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Technique #5 : Logical Effort Path Delay (Equation Derivation) Path

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Technique #5 : Logical Effort Path Delay (Equation Derivation) Path 10/19/2016 Technique #5 : Logical Effort The optimum effective fan-out for a chain of N inverters driving a load C is N L f CL / Cin Set N such that the fan-out per stage is around 4, whenever possible (FO4) Can we generalize this approach (logical effort) to any gate? The inverter equation is Lecture 11, 12: Introduction to Dynamic Logic tp = tp0 (1 + Cext/ Cg) = tp0 (1 + f/) we can generalize it to… Reading: Chap. 9, section 9.2 (esp. 9.2.4) October 17, 19 2016 tp = tp0 (p + g f/) Weste & Harris Prof. R. Iris Bahar tp0 is the intrinsic delay of an inverter f is the effective fan-out (Cext/Cg) – also called the electrical effort p is the ratio of the intrinsic delay of the gate relative to a simple inverter (a function of the gate topology and layout style): parasitic delay g is the logical effort © 2016 R.I. Bahar Portions of these slides taken from Professors J. Rabaey, J. Irwin, V. Narayanan, and S. Reda Path delay (equation derivation) Path delay of logic gate network (cont.) For gate i in the chain, its size is determined by h = figibi The path logical effort, G = gi [1] 1 c Path effective fanout (path electrical effort) is F = CL/Cg1 a b C 5 The branching effort accounts for fan-out to other gates in L the network: b = (Con-path + Coff-path)/Con-path For this network what do we need to compute h? The path branching effort is then B = b i F = CL/Cg1 = 5 …and the total path effort is then H = GFB [1] G = 1 x 5/3 x 5/3 x 1 = 25/9 N B = 1 (no branching) The gate effort that minimizes path delay is h H 4 So, the minimum delay through the path is H = GFB = 125/9, so the optimal stage effort is h= H = 1.93 Fanout factors are computed as fi=h/(gi·bi). Since bi=1 we have: N N N H [1] f = h/g =1.93, f = 1.93 / (5/3) = 1.16, D t p Note the textbook 1 1 2 p0 j swaps the definitions j1 γ f = 1.93/(5/3)=1.16, f = 1.93 of F,H and fi, h 3 4 1 10/19/2016 Path delay of logic gate network (cont.) Path delay of logic gate network (cont.) g Given f for each gate i in the chain, what is the final sizing? So what are the actually scaling sizes of the gates? c c i i g,i g,i 1 h 1 c a b c 1 scal e g,i 1 c 5 a b CL i h 5 CL Consider again the intrinsic capacitance values we calculated f =1.93, f =1.16, f = 1.16, f = 1.93 1 2 3 4 cg,4=5/1.93=2.59, cg,3=2.59/1.16=2.23, cg,2=2.23/1.16=1.93, c =1.93/1.93=1.00 relative to a minimum sized inverter So the gate sizes are (working from outputs to inputs): g,1 We need to adjust to the gate type: c: f =c /c = c /c = 1.93 c =5/1.93=2.59 4 ext,4 g,4 L g,4 g,4 cg,4 = 2.59 gate c is 2.59X size of minimum sized inverter, so S =2.59 since gate c is an inverter as well. b: f3=cext,3/cg,3 = cg,4/cg,3 = 1.16 cg,3=2.59/1.16=2.23 c c = 2.23 gate b is 2.23X size of minimum sized NOR, which is a: f =c /c = c /c = 1.16 c =2.23/1.16=1.93 g,3 2 ext,2 g,2 g,3 g,2 g,2 5/3 as big as min sized INV so Sb = 2.23 X 3/5 = 1.34 NOR is 1.34X size of minimum sized NOR g1: f1=cext,1/cg,1 = cg,2/cg,1 = 1.93 cg,1=1.93/1.93=1.00 cg,2 = 1.93 gate a is 1.93X size of minimum sized NAND3, which is 5/3 as big as min sized NAND so Sa = 1.93X 3/5 = 1.16 NAND is 1.16X size of minimum sized NAND3 Dynamic gate Properties of dynamic gates off Logic function is implemented by the PDN only CLK M CLK M p p on 1 How many transistors are needed to implement an N-input Out Out function in dynamic logic? !((A&B)|C) In1 CL What does this imply about total area and load capacitance, CL? A In2 PDN Full swing outputs (VOL = GND and VOH = VDD) C In Faster switching speeds 3 B Fewer transistors means reduced logical effort. CLK Me off What is the logical effort of a 2-input dynamic NOR gate? CLK Me on no short circuit current, I , so all the current provided by PDN goes Two phase operation sc into discharging CL Precharge (CLK = 0) We assume tpLH = 0, but the presence of the evaluation transistor Evaluate (CLK = 1) slows down the tpHL (extra transistor in series). 