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Penn ESE370 Fall2014 -- DeHon 1 ESE370: Circuit-Level Modeling, Design, and Optimization for Digital Systems Day 19: October 15, 2014 Energy and Power.

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Presentation on theme: "Penn ESE370 Fall2014 -- DeHon 1 ESE370: Circuit-Level Modeling, Design, and Optimization for Digital Systems Day 19: October 15, 2014 Energy and Power."— Presentation transcript:

1 Penn ESE370 Fall2014 -- DeHon 1 ESE370: Circuit-Level Modeling, Design, and Optimization for Digital Systems Day 19: October 15, 2014 Energy and Power Optimization

2 Previously Three components of power –Static –Short circuit –Capacitive switching P tot = P static + P sc + P dyn Penn ESE370 Fall2014 -- DeHon 2

3 Static Penn ESE370 Fall2014 -- DeHon 33 CMOS circuit in steady-state –Leaks subthreshold current determined by off device –P static = V dd × I static

4 Switching (Charging) Penn ESE370 Fall2014 -- DeHon 44 CMOS circuit switching output from 0  1 –Spends energy CV dd 2 charging load

5 Switching (Short Circuit) P=IV When: Vin=Vdd/2 Vth<Vin<Vdd-|Vthp| Penn ESE370 Fall2014 -- DeHon 5

6 Today Power Sources –Static –Short Circuit –Capacitive Switching Reducing Switching Energy Energy-Delay tradeoffs (time permit) Penn ESE370 Fall2014 -- DeHon 6

7 Switching Quick Review Penn ESE370 Fall2014 -- DeHon 7

8 Switching Currents I switch (t) = I sc (t) + I dyn (t) I(t) = I static (t)+I switch (t) Penn ESE370 Fall2014 -- DeHon 8 I sc I static I dyn

9 Charging I dyn (t) –I ds = f(V ds,V gs ) –and V gs, V ds changing Penn ESE370 Fall2014 -- DeHon 9

10 Look at Energy [focus on I dyn (t)] Penn ESE370 Fall2014 -- DeHon 10

11 Energy to Switch Penn ESE370 Fall2014 -- DeHon 11

12 Capacitor Charge Penn ESE370 Fall2014 -- DeHon 12

13 Capacitor Charging Energy Penn ESE370 Fall2014 -- DeHon 13

14 Switching Power Every time output switches 0  1 pay: –E = CV 2 P dyn = (# 0  1 trans) × CV 2 / time # 0  1 trans = ½ # of transitions P dyn = (# trans) × ½CV 2 / time Penn ESE370 Fall2014 -- DeHon 14

15 Short Circuit Power Penn ESE370 Fall2014 -- DeHon 15

16 Preclass 1 Current flow during switching? Penn ESE370 Fall2014 -- DeHon 16

17 Short Circuit Power Between V TN and V dd -V TP –Both N and P devices conducting Roughly: Penn ESE370 Fall2014 -- DeHon 17

18 Peak Current Penn ESE370 Fall2014 -- DeHon 18 I peak around V dd /2 –If |V TN |=|V TP | and sized equal rise/fall

19 Short-Circuit Energy Penn ESE370 Fall2014 -- DeHon 19

20 Short-Circuit Energy Penn ESE370 Fall2014 -- DeHon 20

21 Short Circuit Energy Looks like a capacitance –Q=I×t –Q=CV Penn ESE370 Fall2014 -- DeHon 21

22 Short Circuit Energy and Power Every time switch –Also dissipate short-circuit energy: E = CV 2 –Different C = C sc –C cs “fake” capacitance (for accounting) Largely same dependence as charging Penn ESE370 Fall2014 -- DeHon 22

23 Switching Energy Penn ESE370 Fall2014 -- DeHon 23

24 Reminder Output switches 0  1 charge output from Vdd Output switches 1  0 discharge output to ground – no current from Vdd Penn ESE370 Fall2014 -- DeHon 24

25 Data Dependent Activity Consider an 8b counter –How often do each of the following switch? Low bit? High bit? –Average switching across all 8 output bits? Assuming random inputs –Activity at output of nand4? –Activity at output of xor4? Penn ESE370 Fall2014 -- DeHon 25

26 Gate Output Switching (random inputs) Penn ESE370 Fall2014 -- DeHon 26 Prev. Next 0 1 Pswitch=P(0@i)*P(1@i+1) +P(1@i)*P(0@i+1)

27 Charging Power P dyn = (# trans) × ½CV 2 / time Often like to think about switching frequency Useful to consider per clock cycle –Frequency f = 1/clock-period P dyn = (#trans/clock) ½CV 2 f Penn ESE370 Fall2014 -- DeHon 27

28 Switching Power P dyn = (#trans/clock) ½CV 2 f Let a = activity factor a = average #tran/clock P dyn = a½CV 2 f P sc = aC sc V 2 f Penn ESE370 Fall2014 -- DeHon 28

29 Total Power P tot = P static + P sc + P dyn P dyn + P sc = a(½C load +C sc )V 2 f P tot ≈ a(½C load +C sc )V 2 f+VI ’ s (W/L)e -Vt/(nkT/q) Penn ESE370 Fall2014 -- DeHon 29

30 Dynamic Power Penn ESE370 Fall2014 -- DeHon 30

31 Reduce Dynamic Power? P dyn = a × ½CV 2 f How do we reduce dynamic power? Penn ESE370 Fall2014 -- DeHon 31

32 Slow Down What happens to power contributions as reduce clock frequency? What suggest about V th ? Penn ESE370 Fall2014 -- DeHon 32

33 Reduce V What happens as reduce V? –Delay? –Energy? Static Switching Penn ESE370 Fall2014 -- DeHon 33

34 Penn ESE370 Fall2014 -- DeHon 34 Old Reduce V (no vsat)   gd =Q/I=(CV)/I  I d =(  C OX /2)(W/L)(V gs -V TH ) 2   gd impact?   gd α 1/V

35 Saturation Observe Ignoring leakage Penn ESE370 Fall2014 -- DeHon 35

36 Penn ESE370 Fall2014 -- DeHon 36 Reduce V (velocity saturation)   gd =Q/I=(CV)/I  I ds =( sat C OX )(W)(V gs -V TH -V DSAT /2)

37 Reduce Short-Circuit Power? P sc = aC sc V 2 f Penn ESE370 Fall2014 -- DeHon 37

38 Energy vs. Power? Which do we care about? –Battery operated devices? –Desktops? –Pay for energy by kW-Hr? Penn ESE370 Fall2014 -- DeHon 38

39 Increase V th ? What is impact of increasing threshold on –Delay? –Leakage? Penn ESE370 Fall2014 -- DeHon 39

40 Penn ESE370 Fall2014 -- DeHon 40 Increase V th   gd =Q/I=(CV)/I  I ds =( sat C OX )(W)(V gs -V TH -V DSAT /2)

41 Idea Short circuit energy looks like more capacitance for switching energy Tradeoff –Speed –Switching energy –Leakage energy Energy-Delay tradeoff: E  2 ? E  Penn ESE370 Fall2014 -- DeHon 41

42 Admin HW6 Due tomorrow Project 1 out –Milestone piece due in one week –Full Report in two weeks –That means you need to be starting on it now…and working on it all next week Read assignment today Lecture Friday Penn ESE370 Fall2014 -- DeHon 42


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