Prof. John Nestor ECE Department Lafayette College Easton, Pennsylvania ECE VLSI Circuit Design Lecture 13 - More about Parasitics; Logical Effort Spring 2007

ECE 425 Spring 2007Lecture 13 - Logical Effort2 Announcements  Reading  Book: ,  Weste & Harris Excerpt  Verilog Handout (from ECE 313 last year): 1-4, 5.4  Exam 1: Wed. 3/21  Homework due Wed. 3/28:  Weste & Harris Problems: 4.3, 4.5, 4.9, 4.10, 4.11, 4.12  Logical Effort References:  I. Sutherland, R. Sproull, and D. Harris, Logical Effort: Designing Fast CMOS Circuits, Morgan Kaufmann,  N. Weste and D. Harris, CMOS VLSI Design, 3 rd ed, 2004.

ECE 425 Spring 2007Lecture 13 - Logical Effort3 Where we are  Last Time:  Combinational Network Delay  Today:  Effect of Diffusion Parasitics  Logical Effort

ECE 425 Spring 2007Lecture 13 - Logical Effort4 Impact of Diffusion Parasitics on Delay  The  model ignores parasitics except on output - is this a safe assumption?  Consider a 2-input NAND gate - where are the parastics?

ECE 425 Spring 2007Lecture 13 - Logical Effort5 Parasitics on 2-input NAND  How can we estimate C pdiff and C ndiff ?

ECE 425 Spring 2007Lecture 13 - Logical Effort6 Consider NAND Layout from Lecture 4

ECE 425 Spring 2007Lecture 13 - Logical Effort7 Consider NAND Layout from Lecture 4

ECE 425 Spring 2007Lecture 13 - Logical Effort8 Diffusion Parasitics - Summing Up = 0.725fF = 6.465fF

ECE 425 Spring 2007Lecture 13 - Logical Effort9 Diffusion Parasitics in Large Gates  What is the effect of parasitics in a 4-input NAND?  Use Elmore Delay

ECE 425 Spring 2007Lecture 13 - Logical Effort10 Logical Effort - Making Sizing Systematic  Invented by Ivan Sutherland, Sun Microsystems  Key ideas:  Simple linear model of delay - good for hand calculations  Account for loading in multiples of min-size inverter capacitance  Calculate delays in terms of inverter delay   Account for diffusion parasitics on gate outputs  Use to predict delay of circuits relative to each other instead of estimating  Useful to set sizing from stage to stage  Most accurate when no reconvergent fanout

ECE 425 Spring 2007Lecture 13 - Logical Effort11 Logical Effort - Assumptions  Assume that µ p /µ n = 2 (so R p = 2R n )  Assume all basic gates are sized for equal rise/fall time: t r-inv = t f-inv = t r-nand = t f-nand = t r-nor = t f-nor  Calculate input loading of each basic gate as a multiple of min-size transistor’s gate capacitance  Assume drain diffusion parasitic of a transistor is equal to the transistor’s gate capacitance

ECE 425 Spring 2007Lecture 13 - Logical Effort12 Transistor Sizing for LE assumptions  What are transistor widths as a multiple of min-size?  What are the input loads as a multiple of min-size gate cap? Inverter NAND NOR Load = 3 Load = 4 Load = 5

ECE 425 Spring 2007Lecture 13 - Logical Effort13 Logical Effort - Key definitions   - delay of a minimum-size inverter driving another inverter (without parasitics)  Absolute gate delay: d abs = d *   Values of  for common processes  For 0.6µm process,  = 50ps (approx) [Sutherland 99]  For a 180nm process,  = 15ps (approx) [Weste 04]  Gate delay formula: d = f + p  Effort delay f is related to gate’s load.  Parasitic delay p - due to parasitics in gate itself.

