© r.lauwereins imec 2001 digital design combina- torial circuits sequential circuits fsmd design...

2/1

© R.LauwereinsImec 2001

Digitaldesign

Combina-torialcircuits

Sequentialcircuits

FSMDdesign

VHDL

Course contents

• Digital designCombinatorial circuits: without status• Sequential circuits: with status• FSMD design: hardwired processors• Language based HW design: VHDL

2/2


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Design of Combinatorial Circuits

• Minimization of Boolean functions• Technology mapping• Correct timing behavior• Basic RTL building blocks (Adder, ALU, MUX,

…)

2/3


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• Minimization of Boolean functions Karnaugh map Minimization with the Karnaugh map Don’t care conditions Quine-McCluskey

• Technology mapping• Correct timing behavior• Basic RTL building blocks (Adder, ALU, MUX,

…)

2/4


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL




…)

2/5


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Karnaugh map

• Motivation: Assume: F=xy’z+xy’z’

Cost = (fan-in)complete circuit = (2)+(3)+(3)+(2) = 10

Delay

Assume: relative gate delay NAND or NOR or NOT = 0.6 + fan-in * 0.4

Delay = (gate-delay)critical path = 1 + (1.8+1) + (1.4+1) = 6.2

x y z

F=xy’z+xy’z’

2/6


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Karnaugh map

• Motivation: F =xy’z+xy’z’

=xy’(z+z’)=xy’The value of z hence does not matter

Cost = (fan-in)complete circuit = (1+2) = 3 i.o. 10

Delay

Assume: relative gate delay NAND or NOR or NOT = 0.6 + fan-in * 0.4

Delay = (gate-delay)critical path = 1 + (1.4+1) = 3.4 i.o. 6.2

x y z

F

2/7


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Karnaugh map

• Minimization via manipulation of Boolean expressions is clumsy: no method exists to select the theorems such that we are sure to obtain the minimum cost

• Is it possible to see in the truth table which input value does not matter?

x y z F

0 0 0 0 -0 0 1 0 -

0 1 0 0 -

0 1 1 0 -

1 0 0 1 xy’z’

1 0 1 1 xy’z

1 1 0 0 -

1 1 1 0 -

We indeed see easily that thevalue of F equals 1 for x=1 and y=0

irrespective of the value of z

We however see this easily onlyfor z, since only for z the linesz=0 and z=1 for equal x and y

are consecutive

2/8


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Karnaugh map

• A Karnaugh map contains the same information as a truth table (each square is a minterm), but…

• neighboring squares differ only in the value of 1 variable!!

x’ x

x0 1

x’y’ x’y

xy’ xy

x

0

1

y0 1

x’y’z’ x’y’z

xy’z’ xy’z

x

0

1

yz00 01

x’yz x’yz’

xyz xyz’

11 10

x

y

z

x’z (y does not matter)x’y’z x’yz

xy’z’ xyz’xz’ (y does not matter)

xy’z’ xy’z

xy’ (z does not matter)

