This channel is highly trusted because I found some websites and other channels they don't take data accurate and also don't teach precisely but this channel teach everthing in precisely manner and whenever I study from this channel I don't take any tension for accuracy related problem....
Hi! your video is very informative and easily understandable so I always see the videos related to the digital logic design on your channel. thank you! But I have a question related to this video, what is the previous carry?
(0) We are representing each decimal digit by 4 bits (1) We add 6 because 4 bit includes numbers till 15 and we only want numbers till 9. For the rest of numbers between 10 and 15, we wish to take a carry of 1 to the next set of 4 bits (by adding 6). (2) To identify numbers between 10 and 15, we can look at just the MSB = 1, unfortunately, this includes 8 and 9. So we want all 4 bit numbers such that MSB (S3) is 1, excluding 8 and 9. (3) 8 and 9 have S1 = 0 and S2 = 0. => S1+S2 = 0. For the rest, we want S1+S2 = 1, and also MSB bit as 1. This is S3.(S1+S2) (4) The sum of 2 numbers between 0-9 can also lead to 5 bit numbers, 16, 17 and 18. For these also we need to add 6 as 16 in binary is 1 0000, but we want 1 0110 (similarly for 17 and 18) (5) So the required expression for which we need to add 6 is C + S3.(S1+S2)
@@ikshvaku_allegiance4015 The 1st digit (MSB of the BCD representation) is explicitly calculated by the or gate. So this is correct. Your claim: the carry from the second adder is also correct. Let's check this. Say the sum of the two numbers is 17. Then the binary is 10001. The 0001 is taken and added to 0110 by the second adder to get 0111 = 7. This has no carry. Therefore by your method, the final output will be 00111 (which is wrong). If you use the carry from the or gate, the answer is 10111 (Which is the required answer. See @4:09)
@@RahulMadhavan i have another doubt in video no. 143.Kindly resolve that as well.Thanks. I have posted it below as well. (@Rahul Madhavan .In very first question what happened to the carry outputs after subtraction of the first bit. Cin is 1 but then y0 + X0' must have given a carry which would have then propagated to next bit. How are we handling that?)
@@ikshvaku_allegiance4015 I'm sorry I don't have the context to this question as I don't have video numbers. Furthermore, I saw these videos 3 years back so I don't remember offhand. So maybe someone else can help on that.
@@RahulMadhavan ruclips.net/video/o22PeIImcKc/видео.html this is the video link.First question itself.Nobody else has been able to answer my doubt..kindly help
@@creativestudio5761 If u take the sum of s2* and s1* then you don't need 3no case..Cause it also includes the case..But if you want to take 3 cases then the equations have to be as the comment..
Great explanation. I dont now why we added third case and then removed it through simplification. Even saying s3*(s2*+s1*), it was already covering number 10 - 15.
can you also do videos about verilog there isnt much source, i follow your lectures, thank you , you really give us the best quality lectures and really helpful for all electronic students
thanks a lot sir !!! you just made everything simple i think there are some problem with the subtitles.you are saying something(in the video) and subtitles are listing another
Great and simple explanation sir, can u tell me which board are you using in the video for writing, It would be helpful for me to conduct online lectures for my students.
Watching it first time i didn't understood. But later after three months i understood. First time i say is the time when my sir explained and second time i watched during my semester exam and i finally understood
Hi @7:30 ------> In the (ii) case 12 to 15, instead of using " s*3.(s*2 + s*1)" we can use only "s*3 . s*2" Because 12 to 15 only s*3 and s*2 both is high(1).
Why were 0&3 bits were called min term and maxterm ? Those terms are defined entirely different in some other video! Anyways excellent explanation. Thanks for that 👍
excuse me Professor, unfortunately i could not understand those two options thay you had done, i mean multiplications between S3(S2+S1) and S3(S1), what is the reason for doing that? and how did you get those multiplications? thanks before
NESO ACADEMY ZINDABAD!!! I am passing this semester cause of you guys!!!!
Ur explanation is great ...but u can believe that there is not a single soul in universe who watches ur video in normal speed🤣🤣🤣...#u r awesome
i am that one...
Even i am making notes of his presentation...
@@techvsfake6036you will watch in normal speed if you are making notes. But not if you are reading
@@noobCoder7 i generally watch in 1.25x and take notes after finish. For easy topics, i watch in 1.5 or 1.75x.
@@ImranALI-qh3je i make rough notes and watch in 2x speed always
@@noobCoder7 good
Thanks for all videos Neso Academy. Your videos are easy to understand and also clearly explain the basic concepts.
I love how I can learn all my subjects from Digital Design to Siganls and Systems, Programming, etc with your videos ❤️
This channel is highly trusted because I found some websites and other channels they don't take data accurate and also don't teach precisely but this channel teach everthing in precisely manner and whenever I study from this channel I don't take any tension for accuracy related problem....
Excellent explanation in all the tutorials.
Easy to understand
You are MY Online Guru where i can easily get all the things done thank you sir
the explanation is superb.All the contents are covered
thanks a lot.
Thanks for your explanation! It seems clear to me after listening to your lecture!
Hi! your video is very informative and easily understandable so I always see the videos related to the digital logic design on your channel. thank you! But I have a question related to this video, what is the previous carry?
Thank u so much sir, its really very clear and easy , millions of thank u sir
case 3 is not required
case 2 is valid for both and
Exactly.... Thanks for confirmation 🙌
(0) We are representing each decimal digit by 4 bits
(1) We add 6 because 4 bit includes numbers till 15 and we only want numbers till 9. For the rest of numbers between 10 and 15, we wish to take a carry of 1 to the next set of 4 bits (by adding 6).
