l16_power_dissipation

l16_power_dissipation - L16: Power Dissipation in Digital...

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L16: 6.111 Spring 2007 1 Introductory Digital Systems Laboratory L16: Power Dissipation in Digital Systems L16: Power Dissipation in Digital Systems
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L16: 6.111 Spring 2007 2 Introductory Digital Systems Laboratory Problem #1: Power Dissipation/Heat Problem #1: Power Dissipation/Heat 5KW 18KW 1.5KW 500W 4004 8008 8080 8085 8086 286 386 486 Pentium® proc 0.1 1 10 100 1000 10000 100000 1971 1974 1978 1985 1992 2000 2004 2008 Year Power (Watts) 4004 8008 8080 8085 8086 286 386 486 Pentium® proc P6 1 10 100 1000 10000 1970 1980 1990 2000 2010 Year Power Density (W/cm2) Hot Plate Nuclear Reactor Rocket Nozzle How do you cool these chips?? How do you cool these chips?? chip heat sink Sun’s Surface Courtesy Intel (S. Borkar)
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L16: 6.111 Spring 2007 3 Introductory Digital Systems Laboratory Problem #2: Energy Consumption Problem #2: Energy Consumption (40+ lbs) Battery Year No mi na l Cap acity ( Wa tt -ho ur s/lb ) Nickel-Cadmium Ni-Metal Hydride 65 70 75 80 85 90 95 0 10 20 30 40 50 Rechargable Lithium (from Jon Eager, Gates Inc. , S. Watanabe, Sony Inc.) No Moore’s law for batteries… Today: Understand where power goes and ways to manage it What can One Joule of energy do? Send a 1 Megabyte file over 802.11b Operate a processor for ~ 7s The Energy Problem 7.5 cm 3 AA battery Alkaline: ~10,000J Mow your lawn for 1 ms
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L16: 6.111 Spring 2007 4 Introductory Digital Systems Laboratory Dynamic Energy Dissipation Dynamic Energy Dissipation V DD C L E 0 1 = C L V DD 2 E cap = 1/2C L V DD 2 i DD E diss, RP = 1/2C L V DD 2 V DD C L IN =1 E diss,RN =1/2C L V DD 2 Charging Discharging IN =0 P = C L V DD 2 f clk R N R P R N R P
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L16: 6.111 Spring 2007 5 Introductory Digital Systems Laboratory The Transition Activity Factor The Transition Activity Factor α α 0 −> 1 Output Transition Next Input Current Input 0 −> 0 11 11 0 −> 1 10 11 0 −> 1 01 11 0 −> 1 00 11 1 −> 0 11 10 1 −> 1 10 10 1 −> 1 01 10 1 −> 1 00 10 1 −> 0 11 01 1 −> 1 10 01 1 −> 1 01 01 1 −> 1 00 01 1 −> 0 11 00 1 −> 1 10 00 1 −> 1 01 00 1 −> 1 00 00 α 0 −> 1 = 3/16 Assume inputs (A,B) arrive at f and are uniformly distributed What is the average power dissipation? P = α 0−>1 C L V DD 2 f Z A B
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L16: 6.111 Spring 2007 6 Introductory Digital Systems Laboratory Junction (Silicon) Temperature Junction (Silicon) Temperature Simple Scenario T j -T a =R θ JA P D Silicon R θ JA is the thermal resistance between silicon and Ambient R θ JA P D T j =T a + R θ JA P D Make this as low as possible Realistic Scenario R θ JC P D R θ CA = R θ CS +R θ SA Sink Case Silicon T J T A T J T C T S T A T J T C T S T A R θ CS R θ SA is minimized by facilitating heat transfer (bolt case to extended metal surface – heat sink)
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L16: 6.111 Spring 2007 7
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This note was uploaded on 08/23/2009 for the course EECS 6.111 taught by Professor Prof.ananthachandrakasan during the Spring '04 term at MIT.

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l16_power_dissipation - L16: Power Dissipation in Digital...

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