Introduction toCMOS VLSI

Design

Lecture 8: Combinational Circuits

David Harris

Harvey Mudd CollegeSpring 2004

8: Combinational Circuits Slide 2CMOS VLSI Design

Outlineq Bubble Pushingq Compound Gatesq Logical Effort Exampleq Input Orderingq Asymmetric Gatesq Skewed Gatesq Best P/N ratio

8: Combinational Circuits Slide 3CMOS VLSI Design

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.

8: Combinational Circuits Slide 5CMOS VLSI Design

Example 22) Sketch a design using NAND, NOR, and NOT gates.

Assume ~S is available.

8: Combinational Circuits Slide 8CMOS VLSI Design

Example 33) Sketch a design using one compound gate and one

NOT gate. Assume ~S is available.

8: Combinational Circuits Slide 10CMOS VLSI Design

Compound Gatesq 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= +i Y A B C D= +i i ( )Y A B C D E= + +i iY A=

8: Combinational Circuits Slide 12CMOS VLSI Design

Example 4q 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.

8: Combinational Circuits Slide 14CMOS VLSI Design

NAND Solution

Y

D0S

D1S

8: Combinational Circuits Slide 16CMOS VLSI Design

Compound Solution

Y

D0SD1S

8: Combinational Circuits Slide 18CMOS VLSI Design

Example 5q Annotate your designs with transistor sizes that

achieve this delay.

YY

8: Combinational Circuits Slide 20CMOS VLSI Design

Input Orderq 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

8: Combinational Circuits Slide 22CMOS VLSI Design

Inner & Outer Inputsq Outer input is closest to rail (B)q Inner input is closest to output (A)

q If input arrival time is known– Connect latest input to inner terminal

2

2

22

B

A

Y

8: Combinational Circuits Slide 23CMOS VLSI Design

Asymmetric Gatesq Asymmetric gates favor one input over anotherq 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

q gA =q gB = q gtotal = gA + gB = q Asymmetric gate approaches g = 1 on critical inputq But total logical effort goes up

Areset

Y

4/3

2

reset

AY

8: Combinational Circuits Slide 25CMOS VLSI Design

Symmetric Gatesq Inputs can be made perfectly symmetric

A

B

Y2

1

1

2

1

1

8: Combinational Circuits Slide 26CMOS VLSI Design

Skewed Gatesq Skewed gates favor one edge over anotherq Ex: suppose rising output of inverter is most critical

– Downsize noncritical nMOS transistor

q 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)

8: Combinational Circuits Slide 28CMOS VLSI Design

HI- and LO-Skewq 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.

q Skewed gates reduce size of noncritical transistors– HI-skew gates favor rising output (small nMOS)– LO-skew gates favor falling output (small pMOS)

q Logical effort is smaller for favored directionq But larger for the other direction

8: Combinational Circuits Slide 29CMOS VLSI Design

Catalog of Skewed Gates

1/2

2A Y

Inverter

B

AY

B

A

NAND2 NOR2

HI-skew

LO-skew1

1A Y

B

AY

B

A

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

B

A

11

4

4

unskewedgu = 1gd = 1gavg = 1

gu = 4/3gd = 4/3gavg = 4/3

gu = 5/3gd = 5/3gavg = 5/3

Y

8: Combinational Circuits Slide 32CMOS VLSI Design

Asymmetric Skewq 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

8: Combinational Circuits Slide 33CMOS VLSI Design

Best P/N Ratioq We have selected P/N ratio for unit rise and fall

resistance (µ = 2-3 for an inverter).q Alternative: choose ratio for least average delayq Ex: inverter

– Delay driving identical inverter– tpdf =– tpdr = – tpd =– Differentiate tpd w.r.t. P– Least delay for P =

1

PA

8: Combinational Circuits Slide 35CMOS VLSI Design

P/N Ratiosq 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

B

A

11

2

2

fastestP/N ratio gu =

gd =gavg =

gu =gd =gavg =

gu =gd =gavg =

Y

8: Combinational Circuits Slide 37CMOS VLSI Design

Observationsq For speed:

– NAND vs. NOR– Many simple stages vs. fewer high fan-in stages– Latest-arriving input

q For area and power:– Many simple stages vs. fewer high fan-in stages