Introduction to Multiplexers Video Lecture | Analog and Digital Electronics - Electrical Engineering (EE)

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hello friends my name is Nico Chaudhary
I'm assistant professor at biani girls
college today I am going to deliver my
lecture on multiplexers multiplexer it
is a special combinational circuit we
also call it as Marx okay and it is also
called as MPX so it is a combinational
circuit a combinational circuit is a
combination of different gates here we
have arrangement of different gates they
can be an gate or gate and from the LHS
we have input and RHS we have the output
ok multiplexer is defined as many to 1
that means it can have many inputs but
only single output that's why it is
called many to 1 it is a device which
has many input and single output and
which input will go to the output is
dependent on the selection or select
line here we have select lines okay
so if you give any input it will depend
on these select lines that wait out
which output will you will get so here I
have an example of 2 cross 1 MUX so 2
cross 1 marks in this we have 2 inputs
and 1 output and number of selection
line can be detected by using the
formula 2 to the power m equal to n here
n is the number of input lines M is the
number of select lines so if we have two
inputs then you can see you will get one
select line so by this way you can
calculate for cross 1 8 cross 1 16 cross
1 MUX so next I am going to discuss 4
cross 1 multiplex or cross 1 multiplexer
as the name itself suggests it will have
4 inputs and 1 output so for cross 1 MUX
in this we have 4 inputs that is I not
i1 i2 n i3 and 1 output Y so according
to the formula you can calculate number
of select lines okay 2 to the power M is
equal to n n is number of input lines M
is number of out sorry select lines so
if we have 4 inputs I naught I have an
I 2 and I 3 so number of select lines
will be 2 - the part 2 will give 4 so
select lines will be 2 so here I am
using s 1 and s not so truth table for
this is s 1 s not Y ok s 1 and s not
will take it as inputs in the truth
table and Y is the output so as I told
earlier that input and output will be
dependent on the selection lines so if
you have 0 0 then you'll get I not 0 1 I
1 1 0 I 2 and 1 1 you'll get I 3 so you
will write it soft form so as we know in
soft form 1 is considered as a and 0 is
considered as a bar okay so soft form
for this is using the truth table s 1
and s naught both are 0 so it will be s
1 bar s naught bar I naught plus as I am
using the soft form so it is this sum of
products so it is a product now some
next is 0 1 so it will be s 1 bar s not
i1 plus similarly we have next s 1 s
naught bar I 2 plus both are 1 so it
will be s 1 s naught I think okay so
this is the software next step is to
draw the logic diagram so logic diagram
is as follows you can see the soft form
it is we are using s 1 s not there bar
and simple also so you need s 1 and s 2
and 1 not gate for s 1 and 1 not gate
for s not so they will convert to s 1
bar and here it is s naught again we
need the product of these so for the
product weenies and gates and then sum
of all these so for the sum you take an
or gate okay so I naught here I naught s
1 bar and s naught
so I naught s 1 bar s naught it will
give s 1 bar s naught
I not all the four terms you can get
from this and from the or gate you get
the summation of all these force so this
total is the four cross one multiplexer
using this formula we can calculate
three oh sorry
eight cross one sixteen cross 1
multiplexers also and you can draw this
logic diagram so this completes the
whole question that what is a
multiplexer and what is 4 cross 1
multiplexer and draw its logic diagram
it is used in MSC and for BCA students
also so now we are going to do the
questions what kind of questions do you
get in your net or gate exams for the
multiplexer specially for cross 1
multiplexers like in the question you
have given a 4 cross 1 multiplexers and
inputs are given I naught value is given
1 I 1 is 0 I 2 is 0 and I 3 is 1 and we
have selection line B and a and you have
to calculate the output so how you are
going to solve this question so first of
all from the truth table you will write
the swap expression okay I am going to
write this OP expression y equals 2 as
we have studied earlier swap expression
for this will be B bar a bar I naught
plus B bar a I 1 plus B a bar I 2 plus B
a I 3 ok this is the equation for the
soft form now we have the values for the
I not i1 i2 and i3 just put these values
B bar a bar dot I 1 is 1 plus B bar a I
1 is 0 plus B a bar I 2 is also given as
0 plus B a dot 1 as I 3 is 1 so this
value will be 0 anything multiplied with
0 will be 0 obviously so we are left
with B bar a bar plus B
so this is the equation for the eggs
nugget be ignore so your output will be
your Y value will be B X not a as it is
the value for the X nor gate so in the
previous example we have seen that we
have input 1 0 0 & 1 and you if you
remember that this is the output for the
X nugget so that's what we get X naught
B X naught a similarly if you will get
the input like 1 0 0 1 1 and 0 so that
is the output for the XOR gate so there
we will get B X sorry
ok let's try one more question and here
we have the inputs 0 0 0 & 1 so again
using the steps first write the saw form
okay so B bar a bar I naught plus B bar
a I 1 plus ba ba r i2 plus B a I 3 okay
put the values I not a 0 I 1 is 0 and I
2 is 0 so these terms will be 0 you left
with only term be a dot 1 so your output
will be B or you can write it as a B so
it is a dot B so by this method you can
calculate the value of the Y that is the
output if number of inputs are given and
using the selection lines so you have to
remember the truth table and the logic
diagram also and this diagram oh thank
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