Topic 11
Logical Effort:
Outline
Designing for Speed on the Back of an Envelope
David Harris
David_Harris@hmc.edu
Harvey Mudd College
Claremont, CA
o
o
o
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Introduction
Delay in a Logic Gate
Multi-stage Logic Networks
Choosing the Best Number of Stages
Example
Asymmetric & Skewed Logic Gates
Circuit Families
Summary
Logical Effort
Introduction
How large should the transistors be?
How many stages of logic give least delay?
Decoder specification:
Uses a very simple model of delay
Back of the envelope calculations and tractable optimization
Gives new names to old ideas to emphasize remarkable symmetries
Who cares about logical effort?
o
o
o
Circuit designers waste too much time simulating and tweaking circuits
High speed logic designers need to know where time is going in their logic
CAD engineers need to understand circuits to build better tools
Logical Effort
David Harris
Ben Bitdiddle is the memory designer for the Motoroil 68W86, an embedded
processor for automotive applications. Help Ben design the decoder for a
register file:
32 bits
a<3:0> a<3:0>
Page 3 of 56
o
o
o
o
o
16 word register file
Each word is 32 bits wide
16
Register File
Each bit presents a load of 3 unit-sized transistors
True and complementary inputs of address bits a<3:0> are available
Each input may drive 10 unit-sized transistors
Ben needs to decide:
o
o
o
How many stages to use?
How large should each gate be?
How fast can the decoder operate?
Logical Effort
David Harris
Page 4 of 56
16 words
What is the best circuit topology for a function?
4:16 Decoder
? ? ?
Logical Effort is a method of answering these questions:
o
o
o
Page 2 of 56
Example
Chip designers face a bewildering array of choices.
o
o
o
David Harris
Outline
Let us express delays in a process-independent unit:
Introduction
d abs
d = ---------τ
Delay in a Logic Gate
Multi-stage Logic Networks
Choosing the Best Number of Stages
τ ≈ 12 ps
in 0.18 μm
technology
Delay of logic gate has two components:
Example
effort delay, a.k.a. stage effort
Asymmetric & Skewed Logic Gates
parasitic delay
d = f+p
Circuit Families
Summary
Effort delay again has two components:
f = gh
Logical Effort
David Harris
Page 5 of 56
Logical effort describes relative ability of gate topology to deliver current
(defined to be 1 for an inverter)
o
Electrical effort is the ratio of output to input capacitance
Logical Effort
David Harris
g=
p=
d=
effort
delay
6
How about a
2-input NOR?
1
parasitic delay
4
3
2
AN
D
g = 4/3
p=2
d = (4/3)h + 2
g=1
p=1
d=h+1
inp
ut
N
5
2-
AN
D
Page 6 of 56
Delay Plots
Normalized delay: d
2
in
ve
rte
r
3
g=
p=
d=
inp
ut
N
4
2-
Normalized delay: d
5
electrical effort
is sometimes
called “fanout”
o
Delay Plots
6
logical effort
electrical effort = Cout/Cin
in
ve
rte
r
o
o
o
o
o
o
o
o
Delay in a Logic Gate
effort
delay
1
parasitic delay
1 2 3 4 5
Electrical effort: h = Cout / Cin
1 2 3 4 5
Electrical effort: h = Cout / Cin
o d = f + p = gh + p
o Delay increases with electrical effort
o More complex gates have greater logical effort and parasitic delay
o d = f + p = gh + p
o Delay increases with electrical effort
o More complex gates have greater logical effort and parasitic delay
Logical Effort
Logical Effort
David Harris
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David Harris
Page 45 of 56
Computing Logical Effort
A Catalog of Gates
Table 1: Logical effort of static CMOS gates
DEF: Logical effort is the ratio of the input capacitance of a gate to the input
capacitance of an inverter delivering the same output current.
o
o
Measured from delay vs. fanout plots of simulated or measured gates
inverter
Or estimated, counting capacitance in units of transistor width:
a
b
2
4
2
2
a
Inverter:
Cin = 3
g = 1 (def)
2
2
NAND2:
Cin = 4
g = 4/3
a
b
3
4
4/3
5/3
6/3
7/3
(n+2)/3
NOR
5/3
7/3
9/3
11/3
(2n+1)/3
multiplexer
2
2
2
2
2
XOR, XNOR
4
12
32
Table 2: Parasitic delay of static CMOS gates
Gate type
1
1
Parasitic delay
pinv
inverter
n-input NAND
npinv
n-input NOR
npinv
n-way multiplexer
2npinv
p inv ≈ 1
parasitic delays
depend on diffusion
capacitance
2-input XOR, XNOR 4npinv
Logical Effort
David Harris
Page 8 of 56
Logical Effort
David Harris
Example
Logical Effort:
g =
Electrical Effort:
h =
Parasitic Delay:
p =
Stage Delay:
d =
Estimate the frequency of an N-stage ring oscillator:
Oscillator Frequency: F =
Logical Effort:
g≡1
Electrical Effort:
out
h = --------= 1
Parasitic Delay:
p = p inv ≈ 1
Stage Delay:
d = gh + p = 2
C
C in
1
2Nd abs
1
4N τ
Oscillator Frequency: F = ------------------- = -----------
David Harris
Page 9 of 56
Example
Estimate the frequency of an N-stage ring oscillator:
Logical Effort
n
5
1
x
2
1
NOR2:
