Question

Imagine a new kind of flip-flop called a J N flip-flop which has two inputs J and N. Input J behaves like the J input of the J K flip-flop, and input N behaves like the complement of the K input of a J K flip-flop (N=K’). Derive the characteristic table and excitation table of the J N flip-flop. Show that a D flip-flop can be constructed from a J N flip-flop by connecting its two inputs (J and N) together.

Answer 1

solution:

given data:

Imagine a new kind of flip-flop called a J N flip-flop :

1) Before starting the conversion, let us consider the characteristic table of JK Flipflop

J	K	Qn	Qn+1
0	0	0	0
0	0	1	1
0	1	0	0
0	1	1	0
1	0	0	1
1	0	1	1
1	1	0	1
1	1	1	0

The minimized table will be shown below :

J	K	Output
0	0	Qn
0	1	0
1	0	1
1	1	(Qn )'

Similarly the excitation table of JK-Flipflop is

<-------------------Outputs--------------------><---Inputs-->

Present State(Qn)	Next State(Qn+1)	J	K
0	0	0	Φ
0	1	1	Φ
1	0	Φ	1
1	1	Φ	0

Now coming to new flipflop J-N, given that J=J input of JK Flipflop and N= K' of JK Flipflop. So the characterstic table of J-N Flipflop is as follows:

      J            K'            Qn         Qn+1

J	N	Qn	Qn+1
0	0	0	0
0	0	1	0
0	1	0	0
0	1	1	1
1	0	0	1
1	0	1	0
1	1	0	1
1	1	1	1

Here the characteristic table is filled by taking ( J, N, Qn) as ( J, K', Qn). It means Qn+1 for (0,1,0) of ( J, N, Qn) is same as Qn+1 for (0,0,0) of ( J, K', Qn) because N= K'=1. If K'=1 then K=0. So Qn+1 for (0,1,0) is same as Qn+1 for (0,0,0). In the same way other entries are also filled. The minimized table will be:

J	N	Output
0	0	0
0	1	Qn
1	0	(Qn )'
1	1	1

This can be observed by looking carefully at characteristic table. Consider when J=0 and N=1 output is Qn in minimized table. So for this consider entries in main characteristic table where J=0 and N=1. For these entries if Qn =0, then Qn+1 = 0. If Qn = 1, then Qn+1 = 1. So always same charge is maintained. So Qn+1 = Qn . In this way other entires also can be observed. The excitation table will be as follows:

<-------------------Outputs--------------------><---Inputs-->

Present State(Qn)	Next State(Qn+1)	J	N
0	0	0	Φ
0	1	1	Φ
1	0	Φ	0
1	1	Φ	1

2) Now the obtined JN Flipflop need to be converted to D Flipflop.

Step1: First determine characteristic table of D Flipflop.

D	Qn	Qn+1
0	0	0
0	1	0
1	0	1
1	1	1

Step2: Now this D input need to be connected to J and N inputs in some manner to get the above excitation for D. It means D need to be realised in terms of J and N. It means for this excitation of Qn to Qn+1 , J and N inputs need to be found out.

D	Qn	Qn+1	J	N
0	0	0	0	Φ
0	1	0	Φ	0
1	0	1	1	Φ
1	1	1	Φ	1

Step3: Now J and N need to be realized in terms of D and Qn thorugh K-Map.

The K-Map for J will be as follows:

On solving the above K-Map the expression will be J = D .

The K-Map for N will be as follows:

On solving the above K-Map the expression will be N = D .

So to make JN Flipflop work like D flipflop J and N need to be connected to D irrespective of charge Qn .

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