2/9


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Karnaugh map

m0 m1

x0 1

m0 m1

m2 m3

x

0

1

y0 1

0 1

4 5

x

0

1

yz00 01

3 2

7 6

11 10

x

y

z

2/10


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

x y z w F

0 0 0 0 10 0 0 1 0

0 0 1 0 0

0 0 1 1 0

0 1 0 0 0

0 1 0 1 1

0 1 1 0 0

0 1 1 1 1

1 0 0 0 1

1 0 0 1 0

1 0 1 0 1

1 0 1 1 0

1 1 0 0 0

1 1 0 1 1

1 1 1 0 0

1 1 1 1 1

Karnaugh map

0 1

4 5

xy

00

01

zw00 01

3 2

7 6

11 10

x

z

w

12 13

8 9

15 14

11 10

11

10

y

x y z w F

0 0 0 0 10 0 0 1 0

0 0 1 0 0

0 0 1 1 0

0 1 0 0 0

0 1 0 1 1

0 1 1 0 0

0 1 1 1 1

1 0 0 0 1

1 0 0 1 0

1 0 1 0 1

1 0 1 1 0

1 1 0 0 0

1 1 0 1 1

1 1 1 0 0

1 1 1 1 1

1

x y z w F

0 0 0 0 10 0 0 1 0

0 0 1 0 0

0 0 1 1 0

0 1 0 0 0

0 1 0 1 1

0 1 1 0 0

0 1 1 1 1

1 0 0 0 1

1 0 0 1 0

1 0 1 0 1

1 0 1 1 0

1 1 0 0 0

1 1 0 1 1

1 1 1 0 0

1 1 1 1 1

0

x y z w F

0 0 0 0 10 0 0 1 0

0 0 1 0 0

0 0 1 1 0

0 1 0 0 0

0 1 0 1 1

0 1 1 0 0

0 1 1 1 1

1 0 0 0 1

1 0 0 1 0

1 0 1 0 1

1 0 1 1 0

1 1 0 0 0

1 1 0 1 1

1 1 1 0 0

1 1 1 1 1

0

x y z w F

0 0 0 0 10 0 0 1 0

0 0 1 0 0

0 0 1 1 0

0 1 0 0 0

0 1 0 1 1

0 1 1 0 0

0 1 1 1 1

1 0 0 0 1

1 0 0 1 0

1 0 1 0 1

1 0 1 1 0

1 1 0 0 0

1 1 0 1 1

1 1 1 0 0

1 1 1 1 1

0

x y z w F

0 0 0 0 10 0 0 1 0

0 0 1 0 0

0 0 1 1 0

0 1 0 0 0

0 1 0 1 1

0 1 1 0 0

0 1 1 1 1

1 0 0 0 1

1 0 0 1 0

1 0 1 0 1

1 0 1 1 0

1 1 0 0 0

1 1 0 1 1

1 1 1 0 0

1 1 1 1 1

0 1

x y z w F

0 0 0 0 10 0 0 1 0

0 0 1 0 0

0 0 1 1 0

0 1 0 0 0

0 1 0 1 1

0 1 1 0 0

0 1 1 1 1

1 0 0 0 1

1 0 0 1 0

1 0 1 0 1

1 0 1 1 0

1 1 0 0 0

1 1 0 1 1

1 1 1 0 0

1 1 1 1 1

x y z w F

0 0 0 0 10 0 0 1 0

0 0 1 0 0

0 0 1 1 0

0 1 0 0 0

0 1 0 1 1

0 1 1 0 0

0 1 1 1 1

1 0 0 0 1

1 0 0 1 0

1 0 1 0 1

1 0 1 1 0

1 1 0 0 0

1 1 0 1 1

1 1 1 0 0

1 1 1 1 1

0

x y z w F

0 0 0 0 10 0 0 1 0

0 0 1 0 0

0 0 1 1 0

0 1 0 0 0

0 1 0 1 1

0 1 1 0 0

0 1 1 1 1

1 0 0 0 1

1 0 0 1 0

1 0 1 0 1

1 0 1 1 0

1 1 0 0 0

1 1 0 1 1

1 1 1 0 0

1 1 1 1 1

1

x y z w F

0 0 0 0 10 0 0 1 0

0 0 1 0 0

0 0 1 1 0

0 1 0 0 0

0 1 0 1 1

0 1 1 0 0

0 1 1 1 1

1 0 0 0 1

1 0 0 1 0

1 0 1 0 1

1 0 1 1 0

1 1 0 0 0

1 1 0 1 1

1 1 1 0 0

1 1 1 1 1

1

x y z w F

0 0 0 0 1

0 0 0 1 0

0 0 1 0 0

0 0 1 1 0

0 1 0 0 0

0 1 0 1 1

0 1 1 0 0

0 1 1 1 1

1 0 0 0 1

1 0 0 1 0

1 0 1 0 1

1 0 1 1 0

1 1 0 0 0

1 1 0 1 1

1 1 1 0 0

1 1 1 1 1

0

x y z w F

0 0 0 0 10 0 0 1 0

0 0 1 0 0

0 0 1 1 0

0 1 0 0 0

0 1 0 1 1

0 1 1 0 0

0 1 1 1 1

1 0 0 0 1

1 0 0 1 0

1 0 1 0 1

1 0 1 1 0

1 1 0 0 0

1 1 0 1 1

1 1 1 0 0

1 1 1 1 1

1

x y z w F

0 0 0 0 10 0 0 1 0

0 0 1 0 0

0 0 1 1 0

0 1 0 0 0

0 1 0 1 1

0 1 1 0 0

0 1 1 1 1

1 0 0 0 1

1 0 0 1 0

1 0 1 0 1

1 0 1 1 0

1 1 0 0 0

1 1 0 1 1

1 1 1 0 0

1 1 1 1 1

0

x y z w F

0 0 0 0 10 0 0 1 0

0 0 1 0 0

0 0 1 1 0

0 1 0 0 0

0 1 0 1 1

0 1 1 0 0

0 1 1 1 1

1 0 0 0 1

1 0 0 1 0

1 0 1 0 1

1 0 1 1 0

1 1 0 0 0

1 1 0 1 1

1 1 1 0 0

1 1 1 1 1

0

x y z w F

0 0 0 0 10 0 0 1 0

0 0 1 0 0

0 0 1 1 0

0 1 0 0 0

0 1 0 1 1

0 1 1 0 0

0 1 1 1 1

1 0 0 0 1

1 0 0 1 0

1 0 1 0 1

1 0 1 1 0

1 1 0 0 0

1 1 0 1 1

1 1 1 0 0

1 1 1 1 1

1

x y z w F

0 0 0 0 1

0 0 0 1 0

0 0 1 0 0

0 0 1 1 0

0 1 0 0 0

0 1 0 1 1

0 1 1 0 0

0 1 1 1 1

1 0 0 0 1

1 0 0 1 0

1 0 1 0 1

1 0 1 1 0

1 1 0 0 0

1 1 0 1 1

1 1 1 0 0

1 1 1 1 1

0

x y z w F

0 0 0 0 10 0 0 1 0

0 0 1 0 0

0 0 1 1 0

0 1 0 0 0

0 1 0 1 1

0 1 1 0 0

0 1 1 1 1

1 0 0 0 1

1 0 0 1 0

1 0 1 0 1

1 0 1 1 0

1 1 0 0 0

1 1 0 1 1

1 1 1 0 0

1 1 1 1 1

1

Fill out from truth table

2/11


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Karnaugh map

1 0

0 1

xy

00

01

zw00 01

0 0

1 0

11 10

x

z

w

0 1

1 0

1 0

0 1

11

10

y

MinimizeF=x’y’z’w’+x’yz’w+x’yzw+xy’z’w’+xy’zw’+xyz’w+xyzw

F= yw

+xy’w’

+y’z’w’

2/12


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Karnaugh map

ImplementF=x’y’z’w’+x’yz’w+x’yzw+xy’z’w’+xy’zw’+xyz’w+xyzwx y z w

F

Cost = 4*(1) + 7*(4) + 1*(7)= 39

Delay = 1 + (2.2+1) + (3.4+1)= 8.6

2/13


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Karnaugh map

ImplementF=yw+xy’w’+y’z’w’x y z w

Cost = 3*(1) + {1*(2)+2*(3)} + 1*(3)= 14 i.p.v. 39

Delay = 1 + (1.8+1) + (1.8+1)= 6.6 i.p.v. 8.6

F

2/14


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Karnaugh map

0xy00

01

zw00 01 11 10

1 3 2

4 5 7 6

12 13 15 14

8 9 11 10

18 19 17 16

22 23 21 20

30 31 29 28

26 27 25 24

xy

11

10

10 11 01 00

00

01

11

10x

y

z

w

x

z

w

v

y

Differs from course book

F(v,x,y,z,w)

2/15


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Karnaugh map

0xy00

01

zw00 01 11 10

1 3 2

4 5 7 6

12 13 15 14

8 9 11 10

18 19 17 16

22 23 21 20

30 31 29 28

26 27 25 24

xy

11

10

10 11 01 00

00

01

11

10x

y

z

w

x

z

w

v

y

40 41 43 42

44 45 47 46

36 37 39 38

32 33 35 34

58 59 57 56

62 63 61 60

54 55 53 52

50 51 49 48

10

11

01

00

x

y

10

11

01

00

x

y u

F=(u,v,x,y,z,w)

Differs from course book

2/16


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL




…)

2/17


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Minimization with the Karnaugh map

Determine all prime implicants

Determine all essentialprime implicants

Search for minimal coverage

Create the Karnaugh map

Truth table orcanonical form

2/18


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


Step 1: Create Karnaugh map

w

x

yz

1

1

F=x’y’z’+wz+xyz+w’yF=x’y’z’+wz+xyz+w’yF=x’y’z’+wz+xyz+w’y

1

1

1 1

11

F=x’y’z’+wz+xyz+w’y

1 1

11

1

1

F=x’y’z’+wz+xyz+w’y

1

1

1 1

11

Rule:- Take product term per productterm and indicate where in theKarnaugh map it equals 1

2/19


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


Step 2: Determine all prime implicants

1 1 1

1 1

1 1

1 1 1w

x

yz

11

w’x’z’x’y’z’

1 1w’yyz1 1

1

1

1 11

1

wz

1 11

1

wx’y’

1 11 1Rule:- Analyze each 1-minterm- Determine the largest sub-cube(s)that contain(s) the minterm andadd them to the list of primeimplicants (without adding an

already listed sub-cube)

2/20


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


Step 3: Determine all essential prime implicants

1 1 1

1 1

1 1

1 1 1w

x

yz w’x’z’

x’y’z’w’yyzwz

wx’y’1

1

wz

w’y

Rule:- Search for 1-minterms that areonly contained in 1 prime implicant- Indicate this prime implicant asessential

2/21


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


Step 4: Search minimal coverage

w’x’z’x’y’z’w’yyzwz

wx’y’

1 1 1

1 1

1 1

1 1 1w

x

yz

1 1

1 1

1 1

1 1

12

0

1

1

1

x’y’z’

Fmin=x’y’z’+w’y+wz

Rule:- Goal: search for the smallest set of (as big as possible)prime implicants that contain all 1-minterms- Take all essential prime implicants as initial list- Repeatedly add a prime implicant to the list that containsthe largest number of not yet covered 1-minterms. Whenthere are two that contain the same number of not yetcovered 1-minterms, make a random choice.

