Carry Lookahead Adder I Video Lecture | Digital Electronics - Electrical Engineering (EE)

in this class we are going to discuss
about carry look ahead adder which can
be designed from 4 full adders here we
are going with 4-bit addition only as I
told you in the last classes by using n
bit parallel adder the delay is very
high to reduce it we designed carry look
ahead adder to reduce that delay so now
we will see the procedure here without
using any carries from previous adder
previous full adder we are going to Dale
can so see here so this is the circuit
for carry look ahead adder here you can
see combinational circuit for carry
generation so this circuit always uses
first carry only so that is why no need
of carries from previous adders we are
just using faster carry from this fast
adder so always this adder is a LSB
added that means least significant bit
adder so here this is a must be position
added that means whatever sum here we
got thus that bit is a most significant
bit so here I am taking the example for
this before that so whatever sum is
there for any full adder that sum is
equal to XR b XR C this C is nothing but
C in so that means input carry so
whatever carry that we need to get is CI
plus 1 that is equal into P i CI + GA so
here this P I is nothing but a carry
propagate so that means a i XR be like
this we already discussed in last class
about carry look ahead adder so this GI
is nothing but carry generate so this
carry generate is equal into AI into GA
so B I so here I am taking an example so
that is I am going to add 1 1 0 1 with 1
0 1 1 these are the two 4-bit 4-bit
numbers I am going to add so this is
equivalent to a this I am going to
consider as be so now this big t is
equal into a 1 B 1 so a 1 e 2 e 3 and a
4 I am considering like this and
similarly so this is equivalent to b1 b2
b3 and b4 like this I am considering so
here I am NOT considering a naught if
you want to consider you can consider
that a naught also so now my example is
a is a 4-bit number that 4-bit number I
am considering like W 1 0 1 and Dez
4-bit number that is equal into 1 0 1 1
these two bits I am going to add now I
am going to place the a a values and B
values in this circuit diagram now II 1
B 1 is equal into 1 1 I am going to
place that a 1 B 1 here and now so 0 1 I
am going to place here so 0 1 and then 1
0 so here is 1 0 and last input for this
circuit is 1 1 so 2 hi lines we are
giving so two high voltage lines we are
giving here so now this is the time to
get carry from this combinational
circuit so you can see this
combinational circuit is started from
second full adder onwards for first full
adder we are not having any input carry
or input carry flog is equal into 0 here
you can see I grounded this so whatever
carry we are getting here this carry is
internally connected to all the
combinations or cubes in carry look
ahead adder now we need to define the
value of c1 here so as we know that full
adders sum is equivalent to X are of
inputs and carry is equal into
multiplication and addition of that
inputs so now so here a 1 is equal to 1
B 1 is equal to 1 C naught or carry
input is equal into 0
Shelly as we connected to ground so now
the sum is equal into XOR of all these
three inputs so see these three inputs
here you can observe to number of ones
as XOR gate easy or one detector so the
output here is equal in to 0 that means
X RBX RC is equivalent to 0 why because
you are observing to number of one's is
nothing but even number of ones
so X are gate always going to provide
zero when observes even number of ones
now you are having your carry so this
carry is equaling to a B plus BC plus C
so in all cases at least one case is
going to provide one means obviously the
carry is equal to one so here in a B
case you are getting one into one case
that means you are getting one so
obviously without taking CA or BC values
you are going to get this carry as one
so you know that full adder carry is
equivalent to a B plus BC plus C so in
that addition a B value is equal into
one so that is enough to take C 1 as 1
so now with this C 1 reference we are
going to deal the remaining carries by
using this combinational circuits now
for this combinational circuit what may
be C 2 here so here you are having input
from these combinations are cute as
carrying so here you need to discuss
about C 2 so you just place one in this
CI plus 1 you are going to get C 2 here
so see here I am going to deal with C 2
so C 2 is equal into I am just placing
one in I place so that is equal into P 1
C 1
sg-1 so this is nothing but say - that
means output from this combinational
circuit but we know that P I is equal
equal into AI x RB I so p1 is equal into
even XR beaver so this is P 1 and G 1 is
equal to a 1 into B 1 as GI is equal to
AI into bi and taking G 1 is equal to a
1 into B 1 so now see here you know the
values of a1 and b1 so previously you
are having a 1 and B 1 that values you
need to connect in this combinational
circuit so I am NOT taking that
connection in this circuit I am just
representing that we are having a
connection here so to take out the
complexity of this are cute I am NOT
representing in this diagram but
internally we are having connection from
a1 to here that is why it represented a
1 here so for that combinational circuit
which is providing this second carry so
that is having connection with both a1
and b1 and c1 from here so now see the
c2 value here so when you are exerting
this a1 and b1 1 XR 1 is always equaling
to 0 so p1 value is equal into 0 into C
1 means here you are having c1 plus g1
means a 1 into B 1 so a 1 into B 1 means
again is equal into 1 so as a 1 B 1
values are 1 1 you are getting 1 here
now 0 into 1 you are getting 0 plus 1
means again you are getting 1 whatever
it may be here you are getting C 2 is
equal to 1 from this combinational
circuit so now after getting 1 from this