Cin = 5
g = 5/3
1
NAND
4
x
x
Number of inputs
Gate type
Page 10 of 56
Logical Effort
David Harris
A 31 stage ring
oscillator in a
0.18 μm process
oscillates at about
670 MHz.
Page 46 of 56
Example
Example
Estimate the delay of a fanout-of-4 (FO4) inverter:
Estimate the delay of a fanout-of-4 (FO4) inverter:
d
Logical Effort:
g =
Electrical Effort:
h =
Parasitic Delay:
p =
Stage Delay:
d =
d
Logical Effort:
g≡1
Electrical Effort:
out
h = --------= 4
Parasitic Delay:
p = p inv ≈ 1
Stage Delay:
d = gh + p = 5
C
C in
The FO4 inverter
delay is a useful
metric to characterize
process performance.
1 FO4 delay = 5τ
This is about 60 ps
in a 0.18 μm process.
Logical Effort
David Harris
Page 11 of 56
Logical Effort
David Harris
Outline
o
o
o
o
o
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o
Page 47 of 56
Multi-stage Logic Networks
Logical effort extends to multi-stage networks:
Introduction
Delay in a Logic Gate
10
Multi-stage Logic Networks
Choosing the Best Number of Stages
g1 = 1
h1 = x/10
Example
Asymmetric & Skewed Logic Gates
x
y
g2 = 5/3
h2 = y/x
g3 = 4/3
h3 = z/y
z
20
g4 = 1
h4 = 20/z
Circuit Families
o
Summary
∏ gi
Path Logical Effort:
G =
o
Path Electrical Effort:
C out (path)
H = --------------------C in (path)
o
Path Effort:
F =
∏ fi = ∏ g i h i
Don’t define
H =
∏ hi
because we don’t
know hi until the
design is done
Can we write F = GH ?
Logical Effort
David Harris
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Logical Effort
David Harris
Page 13 of 56
Branching Effort
Branching Effort
No! Consider circuits that branch:
15
G =
H =
GH =
h1 =
h2 =
F =
90
5
15
No! Consider circuits that branch:
90
15
90
5
15
= GH?
G =1
H = 90 / 5 = 18
GH = 18
h1 = (15+15) / 5 = 6
h2 = 90 / 15 = 6
F = 36, not 18!
90
Introduce new kind of effort to account for branching within a network:
C on path + C
o
Branching Effort:
off path
b = ---------------------------------------------
o
Path Branching Effort:
B =
C on path
∏ bi
Now we can compute the path effort:
o
Logical Effort
David Harris
Page 14 of 56
Path Effort:
Logical Effort
Delay in Multi-stage Networks
Path Effort Delay:
o
Path Parasitic Delay:
o
Path Delay:
David Harris
∏ hi = BH ≠ H
in circuits that branch
Page 48 of 56
Determining Gate Sizes
We can now compute the delay of a multi-stage network:
o
F = GBH
Note:
Gate sizes can be found by starting at the end of the path and working backward.
o
∑ fi
PF = ∑ pi
DF = ∑ di = D F + P
DF =
At each gate, apply the capacitance transformation:
C out • g i
i
C in = ---------------------ˆf
i
o
We can prove that delay is minimized when each stage bears the same effort:
Check your work by verifying that the input capacitance specification is satisfied at the beginning of the path.
ˆf = g h = F 1 ⁄ N
i i
Therefore, the minimum delay of an N-stage path is:
NF
o
1⁄N
+P
This is a key result of logical effort. Lowest possible path delay can be found
without even calculating the sizes of each gate in the path.
Logical Effort
David Harris
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Logical Effort
David Harris
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Example
Example
z
Select gate sizes y and z to minimize delay
from A to B
A
Logical Effort:
G =
Electrical Effort:
H =
Branching Effort:
y
C
y
9C
z
z
from A to B
A
9C
B
z
Select gate sizes y and z to minimize delay
3
y
C
9C
z
y
z
9C
B
Logical Effort:
G = (4 ⁄ 3)
B =
Electrical Effort:
C out
H = --------= 9
C in
Path Effort:
F =
Branching Effort:
B = 2•3 = 6
Best Stage Effort:
ˆf =
Path Effort:
F = GHB = 128
Delay:
D =
Best Stage Effort:
ˆf = F 1 ⁄ 3 ≈ 5
Work backward for sizes:
Delay:
D = 3 • 5 + 3 • 2 = 21
• ( 4 ⁄ 3)
y = 3z
---------------------------- = 1.92 C
Work backward for sizes:
z =
y =
Logical Effort
9C
Page 17 of 56
Logical Effort
Outline
o
o
o
o
o
o
o
o
9C • (4 ⁄ 3)
z = ----------------------------- = 2.4 C
5
5
David Harris
David Harris
Page 49 of 56
Choosing the Best Number of Stages
Introduction
How many stages should a path use?