2 10/19/2016 Conditions on output Dynamic behavior CLK Evaluate 2.5 Once the output of a dynamic gate is discharged, it Out cannot be charged again until the next precharge In1 operation. 1.5 In2 What does this imply about the inputs to the gate transitioning during evaluation? In3 In & 0.5 CLK Out Precharge In4 Output can be in the high impedance state during the evaluation stage if the PDN is turned off CLK -0.5 00.51Time, ns State is stored on CL Input/Output Coupling: Bootstrapping Susceptibility to Noise 3 The amount by which the output voltage drops is a strong 2.5 What’s with the overshoot and function of the input voltage and the available evaluation time. 2 undershoot? 1.5 Noise needed to corrupt the signal has to be larger if the evaluation time Notice input changes with steep 1 is short voltage step 0.5 Takes a while for output to react CLK 0 2.5 0 100 200 300 400 to change in input value -0.5 Gate-drain capacitances of the Vout (VG=0.45) 2.5 transistors couple input directly to 1.5 output in the short time before Vout (VG=0.55) transistors have a chance to Vout (VG=0.5) 1.5 switch (V) Voltage 0.5 VG Extra charge must be supplied to 0.5 discharge the bootstrap -0.5 capacitor 0 20 40 60 80 100 See section 4.4.6.6 -0.5 Time (ns) 00.51 3 10/19/2016 Power and energy Power dissipation of dynamic gate Power dissipation: how much energy is consumed per CLK Mp operation and how much heat the circuit dissipates Out Two important components: static and dynamic In1 CL In2 PDN In 2 3 E (joules) = CL Vdd P01 + tsc Vdd Ipeak P01 + Vdd Ileakage CLK Me f01 = P01 * fclock Power only dissipated when previous Out = 0 2 P(watts) = CL Vdd f01 + tscVdd Ipeak f01 + Vdd Ileakage 2 Probability the output E = CL VDD P01 transitions from 0 to 1 Dynamic power is data dependent Properties of dynamic gates (cont.) Dynamic 2-input NOR Gate Power dissipation should be better Assume signal probabilities no short circuit power consumption since the pull-up path is not ABOut on when evaluating PA=1 = 1/2 00 1 PB=1 = 1/2 lower CL 01 0 by construction can have at most one transition per cycle Then transition probability 10 0 no glitching P01 = Pout=0 × Pout=1 11 0 But power dissipation can be significantly higher due to = 3/4 × 1 = 3/4 higher transition probabilities extra load on CLK Switching activity can be higher in dynamic gates! Needs a precharge clock P01 = Pout=0 4 10/19/2016 Cascading dynamic gates Domino Logic V CLK CLK M M CLK p p CLK M Out2 CLK Mp p Out1 1 1 Out1 Out2 In 1 0 In 0 0 In1 0 1 CLK CLK VTn In PDN In4 PDN Me Me Out1 2 In In3 5 V CLK M Out2 CLK Me e t Only a single 0 1 transition allowed at the inputs during the evaluation period! Why Domino? NP-CMOS (Zipper) !CLK M CLK Mp e 1 1 Out1 CLK 1 0 In PUN In1 4 In In1 In2 PDN 5 In PDN In PDN In PDN In PDN In 0 0 Out2 i i i i 3 0 1 Inj Inj Inj Inj (to PDN) CLK M !CLK Mp CLK e to other to other PDN’s PUN’s Only 0 1 transitions allowed at inputs of PDN Like falling dominos! Only 1 0 transitions allowed at inputs of PUN 5 10/19/2016 Dynamic Zero Detector Dynamic Manchester Carry Chain In7 In6 In5 In4 In3 In2 In1 In0 CLK not zero P0 P1 P2 P3 Ci,4 CLK Ci,0 G0 G1 G2 G3 CLK Efficient use of a dynamic gate. Wide NOR gate that uses minimum sized gate !(G0 + P0 Ci,0)!(G1 + P1G0 + P1P0 Ci,0) Low logical effort Charge leakage Impact of charge leakage Output settles to an intermediate voltage determined by a Ideally, if the PDN is off, the output should remain high (VDD) for the duration of the evaluate stage resistive divider of the pull-up and pull-down networks Leakage currents cause charge to degrade over time If output drops below switching threshold, output interprets as low CLK 4 2.5 CLK 3 Mp CLK Out 1 1.5 Out A=0 CL 2 Evaluate V (V) Voltage CLK Out 0.5 Me Precharge -0.5 Leakage sources 02040 Minimum clock rate of a few kHz Time (ms) 6 10/19/2016 A solution to charge leakage Charge sharing Charge stored originally on CL Keeper compensates for the charge lost due to the CLK Mp is redistributed (shared) over C Out L pull-down leakage paths.
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