[0:05]We have already completed all the four types of flip flops and now this is the best time to move to another important topic that is the flip flop conversion. A very important topic for your exams. And what we have to actually do in the flip flop conversion? A flip flop is given to you like in this example JK flip flop is given to you. And you have to convert this JK flip flop to D flip flop. Okay, so a flip flop is given to you and you have to get another flip flop by a small changes in the given flip flop. So what are that small changes? We are going to determine this changes by this five steps and we will implement all these five steps in this example and we will try to convert the JK flip flop to D flip flop. At the end of the presentation, you will find that you can do flip flop conversion very easily. This is a very simple and very important topic. So let's start with it. Our first step is to identify the available and the required flip flop. So let's see in our example what is the available flip flop. The available flip flop. The available flip flop is JK flip flop. So JK is my available flip flop. And what is the required flip flop? What we want? We want D flip flop out of JK flip flop. So my required flip flop is D flip flop. So our first step is now over and it's a very important step because if you switch this things. If you write available flip flop as D flip flop and required flip flop as JK flip flop, everything is going to be changed. Okay, so your answer is 100% wrong in that case. That's why this step is important and you have to take care of this thing. Now the next step is to make characteristic table for required flip flop. That's why I told you to remember the characteristic table and the excitation table of every flip flop. So please open your notes if you have followed my lectures and wrote it down somewhere, then please open it and if you don't, then no problem, we are going to do it right away. So the characteristic table for the required flip flop, it means for D flip flop, we have to make and the inputs to the characteristic table are Qn and D. So I don't know if you remember or not, but I remember that the output Qn + 1 is same as D. So this is a simple characteristic table for us. Let's make it. 0 1 1 0 1 1. Now Qn + 1 is same as D. So it is 0 1 0 1. So our second step is now over. Let's move to our third step. In this, we have to make the excitation table for the available flip flop. So available flip flop is JK and now we need to make its excitation table. So first I'm going to make the excitation table here and then you can follow. Qn Qn + 1 and JK. 0 0 1 1 0 1 0 1. So the values for J and K are 0 1 cross cross cross cross 1 0. So it is way to remember the excitation table for JK flip flop. You can also derive it from the truth table. Now we are going to use it. So let's extend this table.
[3:29]And write here J K. Now when Qn is 0 and Qn + 1 is 0. You can see J is 0, K is don't care. So 0 don't care. When Qn is 0, Qn + 1 is 1, this case, J is 1, K is don't care. So 1 and then don't care. Similarly when Qn is 1, Qn + 1 is 0, J is don't care and K is 1. This case. So don't care and 1 and the last case is don't care 0. So now we have done with our step 3 and now we have to move to our step 4 in which we have to write the boolean expression for the available flip flop. So available flip flop is JK flip flop and we will make the K map for it. So there are four combinations, so definitely we are going to have the four cell K map. Or a two variable K map. So let's make it fast. We are going to require two such maps, one for J and one for K. Okay, and the inputs are definitely Qn and D. So Qn D Qn D. Let's say my first K map is for J and my second K map is for K. Let's fill this map. For J it is 0 1 don't care, don't care. So 0 1 don't care, don't care. And for K it is don't care, don't care 1 0.
[5:11]Fine. Now let's make the groups. Before that, let me write down the values here. 0 0 1 1 0 0 1 1. Now let's do for J first. And my first group is this one. And I have already combined a single one, so no need to make this group. So J is equal to what? Qn is changing from 0 to 1. So J is equal to D. Now let's write down the value for K. K and this is my group and in this you can see Qn is changing from 0 to 1, but D is 0. So K is equal to D complement. Very simple. So we are done with our fourth step in which we found out the values of J and K, the Boolean expression for the available flip flop. Now, let's move to our fifth step that is drawing the circuit and this is very important step. So let's move to it. We have to first draw the available flip flop. Because we are going to do small changes in the available flip flop to get the required flip flop. So the available flip flop is JK. And the two inputs are J K, this is my clock. And the two outputs are Qn Qn complement. So let's see what changes we have to do in this JK flip flop to get the D flip flop. You can see J is D. So if D is my input, so the J is going to be same and K is D complement. So we have to use a not gate here. And it will be given to K. So overall you can see we are having our D flip flop by using a JK flip flop. Only thing you need to do is to find out this expression by following all these five steps and then draw the circuit finally. So it's a very important and very simple presentation. You can do it by yourself. In the next presentation we are going to do some more flip flop conversion. So see you in the next one.
