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L16: Power Dissipation in Digital Systems
L16: 6.111 Spring 2006
Introductory Digital Systems Laboratory
1
 
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Problem #1: Power Dissipation/Heat
100000
10000
Sun’s
Surface
18KW
Rocket
Nozzle
5KW
10000
1.5KW
500W
1000
1000
Nuclear
Reactor
100
Pentium® proc
100
10
400 8008 808 808 8086 286
486
386
10
8086
Hot Plate
4004
8008
8080
P6
1
8085
286
386
486
Pentium® proc
1
0.1
1971 1974 1978 1985 1992 2000 2004 2008
Year
1970
1980
1990
2000
2010
Year
Courtesy Intel (S. Borkar)
How do you cool these chips??
heat sink
chip
L16: 6.111 Spring 2006
Introductory Digital Systems Laboratory
2
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Problem #2: Energy Consumption
The Energy Problem
7.5 cm 3
AA battery
Alkaline:
~10,000J
What can One Joule
of energy do?
(Image by MIT OCW. Adapted from Jon
Eager, Gates Inc. , S. Watanabe, Sony Inc.)
Mow your
lawn for
1 ms
Operate a
processor
for ~ 7s
Send a 1
Megabyte
file over
802.11b
No Moore’s law for batteries…
Today: Understand where power goes
and ways to manage it
Image by MIT OCW.
L16: 6.111 Spring 2006
Introductory Digital Systems Laboratory
2
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Dynamic Energy Dissipation
Charging
Discharging
V DD
V DD
E 0 → 1 = C L V DD 2
i DD
E cap = 1/2C L V DD 2
R P
R P
E diss, RP = 1/2C L V DD 2
E diss,RN =1/2C L V DD 2
IN =1
IN =0
R N
C L
R N
C L
P = C L V DD 2 f clk
L16: 6.111 Spring 2006
Introductory Digital Systems Laboratory
3
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The Transition Activity Factor α 0 −> 1
Current
Next
Output
Input
Input
Transition
A
Z
00
00
1 −> 1
B
00
01
1 −> 1
00
10
1 −> 1
Assume inputs (A,B) arrive
at f and are uniformly
distributed
What is the average
power dissipation?
00
11
1 −> 0
01
00
1 −> 1
01
01
1 −> 1
01
10
1 −> 1
01
11
1 −> 0
10
00
1 −> 1
10
01
1 −> 1
10
10
1 −> 1
α 0 −> 1 = 3/16
10
11
1 −> 0
11
00
0 −> 1
11
01
0 −> 1
11
10
0 −> 1
P = α 0−>1 C L V DD 2 f
11
11
0 −> 0
L16: 6.111 Spring 2006
Introductory Digital Systems Laboratory
4
The Transition Activity Factor
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