- Such a strategy is known as Greedy strategy: at each decision point,take the best choice without looking to future implications- This does not always lead to a global optimum

2/22


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• Original:F=x’y’z’+w’y+xyz+wz

• MinimalFmin=x’y’z’+w’y+wz

wxyz wxyz

Cost=4*1+2*3+2*2+1*4=18 Cost =4*1+1*3+2*2+1*3=14=22% cheaper

Delay =(1)+(.6+3*.4+1)+(.6+4*.4+1)=7

Delay =(1)+(.6+3*.4+1)+(.6+3*.4+1)=6.6=6% faster

2/23


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• Example 2: F(v,w,x,y,z)v w x y z F

0 0 0 0 0 00 0 0 0 1 0

0 0 0 1 0 0

0 0 0 1 1 0

0 0 1 0 0 0

0 0 1 0 1 0

0 0 1 1 0 1

0 0 1 1 1 1

0 1 0 0 0 0

0 1 0 0 1 0

0 1 0 1 0 1

0 1 0 1 1 1

0 1 1 0 0 0

0 1 1 0 1 0

0 1 1 1 0 1

0 1 1 1 1 1

v w x y z F

1 0 0 0 0 0

1 0 0 0 1 0

1 0 0 1 0 0

1 0 0 1 1 0

1 0 1 0 0 0

1 0 1 0 1 1

1 0 1 1 0 0

1 0 1 1 1 1

1 1 0 0 0 0

1 1 0 0 1 1

1 1 0 1 0 0

1 1 0 1 1 1

1 1 1 0 0 0

1 1 1 0 1 1

1 1 1 1 0 0

1 1 1 1 1 1

2/24


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• Realisation as sum of 1-minterms:

F=(6,7,10,11,14,15,21,23,25,27,29,31)vwxyz

Cost=(5*1)+(12*(5+1))+(1*(12+1))=90Delay=(.6+1*.4)+(.6+5*.4+1)+(.6+12*.4+1)=11

2/25


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• MinimisationStep 1: Create Karnaugh map

0 0 0 0

0 0 1 1

0 0 1 1

0 0 1 1

0 0 0 0

0 1 1 0

0 1 1 0

0 1 1 0w

x

y

z zv

2/26


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• MinimisationStep 2: determine all prime implicants

0 0 0 0

0 0 1 1

0 0 1 1

0 0 1 1

0 0 0 0

0 1 1 0

0 1 1 0

0 1 1 0w

x

y

z zv

2/27


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL1 1

11

1 1

11


• MinimisationStep 3: Determine all essential prime implicants

0 0 0 0

0 0

0 0 1 1

0 0

0 0 0 0

0 0

0 1 1 0

0 0w

x

y

z zv

1

1

1

1

F1min2=v’xy+v’wy+vxz+vwz

Is already the minimum coverage

2/28


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• Realisation of F1min2=v’xy+v’wy+vxz+vwz

vwxyz

Cost=1+(4*(3+1))+(1*(4+1))=22 (76% cheaper)Delay=(.6+1*.4)+(.6+3*.4+1)+(.6+4*.4+1)=7 (34% faster)

2/29


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• Realisation in more than two layers F =v’xy+v’wy+vxz+vwz

=v’y(x+w)+vz(x+w)=(x+w)(v’y+vz)

vwxyz

Cost =(1*1)+(5*(2+1))=16(82% cheaper)

Delay =(.6+1*.4)+(.6+2*.4+1)+(.6+2*.4+1)+(.6+2*.4+1)=8.2 (25% faster)

2/30


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• Dual minimisationStep 1: Create the Karnaugh map

0 0 0 0

0 0 1 1

0 0 1 1

0 0 1 1

0 0 0 0

0 1 1 0

0 1 1 0

0 1 1 0w

x

y

z zv

2/31


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• Dual minimisationStep 2: Determine all prime implicants

0 0 0 0

0 0 1 1

0 0 1 1

0 0 1 1

0 0 0 0

0 1 1 0

0 1 1 0

0 1 1 0w

x

y

z zv

2/32


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• Dual minimisationStep 3: Determine all essential prime implicants

0 0 0 0

0 0 1 1

0 0 1 1

0 0 1 1

0 0 0 0

0 1 1 0

0 1 1 0

0 1 1 0w

x

y

z zv

0

0

0

0 0

0

0

0

00

Is already the minimum coverageF0min2=(v+y)(w+x)(v’+z)

2/33


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• Realisation of F0min2=(v+y)(w+x)(v’+z)

vwxyz

Cost=(1*1)+(3*(2+1))+(1*(3+1))=14 (84% cheaper)Delay=(.6+1*.4)+(.6+2*.4+1)+(.6+3*.4+1)=6.2 (44% faster)

2/34


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• Summary

Realisation Cost Rel. cost Delay Rel. delay

Sum of 1-minterms 90 100% 11 100%

Minimal AND-OR 22 24% 7 64%3-layers 16 18% 8.2 75%Minimal OR-AND 14 16% 6.2 56%

Area/time trade-off

We’ll see that, depending on the technology mapping, we will eventually obtain for an ASIC realisation:

OR-AND-INVCost = 11 (Rel. cost=12%)Delay = 4 (Rel. delay=36%)

NORCost = 10 (Rel. cost=11%)Delay = 4.2 (Rel. delay=38%)

2/35


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL




…)

