# 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
26
0
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
5,151
1,266
The state table of a JK FF follows:
$\left[ \begin{array}{c|c|c||c}
J & K & Q_{\small{now}} & Q_{\small{next}} \\
\hline \\
\hline\\
0 &0& 0& 0\\ \hline\\
0 &0 &1 &1 \\ \hline\\
0 &1 &0 &0 \\ \hline \\
0 &1 &1 &0 \\ \hline \\
1 &0 &0 &1 \\ \hline \\
1 &0 &1 &1 \\ \hline \\
1 &1 &0 &1 \\ \hline \\
1 &1 &1 &0 \\ \hline
\end{array} \right]
$

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.

Sep 20, 2010
26
0
4. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
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}
Inputs-JKQ & \text{Next State of the JK-FF} & \text{Needed Input with a T-FF} \\ \hline \\
000 & 0 & 0\\
001 & 1 & 0\\
010 & 0 & 0\\
... & ... & ...
\end{array} \right]
$

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
26
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
5,151
1,266
Can you post it for the rest of the site to see?