Full Adder using Half Adder
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- Опубликовано: 19 окт 2015
- Digital Electronics: Full Adder using Half Adder
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SIR I HAVE COMPLETED LECTURE 103 OUT OF 202. TREMENDOUS LECTURE AND THANKYOU FOR YOUR GUIDANCE !!
this video saved my life, thanks a lot sir! you deserver a medal for your explanation!
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Alas this channel has less subscribers than the gaming or other type of channels. This channel is really helping us
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.
Thank you sir your explanation is so good 🙏
What is the use of Ci in full adder? Does it change from 0-1 automatically?
U saved my 10 Marks thank u
2024! Yet It's like quote which suits for all time.
Very well presented and useful video.
Your channel must compete Khan Academy you have been extreemly helpful :)
Sir you mentioned c input as the carry from the previous sum.
Which previous sum are you talking about.
we use a carry as a input in full adder then how it is a combinational ckt ?? it depends on pervious carry so it should be a sequential ckt
thank you , beautiful work
bro you are doing a very great job
god bless you
Thank you. I learned a lot.
Crystal clear explanation sir 🙏
Afternoon my exam... Tnk god I got ur video 😊 ty sir
thank you sir
very well explained
Great.. Thank you so much.. Such a clarity
TQ sir u r presentation was awesome 🤩🤩🤩🤩
this is really awesome you guy's doing great job seriously😘
bruh did ur meth business get ruined, why r u watching a lecture on digital electronics smh
sir i wanna know that why u add another cin becz we transfer carry from previous aader to next adder
I have the same doubt
thank you Sir You are presentation is good
How is Co2 equal to Cin ( A EXOR B) in the diagram where this is your input to OR gate ?
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).
In love with half adder 😻😻
the theme music sounds a bit like Breaking bad theme. haha. that's awesome! and so are your videos. thank you!
+Raj Kapadia Thanks man !!
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Chintan Meena thanks, why?
Not at all, but it's not bad.
Well, no. But kind of.
It sounds bike race theme, not noticed any1 pointing it out loll
This video made my life easy tq ❤️
My favourite teacher ❤ aapke jaise teacher ho to har bachha topper ho🙏🏻
It's more easy to remmember like this carry of first half adder + carry of second half adder
how fluently u explain the things that is something amazaing
Prashant singh 5:39 Which law is this How can he neglect Xcomplement??
A+(B.C)=(A+B)(A+C)
So similarly,
X+(X'Y)=(X+X')(X+Y)
=(1)(X+Y)
=X+Y. Ans.
Kisika job laga kya?
@@ankitchaturvedi8802 tera laga kya
@@user-qt4ch7uj9z no
Thank you so much sir 😊❤
Hii sir ur videos are more helping to me so I want sum of full adder by using two half adders
Great teacher
The god of teaching
Thank you so much.
Thank you sir
🙏💕Thanks sir this helped me a lot
in which video did you cover 5:46 X+X'Y=X+Y?
if full adder accepts the previous carry then does full adder falls under sequential circuits?
very helpful...thanks
Sir Can You Explain How Can I determine How many HA are needed to Form a 4 bit Parallel Full Adder?
A HA adds two bits
A 4 bit Parallel Adder adds two 4 bit numbers
So a 4 bit Parallel Adder needs 4 HA
thank you
what is the software you are using for lectures to explain sir
Sir , for 4 bit full adder , how many half adder and OR gates required ??
Plz help to solve this qun..🙏
soo good sir
U r just ausome sir 👌.....
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Ur voice is atteractive and ur expilnation
Love you guys ❤
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.
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)..??
The rule is to use the least number of gates. The rule is bent so arrive at a different boolean function
thank u brother
@ 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)
Did u got the answer?
Excellent..
amazing video
Thankss !! :)
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
Can someone explain how the expression of Carry was changed like it was AB+BCin+ACin
Amazing!
instead of giving carry inputs to a 2 input OR gate ...can"t we use another half adder for carry
It is cheaper to just use an or gate rather than an entire half adder
Why in the SR latch when s=0,r=0 we first assume that q complement as 0 or1....
Sir how to obtain how many half adder we'll use?
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
Sir I'm facing problem in making circuit how can I improve it?
Thanks
god bless you man
Sir you have a team for all this study video or you make you do editing and all that things
And please make video on differentiation and interigation with basic
1:24 - why isn't there a proof of this expression being equivalent to AB+BC+CA ?
what there to prove it,its so much easy
@@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 + ?
@@TheViorelFlorea If you express the Function AB+BC+CA in SOP form, it will give you the same result as AB + C (A ㊉ B)
easy to learn
thanks sir
5:39 Which Law is this???
Carry of half adder is A.B then how would we write it A XOR B for Second HA
For the second adder, your inputs are Cin AND A xor B
What is c_in??? From where it is coming???
it's just an input . it's refers to as you got it as a carry from a previous operation
Very Nice explaination sir !!!
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)
Bcz we dont have cin in half adder
Sir why should we neglect the A' & B'?
Cin and C01 are equal?
Yup!! The Cout of the first half adder turns to Cin of the second half adder
why you didn't take the Cout of the first H.A as a Cin of the second H.A rather than bring new Cin??
Why we use 2 half adders to create a full adder?
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)
apply the distributive law to that X+X'Y = (X+X')(X+Y)=X+Y since X+X'=1
varsha Reddy Thanks So Much I was searching for this only
v sai rahul vaddadi Absorption Law is X+XY =X
@@varshanireddypatlolla4887 thanx
Reping Kwame Tech 😂😂😂…Dr Nunoo Dey kill we .
Excellent ❤❤❤
thanksssss
why did we use an OR gate?
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
god bless India
is full adder also a single bit adder..?
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.
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*********************** 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
Tnx