2/36


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Don’t care conditions

• Incompletely specified Boolean function

BCD7-segment

a

b

c

d

e

fg

x y z w a b c d e f g

0 0 0 0 1 1 1 1 1 1 00 0 0 1

0 0 1 0

0 0 1 1

0 1 0 00 1 0 1

0 1 1 0

0 1 1 1

1 0 0 0

1 0 0 1

1 0 1 0

1 0 1 1

1 1 0 0

1 1 0 1

1 1 1 0

1 1 1 1


0 0 0 0 1 1 1 1 1 1 00 0 0 1 0 1 1 0 0 0 0

0 0 1 0

0 0 1 1

0 1 0 00 1 0 1

0 1 1 0

0 1 1 1

1 0 0 0

1 0 0 1

1 0 1 0

1 0 1 1

1 1 0 0

1 1 0 1

1 1 1 0

1 1 1 1


0 0 0 0 1 1 1 1 1 1 00 0 0 1 0 1 1 0 0 0 0

0 0 1 0 1 1 0 1 1 0 1

0 0 1 1

0 1 0 00 1 0 1

0 1 1 0

0 1 1 1

1 0 0 0

1 0 0 1

1 0 1 0

1 0 1 1

1 1 0 0

1 1 0 1

1 1 1 0

1 1 1 1


0 0 0 0 1 1 1 1 1 1 00 0 0 1 0 1 1 0 0 0 0

0 0 1 0 1 1 0 1 1 0 1

0 0 1 1 1 1 1 1 0 0 1

0 1 0 0 0 1 1 0 0 1 10 1 0 1 1 0 1 1 0 1 1

0 1 1 0 1 0 1 1 1 1 1

0 1 1 1 1 1 1 0 0 0 0

1 0 0 0 1 1 1 1 1 1 1

1 0 0 1 1 1 1 1 0 1 1

1 0 1 0 x x x x x x x

1 0 1 1

1 1 0 0

1 1 0 1

1 1 1 0

1 1 1 1


0 0 0 0 1 1 1 1 1 1 00 0 0 1 0 1 1 0 0 0 0

0 0 1 0 1 1 0 1 1 0 1

0 0 1 1 1 1 1 1 0 0 1

0 1 0 0 0 1 1 0 0 1 1

0 1 0 1 1 0 1 1 0 1 1

0 1 1 0 1 0 1 1 1 1 1

0 1 1 1 1 1 1 0 0 0 0

1 0 0 0 1 1 1 1 1 1 1

1 0 0 1 1 1 1 1 0 1 1

1 0 1 0 x x x x x x x

1 0 1 1 x x x x x x x

1 1 0 0 x x x x x x x

1 1 0 1 x x x x x x x

1 1 1 0 x x x x x x x

1 1 1 1 x x x x x x x

2/37


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• Step 1: Create Karnaugh maps

1 0 1 1

0 1 1 1

x x x x

1 1 x x

1 1 1 1

1 0 1 0

x x x x

1 1 x x

1 1 1 0

1 1 1 1

x x x x

1 1 x x

1 0 1 1

0 1 0 1

x x x x

1 1 x x

1 0 0 1

0 0 0 1

x x x x

1 0 x x

1 0 0 0

1 1 0 1

x x x x

1 1 x x

0 0 1 1

1 1 0 1

x x x x

1 1 x x

x

x

y

y

z z z zw w w w

a b c d

e f g

2/38


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• Step 2: determine all prime implicants

1 0 1 1

0 1 1 1

x x x x

1 1 x x

1 1 1 1

1 0 1 0

x x x x

1 1 x x

1 1 1 0

1 1 1 1

x x x x

1 1 x x

1 0 1 1

0 1 0 1

x x x x

1 1 x x

1 0 0 1

0 0 0 1

x x x x

1 0 x x

1 0 0 0

1 1 0 1

x x x x

1 1 x x

0 0 1 1

1 1 0 1

x x x x

1 1 x x

x

x

y

y

z z z zw w w w

a b c d

e f g

2/39


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• Step 3: Determine all essential prime implicants

1 0 1 1

0 1 1 1

x x x x

1 1 x xx

y

zw

a

Completecoverage

1 1 1 0

1 1 1 1

x x x x

1 1 x x

zw

c

Completecoverage

1 0 1 1

0 1 0 1

x x x x

1 1 x x

zw

d

Completecoverage

1 0 0 1

0 0 0 1

x x x x

1 0 x xxy

e

Completecoverage

1 0 0 0

1 1 0 1

x x x x

1 1 x x

f

Completecoverage

0 0 1 1

1 1 0 1

x x x x

1 1 x x

g

Incompletecoverage

1 1 1 1

1 0 1 0

x x x x

1 1 x x

zw

b

Completecoverage

2/40


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• Step 4: Determine minimum coverage

0 0 1 1

1 1 0 1

x x x x

1 1 x x

g Selection of the cube that realisesthe remaining minterm:

- Select all cubes that realisethe minterm and are alreadyessential for another function;in this case, both are alreadyessential

- Select that cube that appearsin the smallest number ofother functions to keep thefan-out as low as possible

2/41


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• Note down the standard form

1 0 1 1

0 1 1 1

x x x x

1 1 x xx

y

zw

a

y’w’zywx

a=y’w’+z+yw+x

2/42


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

1 1 1 1

1 0 1 0

x x x x

1 1 x x

zw

b



y’w’zywx

a=y’w’+z+yw+x

z’w’zw

b=y’+z’w’+zw

y’

2/43


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

1 1 1 0

1 1 1 1

x x x x

1 1 x x

zw

c



y’w’zywx

a=y’w’+z+yw+x

y’z’w’zw

b=y’+z’w’+zw

z’wy

c=z’+w+y

2/44


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

1 0 1 1

0 1 0 1

x x x x

1 1 x x

zw

d



y’w’zywx

a=y’w’+z+yw+x

y’z’w’zw

b=y’+z’w’+zw

z’wy

c=z’+w+y

y’w’

y’zyz’wzw’

x

d=y’w’+y’z+yz’w+zw’+x

2/45


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

1 0 0 1

0 0 0 1

x x x x

1 0 x xxy

e



zywx

a=y’w’+z+yw+x

y’z’w’zw

b=y’+z’w’+zw

z’wy

c=z’+w+y

y’zyz’wzw’

d=y’w’+y’z+yz’w+zw’+x

y’w’y’w’

zw’

e=y’w’+zw’

2/46


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

1 0 0 0

1 1 0 1

x x x x

1 1 x x

f



zywx

a=y’w’+z+yw+x

y’z’w’zw

b=y’+z’w’+zw

z’wy

c=z’+w+y

y’zyz’wzw’

d=y’w’+y’z+yz’w+zw’+xe=y’w’+zw’

z’w’

yz’yw’

x

f=z’w’+yz’+yw’+x

y’w’

2/47


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

0 0 1 1

1 1 0 1

x x x x

1 1 x x

g



zywx

a=y’w’+z+yw+x

y’z’w’zw

b=y’+z’w’+zw

z’wy

c=z’+w+y

y’zyz’wzw’

d=y’w’+y’z+yz’w+zw’+xe=y’w’+zw’

yz’yw’

f=z’w’+yz’+yw’+x

yz’

y’z

yw’

x

g=y’z+yz’+yw’+x

y’w’

