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Design for Low Power potx

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Dynamic Example Static CMOS logic gates: activity factor = 0.1  Memory arrays: activity factor = 0.05 many banks!. Dynamic Example Static CMOS logic gates: activity factor = 0.1  Mem

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Power and Energy

 Power is drawn from a voltage source attached to

the VDD pin(s) of a chip

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Dynamic Power

 Dynamic power is required to charge and discharge

load capacitances when transistors switch

 One cycle involves a rising and falling output

 On rising output, charge Q = CVDD is required

 On falling output, charge is dumped to GND

 This repeats Tfsw times

over an interval of T

C

fsw

iDD(t) VDD

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Dynamic Power Cont.

C

fsw

iDD(t) VDD dynamic

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Dynamic Power Cont.

1

( ) ( )

T

T DD

T V

Tf CV T

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Activity Factor

 Suppose the system clock frequency = f

 Let fsw = f, where  = activity factor

– If the signal is a clock,  = 1

– If the signal switches once per cycle,  = ½

– Dynamic gates:

• Switch either 0 or 2 times per cycle,  = ½– Static gates:

• Depends on design, but typically  = 0.1

 Dynamic power: Pdynamic   CVDD2 f

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Short Circuit Current

 When transistors switch, both nMOS and pMOS

networks may be momentarily ON at once

 Leads to a blip of “short circuit” current

 < 10% of dynamic power if rise/fall times are

comparable for input and output

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 200 Mtransistor chip

– 20M logic transistors

• Average width: 12 – 180M memory transistors

• Average width: 4 – 1.2 V 100 nm process

– Cg = 2 fF/m

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Dynamic Example

 Static CMOS logic gates: activity factor = 0.1

 Memory arrays: activity factor = 0.05 (many banks!)

 Estimate dynamic power consumption per MHz

Neglect wire capacitance and short-circuit current

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Dynamic Example

 Static CMOS logic gates: activity factor = 0.1

 Memory arrays: activity factor = 0.05 (many banks!)

 Estimate dynamic power consumption per MHz

Neglect wire capacitance

 

6 logic

6 mem

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Ratio Example

 The chip contains a 32 word x 48 bit ROM

– Uses pseudo-nMOS decoder and bitline pullups– On average, one wordline and 24 bitlines are high

 Find static power drawn by the ROM

–  = 75 A/V2

– Vtp = -0.4V

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Ratio Example

 The chip contains a 32 word x 48 bit ROM

– Uses pseudo-nMOS decoder and bitline pullups– On average, one wordline and 24 bitlines are high

 Find static power drawn by the ROM

–  = 75 A/V2

– Vtp = -0.4V

pull-up pull-up pull-up

24μAA 2

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Leakage Example

 The process has two threshold voltages and two

oxide thicknesses

 Subthreshold leakage:

– 20 nA/m for low Vt

– 0.02 nA/m for high Vt

 Gate leakage:

– 3 nA/m for thin oxide

– 0.002 nA/m for thick oxide

 Memories use low-leakage transistors everywhere

 Gates use low-leakage transistors on 80% of logic

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Leakage Example Cont.

 Estimate static power:

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Leakage Example Cont.

 Estimate static power:

2.4 10 20 / / 2 3 / 45.6 10 0.02 / / 2 0.002 / 32

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Leakage Example Cont.

 Estimate static power:

2.4 10 20 / / 2 3 / 45.6 10 0.02 / / 2 0.002 / 32

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Low Power Design

 Reduce dynamic power

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Low Power Design

 Reduce dynamic power

– : clock gating, sleep mode

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Low Power Design

 Reduce dynamic power

– : clock gating, sleep mode

– C: small transistors (esp on clock), short wires

– VDD:

– f:

 Reduce static power

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Low Power Design

 Reduce dynamic power

– : clock gating, sleep mode

– C: small transistors (esp on clock), short wires

– VDD: lowest suitable voltage

– f:

 Reduce static power

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Low Power Design

 Reduce dynamic power

– : clock gating, sleep mode

– C: small transistors (esp on clock), short wires

– VDD: lowest suitable voltage

– f: lowest suitable frequency

 Reduce static power

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Low Power Design

 Reduce dynamic power

– : clock gating, sleep mode

– C: small transistors (esp on clock), short wires – VDD: lowest suitable voltage

– f: lowest suitable frequency

 Reduce static power

– Selectively use ratioed circuits

– Selectively use low Vt devices

– Leakage reduction:

stacked devices, body bias, low temperature

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