# Convert a D flip flop into JK

Discussion in 'Homework Help' started by wind_blast942, Nov 19, 2010.

1. ### wind_blast942 Thread Starter New Member

Sep 7, 2010
22
0
I am confused on how to systematically convert a D flip flop into a JK flip flop
J K Q Q+ D
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

I am able to merge the characteristic and excitation table together, but since there are some values where Q to Q+ changes from 1 to 0, and there is no excitation
possible for a d flip flop to effect this change.

In this case what should i do?

2. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
A FF is always excited by its input. As long as the clock is pulsing it will change its state according to the J K and Q (or D and Q) data.

What you need to do is extract the Boolean function Q+=F(J,K,Q) and drive its output in the D input of your FF. Since a D FF will drive its next state where its D input is, you will be able to control the FF by changing that input.

3. ### wind_blast942 Thread Starter New Member

Sep 7, 2010
22
0
By solving Q+ = f(J,K,Q), i get J AND (NOT Q); so does it mean i connect this expression into the D ff?

But in this case wouldn't K not affect the flip flop at all?

4. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
Maybe you need to check again your Karnaugh map solution. Try it once more.

5. ### wind_blast942 Thread Starter New Member

Sep 7, 2010
22
0
My mistake it should be Q+ = (J . |Q) + (|K . Q)

So to confirm will the circuit look like this? With Q as the output and J,K as the input?

6. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
Still no...

Are you sure you 're making this K-map?

I suggest a revision on your K-maps.

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7. ### wind_blast942 Thread Starter New Member

Sep 7, 2010
22
0
But doesn't this K-Map give: (J . |Q) + (|K . Q) ?

8. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
Yes, it does. My bad.
F=JQ'+K'Q is the correct answer.