EE 447 VLSI Design Lecture 7: Combinational Circuits

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