Priority Encoder Video Lecture | Analog and Digital Electronics - Electrical Engineering (EE)

in the last presentation we saw the
basic of the encoders and decoders and I
also told you that we have to study the
four types of encoder in which the first
one was the priority encoder so in this
presentation we are going to complete
the priority encoder you can see a
priority encoder on the right-hand side
of your screen in which there are four
inputs a 0 i1 i2 and i3 and definitely
two outputs y1 and y2 because n is equal
to 4 and we already know to find the
value of output we have to make it equal
to 2 to the power M so solving this M is
equal to 2 fine
now what we have to actually do in this
presentation is to find out this
circuitry inside this block ok and that
we will do by first creating the truth
table for the priority encoder and then
by using the K map we will find out the
logical expression and the
implementation part will be over you you
have to do the implementation part
because it's not very tough thing to do
at this stage of time before moving to
the truth table let us first set the
priority for the inputs I assume a 0 is
having the lowest priority and the i3 on
the other hand is having the highest
priority so I can say that from moving
from top to bottom from eysie row to i3
the priority increases okay the priority
increases when I come from up to down it
means eysie row is having the lowest
priority then I 1
then I 2 then I 3 what actually we have
to do by setting the priority is that if
I 3 is 1 hi let's say then we don't have
to check for eyes 0 I 1 I 2 we will just
set Y 1 equal to 1 and Y 2 equal to 1
because they reflect I 3 1 1 reflect I 3
similarly if I say I 3 0 I 2 is 0 and I
1 is 1 then we don't have to actually
look
for I 0 it is don't care for us whatever
the value of i0 we don't have to do
anything because the priority of a 0 is
less than i1 and the output will be zero
what so this is what we have to do in
the priority encoder so let's quickly
move to the truth table and the once we
are done with our truth table this thing
will be much clearer to you okay so
let's start with our first case when all
the four inputs are zero when all the
four inputs are zero then the outputs
will be simply don't care okay and when
I 3 is 0 I 2 is 0 I 1 is 0 and a 0 is
high which means 1 then Y is 1 is 0 y 0
is 0 very simple let's move to the other
case when I 3 is 0 I 2 is 0 I 1 is 1
it's the case that we just discussed
here this one in this we don't have to
see the value of i0 it simply don't care
if it is 0 doesn't mean anything to us
because the output will still reflect i1
if it is 1 in in that case also the
output will be I 1 so it's don't care
for us and the output will be 0 1 let's
move to the next case in which I 3 is 0
I 2 is 1 again I 2 is having the higher
priority as compared to I 1 and I is 0
so we don't have to look for the values
of i1 and isay Rho because of the
priority of the i2 and output will be 1
0 let's move to the another case in
which I 3 is 1 and i3 is having the
highest priority so again we don't have
to look for the values of I 2 I 1 or a 0
they will be simply don't care to us
whatever their value the priority is
given to i3 and the output will reflect
the i3 so 1 1 ok
so this is a small truth table for the
priority encoder and I hope you have
just learned what is the central idea
for the priority encoder now the next
step is to find out the logical
expression for these two functions y 1
and y 0 and as you can see here there
are
for inputs involved so it is very
obvious that we are going to use a 16
cell km app and also to such key maps
will be required because we are having
two functions y 1 and y 0 so let's move
to it I have already made a 16 cell K
map so the only thing we need to do is
to fill this map ok and you already know
how to do this so let's do it together
ok so the inputs are I 3 I 2 I 1 I 0 so
let's start filling it for the first
cell we are having a don't care
so cross for the second one we are
having a 0 ok and again we are having 0
for this 2 cells because in this we are
having AI 1 as 1 and you can see that I
1 is 1 for this two cells and in that
case we are having Y 1 is 0 very simple
in the next case we are having AI 2 as 1
ok and when I 2 is 1 and the I won and I
0 are the don't cares we are having Y 1
is 1 so that's see for what case is I 2
is 1 I 2 is 1
for this two entire rows you can see
this is your eye - and it is 1 for this
2 rows so I will write one real quick
without any question because it's
clearly mentioned here that output is 1
when I 2 is 1 similarly for the last
case also whatever it be the value of I
2 and I 1 I 0 if I 3 is 1 Y 1 is going
to be 1 so we can see that for this 2
rows I 3 is 1 so we have already put 1
here in this row the only thing left is
to put 1 in all the cells of the last
row it's a very simple key map to do and
let's start grouping it this is our
first group ok and this one is our
second group fine let's ride the
minimized function this is for y 1 so Y
1 is equal to what
let's see for this first group of ones
and let's say this is our second group
of one so for first group of one we are
having simply i2 because all I 1 I 0 is
changing from 0 to 1 and i3 is also
changing from 0 to 1 the only thing
that's not changing is i2 so I will
write I 2 for the first group or for the
second group in the same way i 3 is not
changing its 1 whereas all the three I 2
I 1 and I is 0 are changing from 0 to 1
so I will write i3 so this is your first
function ok y1 we have just found out
let's paste our another key map my 16
cell key map to find out the value of y
0 so let's see for the first cell we are
having a don't here you can match from
the truth table and for the second cell
we are having a 0 okay you can see in
this case it's 0 and when I 1 is 1 which
means for these two cells y 0 is 1 so 1
1 okay and when I 2 is 1 okay for these
2 rows I 2 is 1 the value of y 0 is 0 so
0 0 0 and again 0 for this row and for
the last case when I 3 is 1 the output
is 1 so I will put 1 1 1 1 okay and one
more important thing that you can see
here in this game app is for this I 2
okay if I see for I 2 and it is 1 which
means this row we have to see and this
row we have to see we have put 0 but
when I see I 3 which means again this
row and this row it's 1 so there is a
contradiction in this room it's 1 and 0
both so I will take it as I don't care
for us okay
and we already know how to operate with
the don't cares if it is I don't care I
can take it as one so let's start making
the group I hope you are getting these
things I'm assuming that you know how to
solve the k-map okay so this is my first
group and I have combined these four
ones the two ones are still left so
let's make in the second group this is
my second group this is my first group
this is my second group okay so let's
start writing down the minimized
function y 0 is equal to 4 the first
group I'm having Y 3 with no doubt we
have already done in this case and for
the second group
let's see what's changing and what's not
for the second group you can see that I
1 is 1 and it's 2 1 it's not changing in
this column where is I 0 changing from 1
to 0 so we will neglect eyes ero and I
will write I 1 and as it is a group of
four walls we will eliminate 2 and still
we are left with 2 so y1 is our first
variable let's check for the second out
of Y 3 and I 2 in i3 and I - if I see
for this row and for this row I 2 is not
changing its 0 and again is 0 whereas I
3 changing from 0 to 1 so we will
neglect I 3 and I will write I to
complement because I 2 is down I 2 is
low 0 so this is what we got
after the simplification using the k-map
for the function y 0 now these are the
two function that you got and you can
easily implement them by using the or
and and gates and whatever the circuit
you got will go inside this priority
encoder so this is all for this
presentation a very simple topic and yet
a very important topic there is a great
chances of this question being asked in
your exam so enjoy it and see you in the
next presentation