2/48


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


xyzw

a cb d e f g

2/49


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• Cost when realising as (1-minterms): a: 4*1+8*(4+1)+1*(8+1)=53 b: 4*1+8*(4+1)+1*(8+1)=53 c: 4*1+9*(4+1)+1*(9+1)=59 d: 4*1+7*(4+1)+1*(7+1)=47 e: 4*1+4*(4+1)+1*(4+1)=29 f: 4*1+6*(4+1)+1*(6+1)=41 g: 4*1+7*(4+1)+1*(7+1)=47

• Cost for minimal 2-layer-implementation Invertors: 4*1=4 AND-gates: 8*(2+1)+1*(3+1)=28 OR-gates: 1*(2+1)+2*(3+1)+3*(4+1)

+1*(5+1)=32

329 (100%)

64 (19%)

2/50


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• Delay when realising as (1-minterms): Critical path=c (9-input OR) c: (1)+(.6+4*.4+1)+(.6+9*.4+1)=9.4 (100%)

• Delay for minimal 2-layer-implementation Critical path=d (3-input AND & 5-input OR) d: (1)+(.6+3*.4+1)+(.6+5*.4+1)=7.4 (79%)

2/51


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL




…)

2/52


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Quine-McCluskey

• Method with Karnaugh map OK for human minimisation : visually oriented no guarantee for optimum solution

• Computer method Quine-McCluskey table oriented leads to optimum solution is the basis of all CAD circuit design tools hardly doable by hand

2/53


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• Minimization of Boolean functions• Technology mapping

Gate arrays: NAND, NOR Custom library: AOI, OAI, … PLA FPGA

• Correct timing behavior• Basic RTL building blocks (Adder, ALU, MUX,

…)

2/54


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL





…)

2/55


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Technology Mapping:Gate Arrays

• Properties of the methodology followed: When minimizing 1-minterms: INV-AND-OR When minimizing 0-maxterms: INV-OR-AND Any function can be realized in two layers of logic

with this methodology The fan-in of the gates can become arbitrary large

• Properties of gate arrays: They only contain m-input NAND or m-input NOR

gates

• Technology mapping is: Translating a circuit consisting of INV-AND-OR to

one with only m-input NAND Dual: INV-OR-AND m-input NOR

2/56


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• Design flow:

ConversionReplace each AND and ORby NAND or NOR

Optimisation Eliminate double inversions

Replace each n-input AND(OR) by a few m-inputANDs (ORs), with m<n

Decomposition

RetimingTry to make all input-outputdelays equal

2/57


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• Conversion rules (based on the laws of De Morgan):

= (xy)’ = (x’ + y’)

= (x+y)’ = (x’y’)

• Optimisation rule:

= (x’)’ = x

2/58


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• Conversion rules in practice:

The realisation with only NAND or only NOR is faster:we save an invertor per gate!

2/59


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• Realisation of an invertor:

=

2/60


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• Decomposition:

=

=

2/61


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• Retiming (delay optimisation): try to make the delay from each input to the output equal

• Example:

AND-OR implementation obtained from Karnaughminimisation

2/62


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• Example (first possible decomposition in 3-input NAND):

=

Delay =(.6+3x.4)+(1)+(.6+3x.4)+(.6+3x.4)+(1)+(.6+3x.4)

=9.2

2/63


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• Example (second possible decomposition in 3-input NAND):

=

Delay =(.6+3x.4)+(1)+(.6+3x.4)+(.6+3x.4)

=6.4 (70%)

2/64


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL





…)

2/65


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Technology Mapping:Custom Library

• ASICs have AOI en OAI: small and fast!• For small functions (not in course book):

Realise the inverse function with AND-OR or OR-AND

• Example: realise again following function:

F=(6,7,10,11,14,15,21,23,25,27,29,31) Realisation as sum of 1-minterms: cost = 90; delay

= 11 Minimal realisation as OR-AND: cost = 14 (16%);

delay = 6.2 (56%)

2/66


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• F=(6,7,10,11,14,15,21,23,25,27,29,31)

F’= (0,1,2,3,4,5,8,9,12,13,16,17,18,19,20, 22,24,26,28,30)

1 1 1 1

1 1 0 0

1 1 0 0

1 1 0 0

1 1 1 1

1 0 0 1

1 0 0 1

1 0 0 1w

x

y

z zv

F’ vwxyz

F’

F

wxyz

F

Cost=(5*1)+6=11 (12%)Delay=1+(.6+6*.4)=4 (36%)

2/67


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• For large functions:

Transform to NAND or NOR

Realise as AND-OR or OR-AND

Replace repeatedly 2 layers of gates off critical path by AOI/OAI

Determine critical path

Replace repeatedly 2 layers of gates on critical path by AOI/OAI

2/68


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• F=w’z’+z(w+y)

Realise with AND and OR

y

z

wF

Cost=14Delay=7.2

Transform to NAND and NOR

y

z

wF

Cost=?Delay=?

Transform to NAND and NOR

y

z

wF

Cost=11 (79%)Delay=5.2 (72%)

Determine the critical path

y

z

wF

Cost=11 (79%)Delay=5.2 (72%)

Replace 2 gates on critical path by AOI

y

z

wF

Cost=?Delay=?

Replace 2 gates on critical path by AOI

y

z

wF

Cost=10 (71%)Delay=5.6 (78%)

Replace 2 gates on critical path by AOI(2e possibility)

y

z

wF

Cost=?Delay=?

Replace 2 gates on critical path by AOI(2e possibility)

y

z

wF

Cost=10 (71%)Delay=3.8 (53%)

Analyze the other path

y

z

wF

Cost=10 (71%)Delay=3.8 (53%)

Analyze the other path

y

z

wF

Cost=9 (64%)Delay=3.8 (53%)

2/69


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL





…)

2/70


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Technology Mapping:PLA

• PLA is an AND-plane with large fan-in followed by an OR-plane with large fan-in

• Technology mapping: realisation as AND-OR, without the necessity for decomposition

2/71


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL





…)

2/72


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Technology Mapping:FPGA

• CLB is 2 functions of 4 variables or 1 function of 5 variables

• Technology mapping is similar as for custom design but i.o. AOI/OAI we search for sub-circuits of 4 or 5 variables, first on the critical path, next on the other paths

• For FPGAs technology mapping is done by automatic tools (see next slide); when prototype: no hand optimalisation, when final product: hand optimalisation beneficial. Also for ASICs automatic tools exist, hand optimalisation is beneficial

• BCD7-segment: create 7 truth tables i.f.o. 4 variables: the rest is done by the tools

2/73


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Technology mapping:FPGA

Technologymapping

2/74


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• Minimization of Boolean functions• Technology mappingCorrect timing behavior• Basic RTL building blocks (Adder, ALU, MUX,

…)

2/75


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Correct timing behavior:Hazard-free design

1

1 1 1x

yz

x

y

z

y’a

b

F

x

y

y’

z

a

b

F

1

x

y

z

y’a

b

F

1 1

0 1 2 3 4 5 6

x

y

z

y’a

b

Fx

y

z

y’a

b

Fx

y

z

y’a

b

Fx

y

z

y’a

b

Fx

y

z

y’a

b

Fx

y

z

y’a

b

F

Static 1-hazard

2/76


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• The hazard condition causes an unwanted glitch!

