Dr. Wissam Hasan Mahdi Alagele

1

Dr. Wissam Hasan Mahdi Alagele

e-mail:[email protected]

http://edu-clg.kufauniv.com/staff/Mr.Wesam

العلمي والبحث العالي التعليم وزارةالكوفة الحاسوب- – جامعة علوم قسم التربية كلية

Chapter 3Combinational Design

Digital Logic Design III

Combinational CircuitsOutput is function of input only

i.e. no feedback

When input changes, output may change (after a delay)

CombinationalCircuits

••••••

n inputs m outputs

3

Combinational CircuitsAnalysis

◦Given a circuit, find out its function◦Function may be expressed as:

Boolean function Truth table

Design◦Given a desired function, determine its circuit

◦Function may be expressed as: Boolean function Truth table

CBA

CBA

BA

CA

CB

F1

F2

?

?

?

4

Analysis ProcedureBoolean Expression Approach

CBA

CBA

BA

CA

CB

F1

F2

T2=ABC T1=A+B+

C

F2=AB+AC+BC

F’2=(A’+B’)(A’+C’)(B’+C’)

T3=AB'C'+A'BC'+A'B'C

F1=AB'C'+A'BC'+A'B'C+ABCF2=AB+AC+BC

5

CBA

CBA

BA

CA

CB

F1

F2

Analysis ProcedureTruth Table Approach

A B C F1 F2

0 0 0= 0 = 0= 0= 0= 0= 0= 0= 0

= 0= 0= 0= 0

0

0

0

0

0

0

1

0

00 0

6

CBA

CBA

BA

CA

CB

F1

F2


= 0 = 0= 1= 0= 0= 1= 0= 0

= 0= 1= 0= 1

0

1

0

0

0

0

1

1

1 A B C F1 F2

0 0 0 0 00 0 1 1 0

7

CBA

CBA

BA

CA

CB

F1

F2


= 0 = 1= 0= 0= 1= 0= 0= 1

= 0= 0= 1= 0

0

1

0

0

0

0

1

1

1A B C F1 F2

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

8

CBA

CBA

BA

CA

CB

F1

F2


= 0 = 1= 1= 0= 1= 1= 0= 1

= 0= 1= 1= 1

0

1

0

0

1

1

0

0

0A B C F1 F2

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

9

CBA

CBA

BA

CA

CB

F1

F2


= 1 = 0= 0= 1= 0= 0= 1= 0

= 1= 0= 0= 0

0

1

0

0

0

0

1

1

1A B C F1 F2

0 0 0 0 00 0 1 1 00 1 0 1 00 1 1 0 11 0 0 1 0

10

CBA

CBA

BA

CA

CB

F1

F2

Analysis ProcedureTruth Table Approach= 1 = 0= 1= 1= 0= 1= 1= 0

= 1= 1= 0= 1

0

1

0

1

0

1

0

0

0A B C F1 F2

0 0 0 0 00 0 1 1 00 1 0 1 00 1 1 0 11 0 0 1 01 0 1 0 1

11

CBA

CBA

BA

CA

CB

F1

F2


= 1 = 1= 0= 1= 1= 0= 1= 1

= 1= 0= 1= 0

0

1

1

0

0

1

0

0

0 A B C F1 F2

0 0 0 0 00 0 1 1 00 1 0 1 00 1 1 0 11 0 0 1 01 0 1 0 11 1 0 0 1

12

CBA

CBA

BA

CA

CB

F1

F2


= 1 = 1= 1= 1= 1= 1= 1= 1

= 1= 1= 1= 1

1

1

1

1

1

1

0

0

1A B C F1 F2

0 0 0 0 00 0 1 1 00 1 0 1 00 1 1 0 11 0 0 1 01 0 1 0 11 1 0 0 11 1 1 1 1

B

0 1 0 1A 1 0 1 0

C

B

0 0 1 0A 0 1 1 1

C

F1=AB'C'+A'BC'+A'B'C+ABC F2=AB+AC+BC

13

Design ProcedureGiven a problem statement:

◦Determine the number of inputs and outputs◦Derive the truth table◦Simplify the Boolean expression for each

output◦Produce the required circuit

Example: Design a circuit to convert a “BCD” code to “Excess 3”

code 4-bits 0-9 values

4-bits Value+3

?

