It would be great if you solve one or two sample questions to see how these concepts actually work. Thank you. That would earn you at least 50k more subscribers for sure because that is more convenient.
At 1:34 you have done some mistake as a Xor gate can never be the same as an Or gate, But you have written two expressions with only a difference of an Or gate converting into a Xor gate just because of simplification. That seems confusing.
@ Neso Academy You said in lecture that we can't accept or use Cin(carry of previous sum) in first Half Adder(HA1) then why you use carry of previous sum (Cin) in second half Adder (second half adder is also a half adder then why we use carry of previous sum (Cin) in second half adder)
You said in lecture that we can't accept or use Cin(carry of previous sum) in first Half Adder(HA1) then why you use carry of previous sum (Cin) in second half Adder (second half adder is also a half adder then why we use carry of previous sum (Cin) in second half adder)
Carry of a half adder having inputs X and Y is X.Y ie X and Y In that particular Case, one of the input in HA2 is Ci and the other one is the output of HA1 which is nothing but A exor B. So the carry of HA2 will be (A exor B).(Ci).
Sir, everything is fine except a small mistake is that X+X'Y=X+Y is called as absorption law and not distributive law . distributive law is X+Y.Z=(X+Y)(X+Z)
@@lavishgarg4274 No it's not simply to understand. AB + BC + CA = AB + C (A+B) In the video, he wrote AB + BC + CA = AB + C (A ㊉ B) So... why are they equivalent..since ㊉ does not produce the same output as + ?
are you saying a full adder has 2 XOR gates and 2 AND gates because its made up of 2 half adders and they are connected with an OR PLEASE DRAW A FULL CIRCUTS AND LOCATE THE TWO HALF ADDERS
Full adder is combinational or seq. Circuit? Is it seq ,where is the memory element? Carry iput is taken from where? I am so confused. Pls give me explanation
Thankyou so much sir this vedio is very clear and i got it but i want to know how to impliment full sabtracter using tow half sabtracter i found this type of vedio in your vedio list but i d'nt get any vedio in this topic sir tell me solution of my problem please
by breaking ur own taught rules about k map and implementing carry equation of full adder as "A.B+ Ci (A xor B)" u made me little dubious about k map rule..are we allowed to do what u did in full adder carry k map (selecting single one even though we can select pair)..??
Asma Rahim Ali Jafri To obtain the equation of A.B+Cin(A®B) we had A.B from first half adder and Cin (A®B) from second half adder if add those we get required carry output so we used that OR gate
*********************** DOUBT HERE ******************************* sir y dont we use Co = A.B + ( A+B ) Cin INSTEAD of Co = A.B + ( A XOR B ) Cin AS shown in 6:39
Half adders are single bit adders, but full adders are really just many single bit adders tied together. If I am adding 2 numbers that each have 3 bits, that means I need to do 3 separate 1 bit additions: 110 + 100 = (1+1)+(1+0)+(0+0), where (1+1) is adding the MSB, etc. Because you need 3 separate additions, you will need to make a full adder using 3 half adders, allowing you to complete three 1-bit additions.