Delay in a Logic Gate
o
o
Multi-stage Logic Networks
Choosing the Best Number of Stages
Example
Delay is not always minimized by using as few stages as possible
Example: How to drive 64 bit datapath with unit-sized inverter
1
1
1
1
Initial driver
8
4
2.8
Asymmetric & Skewed Logic Gates
16
Circuit Families
N:
f:
D:
D = NF
Logical Effort
8
22.6
Summary
Datapath load
David Harris
Page 18 of 56
Logical Effort
64
64
1
64
65
1⁄N
+ P = N ( 64 )
64
2
8
18
1⁄N
9C
64
Fastest
3
4
15
4
2.8
15.3
+ N assuming polarity doesn’t matter
David Harris
Page 19 of 56
Derivation of the Best Number of Stages
Suppose we can add inverters to the end of a path without changing its function.
N-n1 extra inverters
Logic Block:
n1 stages
Path effort F
p inv + ρ ( 1 – ln ρ ) = 0 has no closed form solution.
o
o
Neglecting parasitics (i.e. pinv = 0), we get the familiar result that ρ = 2.718 (e)
For pinv = 1, we can solve numerically to obtain ρ = 3.59
How sensitive is the delay to using exactly the best number of stages?
D(N) / D(N)
n1
D = NF
1⁄N
+
∑
p i + ( N – n 1 ) p inv
1 .6
1 .4
1 .2
1 .0
1 .5 1
1 .1 5
1 .2 6
I like to use
ρ=4
1
(ρ = 2 .4 )
1⁄N
1⁄N
1⁄N
∂D
------- = – F
ln ( F
)+F
+ p inv = 0
∂N
Define ρ ≡ F
ˆ
1⁄N
0 .0
to be the best stage effort. Substitute and simplify:
p inv + ρ ( 1 – ln ρ ) = 0
Logical Effort
David Harris
1 .0
1 .4
2 .0
2.4 < ρ < 6 gives delays within 15% of optimal -> we can be sloppy
Logical Effort
Outline
o
o
o
o
o
o
o
o
0 .7
N /N
o
Page 20 of 56
0 .5
David Harris
Page 21 of 56
Example
Let’s revisit Ben Bitdiddle’s decoder problem using logical effort:
32 bits
a<3:0> a<3:0>
Introduction
Delay in a Logic Gate
Multi-stage Logic Networks
Choosing the Best Number of Stages
Decoder specification:
Example
o
o
o
o
o
Asymmetric & Skewed Logic Gates
Circuit Families
Summary
4:16 Decoder
o
(ρ = 6 )
16 word register file
Each word is 32 bits wide
16
Register File
Each bit presents a load of 3 unit-sized transistors
True and complementary inputs of address bits a<3:0> are available
Each input may drive 10 unit-sized transistors
Ben needs to decide:
o
o
o
Logical Effort
David Harris
Page 22 of 56
How many stages to use?
How large should each gate be?
How fast can the decoder operate?
Logical Effort
David Harris
Page 23 of 56
16 words
o
ˆ
How many stages should we use? Let N be the value of N for least delay.
Best Number of Stages (continued)
Example: Number of Stages
Example: Number of Stages
How many stages should Ben use?
How many stages should Ben use?
