Download Advanced static gate design

Introduction to
CMOS VLSI
Design
Combinational Circuits
Outline







Bubble Pushing
Compound Gates
Logical Effort Example
Input Ordering
Asymmetric Gates
Skewed Gates
Best P/N ratio
Combinational Circuits
CMOS VLSI Design
Slide 2
Example 1
module mux(input s, d0, d1,
output y);
assign y = s ? d1 : d0;
endmodule
1) Sketch a design using AND, OR, and NOT gates.
Combinational Circuits
CMOS VLSI Design
Slide 3
Example 1
module mux(input s, d0, d1,
output y);
assign y = s ? d1 : d0;
endmodule
1) Sketch a design using AND, OR, and NOT gates.
D0
S
Y
D1
S
Combinational Circuits
CMOS VLSI Design
Slide 4
Example 2
2) Sketch a design using NAND, NOR, and NOT gates.
Assume ~S is available.
Combinational Circuits
CMOS VLSI Design
Slide 5
Example 2
2) Sketch a design using NAND, NOR, and NOT gates.
Assume ~S is available.
D0
S
Y
D1
S
Combinational Circuits
CMOS VLSI Design
Slide 6
Bubble Pushing
 Start with network of AND / OR gates
 Convert to NAND / NOR + inverters
 Push bubbles around to simplify logic
– Remember DeMorgan’s Law
Y
Y
(a)
(b)
Y
(c)
D
Combinational Circuits
Y
(d)
CMOS VLSI Design
Slide 7
Example 3
3) Sketch a design using one compound gate and one
NOT gate. Assume ~S is available.
Combinational Circuits
CMOS VLSI Design
Slide 8
Example 3
3) Sketch a design using one compound gate and one
NOT gate. Assume ~S is available.
D0
S
D1
S
Combinational Circuits
Y
CMOS VLSI Design
Slide 9
Compound Gates
 Logical Effort of compound gates
unit inverter
AOI21
YA
Y  A BC
Y  A BC D
A
B
C
A
B
C
D
A
Y
Y
A
A
2
1
Y
AOI22
4 B
C
A
2
B
2
4
4
C
Y
1
Complex AOI
Y
A
4 B
4
C
4 D
4
A
2 C
2
B
2 D
2
Y
Y  A B  C  D E
D
E
A
B
C
Y
B
6
C
6
A
3
D
6
E
6
E
2
A
2
D
2
2
C
gA = 3/3
gA = 6/3
gA =
gA =
p = 3/3
gB = 6/3
gB =
gB =
gC = 5/3
gC =
gC =
p = 7/3
gD =
gD =
p=
gE =
B
Y
2
p=
Combinational Circuits
CMOS VLSI Design
Slide 10
Inner & Outer Inputs
 Outer input is closest to rail (B)
 Inner input is closest to output (A)
2
A
B
 If input arrival time is known
– Connect latest input to inner terminal
Combinational Circuits
CMOS VLSI Design
2
Y
2
2
Slide 22
Asymmetric Gates
 Asymmetric gates favor one input over another
 Ex: suppose input A of a NAND gate is most critical
– Use smaller transistor on A (less capacitance)
A
– Boost size of noncritical input
reset
– So total resistance is same
 gA =
2
Y
A
4/3
 gB =
reset
 gtotal = gA + gB =
 Asymmetric gate approaches g = 1 on critical input
 But total logical effort goes up
Combinational Circuits
CMOS VLSI Design
Y
Slide 23
Asymmetric Gates
 Asymmetric gates favor one input over another
 Ex: suppose input A of a NAND gate is most critical
– Use smaller transistor on A (less capacitance)
A
– Boost size of noncritical input
reset
– So total resistance is same
 gA = 10/9
2
2
Y
A
4/3
 gB = 2
4
reset
 gtotal = gA + gB = 28/9
 Asymmetric gate approaches g = 1 on critical input
 But total logical effort goes up
Combinational Circuits
CMOS VLSI Design
Y
Slide 24
Symmetric Gates
 Inputs can be made perfectly symmetric
2
2
A
1
1
B
1
1
Combinational Circuits
CMOS VLSI Design
Y
Slide 25
Skewed Gates
 Skewed gates favor one edge over another
 Ex: suppose rising output of inverter is most critical
– Downsize noncritical nMOS transistor
HI-skew
inverter
unskewed inverter
(equal rise resistance)
2
A
unskewed inverter
(equal fall resistance)
2
Y
1/2
A
1
Y
A
1
Y
1/2
 Calculate logical effort by comparing to unskewed
inverter with same effective resistance on that edge.
– gu =
– gd =
Combinational Circuits
CMOS VLSI Design
Slide 26
Skewed Gates
 Skewed gates favor one edge over another
 Ex: suppose rising output of inverter is most critical
– Downsize noncritical nMOS transistor
HI-skew
inverter
unskewed inverter
(equal rise resistance)
2
A
unskewed inverter
(equal fall resistance)
2
Y
1/2
A
1
Y
A
1
Y
1/2
 Calculate logical effort by comparing to unskewed
inverter with same effective resistance on that edge.
– gu = 2.5 / 3 = 5/6
– gd = 2.5 / 1.5 = 5/3
Combinational Circuits
CMOS VLSI Design
Slide 27
HI- and LO-Skew
 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.
