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## How to solve questions on conversion of Flip-Flops

For conversion the steps needed are

1. Identify available and required flipflop .
2. Make the characteristic table for required flipFlop
3. Make excitation table for available flipflop
4. Write Boolean expression for available flipflop

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In order to solve questions on flipflops, one must know the characteristics and excitation tables of Flip Flops.

From truth table, we can derive characteristics table and from characteristics table we can derive excitation table.

## For creating characteristics table

For creating characteristics table, following equations can be used.

SR-

Qnext= S+RlQ, for SR=0

For S=1, R=1,Qnext is Don’t care

JK -

Qnext= JQl + KlQ

D -

Qnext= D

T -

Qnext= TQl + TlQ

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## Characteristics of Various FlipFlops

The characteristics can be summed as:

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## Creating an excitation Table from characteristics table

We can create excitable table from characteristics table .  Lets take an example for creating excitation table for JK Flip Flop.

Characteristics of  J K Flip Flop :

Excitation table of J K Flip Flop :

Here if we combine  2 rows of 0 ,0 for Qn and Qn+1  respectively, we can see J gets constant output of 0 and K can be 1 or 0, so we place don’t care. Similarly for 2 rows of 0,1   for Qn and Qn+1  , we get J as constant output of 1 and k can be 1 or 0 so we place don’t care in K. Similarly for other 2 cases.

Now for conversion of 1 Flip Flop to another. Lets understand this by an example

## Example : Suppose we convert JK Flip Flop to D Flip Flop

Available Flip Flop : JK Flip Flop

Required : D Flip Flop

We need characteristic table of D(required ) and Excitation Table of JK (available)

So J=D and K= D'

### Important tips To learn

a) SR to

JK :-           S=JQl                R=KQ

D :-           S= D              R= Dl

T :-            S=TQl                     R=TQ

b) JK to

SR :-          J= S                K=R

D :-            J= D                K=Dl

T :-            J=T                 K=T

c) D to

SR :-           D=S+RlQ

JK :-            D=JQl+KlQ

T :-              D=T xnor Q

d) T to

SR :-       T=SQl+RQ

JK :-        T=JQl+KQ

D :-          T=D xnorQ

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Example from Gate 2000: A sequential circuit using D flip flop and logic gates is shown in figure, where X and Y are the inputs and Z is output. The circuit is

1. S-R flip flop with inputs X = R and Y = S
2. S-R flip flop with inputs X = S and Y = R
3. J-K flip flop with inputs X = J and Y = K
4. J-K flip flop with inputs X = K and Y = J

Solution

We can say, D=XlZ+YlZ

D=Qn+1   , for  D Flip Flop

Truth Table

 X Y Z Q/Qn+1 0 0 0 0 0 0 1 1 0 1 0 1 1 0 1 1 1 0 0 0 1 0 1 0 1 1 0 1 1 1 1 0

We can sum it as :

 X Y Qn+1 0 0 Zl 0 1 1 1 0 0 1 1 Zl

This is an Jk Flop characterstics  X = K and Y = J, Hence d.

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