1)
2
 k  2  0.5N
k 0
N
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Penn ESE370 Fall2011 -- DeHon
Wire Delay
• Wire N units long:
Runit*(N*Cunit)+Runit((N-1)*Cunit
+Runit*(N-2)*Cunit+…+Runit*Cunit
=(Runit*Cunit)*(N+N-1+N-2+….1)
=Runit*Cunit*N2/2
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Penn ESE370 Fall2011 -- DeHon
Wire Delay
•
•
•
•
Rwire = N*Runit
Cwire=N*Cunit
Wire delay = Runit*Cunit*N2/2
Wire delay = 0.5 * Rwire*Cwire
• Half the delay of lumped RC product
• Quadratic in length of wire
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Penn ESE370 Fall2011 -- DeHon
Branching Wire
• What is delay of:
Drive
L/2
L/2
Receive
L/2
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Penn ESE370 Fall2011 -- DeHon
Branching Wire
• Drive Wire of length L
– L2/2 (Runit*Cunit)
• Charge L/2 Cunit through R=L/2 Runit
– L2/4 (Runit*Cunit)
• Total 0.75 L2
Drive
L/2
L/2
Receive
L/2
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Penn ESE370 Fall2011 -- DeHon
Branching Wire
• Direct calc like ladder for first half
 3L 2
 5 
3L 3L 

RC   1 ...L RC0.5   0.5L2   RCL2

 2  2

  2 
 8 
• Then wire of length L/2

– 0.5(L/2)2=L2/8
Total 0.75 L2
Drive
L/2
L/2
Receive
L/2
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Penn ESE370 Fall2011 -- DeHon
Admin
• Project
– Due next Friday
– Baseline done; many ideas on what to do
– Starting to optimize….
– Now dig into optimization and designspace exploration
• Midterm 2: Nov. 9th
– Wednesday, 1.5 week, in evening
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Penn ESE370 Fall2011 -- DeHon
Idea
• Lumped wiring calculation is pessimistic
– Not all capacitance at end of wire
• Elmore delay calculation allows us to
estimate
• Wires are distributed RC
– Half delay lumped calculation
– Still quadratic in length
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Penn ESE370 Fall2011 -- DeHon