 Skewed gates reduce size of noncritical transistors
– HI-skew gates favor rising output (small nMOS)
– LO-skew gates favor falling output (small pMOS)
 Logical effort is smaller for favored direction
 But larger for the other direction
Combinational Circuits
CMOS VLSI Design
Slide 28
Catalog of Skewed Gates
Inverter
NAND2
2
unskewed
1
2
Y
2
A
NOR2
Y
gu = 1
gd = 1
gavg = 1
A
2
B
2
B
4
A
4
Y
1
gu = 4/3
gd = 4/3
gavg = 4/3
1
gu = 5/3
gd = 5/3
gavg = 5/3
B
HI-skew
Y
2
A
1/2
Y
gu = 5/6
gd = 5/3
gavg = 5/4
A
A
Y
B
gu =
gd =
gavg =
gu =
gd =
gavg =
B
LO-skew
Y
1
A
1
Combinational Circuits
Y
gu = 4/3
gd = 2/3
gavg = 1
A
A
B
Y
gu =
gd =
gavg =
CMOS VLSI Design
gu =
gd =
gavg =
Slide 29
Catalog of Skewed Gates
Inverter
NAND2
2
unskewed
1
Y
gu = 1
gd = 1
gavg = 1
A
2
B
2
2
HI-skew
1/2
Y
gu = 5/6
gd = 5/3
gavg = 5/4
1
B
1
1
LO-skew
1
Combinational Circuits
Y
gu = 4/3
gd = 2/3
gavg = 1
2
B
2
4
1
1
B
4
A
4
1/2
gu =
gd =
gavg =
1
A
A
gu = 5/3
gd = 5/3
gavg = 5/3
Y
Y
1
A
4
gu = 4/3
gd = 4/3
gavg = 4/3
2
A
B
Y
Y
2
A
2
Y
2
A
NOR2
1/2
B
2
A
2
gu =
gd =
gavg =
Y
gu =
gd =
gavg =
CMOS VLSI Design
1
1
gu =
gd =
gavg =
Slide 30
Catalog of Skewed Gates
Inverter
NAND2
2
unskewed
1
Y
gu = 1
gd = 1
gavg = 1
A
2
B
2
2
HI-skew
1/2
Y
gu = 5/6
gd = 5/3
gavg = 5/4
1
B
1
1
LO-skew
1
Combinational Circuits
Y
gu = 4/3
gd = 2/3
gavg = 1
2
B
2
4
1
1
B
4
A
4
1/2
gu = 1
gd = 2
gavg = 3/2
1
A
A
gu = 5/3
gd = 5/3
gavg = 5/3
Y
Y
1
A
4
gu = 4/3
gd = 4/3
gavg = 4/3
2
A
B
Y
Y
2
A
2
Y
2
A
NOR2
1/2
B
2
A
2
gu = 3/2
gd = 3
gavg = 9/4
Y
gu = 2
gd = 1
gavg = 3/2
CMOS VLSI Design
1
1
gu = 2
gd = 1
gavg = 3/2
Slide 31
Asymmetric Skew
 Combine asymmetric and skewed gates
– Downsize noncritical transistor on unimportant
input
– Reduces parasitic delay for critical input
A
reset
Y
1
A
reset
Combinational Circuits
2
Y
4/3
4
CMOS VLSI Design
Slide 32
Best P/N Ratio
 We have selected P/N ratio for unit rise and fall
resistance (m = 2-3 for an inverter).
 Alternative: choose ratio for least average delay
 Ex: inverter
P
– Delay driving identical inverter
A
1
– tpdf =
– tpdr =
– tpd =
– Differentiate tpd w.r.t. P
– Least delay for P =
Combinational Circuits
CMOS VLSI Design
Slide 33
Best P/N Ratio
 We have selected P/N ratio for unit rise and fall
resistance (m = 2-3 for an inverter).
 Alternative: choose ratio for least average delay
 Ex: inverter
P
– Delay driving identical inverter
A
1
– tpdf = (P+1)
– tpdr = (P+1)(m/P)
– tpd = (P+1)(1+m/P)/2 = (P + 1 + m + m/P)/2
– Differentiate tpd w.r.t. P
– Least delay for P = m
Combinational Circuits
CMOS VLSI Design
Slide 34
P/N Ratios
 In general, best P/N ratio is sqrt of equal delay ratio.
– Only improves average delay slightly for inverters
– But significantly decreases area and power
Inverter
NAND2
2
fastest
P/N ratio
A
1.414
Y
1
gu =
gd =
gavg =
Combinational Circuits
NOR2
2
Y
A
2
B
2
B
2
A
2
Y
gu =
gd =
gavg =
CMOS VLSI Design
1
1
gu =
gd =
gavg =
Slide 35
P/N Ratios
 In general, best P/N ratio is sqrt of that giving equal
delay.
– Only improves average delay slightly for inverters
– But significantly decreases area and power
Inverter
NAND2
2
fastest
P/N ratio
A
1.414
Y
1
gu = 1.15
gd = 0.81
gavg = 0.98
Combinational Circuits
NOR2
2
Y
A
2
B
2
B
2
A
2
Y
gu = 4/3
gd = 4/3
gavg = 4/3
CMOS VLSI Design
1
1
gu = 2
gd = 1
gavg = 3/2
Slide 36
Observations
 For speed:
– NAND vs. NOR
– Many simple stages vs. fewer high fan-in stages
– Latest-arriving input
 For area and power:
– Many simple stages vs. fewer high fan-in stages
Combinational Circuits
CMOS VLSI Design
Slide 37