Working of a basic sequential circuit

In this article we will explain how a basic sequential circuit works and how its output is dependent on the previous inputs.Consider the figure below

In the figure we have two NOR gate connected such that the output of one NOR gate becomes one of the input of the other gate. The two inputs R and S are the primary inputs and can be changed when we need. But this circuit shown above is not a practical one because it has no delay. In real world every nor gate will be associated with some delay . The diagram along with delay is shown in figure below.

Here the two gates have a delay of “d”. Their outputs x and y becomes X and Y after a delay of d. By the examination of equivalent circuit we found that the output x and y follow changes in input R,S,X and Y instantaneously. From this observation we write the combinatorial relationships as

X and Y are nothing but x and y delayed for some time. So we can write

X(t) =x(t-d) Y(t) =y(t-d)

We can represent x and y (eq.1) in K-maps as shown below.

For simplicity we combine the two k-map into one as shown below.

With K-map we cannot find the sequence of events ,but we can find out the state of the circuit at a particular time using K-Map. Ie we can determine what happens after a time delay d just by saying X becomes x and Y becomes y after a time delay d.

The circuit is stable when x= X and y=Y. In all other cases the circuit is unstable and further changes must be take place to make the circuit stable . In a K- map stable states can be identified by comparing the XY values of each row and the entries of xy in the same row.

Sequence of events or cycles by which the sequential switching circuit come to a stable state.

Consider that the circuit is in a unstable condition where RS= 10 and XY =11 (4^th column 3^rd row). We can see that the value of xy at this position 00 and is not equal to the value of XY. But XY assumes the value of xy(as it is given as feed back) after some time and now becomes 00. So our point of attention moves to 4^th column 1^st row . Here also its unstable since xy =01 and is not equal to XY. So after some time XY becomes 01 and our point of attention moves to 2^nd row 1^st column. Here xy = 01 which is equal to XY . This is a stable situation.

Proving that sequential switching circuit exhibit memory.

When the primary input RS has value 00 it can assume two stable states where XY=01(A) or 10 (B). But what was RS before it became 00. It could have been 01 or 10 from which they changed one digit to become 00.

Suppose that RS was equal to 01 and the circuit was in stable state C. We now make S=0 making RS=00 making us move to the first column to reach the stable state B.

Suppose RS was equal to to 10 and the circuit was in stable state E. We now make R=0 making RS=00 making us move to the first column to reach the stable state A.

Thus we clearly prove that sequential switching circuit ie it outputs depend on previous inputs.