14

Design ProcedureBCD-to-Excess 3 Converter

A B C D w x y z0 0 0 0 0 0 1 10 0 0 1 0 1 0 00 0 1 0 0 1 0 10 0 1 1 0 1 1 00 1 0 0 0 1 1 10 1 0 1 1 0 0 00 1 1 0 1 0 0 10 1 1 1 1 0 1 01 0 0 0 1 0 1 11 0 0 1 1 1 0 01 0 1 0 x x x x1 0 1 1 x x x x1 1 0 0 x x x x1 1 0 1 x x x x1 1 1 0 x x x x1 1 1 1 x x x x

C

1 1 1 BA

x x x x1 1 x x

D

C

1 1 11 B

Ax x x x

1 x xD

C

1 11 1 B

Ax x x x1 x x

D

C

1 11 1 B

Ax x x x1 x x

D

w = A+BC+BD x = B’C+B’D+BC’D’

y = C’D’+CD z = D’

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Design ProcedureBCD-to-Excess 3 Converter

A B C D w x y z0 0 0 0 0 0 1 10 0 0 1 0 1 0 00 0 1 0 0 1 0 10 0 1 1 0 1 1 00 1 0 0 0 1 1 10 1 0 1 1 0 0 00 1 1 0 1 0 0 10 1 1 1 1 0 1 01 0 0 0 1 0 1 11 0 0 1 1 1 0 01 0 1 0 x x x x1 0 1 1 x x x x1 1 0 0 x x x x1 1 0 1 x x x x1 1 1 0 x x x x1 1 1 1 x x x x

w

x

D

C

z

y

B

A

w = A + B(C+D)x = B’(C+D) + B(C+D)’

y = (C+D)’ + CDz = D’

16

Seven-Segment Decoderw x y z a b c d e f g

0 0 0 0 1 1 1 1 1 1 00 0 0 1 0 1 1 0 0 0 00 0 1 0 1 1 0 1 1 0 10 0 1 1 1 1 1 1 0 0 10 1 0 0 0 1 1 0 0 1 10 1 0 1 1 0 1 1 0 1 10 1 1 0 1 0 1 1 1 1 10 1 1 1 1 1 1 0 0 0 01 0 0 0 1 1 1 1 1 1 11 0 0 1 1 1 1 1 0 1 11 0 1 0 x x x x x x x1 0 1 1 x x x x x x x1 1 0 0 x x x x x x x1 1 0 1 x x x x x x x1 1 1 0 x x x x x x x1 1 1 1 x x x x x x x

1 1 11 1 1

x x x x1 1 x x

?

wxyz

abcdefg

BCD code

a = w + y + xz + x’z’

b = x’+yz+y’z’c = x+y’+zd = x’z’+yz’+x’y+xy’z

e = x’z’+yz’

f = w+x+y’z’g = w+x’y+xy’+x’y+xz’

f

e

g

d

a

b

c

17

18

19

20

Problem Design a circuit which displays the letters A through J on a seven-segment indicator. The circuit has four inputs W, X, Y, Z which represent the last 4 bits of the ASCII code for the letter to be displayed. For example, if WXYZ = 0001, “A” will be displayed. The letters should be displayed in the following form:

Design your circuit using only two-, three-, and four-input NOR gates and inverters. Any solution with 22 or fewer gates and inverters (not counting the four inverters for the inputs) is acceptable.

21

Binary AdderHalf Adder

◦Adds 1-bit plus 1-bit◦Produces Sum and Carry

HAxy

SC

x y C S0 0 0 00 1 0 11 0 0 11 1 1 0

x+ y───C S

x

y

S

C

22

Binary AdderFull Adder

◦Adds 1-bit plus 1-bit plus 1-bit◦Produces Sum and Carry

x y z C S0 0 0 0 00 0 1 0 10 1 0 0 10 1 1 1 01 0 0 0 11 0 1 1 01 1 0 1 01 1 1 1 1

x+ y+ z───C S

FAxyz

SC

y

0 1 0 1x 1 0 1 0

z

y

0 0 1 0x 0 1 1 1

z

S = xy'z'+x'yz'+x'y'z+xyz = z’(xy’+x’y)+z(x’y’+xy) =z’(x y) + z(x y)’ = x y z

C = xy + xz + yz

23


xyzxyzxyzxyz

x

y

z

S

C

xy

xzyz

xyzx

y

z

xy

xzyz

S

C

S = xy'z'+x'yz'+x'y'z+xyz=x y z

C = xy + xz + yz

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x

y

z

S

C

HAxy

z

HA S

C

QUIZ

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?