ECE 425 Spring 2007Lecture 13 - Logical Effort14 Logical Effort - Continued  Delay formula (last slide) d = f + p  Effort Delay  Effort delay has two components: f = g*h  Electrical effort h is determined by gate’s load: h = C out /C in  Logical effort g is determined by gate’s structure (see next slide)  Modified Delay Formula: d = g*h + p

ECE 425 Spring 2007Lecture 13 - Logical Effort15 Logical Effort (g) of Common Gates  Definition: the ratio of input capacitance of the gate to the input capacitance of an inverter that can deliver the same current

ECE 425 Spring 2007Lecture 13 - Logical Effort16 Logical Effort of Basic Gates  Definition: the ratio of input capacitance of the gate to the input capacitance of an inverter that can deliver the same current NOR Inverter NAND g = 3/3 = 1 g = 4/3 g = 5/3

ECE 425 Spring 2007Lecture 13 - Logical Effort17 Parasitic Delay (p) of common gates  Assumption for one transistor: diffusion capacitance = gate capacitance  Delay due to parasitics on min-size inverter Parastic loading 3C * R = 3*RC =   Count diffusion capacitance on output nodes only

ECE 425 Spring 2007Lecture 13 - Logical Effort18 Logical Effort: Normalized Delay vs. Fanout Inverter 2-Input NAND Electrical Effort: h = C out /C in Parasitic Delay: p Effort Delay: f = gh g = 1 p = 1 d= h + 1 Normallized Delay: d g = 4/3 p = 2 d= (4/3)h + 2 *From Weste & Harris, CMOS VLSI Design d = gh + p

ECE 425 Spring 2007Lecture 13 - Logical Effort19 Example - Delay Calculations w/ NAND g = 4/3 (from table - p. 14) h = C out /C in = 20/4 = 5 p = 2 (from table - p. 15) d = gh + p = (4/3)*5 + 2 = 8.67 g = 4/3 (unchanged!) h = C out /C in = 20/8 = 2.5 p = 2 (unchanged - why?) d = gh + p = (4/3)* = 5.33

ECE 425 Spring 2007Lecture 13 - Logical Effort20 Example * : FO4 Delay  What is the delay of a fanout-of-4 (FO4) inverter? g = 1(from table - p. 14) p = 1(from table - p. 15) h = C out /C in =12/3=4 d = g*h + p = 1*4 + 1 = 5 If  =15ps in a 0.18µm technology, d abs = d*  = 75ps  Note: FO4 delay is often used to characterize the speed of a process d=? *From Weste & Harris, CMOS VLSI Design

ECE 425 Spring 2007Lecture 13 - Logical Effort21 Example * : Ring Oscillator  What is the frequency of an N-stage ring oscillator?  g = 1from table - p. 14  p = 1from table - p. 15  h = C out /C in = 1/1 =1  d = g*h + p = 1*2 + 1 = 2  T osc = 2*N*d = 4*Nf osc = 1/T osc = 1/(4*N)  Given  =15ps in a 0.18µm technology, and N=31 T osc(s) = T osc *  = 4*31*15ps = 1.86ns f osc = 1/T osc = 536MHz  Ring oscillators often used as process monitors *From Weste & Harris, CMOS VLSI Design N Stages

ECE 425 Spring 2007Lecture 13 - Logical Effort22 Example * : NOR Driving 10 NORs  What is a NOR gate with input capacitance x driving 10 identical NOR gates? g = 9/3from table - p. 14 p = 4from table - p. 15 h = C out /C in = 10x/x = 10 d= g*h + p = 9/3 * = 34 If  =15ps in a 0.18µm technology, d abs = d*  = 465ps  We don’t need it here, but what is the input loading x? *From Sutherland et. al., Logical Effort

