2. Ch10L3-"Digital Principles and Design", Raj Kamal, Pearson Education, 2006 2
Lesson 3
Logic Design and Boolean-
Function Implementation Using
Decoders
3. Ch10L3-"Digital Principles and Design", Raj Kamal, Pearson Education, 2006 3
Outline
• Three variable Boolean Function
implementation
• Four variable Boolean Function
implementation
4. Ch10L3-"Digital Principles and Design", Raj Kamal, Pearson Education, 2006 4
Decoder of active output is 1
• A decoder circuit can be used to implement
AND-OR circuit SOP Boolean expression
when decoder active state output is 1 and
inactive 0
• Number of binary inputs = n
• Number of binary outputs = 2n = Maximum
number of miniterms, where n is the
number of literals in F
• Its outputs reflect the Mini-terms with one
term each at each of the output
5. Ch10L3-"Digital Principles and Design", Raj Kamal, Pearson Education, 2006 5
1 of 8 Decoder with active state = 1
m1
m0
m2
m3
m4
m5
m6
m7
A0
A1
A2
7. Ch10L3-"Digital Principles and Design", Raj Kamal, Pearson Education, 2006 7
Decoder of active output is 0
• A decoder circuit can be used to implement
OR-AND circuit SOP Boolean expression
when decoder active state output is 0 and
inactive 1
• Number of binary inputs = n
• Number of binary outputs = 2n = Maximum
number of Maxterms, where n is the
number of literals in F
• Its outputs reflect the Maxterms with one
term each at each of the output
8. Ch10L3-"Digital Principles and Design", Raj Kamal, Pearson Education, 2006 8
1 of 8 Decoder with active 0 output
M1
M8
M2
M3
M4
M5
M6
M6
A0
A1
A2
9. Ch10L3-"Digital Principles and Design", Raj Kamal, Pearson Education, 2006 9
Implement F= ΠΠΠΠ M (2, 5)
M1
M0
M2
M3
M4
M5
M6
M7
A0
A1
A2 F
10. Ch10L3-"Digital Principles and Design", Raj Kamal, Pearson Education, 2006 10
Outline
• Three variable Boolean Function
implementation
• Four variable Boolean Function
implementation
11. Ch10L3-"Digital Principles and Design", Raj Kamal, Pearson Education, 2006 11
Decoder of active output is 1
• Number of binary inputs = 4
• Number of binary outputs = 16=
Maximum number of miniterms
• Number of literals in F = 4
• Its outputs reflect the Mini-terms with
one term each at each of the input of an
OR to get F
12. Ch10L3-"Digital Principles and Design", Raj Kamal, Pearson Education, 2006 12
1 of 16 Decoder with active state = 1
m1
m0
m2
.
.
.
m14
m15
A0
A1
A2
A3
14. Ch10L3-"Digital Principles and Design", Raj Kamal, Pearson Education, 2006 14
Decoder of active output is 0
• Number of binary inputs = 4
• Number of binary outputs = 16 =
Maximum number of Maxterms.
Number of literals in F = 4
• Its outputs reflect the Maxterms with
one term each at each of the AND
input
15. Ch10L3-"Digital Principles and Design", Raj Kamal, Pearson Education, 2006 15
1 of 16 Decoder with active state = 0
M1
M0
M2
.
.
.
M14
M15
A0
A1
A2
A3
18. Ch10L3-"Digital Principles and Design", Raj Kamal, Pearson Education, 2006 18
Decoder
• Decoder active one output pin is also a
miniterm at one of the 2n output pins
• For n terms, use active 1 line decoder
and m-input OR gate for m terms in an
SOP Boolean function
19. Ch10L3-"Digital Principles and Design", Raj Kamal, Pearson Education, 2006 19
Decoder
• Decoder active 0 output pin is also a
Maxterm at one of the 2n output pins
• For n terms, use active 0 line decoder
and m-input AND gate for m terms in a
POS Boolean function
20. Ch10L3-"Digital Principles and Design", Raj Kamal, Pearson Education, 2006 20
End of Lesson 3
Logic Design and Boolean-
Function Implementation
Using Decoders