Boolean Expression To Logic Circuit Solved Examples

Boolean Expression/Function to Logic Circuit and Truth Table Evaluation

This topic is important from the examination point of view. It is easy and you can do it in a couple of minutes. You are given an expression and you have to draw a logic circuit for this expression. I will explain with the help of examples. Some points to follow,

  • Look for parentheses, draw them first

  • Proceed from left to right

There are no other rules or suggestions for solving these exercises. Just do some examples.


Example # 1: \[(A+B).A\]

Convert Boolean expression to logic circuit



A

B

Output

0

0

0

0

1

0

1

0

1

1

1

1


Example # 2: \[(\bar A+B).B\]

Convert Boolean expression to logic circuit



A

B

Output

0

0

1

0

1

1

1

0

0

1

1

0



Example # 3: \[(\bar A. \bar B)+ B\]

Convert Boolean expression to logic circuit


A

B

Output

0

0

1

0

1

1

1

0

0

1

1

1


Example # 4: \[(A.B) \oplus C\]

Convert Boolean expression to logic circuit

A

B

C

\[A.B\]

\[(A.B) \oplus C\]

0

0

0

0

0

0

0

1

0

1

0

1

0

0

0

0

1

1

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1

1

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1

0


Example # 5: \[A.B + B.C\]

Convert Boolean expression to logic circuit


A

B

C

\[A.B\]

\[B.C\]

\[A.B + B.C\]

0

0

0

0

0

0

0

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1

0

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1

Example # 6: \[(A+B).(A+C)\]

Convert Boolean expression to logic circuit

A

B

C

\[A+B\]

\[A+C\]

\[(A+B).(A+C)\]

0

0

0

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1

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1

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1


Example # 7: \[(\bar A \oplus B).(A \oplus C)\]

Convert Boolean expression to logic circuit



A

B

C

\[\bar A \oplus B\]


\[A \oplus C\]

\[(\bar A \oplus B).(A \oplus C)\]

0

0

0

1

0

0

0

0

1

1

1

1

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1

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1



Example # 8: \[A . B + A . C \oplus C . D\]

Convert Boolean expression to logic circuit



A

B

C

D

\[A.B\]

\[A.C\]

\[C.D\]

\[A.B+A.C\]

\[A . B + A . C \oplus C . D\]

0

0

0

0

0

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0

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1

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1

0


Example # 9: \[(A + B) \oplus (A . C)\]

Convert Boolean expression to logic circuit


A

B

C

\[A+B\]


\[A.C\]

\[(\bar A \oplus B).(A \oplus C)\]

0

0

0

0

0

0

0

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1

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1

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0


Example # 10:  \[A . B + B . C + (C \oplus D)\]

Convert Boolean expression to logic circuit


A

B

C

D

\[A.B\]

\[B.C\]

\[C \oplus D\]

\[A.B + B.C\]

\[A . B + B . C + (C \oplus D)\]


0

0

0

0

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1



Example # 11: \[(\bar A . \bar B) \oplus (A + \bar B)\]


Convert Boolean expression to logic circuit

A

B

\[\bar A. \bar B\]

\[A + \bar B\]

\[(\bar A . \bar B) \oplus (A + \bar B)\]

0

0

1

1

0

0

1

0

0

0

1

0

0

1

1

1

1

0

1

1


Example # 12: \[(\bar A . B) + \bar C\]

Convert Boolean expression to logic circuit


A

B

C

\[\bar A.B\]


\[(\bar A . B) + \bar C\]


0

0

0

0

1

0

0

1

0

0

0

1

0

1

1

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1

1

1

1

1

0

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1

1

1

0

0


Example # 13: \[\overline {(\bar A +\bar B) \oplus (B+C)}\]

Convert Boolean expression to logic circuit


A

B

C

\[\bar A + \bar B\]

\[B+C\]

\[\overline {(\bar A+\bar B) \oplus (B+C)}\]


0

0

0

1

0

0

0

0

1

1

1

1

0

1

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1

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1

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1

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1

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0


Example # 14: \[\overline {A.B} + \overline{C+B}\]


Convert Boolean expression to logic circuit



A

B

C

\[\overline {A.B}\]

\[\overline {C+B}\]

\[\overline {A.B} + \overline{C+B}\]

0

0

0

1

1

1

0

0

1

1

0

1

0

1

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1

0

1

0

1

1

1

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1

1

0

0

1

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1

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1

1

1

0

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0


Example # 15: \[\overline {(A.\bar C) +(\overline{C.B})}\]

Convert Boolean expression to logic circuit



A

B

C

\[A. \bar C\]

\[\overline {C.B}\]

\[\overline {(A.\bar C) +(\overline{C.B})}\]


0

0

0

0

1

0

0

0

1

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1

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1

0

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1


Example # 16: \[\overline {(\overline {B+C}).(\overline {B.C})}\]

Convert Boolean expression to logic circuit


B

C

\[\overline {B+C}\]

\[\overline {B.C}\]

\[\overline {(\overline {B+C}).(\overline {B.C})}\]


0

0

1

1

0

0

1

0

1

1

1

0

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1

1

1

1

0

0

1



Example # 17: \[(\bar A+B).(A+\bar B)\]



A

B

\[bar A + B\]

\[A + \bar B\]

\[\overline {(\bar A+B).(A+\bar B)}\]

0

0

1

1

0

0

1

1

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1

0

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1

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0



Example # 18: \[\overline {(A \oplus B) + (\bar A. B)} + \overline{(B+C)}\]


A

B

C

\[\overline {A \oplus B}\]

\[\bar A.B\]

\[\overline {\overline {(A \oplus B)} + (\bar A. B)}\]

\[\overline {B+C}\]

\[\overline {\overline {(A \oplus B)} + (\bar A. B)} + \overline{(B+C)}\]

0

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