ECE 425 Spring 2007Lecture 13 - Logical Effort23 Delay in Multiple Stages  Path Logical Effort G - LE along a chain of gates: G =  g i  Path Electrical Effort H - depends on ratio of first and last stage capacitance: H = C out /C in x yz g 1 =1 h 1 =x/10 g 2 =5/3 h 1 =y/x g 3 =4/3 h 3 =z/y g 4 =1 h 1 =20/z G = g 1 *g 2 *g 3 *g 4 = 1*5/3*4/3*1 = 20/9H = 20/10 = 2 *From Weste & Harris, CMOS VLSI Design

ECE 425 Spring 2007Lecture 13 - Logical Effort Branching Effort  Takes into account fanout  Branching effort at one stage: b = (C onpath + C offpath )/ C onpath  Branching effort along path: B =  b i Path b= (15+15/15) = 2 *From Weste & Harris, CMOS VLSI Design

ECE 425 Spring 2007Lecture 13 - Logical Effort25 Path Delay  Path effort - product of stage efforts F =  f i = GBH  Path delay - sum of delays of gates along the path: D =  g i h i +  p i = D F + P D F =  f i P =  p i

ECE 425 Spring 2007Lecture 13 - Logical Effort26 Minimizing Path Delay  Path effort: F =  f i = GBH - independent of sizes  Path delay is sum of delays of gates along the path: D =  g i h i +  p i = D F + P D F =  f i P =  p i  To minimize  f i make f i equal in each stage  Minimum possible delay in an N-stage path Key Result of Logical Effort. Min delay can be found without calculating the sizes of each gate in path

ECE 425 Spring 2007Lecture 13 - Logical Effort27 Sizing the Transistors  Minimum possible delay in an N-stage path: D=N*F 1/N + P  Determine W/L of each gate on path by working backward from the final gate: Capacitance Transformation

ECE 425 Spring 2007Lecture 13 - Logical Effort28 Example - Transistor Sizing  Estimate minimum delay of path from A to B  Choose transistor sizes to achieve delay 8 Path x x x y y 45

ECE 425 Spring 2007Lecture 13 - Logical Effort29 Example - Transistor Sizing  Path Logical Effort  Branching Effort  Path Electrical Effort  Path Effort  Best Stage Effort  Min Delay 8 Path x x x y y 45 H = C out /C in = 45/8 g 1 = 4/3 g 2 = 5/3 g 3 = 5/3 b 1 = 3 p 2 = 3 p 3 = 2 B = b 1 *b 2 *b 1 = 3 * 2 * 1 = 6 G = g 1 *g 2 *g 3 = 4/3 * 4/3 * 5/3 = 100/27 F = G*B*H = D = N*F 1/N + P = 3*5 + (2+3+2) = 22 p 1 = 2 b 2 = 2 b 3 = 1

ECE 425 Spring 2007Lecture 13 - Logical Effort30 Example - Transistor Sizing  Now assign transistor sizes using capacitance transformation 8 Path x x x y y 45 g 1 = 4/3 g 2 = 5/3 g 3 = 5/3 p 2 = 3 p 3 = 2 45 p 1 = 2 P 3 = 12 N 3 = 3 P 2 = 4 N 2 = 6 P 2 = 4 N 2 = 4 y = 15 x = 10

ECE 425 Spring 2007Lecture 13 - Logical Effort31 Example - Transistor Sizing  Double-Check: Calculate Delay of Each Stage 8 Path g 1 = 4/3 g 2 = 5/3 g 3 = 5/3 p 2 = 3 p 3 = 2 45 p 1 = 2 h 1 = ( )/8 = 3.75 h 2 = 2 h 3 = 2 d i = g i *h i + p i d 1 = (4/3)* = 7 h 2 = (15+15)/10 = 3 d 2 = (5/3)*3 + 3 = 8 h 3 = 45/15 = 3 d 3 = (5/3)*3 + 2 = 7 D = d 1 + d 2 + d 3 = 22

ECE 425 Spring 2007Lecture 13 - Logical Effort32 Choosing the Best Number of Stages  Which buffer circuit is fastest? NfD x 4x16x 2.8x 8x23x

ECE 425 Spring 2007Lecture 13 - Logical Effort33 Choosing the Best Number of Stages  The General Case - What should N be?