o
o
o
o
Effort of decoders is dominated by electrical and branching portions
o
Electrical Effort:
o
Branching Effort:
Effort of decoders is dominated by electrical and branching portions
Electrical Effort:
Branching Effort:
H =
B =
32 • 3- = 9.6
H = -------------10
B = 8
because each address input
controls half the outputs
If we neglect logical effort (assume G = 1),
o
F =
Path Effort:
If we neglect logical effort,
o
Remember that the best stage effort is about ρ = 4
o
Hence, the best number of stages is: N =
Remember that the best stage effort is about ρ = 4
o
o
Logical Effort
David Harris
Page 24 of 56
Hence, the best number of stages is: N = log 476.8 = 3.1
Let’s try a 3-stage design
Logical Effort
Example: Gate Sizes & Delay
z
y
o
o
o
o
Actual path effort is:
Therefore, stage effort should be:
Gate sizes:
Path delay:
Logical Effort
z
F =
f =
z =
D =
David Harris
David Harris
Page 50 of 56
Example: Gate Sizes & Delay
Lets try a 3-stage design using 16 4-input NAND gates with G =
a0a0 a1a1 a2a2 a3a3
10 unit input capacitance
y
F = GBH = 8 • 9.6 = 76.8
Path Effort:
Lets try a 3-stage design using 16 4-input NAND gates with G = 1 • 2 • 1 = 2
a0a0 a1a1 a2a2 a3a3
10 unit input capacitance
out0
y
z
96 unit wordline
capacitance
y
out15
o
Actual path effort is:
z
out0
96 unit wordline
capacitance
out15
F = 2 • 8 • 9.6 = 154
1⁄3
o Therefore, stage effort should be: f = ( 154 )
= 5.36
o z = 96 • 1 ⁄ 5.36 = 18
y = 18 • 2 ⁄ 5.36 = 6.7
o D = 3f + P = 3 • 5.36 + 1 + 4 + 1 = 22.1
y =
Page 25 of 56
Logical Effort
David Harris
Close to
4, so f is
reasonable
Page 51 of 56
Example: Alternative Decoders
Outline
Table 3: Comparison of Decoder Designs
Design
Stages
G
P
D
NAND4; INV
2
2
5
29.8
INV; NAND4; INV
3
2
6
22.1
INV; NAND4; INV; INV
4
2
7
21.1
NAND2; INV; NAND2; INV
4
16/9
6
19.7
INV; NAND2; INV; NAND2; INV
5
16/9
7
20.4
NAND2; INV; NAND2; INV; INV; INV
6
16/9
8
21.6
INV; NAND2; INV; NAND2; INV; INV; INV
7
16/9
9
23.1
NAND2; INV; NAND2; INV; INV; INV; INV; INV 8
16/9
10
24.8
o
o
o
o
o
o
o
o
Introduction
Delay in a Logic Gate
Multi-stage Logic Networks
Choosing the Best Number of Stages
Example
Asymmetric & Skewed Logic Gates
Circuit Families
Summary
We underestimated the best number of stages by neglecting the logical effort.
o
o
o
Logical effort facilitates comparing different designs before selecting sizes
Using more stages also reduces G and P by using multiple 2-input gates
Our design was about 10% slower than the best
Logical Effort
David Harris
Page 26 of 56
Logical Effort
Asymmetric Gates
David Harris
Page 27 of 56
Asymmetric Gates
Asymmetric logic gates favor one input over another.
Asymmetric logic gates favor one input over another.
Example: suppose input A of a NAND gate is most critical.
Example: Suppose input A of a NAND gate is most critical:
o
o
o
o
Select sizes so pullup and pulldown still match unit inverter
Place critical input closest to output
2
Select sizes so pullup and pulldown still match unit inverter
Place critical input closest to output
2
2
x
x
a
o
o
o
Logical Effort on input A:
gA =
Logical Effort on input B:
gB =
Total Logical Effort:
g tot = g A + g B
Logical Effort
a
4/3
4
David Harris
2
4/3
4
b
o
o
o
Page 28 of 56
b
Logical Effort on input A:
g A = 10 ⁄ 9
Logical Effort on input B:
gB = 2
Total Logical Effort:
g tot = g A + g B = 28 ⁄ 9
Logical Effort
David Harris
Effort on A
goes down at
expense of
effort on B and
total gate effort
Page 52 of 56
Symmetry Factor
Skewed Gates
In general, consider gates with arbitrary symmetry factor s:
o
o
s = 1/2 in symmetric gate with equal sizes
Skewed gates favor one edge over the other.
2
s = 1/4 in previous example
Example: suppose rising output of inverter is most critical.
2
x
a
o
Downsize noncritical NMOS transistor to reduce total input capacitance
1/(1-s)
2
b
1/s
Logical effort of inputs:
gA
o
o
1
------------ + 2
1
–s
= ---------------------3
a
gB
1
--- + 2
s
= ------------3
g tot
1
-------------------- + 4
s
(1 – s)
= -----------------------------3
o
o
o
x
a
1
1/2
Page 29 of 56
Logical Effort for rising (up) output:
gu =
Logical Effort for falling (down) output:
gd =
Average Logical Effort:
g avg = ( g u + g d ) ⁄ 2
Logical Effort
Skewed Gates
David Harris
Page 30 of 56
HI- and LO-Skewed Gates
Skewed gates favor one edge over the other.
Example: suppose rising output of inverter is most important.
o
a
Unskewed w/
Unskewed w/
equal rise
equal fall
Compare with unskewed inverter of the same rise/fall time to compute effort.
