EE 447 VLSI Design
Lecture 7: Combinational Circuits
Outline Bubble Pushing Compound Gates Logical Effort Example Input Ordering Asymmetric Gates Skewed Gates Best P/N ratio
Example 1module mux(input s, d0, d1,
output y);
assign y = s ? d1 : d0;endmodule
1) Sketch a design using AND, OR, and NOT gates.
Example 1module mux(input s, d0, d1,
output y);
assign y = s ? d1 : d0;endmodule
1) Sketch a design using AND, OR, and NOT gates.
D0S
D1S
Y
Example 2
2) Sketch a design using NAND, NOR, and NOT gates. Assume ~S is available.
Example 2
2) Sketch a design using NAND, NOR, and NOT gates. Assume ~S is available.
Y
D0S
D1S
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
Y
D
Y
(a) (b)
(c) (d)
Example 3
3) Sketch a design using one compound gate and one NOT gate. Assume ~S is available.
Example 3
3) Sketch a design using one compound gate and one NOT gate. Assume ~S is available.
Y
D0SD1S
Compound Gates Logical Effort of compound gates
ABCD
Y
ABC Y
A
BC
C
A B
A
B
C
D
A
C
B
D
2
21
4
44
2
2 2
2
4
4 4
4
gA = 6/3
gB = 6/3
gC = 5/3
p = 7/3
gA =
gB =
gC =
p =
gD =
YA
A Y
gA = 3/3
p = 3/3
2
1YY
unit inverter AOI21 AOI22
A
C
DE Y
B
Y
B C
A
D
E
A
B
C
D E
gA =
gB =
gC =
gD =
2
2 2
22
6
6
6 6
3
p =
gE =
Complex AOI
Y A B C Y A B C D Y A B C D E Y A
Compound Gates Logical Effort of compound gates
ABCD
Y
ABC Y
A
BC
C
A B
A
B
C
D
A
C
B
D
2
21
4
44
2
2 2
2
4
4 4
4
gA = 6/3
gB = 6/3
gC = 5/3
p = 7/3
gA = 6/3
gB = 6/3
gC = 6/3
p = 12/3
gD = 6/3
YA
A Y
gA = 3/3
p = 3/3
2
1YY
unit inverter AOI21 AOI22
A
C
DE Y
B
Y
B C
A
D
E
A
B
C
D E
gA = 5/3
gB = 8/3
gC = 8/3
gD = 8/3
2
2 2
22
6
6
6 6
3
p = 16/3
gE = 8/3
Complex AOI
Y A B C Y A B C D Y A B C D E Y A
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 NAND and compound gate designs.
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 NAND and compound gate designs.
Y
D0S
D1S
Y
D0SD1S
H = 160 / 16 = 10B = 1N = 2
NAND Solution
Y
D0S
D1S
NAND Solution
Y
D0S
D1S
2 2 4(4 / 3) (4 / 3) 16 / 9
160 / 9ˆ 4.2
ˆ 12.4
N
PGF GBH
f F
D Nf P
Compound Solution
Y
D0SD1S
Compound Solution
4 1 5(6 / 3) (1) 2
20ˆ 4.5
ˆ 14
N
PGF GBH
f F
D Nf P
Y
D0SD1S
Example 5 Annotate your designs with transistor sizes
that achieve this delay.
YY
Example 5 Annotate your designs with transistor sizes
that achieve this delay.
6
6 6
6
10
10Y
24
12
10
10
8
8
88
8
8
88
25
25
2525Y
16 16160 * (4/3) / 4.2 = 50 160 * 1 / 4.5 = 36
Input Order Our parasitic delay model was too simple
Calculate parasitic delay for Y falling If A arrives latest? If B arrives latest?
