NOR Gate as Universal Gate
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- Опубликовано: 18 мар 2016
- Digital Electronics: NOR Gate as Universal Gate
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Your explanation is awesome... Sometime I want to like the video and found that video is already liked by me.😊😊
😆
🤣🤣Joke
Then first time you wasn't listen the lecture clearly......
Then you should try taking notes on the first go itself
Thank you sir for giving great concept with no nonsense talks
1. 5
2. 7
#Neso Academy homework makes the concept concrete in the mind...
5&7
@pradeep kumar i dont understand h. W que plz explain me
Hw Problm 1 explained properly😊:
We use complement of A and B asitis as inputs for nor gate. And for other we use complement of C and D asitis as input for other nor gate. We use output of both nor gates as input of one more not gate and hence we get required result. :D
So, in total we use 5 nor gates.
What about the second question?
Sir, you are the best, thank you very much for your effort.
thank you for saving me the stress to learn these on my own
5 and 7
Great lecture :)
just amazing explanation......
Thanks a lot man. Really helpful
Thank u for explaining !
5, and 7.,very well explained sir thank you
Thanks for the video
VERY HELPFUL
Thank you very much sir...
Your teaching skills are too good.
The answers for h/w
1. 5
2. 7
someone can explain me why for Question 1 of the he it is 5 and not 7?
@@MrPotaatooo a nor a=a' and this a' nor with B gives (a'+b)'=a.b'
Then c nor c=c' and this c' nor with D gives (c'+d)'=c.d' then nor both the output I.e a.b' nor c.d' this will give the final output (a.b'+c.d')'=(a.b')'.(c.d')'=(a'+b).(c'+d)
My 2nd answer is also coming as 5.
@@MrPotaatooo 1st question can be rewritten as ((A'+B)' + (c' +d)')' . Do this and you will also get 5
Great lecture sir. Now I am confident that I can find minimum gates required to design any boolean circuit efficiently.
😉😇
thank you very much 🌸
thanks for making this video
Prerequisite for this chapter, you should know De-morgans law of boolean algebra.
sir ...what is the process to implement the given functions using minimum no. of nand gate
Thanks it help me a lot
if the -ve logic is used the diode gate will represents which gate ?
Great video
@Neso Academy do we have to get Paid Neso to get answer of homework problem?
Thank You Sir
Neso, your page always shows error. is the site under maintainence ?
Number of NOR gates
NOT-1
AND-3
OR-2
X-OR=5
X-NOR=4
NAND-4
If y=A.B.C.D.E.F......n variables then,
Min no of NOR gates=3(n-1)-k
If y=A+B+C+D+E+...+n variables
Min no.of nor gates=2(n-1)+k
Where n= total no.of variables
K= no.of complement variables
this is very helpful in 2019 :)
2022 too😂
thank u sir
Thanks a lot
PLZZ do something regarding this problem
H. W problems answers
1)5 NOR gates r required to get (A' +B). (C' + D)
2)7 NOR gate r required to get A. B + C. D'
Thanks Sir😃😃
could plz explain one example problem
Thanx sir ji
Your page show error on finding solution for these problems. I request u to fix it.
Yes same problem faced by me
"And check for it OUTPUT "--- i like way the you speak 💝😋
There is an error shows when I open solutions for digital electronics
great lectures sir! can u plz explain how you derive the form AB + (A+B)' from (A+(A+B)')*(B+(A+B)') at exactly 11:53?? i can't grasp it ...thank's in advance
Answer:A+B'
Bro it's distributive law a+(bc)=(a+b).(a+c)
Please add ppt on your website of digital electronics
🔴H/W problem 1:
✔️Ans: 5 NOR Gates are required.
..............................................................
🔴H/W problem 2:
✔️Ans: 7 NOR Gates are required.
How it is coming
Thank you so much. But how many gates we need to verify using NOR gates to prove it as a universal gate? Please reply asap cuz i have an exam
'or', 'and', 'not' Hope you scored well in your exam 5 years back😉
during driving 5.XNOR === how did you drive (A+(A+B)').(B+(A+B)') to (A+B)' +AB ?
by using the distributive propriety of addition
x1 + (x2 * x3) = (x1 + x2)(x1 + x3)
(A+B) is a common term for both, so it's like x1
I Love Neso Academy ❤️
Pls explain me the answer I am not getting 5
1. 5
2. 7
Ravi sodhi plz explain both the ans
Sir is it possible to know how many nor gate require by watching the x-nor gate I. E (ab+a'b')
6 nor gates are needed...
Actually 10 nor gates are there...but ignoring the redundant gates we get 6 gates as the final answer...
Q1 : 3 (2 i/p NOR gate)
Q2 : 3 (2 i/p NOR gate)
Totally different from everyone .... Please explain.... Dont write anything. To confuse everyone ..
Infinity x her first answer is right acc. To me my ans is also 3
I m not able to find 2 one
Plz hell shanmuga
@@infinityx9937 and yes just bcz uska ans different hai iska mtlb ye ni ki galt hai🙄
@@pakhisingh1470 please explain, how you got 3 for 1st question.
thank you sir ji thoda zoom
Q1-> 7 Gates
Q2-> 4 Gates by modifying the expression: ( ( A' + B' ) . ( C' + D ) )'
You will get 5 in q2
i got 5 for q1
5 & 7 resp.
how to implement NAND gate from NAND gate and NOR gate from NOR gate?