• Static 1-hazard: the output had to stay 1 but became briefly 0

• Cause: different delay in two paths

• Solution: see next slide

2/77


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


1

1 1 1x

yz

x

y

z

y’a

bF

c1

x

y

y’

z

a

b

F

0 1 2 3 4 5 6

c

x

y

z

y’a

bF

c

x

y

z

y’a

bF

c11

x

y

z

y’a

bF

c

x

y

z

y’a

bF

c

x

y

z

y’a

bF

c

2/78


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• Dynamic hazard: the output had to switch (eg. from 1 to 0) but switched several times (bvb. 1 0 1 0)

• Cause: different delay in multiple paths

• Example: see next slide

2/79


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

x’ x’’a

F

x’ x’’a

Fx


Statically equivalent to:

x F

x’’

0 1 2 3 4 5 6 7 8 9

x

x’

a

F

x’ x’’a

F

x’ x’’a

F

x’ x’’a

F

x’ x’’a

F

x’ x’’a

F

x’ x’’a

F

x’ x’’a

F

x’ x’’a

F

x’ x’’a

F

x’ x’’a

F

2/80


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• Hazards are hard to detect by hand: importance of simulation

• The danger for hazards increases when rise times and fall times are not equal

• Are hazards a problem? For synchronous circuits, they are not Unless they control the clock of a memory element For asynchronous circuits, they always are a

problem This is why asynchronous design is heavily

demotivated for FPGAs (the delay of the different paths is only known after APR-Automatic Placement and Routing)

2/81


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Design of Combinatorial Circuits• Minimization of Boolean functions• Technology mapping• Correct timing behavior• Basic RTL building blocks

Ripple-carry adders Carry-look-ahead adders Adder/subtractors Multipliers Logic units Arithmetic-logic units Decoders Selectors Buses Priority encoders Magnitude comparators Shifters and rotators

2/82


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL



2/83


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Ripple-carry adders

• Half adder

xi yi c i+1 si

0 0 0 00 1 0 1

1 0 0 1

1 1 1 0

0 0

0 1xi

yici+1

0 1

1 0xi

yisi

xi yi

ci+1 si

HA

xi yi

ci+1

si

1 CLB

2/84


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Ripple-carry adders

• Full adderxi yi c i c i+1 si

0 0 0 0 00 0 1 0 1

0 1 0 0 1

0 1 1 1 0

1 0 0 0 1

1 0 1 1 0

1 1 0 1 0

1 1 1 1 1

1

1 1 1ci

xi

yi

ci+1

1 1

1 1ci

xi

yi

si

xiyi

ci

ci+1 si

1 CLB

2/85


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Ripple-carry adders

• Full adder: alternative implementationxi yi c i c i+1 si

0 0 0 0 00 0 1 0 1

0 1 0 0 1

0 1 1 1 0

1 0 0 0 1

1 0 1 1 0

1 1 0 1 0

1 1 1 1 1

yi

1 1

1 1ci

xi

yi

si

1

1 1 1ci

xici+1

xiyi

ci

ci+1

si

1 gate less, larger delayfrom xi&yi to ci+1, same delayfrom ci to ci+1

FA

xi yi

cici+1

si

2/86


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Ripple-carry adders

• 4-bit ripple-carry adder

FA

x0 y0

c0=0c1

s0

FA

x1 y1

c2

s1

FA

x2 y2

c3

s2

FA

x3 y3

c4

s3

Critical path: x0 or y0 to c4: 1 XOR + 4 AND + 4 OR

In principal 1 CLB per bitBecause of special circuitry (dedicated carry chain):

1 CLB per 2 bits

2/87


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL



2/88


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Carry-look-ahead adders

• Ripple-carry adder is slow because the critical path x0 to cn+1 is long

• Speed-up is possible by computing for example c4 directly (in principle in 2 layers of logic) from c0, x0…x3 en y0…y3. Hence the name Carry-look-ahead

• How is this done? See exercises and course book

4-bit CLA ci

xi…i+3 yi…i+3

si…i+3

ci+4

2/89


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL



2/90


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Adder-subtractors

FA

x0 y0

c0c1

f0

FA

x1 y1

c2

f1

FA

x2 y2

c3

f2

FA

x3 y3

c4

f3

S

S Function Note

0 X+Y Addition

1 X-Y =X+Y *=X+Y ’+1 SubtractionAdder/

subtractor

X Y

F

SCout

overflow Only for 2-complement!!!

2/91


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL



2/92


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Multipliers

• A 1-bit by 1-bit multiplier:C = B • A

b0

a0

a0b0

b0 a0 c0=a0.b0

0 0 0

b0 a0 c0=a0.b0

0 0 0

0 1 0

b0 a0 c0=a0.b0

0 0 0

0 1 0

1 0 0

b0 a0 c0=a0.b0

0 0 0

0 1 0

1 0 0

1 1 1

b0 a0 c0=a0.b0

0 0 0

0 1 0

1 0 0

1 1 1

A 1-bit by 1-bit multiplier is hence an AND gate

a0 b0

c0

2/93


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Multipliers

• A 2-bit by 2-bit multiplier:C = B • A

a0b0

b0

a0

b1

a1

a0b1

a1b0a1b1

c0c1c2c3

Each of theseterms is a

1-bit by 1-bitmultiplier: AND

b0b1

a0

b0b1

a1

HAHA

c0c1c2c3

2/94


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Multipliers

• A 4-bit by 3-bit multiplier: cost=O(n2)b0

b1

a0b2b3

b0b1

a1 b2b3

4-bit adder

0

b0b1

a2 b2b3

4-bit adder

c0c1c2c3c4c5c6

s0s1s2s3cout

s0s1s2s3cout

2/95


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL



2/96


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Logic units

• Goal: implement a unit that can realise all 16 Boolean functions of 2 bits

• The unit has two inputs X and Y and 4 select bits S3S2S1S0 that select the wanted function

• The coding of the select bits is identical to the function number in the table of possible Boolean functions