6C
2C2
2
22
B
Ax
Y
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
6C
2C2
2
22
B
Ax
Y
Inner & Outer Inputs Outer input is closest to rail (B) Inner input is closest to output (A)
If input arrival time is known Connect latest input to inner terminal
2
2
22
B
A
Y
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) Boost size of noncritical input So total resistance is same
gA = gB = gtotal = gA + gB = Asymmetric gate approaches g = 1 on critical input But total logical effort goes up
Areset Y
4/3
2
reset
AY
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) Boost size of noncritical input So total resistance is same
gA = 10/9 gB = 2 gtotal = gA + gB = 28/9 Asymmetric gate approaches g = 1 on critical input But total logical effort goes up
Areset Y
4
4/3
22
reset
AY
Symmetric Gates Inputs can be made perfectly symmetric
A
B
Y2
1
1
2
1
1
Skewed Gates Skewed gates favor one edge over another Ex: suppose rising output of inverter is most critical
Downsize noncritical nMOS transistor
Calculate logical effort by comparing to unskewed inverter with same effective resistance on that edge. gu = gd =
1/2
2A Y
1
2A Y
1/2
1A Y
HI-skewinverter
unskewed inverter(equal rise resistance)
unskewed inverter(equal fall resistance)
Skewed Gates Skewed gates favor one edge over another Ex: suppose rising output of inverter is most critical
Downsize noncritical nMOS transistor
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
1/2
2A Y
1
2A Y
1/2
1A Y
HI-skewinverter
unskewed inverter(equal rise resistance)
unskewed inverter(equal fall resistance)
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
Catalog of Skewed Gates
1/2
2A Y
Inverter
B
AY
BA
NAND2 NOR2
HI-skew
LO-skew1
1A Y
B
AY
BA
gu = 5/6gd = 5/3gavg = 5/4
gu = 4/3gd = 2/3gavg = 1
gu =gd =gavg =
gu =gd =gavg =
gu =gd =gavg =
gu =gd =gavg =
Y
Y
1
2A Y
2
2
22
B
AY
BA
11
4
4
unskewedgu = 1gd = 1gavg = 1
gu = 4/3gd = 4/3gavg = 4/3
gu = 5/3gd = 5/3gavg = 5/3
Y
Catalog of Skewed Gates
1/2
2A Y
Inverter
1
1
22
B
AY
BA
NAND2 NOR2
1/21/2
4
4
HI-skew
LO-skew1
1A Y
2
2
11
B
AY
BA
11
2
2
gu = 5/6gd = 5/3gavg = 5/4
gu = 4/3gd = 2/3gavg = 1
gu =gd =gavg =
gu =gd =gavg =
gu =gd =gavg =
gu =gd =gavg =
Y
Y
1
2A Y
2
2
22
B
AY
BA
11
4
4
unskewedgu = 1gd = 1gavg = 1
gu = 4/3gd = 4/3gavg = 4/3
gu = 5/3gd = 5/3gavg = 5/3
Y
Catalog of Skewed Gates
1/2
2A Y
Inverter
1
1
22
B
AY
BA
NAND2 NOR2
1/21/2
4
4
HI-skew
LO-skew1
1A Y
2
2
11
B
AY
BA
11
2
2
gu = 5/6gd = 5/3gavg = 5/4
gu = 4/3gd = 2/3gavg = 1
gu = 1gd = 2gavg = 3/2
gu = 2gd = 1gavg = 3/2
gu = 3/2gd = 3gavg = 9/4
gu = 2gd = 1gavg = 3/2
Y
Y
1
2A Y
2
2
22
B
AY
BA
11
4
4
unskewedgu = 1gd = 1gavg = 1
gu = 4/3gd = 4/3gavg = 4/3
gu = 5/3gd = 5/3gavg = 5/3
Y
Asymmetric Skew Combine asymmetric and skewed gates
Downsize noncritical transistor on unimportant input
Reduces parasitic delay for critical input
Areset Y
4
4/3
21
reset
AY
Best P/N Ratio We have selected P/N ratio for unit rise and fall
resistance ( = 2-3 for an inverter). Alternative: choose ratio for least average delay Ex: inverter
Delay driving identical inverter tpdf = tpdr = tpd = Differentiate tpd w.r.t. P Least delay for P =
1
PA
Best P/N Ratio We have selected P/N ratio for unit rise and fall
resistance ( = 2-3 for an inverter). Alternative: choose ratio for least average delay Ex: inverter
Delay driving identical inverter tpdf = (P+1) tpdr = (P+1)(/P) tpd = (P+1)(1+/P)/2 = (P + 1 + + /P)/2 Differentiate tpd w.r.t. P Least delay for P =
1
PA
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 NOR2
1
1.414A Y
2
2
22
B
AY
BA
11
2
2
fastestP/N ratio gu =
gd =gavg =
gu =gd =gavg =
gu =gd =gavg =
Y
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 NOR2
1
1.414A Y
2
2
22
B
AY
BA
11
2
2
fastestP/N ratio gu = 1.15
gd = 0.81gavg = 0.98
gu = 4/3gd = 4/3gavg = 4/3
gu = 2gd = 1gavg = 3/2
Y
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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