Realizing Full Adder using NAND Gates only
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- Опубликовано: 20 окт 2014
- Digital Electronics: Realizing Full Adder using NAND Gates only.
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Thanks Neso You Simplified This ❤️
thank u neso. i was struggling with this.
superb and simple explanation..
u r the best man!
Thank you so much
sir u r just awesome!!! Thank u sooooo much :)
Thanks for the video
What is the need to implement full adder using nand gate, when we can easily implement it using EX-OR gate
maybe because they're universal gates-
Thank You
Thank u so much sir🙏🙏🙏
QUITE SATISFACTORY EXPLANTION THANK U VERY MUCH
Waste
Sir did you made lectures on series and parallel adders..please reply
Could you explain full adder using NOR gate only??
God help me
keep it up Bro....
Thanks a lot ❤❤
Thank you good explanation
Sir pls add a video of full adder using NOR gates if possible
Proper explanation
Thankyou sir ❤️
Osm bro....👌👌
where is full adder by nor gate?
Thanks a lot
Thanks Sir ☺️☺️
Plz explain full adder using nor gate
thanks
Impressive
How can a nand make the sum? I mean how do you arrived to the fact [(A*B)(A+B)Ci] = (A*B)+Ci(A+B)?
COUT = A B + A CIN + B CIN
= A B + A CIN + B CIN (A + A)
= A B + A CIN + A B CIN + A B CIN
= A B (1 + CIN)+ A CIN + A B CIN
= A B + A CIN + A B CIN
= A B + A CIN (B + B) + A B CIN
= A B + A B CIN + A B CIN + A B CIN
= A B (1 + CIN )+ CIN (A B + A B)
= A B + CIN (A B + A B)
= A B + CIN (A Ex-OR B)
Therefore, COUT = A B + CIN (A ⊕ B)
I have same question sir please reply ?
[(A*B)' . ( (A xor B)*C ) ' ] '
Using De-Morgans law ,
A*B + (A XOR B)*C
Which is the Carry of Full Adder
It’s a nand gate bro ,first sir did the multiplication of the two input and then complimented the both the input
De Morgans
saved my grade, thx
Expression plz
super sir❤
Why we use and gate at last why not just or... To add them ....
Coz we have both of them why did we just not or them
Can we design it more easily by giving input A' with B, B' with A??
It will form a different circuit diagram but the output is same🤔
no , we only have inputs and nand gate. In order to get A" we need to use NOT gate which we can't use(not given)
i need proof for carry from nand gates... ? will the expression included above is same as AB + BC + CA ??
isn't the carry output supposed to be AB'+BC'+CA' ?
correct me if I'm wrong please.
it is supposed to be A.B + B.Ci + A.Ci which is same as A.B + Ci(A XOR B). Hope that helped.
U r right, actually these two expressions are equivalent. It depends on convenience.
@@NakshatraSaxena Hi. I know it's three years late.😂 But please let me know how AB+BCi+ACi is equivalent to AB+Ci(A xor B)? Isn't it supposed to be AB+Ci(A or B)??? What am I missing?
@@AnoNymous-po5sx It means that carry is only generated whenever either a and b are 's1 means AB or either A or B are once means A Xor B and c is 1 means C(A xor B).
So ans is AB + C(A xor B)
How do you get the idea of xor gate to be like that? If iam doing it iam getting one gate extra everywhere.iam writing the expression using and,or gates then converting them using nand gates.is there any other way? Or just we muat remember?
It's actually an optimised version.Needs some practice to understand!
@@nishanth1828 I am having assignments on implementing function using only universal gates! Where can I read some examples of these?
Maybe this link gives you a bit of confidence while implementing xor gate.
Are A and B two 8 bit numbers?
No. They are 1 bit numbers
Why did you don't do Full adder using NOR gate
Bhai jbh figure ka ss lena hota aapki video ajati hai end mein
kuch samjh nhi aaya
❣️❣️
where's the goddamn boolean expression of HA using nand gate???
Akshay Das M.K khud b pdh liya kro kuch to 😝😂😂
In the title, using the word "Gate" in plural form ("Gates") would have been more clear.
Full Adder NOR Gate Please 🥹