combinational circuit what may be the
output of this full adder now again for
some you are getting XR of all inputs so
if you take in
Sahra for input see here 0 1 1 or given
as input for this second full adder so
this full adder to get the sum if you
take in XR of all those inputs again you
are getting 0 why because here even
number of ones are present so again is
to place you are getting 0 so now come
to third step no need to get carry here
why because no need of that carry to
give this for this full adder because we
are not taking reference of previous
carry we are generating ourself for this
full adder so now I am generating carry
for this full adder by using second
combinational circuit so here to get
this c3 so I am just placing I value as
2 so here I am placing I value as 2 and
P 2 C 2 plus G 2 I am getting so P 2 G 2
plus C 2 means so C 3 is equal to P 2 C
2 plus G 2 so now here this is equaling
2 P 2 in place of C 2 as we know that so
P 1 C 1 plus G 1 we can place this plus
G 2 I am writing here now the value is
equal to P 1 P 2 C 1 plus P 2 G 1 plus G
2 this is the value of C 3 now we are
going to take that value by using given
values so now P 1 P 2 I am going to take
P 1 is nothing but a 1 X of B 1 a 1 X R
B 1 is equivalent to 0 so if anything in
this
is zero total value in this is
equivalent to zero so now coming to P 2
so P 2 is nothing but a 2 XR b 2 so 0 XR
1 is equivalent to always 1 but when
coming to g 1 you are going to multiply
a 1 into B 1 so G 1 is always
multiplication of a and B P is always XR
of a and B so reference is added to the
a and B values so here G 1 means a 1
into B 1 so 1 into 1 means here you are
taking 1 as I told you in addition any
one value is equivalent to 1 means total
value you are getting 1 so see here this
multiplication term is 0 but this
multiplication term is going to provide
1 so in these three terms if any one is
equivalent to 1 then total value is
going to be 1 but according to boolean
algebra theorems so now according to
this formula we generated C 3 as 1 so
that means here you are having 1 from
this combinational circuit now see here
what may be the output of this full
adder so now the output of this full
adder is also equivalent to 0 because
you are observing two ones as input as X
our gate is a odd one detector if you
observed any even number of ones then
your some output is equivalent to 0 no
need to calculate carry here because
there is no use of that carry to give to
the next full adder now we are in the
final step so that is equivalent to
fourth carry that is C 4 now if you want
to derive for the carry what may be the
equation means here you need to take 3
in place of Phi so here I am taking T in
place of Phi so I am getting C 4 here so
P 4 and
P 3 and C 3 plus G 3 maybe the value of
this seafood as you know that P 3 is
equal in to some value equal into a 3x
ax B 3 now I am going to take the value
of C 3 here so C 3 is equal to I am
taking C 3 value again so that means P 2
C 2 plus G 2 as C 2 is equal into P 1 C
1 plus G 1 plus G 2 so this is
equivalent we know the C 3 value same C
3 value I am going to take here so that
means P 3 into c 3 value i am going to
consider here P 1 P 2 C 1 plus G 1 so
again you are having this multiplication
factor and G 2 so this total value I am
considering here see here so now I
multiplied inside and I considered this
G 2 total value I considered as C 3 and
plus G 3 you are having so what may be
the total value here means here you are
having P 1 P 2 C 1 and P 3 again I'm
multiplying the inside so here G 1 into
P 2 into P 3 so here G 2 and P 3 finally
you are having GT now you need to
consider the values for each and
everything now you know that P 1 that
means XRF a 1 and B 1 is equal into zero
if any one value in this multiplication
is zero total multiplication term is
going to be zero now come to P 2 and P 3
so now P 2 is equal in two so a 2 XR b 2
this value is 1
but coming to p3 so this is also
equivalent to one a 3 x RB 3 this is
also equal into one coming to G 1 so G 1
means so a 1 into B 1 so total all these
are ones
all these are ones means when we
multiplied three ones you are getting
one so if any one term in this addition
is equivalent one obviously you are
going to get c4 as one so now you are
getting your combinational circuit
output as one now see here to get is 4
that means final sum you need to take
XOR of all these three inputs so if you
observe three ones in a full adder input
you can take this sum as one because as
X our gate is a odd one detector here if
you observe odd number of ones
definitely the output of this s4 is
equivalent to one now this is the time
to calculate the carry now Kari is
equalling to a B plus BC let's see in
all cases you are getting one that means
so here carry is equal into one now so
in all the cases you are not depending
upon the previous carries you just
depend upon the first carry which is
generated by first full adder so by this
process you feel some you are taking
some delay but when you connected the
circuit when you switch on the secure
all the inputs are going in the same
time and you are getting output in the
same time because any carry output from
any combinational circuit is not
dependent upon the previous circuits
only dependent upon the first carry so
here will cross check the result now so
we take in two numbers addition a and B
so I am taking normal addition 1 plus 1
0 with carry 1 again 1 plus 1 you are
adding 0 with carry 1 again 1 plus 1
that means 0 with
one you are adding three ones in last
step so definitely are going to get one
with carry one you need to get one one
triple zero as output you can see carry
out always you need to take as another
big one one triple zero you want as your
circuit output this is the carry look
ahead adder by using different
combinational circuit four carry
generation