But total logical effort is higher for asymmetric gates
David Harris
1/2
1
x
HI-Skewed inverter
Critical input approaches logical effort of inverter = 1 for small s
Logical Effort
2
x
Downsize noncritical NMOS transistor to reduce total input capacitance
DEF: Logical effort of a skewed gate for a particular transition is the ratio of the
input capacitance of that gate to the input capacitance of an unskewed
inverter delivering the same output current for the same transition.
Skew gates by reducing size of noncritical transistors.
2
2
x
a
1/2
1
x
a
1
x
a
1/2
Skewed inverter
Unskewed w/
Unskewed w/
equal rise
equal fall
Critical rising
Compare with unskewed inverter of the same rise/fall time
effort goes down
o Logical Effort for rising (up) output:
gu = 5 ⁄ 6
at expense of
noncritical and
o Logical Effort for falling (down) output: g d = 5 ⁄ 3
average effort
o
HI-Skewed gates favor rising outputs by downsizing NMOS transistors
LO-Skewed gates favor falling outputs by downsizing PMOS transistors
Logical effort is smaller for the favored input due to lower input capacitance
Logical effort is larger for the other input
g avg = ( g u + g d ) ⁄ 2 = 5 ⁄ 4
Average Logical Effort:
Logical Effort
o
o
o
o
David Harris
Page 53 of 56
Logical Effort
David Harris
Page 31 of 56
Catalog of Skewed Gates
Inverter
NAND2
4
2
2
½
4
gu = 5/6
gd = 5/3
gavg = 5/4
1
1
gu = 1
gd = 2
gavg = 3/2
LO-Skew
1
½
1
2
gu = 4/3
gd = 2/3
gavg = 1
Logical Effort
2
2
gu = 2
gd = 1
gavg = 3/2
gu = 3/2
gd = 3
gavg = 9/4
½
2
1
1
o
o
o
o
o
o
o
o
NOR2
2
HI-Skew
Outline
Introduction
Delay in a Logic Gate
Multi-stage Logic Networks
Choosing the Best Number of Stages
Example
Asymmetric & Skewed Logic Gates
Circuit Families
Summary
gu = 2
gd = 1
gavg = 3/2
1
1
David Harris
Page 32 of 56
Logical Effort
Pseudo-NMOS
David Harris
Page 33 of 56
Pseudo-NMOS
Pseudo-NMOS gates replace fat PMOS pullups on inputs with a resistive pullup.
Pseudo-NMOS gates replace fat PMOS pullups on inputs with a resistive pullup.
o
o
o
o
o
o
o
o
o
o
Resistive pullup must be much weaker than pulldown stack (e.g. 4x)
Reduces logical effort because inputs must only drive the NMOS transistors
However, NMOS current reduced by contention with pullup
Unequal rising and falling efforts
Quiescent power dissipation when output is low
2/3
Example: Pseudo-NMOS inverter
o
o
o
x
Logical Effort for falling (down) output:
gd =
Logical Effort for rising (up) output:
gu =
Average Logical Effort:
g avg = ( g u + g d ) ⁄ 2
Logical Effort
David Harris
a
Page 34 of 56
4/3
Resistive pullup must be much weaker than pulldown stack (e.g. 4x)
Reduces logical effort because inputs must only drive the NMOS transistors
However, NMOS current reduced by contention with pullup
Unequal rising and falling efforts
Logical effort can be applied to domino, pseudo-NMOS, and other logic families
2/3
Example: Pseudo-NMOS inverter
o
o
o
x
Logical Effort for falling (down) output:
gd = 4 ⁄ 9
Logical Effort for rising (up) output:
gu = 4 ⁄ 3
Average Logical Effort:
g avg = ( g u + g d ) ⁄ 2 = 8 ⁄ 9
Logical Effort
David Harris
a
Page 54 of 56
4/3
Pseudo-NMOS Gates
Dynamic Logic
Dynamic logic replace fat PMOS pullups on inputs with a clocked precharge.
Inverter
NAND2
NOR2
2/3
2/3
2/3
x
a
4/3
gd = 4/9
gu = 4/3
gavg = 8/9
x
a
8/3
b
8/3
o
o
o
o
x
a
4/3
4/3
b
Reduces logical effort because inputs must only drive the NMOS transistors
Eliminates pseudo-NMOS contention current and power dissipation
Only the falling (“evaluation”) delay is critical
Downsize noncritical precharge transistors to reduce clock load and power
φ
Example: Footless dynamic inverter
gd = 8/9
gu = 8/3
gavg = 16/9
o
gd = 4/9
gu = 4/3
gavg = 8/9
x
gd =
Logical Effort for falling (down) output:
1
a
1
Robust gates may require keepers and clocked pulldown transistors (“feet”).
o
o
Tradeoffs exist between power and effort by varying P/N ratio.
