Transcript Lecture 9: Combinational Circuit Design

Lecture 9:
Combinational
Circuit Design
Outline



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Bubble Pushing
Compound Gates
Logical Effort Example
Input Ordering
Asymmetric Gates
Skewed Gates
Best P/N ratio
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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
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Example 2
2) Sketch a design using NAND, NOR, and NOT gates.
Assume ~S is available.
D0
S
Y
D1
S
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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
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Y
(d)
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Example 3
3) Sketch a design using one compound gate and one
NOT gate. Assume ~S is available.
D0
S
D1
S
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Y
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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
A
A
2
1
Y
AOI22
Y
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
B
gA = 3/3
gA = 6/3
gA = 6/3
gA = 5/3
p = 3/3
gB = 6/3
gB = 6/3
gB = 8/3
gC = 5/3
gC = 6/3
gC = 8/3
p = 7/3
gD = 6/3
gD = 8/3
p = 12/3
gE = 8/3
Y
2
p = 16/3
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Example 4
 The multiplexer has a maximum input capacitance of
16 units on each input. It must drive a load of 160
units. Estimate the delay of the two designs.
H = 160 / 16 = 10 B = 1 N = 2
D0
S
Y
D1
S
D0
S
D1
S
P 224
G  (4 / 3) (4 / 3)  16 / 9
F  GBH  160 / 9
fˆ  N F  4.2
P  4 1  5
G  (6 / 3) (1)  2
F  GBH  20
fˆ  N F  4.5
D  Nfˆ  P  12.4
D  Nfˆ  P  14
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Example 5
 Annotate your designs with transistor sizes that
achieve this delay.
8
8
8
8
25
25
25
8
8
Y
25
10
10
10
10
24
6
6
12
6
6
Y
8
8
16
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160 * (4/3) / 4.2 = 50
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160 * 1 / 4.5 = 36
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Input Order
 Our parasitic delay model was too simple
– Calculate parasitic delay for Y falling
• If A arrives latest? 2
• If B arrives latest? 2.33
2
2
A
2
B
2x
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Y
6C
2C
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Inner & Outer Inputs
 Inner input is closest to output (A)
 Outer input is closest to rail (B)
2
A
B
 If input arrival time is known
– Connect latest input to inner terminal
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Y
2
2
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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
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Symmetric Gates
 Inputs can be made perfectly symmetric
2
2
A
1
1
B
1
1
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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
1
A
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
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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
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Catalog of Skewed Gates
Inverter
NAND2
2
unskewed
1
Y
guu = 1
gdd = 1
gavg
=1
avg
A
2
B
2
2
HI-skew
2
A
Y
1/2 gu
u
gdd
gavg
avg
= 5/6
= 5/3
= 5/4
B
1
A
1
1
Y
guu = 4/3
gdd = 2/3
gavg
=1
avg
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1
A
2
B
2
4
A
4
1
guu = 4/3
gdd = 4/3
gavg
= 4/3
avg
Y
1
B
Y
2
A
1
LO-skew
2
Y
2
A
NOR2
1
B
4
A
4
guu = 5/3
gdd = 5/3
gavg
= 5/3
avg
Y
guu
gdd
gavg
avg
1/2
=1
=2
= 3/2
guu
gdd
gavg
avg
= 3/2
=3
= 9/4
2
B
A
Y
1/2
2
Y
guu = 2
gdd = 1
gavg
= 3/2
avg
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1
guu = 2
gdd = 1
gavg
= 3/2
avg
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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
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2
Y
4/3
4
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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
– dtpd / dP = (1- m/P2)/2 = 0
– Least delay for P = m
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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 = 1.15
gd = 0.81
gavg = 0.98
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NOR2
2
Y
A
2
B
2
B
2
A
2
Y
gu = 4/3
gd = 4/3
gavg = 4/3
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1
gu = 2
gd = 1
gavg = 3/2
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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
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