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  #1  
Old 11-19-2010, 05:28 AM
wind_blast942 wind_blast942 is offline
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Default Convert a D flip flop into JK

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?
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Old 11-19-2010, 08:27 AM
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Georacer Georacer is offline
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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.
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Old 11-19-2010, 08:53 AM
wind_blast942 wind_blast942 is offline
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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?

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Old 11-19-2010, 11:36 AM
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Maybe you need to check again your Karnaugh map solution. Try it once more.
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Old 11-19-2010, 11:49 AM
wind_blast942 wind_blast942 is offline
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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?

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Old 11-19-2010, 12:27 PM
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Still no...

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

I suggest a revision on your K-maps.
Attached Images
File Type: png K-Map.png (17.0 KB, 62 views)
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Old 11-19-2010, 01:12 PM
wind_blast942 wind_blast942 is offline
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But doesn't this K-Map give: (J . |Q) + (|K . Q) ?
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Old 11-19-2010, 03:38 PM
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Yes, it does. My bad.
F=JQ'+K'Q is the correct answer.
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