Logical Effort
David Harris
Page 35 of 56
Feet prevent contention during precharge but increase logical effort
Weak keepers prevent floating output at cost of slight contention during eval
Logical Effort
David Harris
Dynamic Logic
Dynamic Gates
Dynamic logic replace fat PMOS pullups on inputs with a clocked precharge.
o
o
o
o
Inverter
Reduces logical effort because inputs must only drive the NMOS transistors
Footless
Critical pulldown (“evaluation”) delay independent of precharge size
φ
Logical Effort for falling (down) output:
gd = 1 ⁄ 3
a
1
1
x
2
b
2
φ
1
x
a
1
1
b
gd = 1/3
gd = 2/3
x
a
1
φ
Footed
Page 55 of 56
1
a
a
2
2
gd = 2/3
b
φ
Logical Effort
φ
1
x
x
φ
Weak keepers prevent floating output at cost of slight contention during eval
David Harris
a
NOR2
φ
1
x
1
Feet prevent contention during precharge but increase logical effort
Logical Effort
NAND2
φ
1
gd = 1/3
Robust gates may require keepers and clocked pulldown transistors (“feet”).
o
o
φ
Eliminates pseudo-NMOS contention current and power dissipation
Example: Footless dynamic inverter
Page 36 of 56
3
x
a
3
gd = 1
3
David Harris
φ
2
2
b
gd = 2/3
2
Page 37 of 56
Domino Gates
Domino Gates
Dynamic gates require monotonically rising inputs.
Dynamic gates require monotonically rising inputs.
o
o
o
o
o
o
However, they generate monotonically falling outputs
Alternate dynamic gates with HI-skew inverting static gates
Dynamic / static pair is called a domino gate
However, they generate monotonically falling outputs
Alternate dynamic gates with HI-skew inverting static gates
Dynamic / static pair is called a domino gate
Example: Domino Buffer
Example: Domino Buffer
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
Constraints: maximum input capacitance = 3, load = 54
Logical Effort:
G=
Branching Effort: B =
Electrical Effort:
H=
Path Effort:
F=
Stage Effort:
f=
φ
3
p HI-Skew
a
3
n
g1
g2
54
HI-Skew Inverter: size =
Transistor Sizes: n = p =
Logical Effort
David Harris
Page 38 of 56
Constraints: maximum input capacitance = 3, load = 54
Logical Effort:
G = (1/3) * (5/6) = 5/18
Branching Effort: B = 1
Electrical Effort:
H = 54/3 = 18
Path Effort:
F = (5/18) * 1 * 18 = 5
Stage Effort:
f=
5 = 2.2
HI-Skew Inverter: size =54 * (5/6) / 2.2 = 20
Transistor Sizes: n = 4 p = 16
Logical Effort
Comparison of Circuit Families
PMOS transistors have half the drive of NMOS transistors
Skewed gates downsize noncritical transistors by factor of two
Pseudo-NMOS gates have 1/4 strength pullups
Table 4: Summary of Logical Efforts
Circuit Style
Inverter g
gd
gu
Static CMOS
n-input NAND g
gu
1
gd
n-input NOR g
gu
(n+2)/3
3
p HI-Skew
a
3
n
54
g1 = 1/3
g2 = 5/6
Page 56 of 56
Outline
Assumptions:
o
o
o
David Harris
φ
gd
(2n+1)/3
HI-Skew
5/6
5/3
(n/2+2)/3
(n+4)/3
(2n+.5)/3
(4n+1)/3
LO-Skew
4/3
2/3
2(n+1)/3
(n+1)/3
2(n+1)/3
(n+1)/3
Pseudo-NMOS
4/3
4/9
4n/3
4n/9
4/3
o
o
o
o
o
o
o
o
Introduction
Delay in a Logic Gate
Multi-stage Logic Networks
Choosing the Best Number of Stages
Example
Asymmetric & Skewed Logic Gates
Circuit Families
Summary
4/9
Footed Dynamic
2/3
(n+1)/3
2/3
Footless Dynamic
1/3
n/3
1/3
Adjust these numbers as you change your assumptions.
