Help building one flip flop using another?

Discussion in 'Homework Help' started by TripleDeuce, Dec 4, 2010.

  1. TripleDeuce

    Thread Starter New Member

    Sep 20, 2010
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    Can anyone show me the proper procedure to construct a flip flop using another flip flop? My Teaching Assistant gave us these steps you can use on any occasion but I cannot seem to apply it in this certain case.

    How do you build a JK flip flop using a Toggle Flip Flop with an Enable?

    The steps my Teaching Assistant said was to

    first construct a Transition Diagram (For the JK flip flop)
    Write down the Excitation Table for the Toggle
    construct a State Table
    make a K-map so you know what gates to connect with what inputs

    I am stuck at how to make the state table. Can someone show me what it would look like?

    I need to learn this before my exam Tuesday...
     
  2. Georacer

    Moderator

    Nov 25, 2009
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    1,266
    The state table of a JK FF follows:
    \left[ \begin{array}{c|c|c||c}<br />
J & K & Q_{\small{now}} & Q_{\small{next}} \\<br />
\hline \\<br />
\hline\\<br />
0 &0& 0& 0\\ \hline\\<br />
0 &0 &1 &1 \\ \hline\\<br />
0 &1 &0 &0 \\ \hline \\<br />
0 &1 &1 &0 \\ \hline \\<br />
1 &0 &0 &1 \\ \hline \\<br />
1 &0 &1 &1 \\ \hline \\<br />
1 &1 &0 &1 \\ \hline \\<br />
1 &1 &1 &0 \\ \hline <br />
\end{array} \right]<br />

    The 3 first columns describe its current state and the 4th the state it will end to depending on its current states.

    Have you written the excitation diagram for the T-FF? If so, then you know what are your inputs and your outputs. What you must do is use the T-FF and some gates to convert the T-FF into a JK-FF.

    A Karnaugh Map will help you at that. Use it to construct a boolean
    function that will have J,K and Qnow as inputs and Qnext as output.
    After you do that, use the excitation table of the T-FF to find what gates you must also intervene in order to go from the desired Qnext to the actual Qnext of the T-FF.

    It's not that long since I woke up and I 'm starvng too, so be cauitious about what I write. I 'll check back later to correct any mistakes or inaccuracies I have written.
     
  3. TripleDeuce

    Thread Starter New Member

    Sep 20, 2010
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  4. Georacer

    Moderator

    Nov 25, 2009
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    Think of the problem this way: The variables you have at hand are the J and K inputs of your circuit and the current state Q. What you must do is translate the inputs in order to achieve a JK-FF behaviour with a T-FF.

    \left[ \begin{array}{c|c|c}<br />
Inputs-JKQ & \text{Next State of the JK-FF} & \text{Needed Input with a T-FF} \\ \hline \\<br />
000 & 0 & 0\\<br />
001 & 1 & 0\\<br />
010 & 0 & 0\\<br />
... & ... & ...<br />
\end{array} \right]<br />

    What I have written above is a partial state table that will facilitate the solution of your problem. I will describe its logic that is applied on every row:
    My inputs are the J,K and Q signals. I see that I have Q=0 and J,K=0,0. That means that the virtual JK-FF must have Qnext=0 (second column). The actual T-FF must get an input T=0 in order to have Qnext=0 (third column).
    Do the above for all the possible inputs from 000 'till 111.

    That way you will have a truth table with input variable the JKQ and output variable the last column T. Solve it with a Karnaugh map and create the matching circuit.

    The Enable control can easilly be applied with an AND gate right before the T input of the T-FF.
     
  5. TripleDeuce

    Thread Starter New Member

    Sep 20, 2010
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    0
    At first I did not know where you got your next states from but then I see you used Characteristic Tables. In class we were just given the Characteristic Equations and the book is so bad that those tables weren't even provided.

    I have obtained the correct answer that was posted on my HW solutions. Thanks,
     
  6. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    Can you post it for the rest of the site to see?
     
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