Now when we already know how to
design a control logic for a 2 switch input let us proceed further now taking 3
input switches and solving some questions on them;
Ex 1:- Given three switches
I1, I2, I3 to control the O/P, it is mandatory
that all the three switches are pressed together to activate the O/P.
I1
|
I2
|
I3
|
O/P
|
0
|
0
|
0
|
0
|
0
|
0
|
1
|
0
|
0
|
1
|
0
|
0
|
0
|
1
|
1
|
0
|
1
|
0
|
0
|
0
|
1
|
0
|
1
|
0
|
1
|
1
|
0
|
0
|
1
|
1
|
1
|
1
|
Let us check out the
question first so that we can pull out a truth table from it which will tell us
that on what condition of the switches the output was ON or the relay coil was
ON.
Just like a general truth
table we put the inputs on one side and output on another one.
Each input will have only
two conditions but when connected to each other that might raise upto 8
Now keeping in mind,
1 = pressed
0 = released (no action)
We
need to find out on which situation the output turn ON(high).
Recalling the question we
find out a specific line “all the three switches are pressed together
to activate the O/P” that
verifies that the output will turn ON when all the switches (inputs) are
pressed(1).
1
|
1
|
1
|
1
|
Hence forth the output
will turn on when I1 = 1, I2 = 1,
and I3
= 1.
Checking
out the status of the switches and will give us there types and as the situation
in which the output turns ON is when I1 is pressed(1) and
I2 is pressed(1), and I3 is pressed(1),
it
provides us the relation between the inputs. i.e. AND gate.
Hence forth the Boolean equation can be
deduced, i.e.
I1 . I2 . I3 = O/P
Now
let’s pen down the relay logic diagram for the given Boolean equation that is
easy just keep in mind the simple logics that we have created.
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