Logical Effort
David Harris
Page 39 of 56
Logical Effort
David Harris
Page 40 of 56
Summary
Method of Logical Effort
Table 5: Key Definitions of Logical Effort
Term
Logical effort
Electrical effort
Branching effort
Stage expression
g
(seeTable 1)
C
C in
out
h = --------n/a
Logical effort helps you find the best number of stages, the best size of each gate,
and the minimum delay of a circuit with the following procedure:
Path expression
∏ gi
G =
C
C in (path)
out (path)
H = ---------------------
B =
∏ bi
o
Compute the path effort:
o
Estimate the best number of stages:
F = GBH
ˆ ≈ log F
N
o
Estimate the minimum delay:
D = N̂F
+P
F = GBH
o
Sketch your path using the number of stages computed above
∑ fi
Compute the stage effort:
Number of stages
1
N
o
o
Parasitic delay
p
Delay
d = f+p
Effort delay
(seeTable 2)
Logical Effort
DF =
P =
∑
ˆf = F 1 ⁄ N
Starting at the end, work backward to find transistor sizes:
C out • g i
i
C in = ---------------------ˆf
i
pi
D = DF + P
David Harris
Page 41 of 56
Logical Effort
Limitations of Logical Effort
David Harris
Page 42 of 56
Conclusion
Logical effort is not a panacea. Some limitations include:
Logical effort is a useful concept for thinking about delay in circuits:
o
Chicken & egg problem
how to estimate G and best number of stages before the path is designed
o
Simplistic delay model
neglects effects of input slopes
o
ˆ
1⁄N
f = gh
f
Effort
o
4
Interconnect
iteration required in designs with branching and non-negligible wire C or RC
o
o
o
o
o
o
Facilitates comparison of different circuit topologies
Easily select gate sizes for minimum delay
Circuits are fastest when effort delays of each stage are equal and about 4
Path delay is insensitive to modest deviations from optimal sizes
Logic gates can be skewed to favor one input or edge at the cost of another
Logical effort can be applied to domino, pseudo-NMOS, and other logic families
same convergence difficulties as in synthesis / placement problem
Logical effort provides a language for engineers to discuss why circuits are fast.
Maximum speed only
optimizes circuits for speed, not area or power under a fixed speed constraint
o
Like any language, requires practice to master
A book on Logical Effort is available from Morgan Kaufmann Publishers
o http://www.mkp.com/Logical_Effort
o Discusses P/N ratios, gate characterization, pass gate logic, forks, wires, etc.
Logical Effort
David Harris
Page 43 of 56
Logical Effort
David Harris
Page 44 of 56
Delay Plots
Estimate the frequency of an N-stage ring oscillator:
AN
D
g=1
p=1
d=h+1
in
ve
rte
r
3
g = 4/3
p=2
d = (4/3)h + 2
inp
ut
N
5
2-
Normalized delay: d
6
4
Example
effort
delay
2
1
parasitic delay
1 2 3 4 5
Electrical effort: h = Cout / Cin
o d = f + p = gh + p
o Delay increases with electrical effort
o More complex gates have greater logical effort and parasitic delay
Logical Effort
David Harris
Page 45 of 56
Logical Effort:
g≡1
Electrical Effort:
out
h = --------= 1
Parasitic Delay:
p = p inv ≈ 1
Stage Delay:
d = gh + p = 2
C
C in
1
2Nd abs
Logical Effort
David Harris
Example
No! Consider circuits that branch:
d
15
15
Electrical Effort:
out
h = --------= 4
C
C in
Parasitic Delay:
p = p inv ≈ 1
Stage Delay:
d = gh + p = 5
Logical Effort
David Harris
The FO4 inverter
delay is a useful
metric to characterize
process performance.
G =1
H = 90 / 5 = 18
GH = 18
h1 = (15+15) / 5 = 6
h2 = 90 / 15 = 6
F = 36, not 18!
90
5
g≡1
Page 46 of 56
Branching Effort
Estimate the delay of a fanout-of-4 (FO4) inverter:
Logical Effort:
1
4N τ
Oscillator Frequency: F = ------------------- = -----------
A 31 stage ring
oscillator in a
0.18 μm process
oscillates at about
670 MHz.
90
Introduce new kind of effort to account for branching within a network:
C on path + C
o
Branching Effort:
off path
b = ---------------------------------------------
1 FO4 delay = 5τ
o
Path Branching Effort:
B =
This is about 60 ps
in a 0.18 μm process.
Now we can compute the path effort:
Page 47 of 56
o
Path Effort:
Logical Effort
C on path
∏ bi
F = GBH
David Harris
Note:
∏ hi = BH ≠ H
in circuits that branch
Page 48 of 56
Example
Example: Number of Stages
z
Select gate sizes y and z to minimize delay
from A to B
A
3
y
C
9C
z
y
9C
z
B
How many stages should Ben use?
o
Effort of decoders is dominated by electrical and branching portions
o
Electrical Effort:
o
Branching Effort:
Logical Effort:
G = (4 ⁄ 3)
Electrical Effort:
C out
H = --------= 9
C in
Branching Effort:
B = 2•3 = 6
If we neglect logical effort,
Path Effort:
F = GHB = 128
o
Best Stage Effort:
ˆf = F 1 ⁄ 3 ≈ 5
Delay:
9C
Work backward for sizes:
9C • (4 ⁄ 3)
z = ----------------------------- = 2.4 C
5
D = 3 • 5 + 3 • 2 = 21
• ( 4 ⁄ 3)
y = 3z
---------------------------- = 1.92 C
5
Logical Effort
David Harris
Page 49 of 56
controls half the outputs
Path Effort:
Hence, the best number of stages is: N = log 476.8 = 3.1
Let’s try a 3-stage design
Logical Effort
y
o
Actual path effort is:
z
Example: Suppose input A of a NAND gate is most critical:
o
o
Select sizes so pullup and pulldown still match unit inverter
Place critical input closest to output
2
96 unit wordline
capacitance
2
x
a
4/3
4
out15
F = 2 • 8 • 9.6 = 154
David Harris
Page 50 of 56
Asymmetric logic gates favor one input over another.