ECE 425 Spring 2007Lecture 13 - Logical Effort34 Example: Decoder Design  Ben Bitdiddle is the memory designer for the Motoroil 68W86, an embedded automotive processor. Help Ben design the decoder for a register file. Decoder Register File 16 A[3:0] A’[3:0] 32 bits 16 words

ECE 425 Spring 2007Lecture 13 - Logical Effort35 Example: Decoder Design  Decoder specifications:  16 word register file  Each word is 32 bits wide  Each bit presents load of 3 unit-sized transistors  True and complementary address inputs A[3:0]  Each input may drive 10 unit-sized transistors  Ben needs to decide:  How many stages to use?  How large should each gate be?  How fast can decoder operate? DecoderRegister File 16 A[3:0] A’[3:0] 32 bits 16 words

ECE 425 Spring 2007Lecture 13 - Logical Effort36 One Possible Design - 3 Stages  Decoder effort is mostly electrical and branching  If we neglect logical effort (assume G = 1)  Try a 3-Stage Design Electrical Effort:H = (32*3)/10 = 9.6 Branching Effort:B = 8 Path Effort: F = GBH = 76.8 Number of StagesN = log 4 F = 3.1

ECE 425 Spring 2007Lecture 13 - Logical Effort37 3-Stage Decoder Design b 1 =8 b 2 =1b 3 =8 g 2 =6/3

ECE 425 Spring 2007Lecture 13 - Logical Effort38 Gate Sizes - 3-Stage Decoder Electrical Effort:H = (32*3)/10 = 9.6 Branching Effort:B = 8 Path Logical Effort:G = (1*6/3*1) = 2 Path Effort:F = GBH = 2*9.6*8 = Stage Effort:

ECE 425 Spring 2007Lecture 13 - Logical Effort39 Decoder - Alternative Designs /96NAND2-INV-NAND2-INV-INV-INV /95INV-NAND2-INV-NAND2-INV /94NAND2-INV-NAND2-INV /94NAND2-NOR2-INV-INV NAND4-INV-INV-INV INV-NAND4-INV /92NAND2-NOR NAND4-INV DPGNDesign

ECE 425 Spring 2007Lecture 13 - Logical Effort40 Summary of Notation D =  d i = D F + P d = f + pdelay P =  p i p (see table)parasitic delay D F =  f i feffort delay F = GBHf = gheffort B =  b i branching effort electrical effort G =  g i g (see table)logical effort N1number of stages PathStageTerm

ECE 425 Spring 2007Lecture 13 - Logical Effort41 Summary - The Method of Logical Effort  Compute the path effortF = GBH  Estimate the best number of stages  Sketch a path using:  Estimate the minimum delay: D=N*F 1/N + P  Determine the best stage effort:  Starting with output, work backward to find sizes:

ECE 425 Spring 2007Lecture 13 - Logical Effort42 Limitations of Logical Effort  Chicken and egg problem  Need path to compute G  But don’t know number of stages without G  Simplistic delay model  Neglects input rise time effects  Neglects parasitics not on output node  Interconnect  Iteration required in designs with wire  Maximum speed only  Not minimum area/power for constrained delay

ECE 425 Spring 2007Lecture 13 - Logical Effort43 Summary  Logical effort is useful for thinking of delay in circuits  Numeric logical effort characterizes gates  NANDs are faster than NORs in CMOS  Paths are fastest when effort delays are ~4  Path delay is weakly sensitive to stages, sizes  But using fewer stages doesn’t mean faster paths  Delay of path is about log 4 F FO4 inverter delays  Inverters and NAND2 best for driving large caps  Provides language for discussing fast circuits  But requires practice to master

ECE 425 Spring 2007Lecture 13 - Logical Effort44 Coming Up  Testing  Comb. Logic Design using Standard Cells  Sequential Logic