2/97


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Logic units

Function valuefor x,y

1-minterms 00 01 10 11 Expression

- 0 0 0 0 F0=0m3 0 0 0 1 F1=xym2 0 0 1 0 F2=xy’

m2+m3 0 0 1 1 F3=xm1 0 1 0 0 F4=x’y

m1+m3 0 1 0 1 F5=ym1+m2 0 1 1 0 F6=xy’+x’y

m1+m2+m3 0 1 1 1 F7=x+y

m0 1 0 0 0 F8=(x+y)’m0+m3 1 0 0 1 F9=xy+x’y’m0+m2 1 0 1 0 F10=y’

m0+m2+m3 1 0 1 1 F11=x+y’m0+m1 1 1 0 0 F12=x’

m0+m1+m3 1 1 0 1 F13=x’+ym0+m1+m2 1 1 1 0 F14=(xy)’

m0+m1+m2+m3 1 1 1 1 F15=1

2/98


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Logic units

S0S1S2S3

xiyi

fi

LU

xi yi

fi

S0..3

2/99


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL



2/100


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Arithmetic-logic units

• Goal: build a unit that realises 4 arithmetic operations (addition, subtraction, increment and decrement) and 4 logic operations (AND, OR, INV, identity)

• Realisation principle: use an adder in front of which we place a modifier circuit (Arithmetic-Logic Extender)

• This principle has already been applied for the 2-complement adder/subtractor: this was an adder in front of which we placed an exor circuit to allow for subtraction

2/101


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


FAFAFAFAFA

a0 b0a1 b1a2 b2a3 b3a4 b4

f0f1f2f3f4

S

Cout

S selects the function to be executed: 0=addition, 1=subtraction

2/102


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


M

M selects the type of operation: 0=logic, 1=arithmeticS0 and S1 select the operation

FA

a0 b0

f0

ALE

FA

a1 b1

f1

ALE

FA

a2 b2

f2

ALE

FA

a3 b3

f3

ALE

FA

a4 b4

f4

ALE

Cout

S01

X Y

2/103


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


M S1 S0 Function F X Y C0

0 0 0 Complement A’ A’ 0 0



0 0 1 AND A AND B A AND B 0 0



0 0 1 AND A AND B A AND B 0 00 1 0 Identity A A 0 0



0 0 1 AND A AND B A AND B 0 00 1 0 Identity A A 0 00 1 1 OR A OR B A OR B 0 0



0 0 1 AND A AND B A AND B 0 00 1 0 Identity A A 0 00 1 1 OR A OR B A OR B 0 01 0 0 Decrement A-1 A all 1 0



0 0 1 AND A AND B A AND B 0 00 1 0 Identity A A 0 00 1 1 OR A OR B A OR B 0 01 0 0 Decrement A-1 A all 1 01 0 1 Add A+B A B 0



0 0 1 AND A AND B A AND B 0 00 1 0 Identity A A 0 00 1 1 OR A OR B A OR B 0 01 0 0 Decrement A-1 A all 1 01 0 1 Add A+B A B 01 1 0 Subtract A-B A B’ 1



0 0 1 AND A AND B A AND B 0 00 1 0 Identity A A 0 00 1 1 OR A OR B A OR B 0 01 0 0 Decrement A-1 A all 1 01 0 1 Add A+B A B 01 1 0 Subtract A-B A B’ 11 1 1 Increment A+1 A all 0 1

2/104


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Arithmetic-logic unitsM S1 S0 Function F X Y C0



1 1M

S1

S0c0

M S1

c0

2/105


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL




1

1 1

M

S1

S0X

1 1 1

1 1ai

bi

1 1 1 1

1 1 1 1

S1

S0

ba

S0S1

M

X

2/106


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL




bi

M

S1

S0Y

ai

1 1

1 1

1 1

1 1

S1

S0

ba

S0S1

M

Y

2/107


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


M

FA

a0 b0

f0

ALE

FA

a1 b1

f1

ALE

FA

a2 b2

f2

ALE

FA

a3 b3

f3

ALE

FA

a4 b4

f4

ALE

Cout

S01

X Y

2/108


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL



2/109


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Decoders

E A1 A0 C3 C2 C1 C0

0 0 0 0 0 0 00 0 1 0 0 0 0

0 1 0 0 0 0 0

0 1 1 0 0 0 0

1 0 0 0 0 0 1

1 0 1 0 0 1 0

1 1 0 0 1 0 0

1 1 1 1 0 0 0

EA1

A0

C3 C2 C1 C0

Decoder

C3..0

A1..0

E

2 CLB

2/110


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Decoders

Decoder

C3..0

A1..0

EDecoder

C7..4

A1..0

EDecoder

C11..8

A1..0

EDecoder

C15..12

A1..0

E

Decoder

A3..2

E

2/111


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL



2/112


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Selectors

S1 S0 Y

0 0 D0

0 1 D1

1 0 D2

1 1 D3

D3 D2 D1 D0

S1

S0

Y

4-to-1MUX

D3..0

S1..0

Y

In principle:2-to-1 MUX is 1/2 CLBDue to special provisions:4-to-1 MUX is 1 CLB

2/113


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Selectors

S1 S0 Y

0 0 D0

0 1 D1

1 0 D2

1 1 D3

D3 D2 D1 D0

Y

Deco

der

S1

S0

Alternative implementation

2/114


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Selectors

4-to-1selector

S3..2

4-to-1selector

D7..4

S1..0

4-to-1selector

D11..8

S1..0

4-to-1selector

D15..12

S1..0

4-to-1selector

D3..0

S1..0

Y

2/115


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL



2/116


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Buses

• Problem with high fan-in MUX: fan-in OR gate too big all inputs have to be routed to 1 central location:

substantial routing delay and difficult routing

• Solution: bus with tristate drivers

Deco

der

D3 D2 D1 D0 S1 S0

E Y

0 Z1 D

Y

2/117


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Buses

• In an FPGA (lab session) a limited number of tristate buffers is foreseen, connected to horizontal long lines. It is possible to indicate for a certain signal that we prefer to map it to a long line.