out0
o Therefore, stage effort should be: f = ( 154 )1 ⁄ 3 = 5.36
o z = 96 • 1 ⁄ 5.36 = 18
y = 18 • 2 ⁄ 5.36 = 6.7
o D = 3f + P = 3 • 5.36 + 1 + 4 + 1 = 22.1
Logical Effort
David Harris
Asymmetric Gates
Lets try a 3-stage design using 16 4-input NAND gates with G = 1 • 2 • 1 = 2
a0a0 a1a1 a2a2 a3a3
10 unit input capacitance
z
F = GBH = 8 • 9.6 = 76.8
Remember that the best stage effort is about ρ = 4
o
o
Example: Gate Sizes & Delay
y
32 • 3- = 9.6
H = -------------10
B = 8
because each address input
Close to
4, so f is
reasonable
Page 51 of 56
o
o
o
b
Logical Effort on input A:
g A = 10 ⁄ 9
Logical Effort on input B:
gB = 2
Total Logical Effort:
g tot = g A + g B = 28 ⁄ 9
Logical Effort
David Harris
Effort on A
goes down at
expense of
effort on B and
total gate effort
Page 52 of 56
Skewed Gates
Pseudo-NMOS
Skewed gates favor one edge over the other.
Pseudo-NMOS gates replace fat PMOS pullups on inputs with a resistive pullup.
Example: suppose rising output of inverter is most important.
o
o
o
o
o
o
Downsize noncritical NMOS transistor to reduce total input capacitance
2
2
x
a
1
x
a
1/2
x
a
1
Reduces logical effort because inputs must only drive the NMOS transistors
However, NMOS current reduced by contention with pullup
Unequal rising and falling efforts
Logical effort can be applied to domino, pseudo-NMOS, and other logic families
1/2
Skewed inverter
Unskewed w/
Unskewed w/
equal rise
equal fall
Critical rising
Compare with unskewed inverter of the same rise/fall time
effort goes down
o Logical Effort for rising (up) output:
gu = 5 ⁄ 6
at expense of
noncritical and
o Logical Effort for falling (down) output: g d = 5 ⁄ 3
average effort
o
Resistive pullup must be much weaker than pulldown stack (e.g. 4x)
2/3
Example: Pseudo-NMOS inverter
o
o
o
x
Logical Effort for falling (down) output:
gd = 4 ⁄ 9
Logical Effort for rising (up) output:
gu = 4 ⁄ 3
Average Logical Effort:
g avg = ( g u + g d ) ⁄ 2 = 8 ⁄ 9
David Harris
Page 53 of 56
Logical Effort
David Harris
Dynamic Logic
Dynamic gates require monotonically rising inputs.
o
o
o
o
o
o
Reduces logical effort because inputs must only drive the NMOS transistors
Eliminates pseudo-NMOS contention current and power dissipation
Critical pulldown (“evaluation”) delay independent of precharge size
φ
Example: Footless dynamic inverter
Logical Effort for falling (down) output:
gd = 1 ⁄ 3
x
a
1
Feet prevent contention during precharge but increase logical effort
Weak keepers prevent floating output at cost of slight contention during eval
David Harris
Page 55 of 56
However, they generate monotonically falling outputs
Alternate dynamic gates with HI-skew inverting static gates
Dynamic / static pair is called a domino gate
Example: Domino Buffer
1
Robust gates may require keepers and clocked pulldown transistors (“feet”).
Logical Effort
Page 54 of 56
Domino Gates
Dynamic logic replace fat PMOS pullups on inputs with a clocked precharge.
o
o
4/3
g avg = ( g u + g d ) ⁄ 2 = 5 ⁄ 4
Average Logical Effort:
Logical Effort
o
a
o
o
o
o
o
o
o
o
Constraints: maximum input capacitance = 3, load = 54
Logical Effort:
G = (1/3) * (5/6) = 5/18
Branching Effort: B = 1
Electrical Effort:
H = 54/3 = 18
Path Effort:
F = (5/18) * 1 * 18 = 5
Stage Effort:
f=
5 = 2.2
HI-Skew Inverter: size =54 * (5/6) / 2.2 = 20
Transistor Sizes: n = 4 p = 16
Logical Effort
David Harris
φ
3
p HI-Skew
a
3
n
54
g1 = 1/3
g2 = 5/6
Page 56 of 56