Q#1- This combinational logic circuit is described as a(n) ___.

ANS: full-adder

Q#2- What are the sum and carry out outputs of thisfull-adder circuit?

Cin = 0

A = 0

B = 0

ANS: Sum=0, Carry out=0


Cin = 0

A = 0

B = 1



Cin = 1

A = 0

B = 1



Cin = 1

A = 1

B = 1



Cin = 0

A = 1

B = 1


26

Binary Adderc3 c2 c1 .

+ x3 x2 x1 x0

+ y3 y2 y1 y0

────────Cy S3 S2 S1

S0

FA

x3 x2 x1 x0

FAFAFA

y3 y2 y1 y0

S3 S2 S1 S0

C4 C3 C2 C1

0

Binary Adder

x3x2x1x0 y3y2y1y0

S3S2S1S0

C0CyCarry Propagate Addition

74x2834-bit adder

Uses carry look-ahead internally

28

Binary AdderCarry Propagate Adder

CPA

A3 A2 A1 A0 B3 B2 B1 B0

S3 S2 S1 S0

C0CyCPA

A3 A2 A1 A0 B3 B2 B1 B0

S3 S2 S1 S0

C0Cy

x3 x2 x1 x0y3 y2 y1 y0

x7 x6 x5 x4y7 y6 y5 y4

S3 S2 S1 S0S7 S6 S5 S4

0

c3 c2 c1 . x3 x2 x1 x0

y3 y2 y1 y0

────────Cy S3 S2 S1 S0

c7 c6 c5 Cy .+ x7 x6 x5 x4

+ y7 y6 y5 y4

────────Cy S7 S6 S5 S4

Subtractors

Half SubtractorFull SubtractorAdder/Subtractor - 1Adder/Subtractor - 2

Half Subtractor

A

B

Di (difference)

B0 (borrow out)

HalfSubtractor

Input Logic

Symbol:

Logic Diagram:

A B Di

Bo

Bo A B Di 0 0 0 00 1 1 11 0 1 01 1 0 0

0-1 1

21

Output

QUIZ

?

?

Q#1- What is the difference and borrow outputs fromThis half-subtractor circuit?

(A – B)

A = 0

B = 0

ANS: Di= 0, Bo= 0

Q#2- What is the difference and borrow outputs fromthis half-subtractor circuit?

(A – B)

A = 1

B = 0

ANS: Di= 1, Bo= 0


(A – B)

A = 1

B = 1

ANS: Di= 0, Bo= 0


(A – B)

A = 0

B = 1

ANS: Di= 1, Bo= 1

Full Subtractor

Logic Diagram:

Logic Symbol: A

B

Di (difference)

B0 (borrow out)

FullSubtractor

Input Output

Bin

AB

DiB0H. S.

H. S.Bin

A B

D

C

C i+1

i i

i

i

half subtractor

half subtractorBo

Full Subtractor

0 0 0 0 00 0 1 1 10 1 0 1 00 1 1 0 01 0 0 1 11 0 1 0 11 1 0 0 01 1 1 1 1

Ci A B Di Bo

Di Same as S in full adder

0 1 0 01 1 1 0

Bo

00 01 11 1001

1 1

1 1

Ci

A B 00 01 11 10

0

1

Di

QUIZ

?

?

Q#1- What are the Difference and Borrow out output fromthis full-subtractor circuit?

(A – B - Bin)Bin = 0

A = 0

B = 0

HINT: truth table from textbook (Fig. 10-10) is helpfulANSWER: Di = 0, Bo = 0



A = 0

B = 0

ANSWER: Di = 1, Bo = 1



A = 1

B = 1




A = 0

B = 1




A = 1

B = 0




A = 1

B = 1


Adder/Subtractor - 1

A B Di

Bo

A B S

C

A 0 B 0 0

CB1 E

SD

Half adder Half subtractor

E = 0: Half adderE = 1: Half subtractor

i

o

Adder/Subtractor-1

A B

Di C

Bo

i

E

E = 0: Full adderE = 1: Full subtractor

Adder/Subtractor-2

Full Adder

A B

C

0 0

1

0

Full Adder

A B

C

1 1

2

1

Full Adder

A B

C

2 2

3

2

Full Adder

A B

C SD

3 3

4 3 SD SD SD

E

E = 0: 4-bit adderE = 1: 4-bit subtractor

Dr. Wissam Hasan Mahdi Alagele

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