Bruh what ???!!!! 🧠📉
why mostly NAND gates are to be used in layout designing?
u should read about CMOS logic circuit, in NAND gates, it uses 4 transistor only for each gate, and it will be efficient to apply it on TTL IC 7400 series (common uses for this chapter) its not present in this video btw, cmiiw
coz its a universal gate
Sir your site is not working
someone can explain me why for Question 1 of the he it is 5 and not 7?
5&7
In this video xor needs 5 nor gate amd xnor needs 4 nor gate..is it correct.pls anyone clear my doubt
mera v yahi aa raha h, byt everyone has written 7 for 2nd question.
5 and 7
1) 5
2) 7
can explain me why for Question 1 of the he it is 5 and not 7?
@@MrPotaatooo There were total 9 nor gates used in answer of first question and 4 of them were redundant as they were serving as not gates and eliminating them doesn't effect the output. So the answer is 9-4 which is 5.
@@oggy107 thank you so much sir..now I got cleared 🙏🙏
1: You need 5
2: You need 3...... not 7 NAND gates because the 4 inverters are redundant
(X&Y) to one NAND and (W&Z) to another NAND and finally a NAND which has outputs of the 2 NANDS used earlier as i/p.
we are using NOR gates here
@@srinityapadma5125yes 😂
thank u brother
I think XOR + 1 NOR gate gives XNOR ...But you did the opposite .Correct me if I'm Wrong
same doubt
H.w problem answers
1). 5 NOR gate
2). 7 NOR gate
1) 5 gates
2) 4 gates (if we first take dual of expression and then solve)
How answer of 1st question is 5 ?
Is the second one a self dual function ?? I don't think so. So it's not possible with 4 nor gates. Normal simplification leads to 7 nor gates.
U r right.. It is self dual....
But how you combine (ab) and (c. d') .U need to put one extra gate to get the output expression. .It will be an OR gate.... But here is the condition we must only use nor gate ....so in this case there is no other way to get the output .
So we require 7 gates to get the output....
1)correct
2) 7 gates
You are Right!
Q2-> 4 Gates by modifying the expression: ( ( A' + B' ) . ( C' + D ) )'
🔥🔥🔥
1) 5 and 2) 7
Ans 1) 5
2) 5
5 nor gates
1st=5min gates
Detailed Electronics lectures available in Urdu language
FROM WHERE i will get video on morgan's law & other law like a'+a=1,a*a'=0.,etc..???
thaks
if anybody who knows how to solve hw problem please explain
🙏🙏
1.{A Nor1->A(COM)} B nor2
{C Nor3->c(com)} D nor4
combining nor 2 ,4---->nor5
2. A Nor1-->A(com),Bnor2-->B(comp) combine nor 1,2 obtain nor3 (AB), C Nor4-->c(comp),D combine c(comp),D by Nor5 we obtain (CDbar) .then combine Nor3,5 .we got(( AB+CDbar)whole bar) in NOR6 .to obtain result again NOR7 ----result AB+CDbar
ഗ്ഗ്ഗ്
@@brindhabrindha2884 wowwwwww what an explanation thanks 😇
answer
1) 5
2) 3
Mera ans first ka 3 aur second ka 8 aa rha....
First ka mere according mera correct hai......
second ka koi bata do kaise kru🙏🙏
i dont understand what demorgands law
1- 5
5 nor gates are minimum requirment......
Can someone explain how for second question it is 7
1 A ---> A' , B----> B' ( using nor as not) it needs two nor gates
A' nor B' = (A' + B') ' = A. B (demorgans law) [ 1 nor]
2 C ----> C' ( need one nor gate)
C' nor D = (C'+ D) ' =
C. D' ( demorgans law) [1 nor]
3 A. B nor C. D' ( one nor)
= (A. B + C. D' ) '
4 again take step 3 answer as input like nor as not
(A. B + C. D' ) ' nor (A. B + C. D' ) ' [1 nor]
=( (A. B + C. D' ) ') '
= A. B + C. D'
So total gates is equal to 7
@@artisharma9204 For minimum, you'll need 4 nor gates. (a nor b) nor (c nor d'). For d' you'll need 1 nor too. So in total 4.
@@jugnugill ok
Thanks
sir please upload the homework problems
9 nor gate
4:06
5:13
6:21
How A+A complement is only A complement?🇳🇵
Because A+A is considered as A
1---5
2---7
1)5
2)5
5
6
These is confused me 😶
It will be more better if you talk abit louder
57
Answers to h.w qns
(1). 9
(2). 7
(i) 5
(ii) 7
5
5
7
Hindi nahi useko
The answers for homework are -
1- 5
2 - 5
How?
Mine is also coming 5 for both the questions.
Mine also
Mine ans is also 5 for both
@@bhanupratap-qj4lp my answer is 7 gates not 5
1-) 5NOR or 6NAND
2-) 7NOR or 4NAND
First one i got 9
I got now 1)5
2)7
Please solve 404 error problem in the solution of homework asap
Hindi may bolo
3
7
how come 3 ? Its coming 5. Please check and explain!
5 & 7