MOD 12 Synchronous Counter

MOD 12 Synchronous Counter

How to draw circuit for MOD-12 synchronous counter using flip flops?

Drawing synchronous counter is a very interesting and easy task. Here, we are going to discuss and learn the techniques to draw circuit diagram for any synchronous counter.

Step I : Draw a state diagram for the required synchronous counter. For example, if we are making synchronous counter that counts from 0 to 15, then

State diagram in decimal

0→1→2→3→4→5→6→7→8→9→10→11→0

State diagram in binary

0000→0001→0010→0011→0100→0101→0110→0111→1000→1001→1010→1011→0000

Step 2: Decide no. of flip flops and name of flip flop

In the above state diagram, we can see that we need 4 bits to represents the no. in binary. Thus, we need to use 4 no. of flip flops. Also, we are using t flip flop for this.  You can use any flip flop of your choice.

Step 3: Draw/Write excitation table of flip flop

We need excitation table of the flip flop we are using. Since, we are using T-flip flop for this,

Excitation Table of T-Flip flop

Present State

Next State

T

Qt

Qt

0

Qt

Qt

1

 

Step 4: Draw excitation table of state diagram

Present State

Next State

T

Qd

Qc

Qb

Qa

Qd

Qc

Qb

Qa

Ta

Tb

Tc

Td

0

0

0

0

0

0

0

1

1

0

0

0

0

0

0

1

0

0

1

0

1

1

0

0

0

0

1

0

0

0

1

1

1

0

0

0

0

0

1

1

0

1

0

0

1

1

1

0

0

1

0

0

0

1

0

1

1

0

0

0

0

1

0

1

0

1

1

0

1

1

0

0

0

1

1

0

0

1

1

1

1

0

0

0

0

1

1

1

1

0

0

0

1

1

1

1

1

0

0

0

1

0

0

1

1

0

0

0

1

0

0

1

1

0

1

0

1

1

0

0

1

0

1

0

1

0

1

1

1

0

0

0

1

0

1

1

0

0

0

0

1

1

0

1

 

Step 4: Find the expression

To find the expression for Ta , Tb Tc and Td , we are using k map

For Ta

 

                                                                 QbQa

QdQc

00

01

11

10

00

 

1

 

1

 

1

 

1

01

 

1

 

1

 

1

 

1

11

 

*

 

*

 

*

 

*

10

 

1

 

1

 

1

 

1

 

Solving k map we get, Ta= 1

For Tb

                                             QbQa

QdQc

00

01

11

10

00

 

0

 

1

 

1

 

0

01

 

0

 

1

 

1

 

0

11

 

*

 

*

 

*

 

*

10

 

0

 

1

 

*

 

0

 

Solving k map we get, Tb=Qa

 

For Tc

                                             QbQa

QdQc

00

01

11

10

00

 

0

 

0

 

1

 

0

01

 

0

 

0

 

1

 

0

11

 

*

 

*

 

*

 

*

10

 

0

 

0

 

0

 

0

 

Solving k map we get, Tc= Qb.Qa Qd

For Td

                                             QbQa

QdQc

00

01

11

10

00

 

0

 

0

 

0

 

0

01

 

0

 

0

 

1

 

0

11

 

*

 

*

 

*

 

*

10

 

0

 

0

 

1

 

0

 

Solving the above k map we get,

Td=Qb.Qa.Qc Qd’+ Qb.Qa.Qc ‘Qd
 =Qb.Qa.(Qc Qd’+ Qc ‘Qd )
=Qb.Qa.(Qc Ꚛ Qd )

Step 5 : Write the expression for Ta, Tb Tc and Td.

Ta=1, Tb= Qa , Tc= Qb.Qa Qd’and Td= Qb.Qa.(Qc Ꚛ Qd )

Step 6: Draw the circuit diagram