• Note that a Boolean signal already can have 4 different values: 0: the logical signal “0” 1: the logical signal “1” x: don’t care Z: high-impedant

• Simulations will allow to visualize each of the 4 different values

2/118


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL



2/119


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Priority encoders

D3 D2 D1 D0 A1 A0 Any

0 0 0 0 0 0 0


0 0 0 0 0 0 0

0 0 0 1 0 0 1


0 0 0 0 0 0 0

0 0 0 1 0 0 1

0 0 1 0 0 1 1

0 0 1 1 0 1 1


0 0 0 0 0 0 0

0 0 0 1 0 0 1

0 0 1 0 0 1 1

0 0 1 1 0 1 1

0 1 0 0 1 0 1

0 1 0 1 1 0 1

0 1 1 0 1 0 1

0 1 1 1 1 0 1


0 0 0 0 0 0 0

0 0 0 1 0 0 1

0 0 1 0 0 1 1

0 0 1 1 0 1 1

0 1 0 0 1 0 1

0 1 0 1 1 0 1

0 1 1 0 1 0 1

0 1 1 1 1 0 1

1 0 0 0 1 1 1

1 0 0 1 1 1 1

1 0 1 0 1 1 1

1 0 1 1 1 1 1

1 1 0 0 1 1 1

1 1 0 1 1 1 1

1 1 1 0 1 1 1

1 1 1 1 1 1 1


0 0 0 0 0 0 00 0 0 1 0 0 1

0 0 1 0 0 1 1

0 0 1 1 0 1 1

0 1 0 0 1 0 1

0 1 0 1 1 0 1

0 1 1 0 1 0 1

0 1 1 1 1 0 1

1 0 0 0 1 1 1

1 0 0 1 1 1 1

1 0 1 0 1 1 1

1 0 1 1 1 1 1

1 1 0 0 1 1 1

1 1 0 1 1 1 1

1 1 1 0 1 1 1

1 1 1 1 1 1 1

0

D2

D3

D1

D0

Any

0 0 0 0

D2

D3

D1

D0

A1

1 1

1 1 1 11 1 1 1

D2

D3

D1

D0

A0

2/120


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Priority encoders

0

D2

D3

D1

D0

Any

0 0 0 0

D2

D3

D1

D0

A1

1 1

1 1 1 11 1 1 1

D2

D3

D1

D0

A0

D3

D0

Any A1 A0

Priorityencoder

D3..0

A1..0

Any1 1/2 CLB

2/121


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Priority encoders

Priorityencoder

D3..0

Priorityencoder

D7..4

Priorityencoder

D11..8

Priorityencoder

D15..12

Priorityencoder

A3..2

Any 4-to-1MUX

A1

4-to-1MUX

A0

2/122


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL



2/123


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Magnitude comparators

x1 y1 x0 y0 G (X>Y) L (X<Y)

0 0 0 0 0 0

0 0 0 1 0 1

0 0 1 0 1 0

0 0 1 1 0 0

x1 y1 x0 y0 G (X>Y) L (X<Y)

0 0 0 0 0 00 0 0 1 0 1

0 0 1 0 1 0

0 0 1 1 0 0

0 1 0 0 0 1

0 1 0 1 0 1

0 1 1 0 0 1

0 1 1 1 0 1

x1 y1 x0 y0 G (X>Y) L (X<Y)

0 0 0 0 0 00 0 0 1 0 1

0 0 1 0 1 0

0 0 1 1 0 0

0 1 0 0 0 1

0 1 0 1 0 1

0 1 1 0 0 1

0 1 1 1 0 1

1 0 0 0 1 0

1 0 0 1 1 0

1 0 1 0 1 0

1 0 1 1 1 0

x1 y1 x0 y0 G (X>Y) L (X<Y)

0 0 0 0 0 00 0 0 1 0 1

0 0 1 0 1 0

0 0 1 1 0 0

0 1 0 0 0 1

0 1 0 1 0 1

0 1 1 0 0 1

0 1 1 1 0 1

1 0 0 0 1 0

1 0 0 1 1 0

1 0 1 0 1 0

1 0 1 1 1 0

1 1 0 0 0 0

1 1 0 1 0 1

1 1 1 0 1 0

1 1 1 1 0 0

x1 y1 x0 y0 G (X>Y) L (X<Y)

0 0 0 0 0 00 0 0 1 0 1

0 0 1 0 1 0

0 0 1 1 0 0

0 1 0 0 0 1

0 1 0 1 0 1

0 1 1 0 0 1

0 1 1 1 0 1

1 0 0 0 1 0

1 0 0 1 1 0

1 0 1 0 1 0

1 0 1 1 1 0

1 1 0 0 0 0

1 1 0 1 0 1

1 1 1 0 1 0

1 1 1 1 0 0

1

11 1 1 1

y1

x1

x0

y0

G

11 1 1 1

1y1

x1

x0

y0

L

2/124


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

x1x0y1y0


1

11 1 1 1

y1

x1

x0

y0

G

11 1 1 1

1y1

x1

x0

y0

L

G L

Comp

Xi Yi

Gi

Li

Gi+1

Li+1 1 CLB

2/125


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


Comp

x1 y1

Comp

x2 y2

Comp

x3 y3

Comp

x4 y4

Comp

x5 y5

Comp

x6 y6

Comp

x7 y7 x0 y0

G

L

Comp

x1 y1x2 y2

Comp

x3 y3x4 y4

Comp

x5 y5x6 y6

Comp

x7 y7 x0 y0

Comp

Comp

Comp

G L

2/126


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


• Simpler circuits are used for comparison with constants!!

y=1 when X=0

X

y

y=1 when X=255

y=1 when X>=64

X

y

y=1 when X<192

x7 x6

y

y=1 when X is even

x0

y

x7 x6

y

X is8 bits

2/127


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL



2/128


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Shifters and rotators• Shifter:

An input word is shifted m positions to the left or to the right

m bits disappear at one side m bits are created at the other side For an arithmetic shift (word = 2-complement)

For a left shift m zeros are shifted in from the right

For a right shift m times the MSB is shifted in from the left (for 2-complement)

For a logic shift

It is possible to indicate which value is shifted in

m bit left shift is multiplication with 2m

m bit right shift is division by 2m

2/129


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL

Shifters and rotators

• Rotator: An input word is shifted m positions to the left or to

the right The bits that drop-off at one side, are shifted back

in at the other side

2/130


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


d0d1d2d3S2S1S0L-inR-in

4-to-1MUX

4-to-1MUX

4-to-1MUX

4-to-1MUX

MM

y3 y2 y1 y0

S2: 0=no shift,1=shift S1: 0=left,1=right S0: 0=shift,1=rotate

4-to-1MUX

4-to-1MUX

4-to-1MUX

4-to-1MUX

MM

y3 y2 y1 y0


4-to-1MUX

4-to-1MUX

4-to-1MUX

4-to-1MUX

MM

y3 y2 y1 y0


4-to-1MUX

4-to-1MUX

4-to-1MUX

4-to-1MUX

MM

y3 y2 y1 y0


4-to-1MUX

4-to-1MUX

4-to-1MUX

4-to-1MUX

MM

y3 y2 y1 y0


4-to-1MUX

4-to-1MUX

4-to-1MUX

4-to-1MUX

MM

y3 y2 y1 y0


2/131


Digitaldesign


Sequentialcircuits

FSMDdesign

VHDL


MMMMMMMM

MMMMMMMM

MMMMMMMM

S0

S1

S2

8-bit barrel left rotator

© r.lauwereins imec 2001 digital design combina- torial circuits sequential circuits fsmd design...

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