(2) To identify numbers between 10 and 15, we can look at just the MSB = 1, unfortunately, this includes 8 and 9. So we want all 4 bit numbers such that MSB (S3) is 1, excluding 8 and 9.
(3) 8 and 9 have S1 = 0 and S2 = 0. => S1+S2 = 0. For the rest, we want S1+S2 = 1, and also MSB bit as 1. This is S3.(S1+S2)
(4) The sum of 2 numbers between 0-9 can also lead to 5 bit numbers, 16, 17 and 18. For these also we need to add 6 as 16 in binary is 1 0000, but we want 1 0110 (similarly for 17 and 18)
(5) So the required expression for which we need to add 6 is C + S3.(S1+S2)
can you tell why carry from or gate comes straight as is rather than coming from the second adder
@@ikshvaku_allegiance4015 The 1st digit (MSB of the BCD representation) is explicitly calculated by the or gate. So this is correct.
Your claim: the carry from the second adder is also correct. Let's check this.
Say the sum of the two numbers is 17. Then the binary is 10001. The 0001 is taken and added to 0110 by the second adder to get 0111 = 7. This has no carry.
Therefore by your method, the final output will be 00111 (which is wrong).
If you use the carry from the or gate, the answer is 10111 (Which is the required answer. See @4:09)
@@RahulMadhavan i have another doubt in video no. 143.Kindly resolve that as well.Thanks.
I have posted it below as well.
(@Rahul Madhavan .In very first question what happened to the carry outputs after subtraction of the first bit. Cin is 1 but then y0 + X0' must have given a carry which would have then propagated to next bit. How are we handling that?)
@@ikshvaku_allegiance4015 I'm sorry I don't have the context to this question as I don't have video numbers. Furthermore, I saw these videos 3 years back so I don't remember offhand. So maybe someone else can help on that.
@@RahulMadhavan ruclips.net/video/o22PeIImcKc/видео.html
this is the video link.First question itself.Nobody else has been able to answer my doubt..kindly help
Cases should be like
1. c*=1 (for 16-19)
2. s3*.s2* (for 12-15)
3. s3*.s1* (for 10-11)
Yes you are right.
It doesn't matter either way, because when you take the sum of cases both answers are same.
Yes
@@creativestudio5761 If u take the sum of s2* and s1* then you don't need 3no case..Cause it also includes the case..But if you want to take 3 cases then the equations have to be as the comment..
No one can explain such a complex method in one go,just understand his efforts
Great explanation. I dont now why we added third case and then removed it through simplification. Even saying s3*(s2*+s1*), it was already covering number 10 - 15.
Don't forget + c*. But, yeah, what you're saying is ultimately correct.
I find using the K-map to find the equation simpler.
can you also do videos about verilog there isnt much source, i follow your lectures, thank you , you really give us the best quality lectures and really helpful for all electronic students
Thank you ,you really saved someone's life. really helping
Such easy after your explanation...and how genius ppl who created this
awesome....extremely awesome...he made it so easy for me to understand...thanks
Case 3 can be obtained from case 2 itself. Instead of eliminating s3*. s1* during simplification we can just have two cases.
Yes I thought the same I expect the creator to respond to this
the lecture was very very helpful...thank you
thanks a lot sir !!! you just made everything simple
i think there are some problem with the subtitles.you are saying something(in the video) and subtitles are listing another
I think you can group 10 and 11 in the group of 12,13,14,15 as well since their expression holds true for 10 and 11 as well
Thanks for the awesome explanation otherwise its so hard to understand👍
Awesome explanation. Thank you. ❤
sir u r lecture is awesome !! I like it sir
Thank You so much for the great content !!! 😊
one like is not enough ,if we could get such crystal clear concepts in all subjects
Then give two.
7:10 y r u considering 12-15 and not 10-15?
such a great teacher👍
really,,it's very much helpful for me,thank you
thnx,earlier i memorized it but now i understand each component of bcd adder completely
Nailed it buddy .....!!!
Really nice explanation
Great and simple explanation sir, can u tell me which board are you using in the video for writing, It would be helpful for me to conduct online lectures for my students.
Nicely explained!
Neso Academy you are the best...
Excellent explanation thank you so much for making this video 😊😊😊
Great explanation sir!!
Thanks,that was really helpful
Omggg, thanks for existing!! Excellent video!
You look very beautiful. Beauty with brain
damn Indians are the best!!!
😂😂😂makes me proud!
Lol
Indians with fake names
...we would be if we were all as smart as him
FAX
Your videos are super easily understandable.
Clear explanation Tq sir I makes Me easy to understand the topic
just wow i got it within 1 view.great explanation .love from Nepal.Thank u
nepaad
Explained very nicely😎
Excellent Sir!
Watching it first time i didn't understood. But later after three months i understood. First time i say is the time when my sir explained and second time i watched during my semester exam and i finally understood
Soooooooooooo much great explanation sir...........
You made it really easy
Thank you so much sir for posting such amazing videos !
Awesome explanation....great
Excellent Explanation.
Superbb explanation..
Sir, I really like your video. If you make a lot of videos like this, I would be very happy.
Nigga nobody cares about u
Good explained.
sir in 8:37 m, (ii) case mei S2*=1 than 11 to 15 cover bhi hogye or 8 and 9 se disting bhi
Thank you so much for the video!
Very well explained....sir
Hi
@7:30 ------> In the (ii) case 12 to 15, instead of using " s*3.(s*2 + s*1)" we can use only "s*3 . s*2" Because 12 to 15 only s*3 and s*2 both is high(1).
agreed
www.geeksforgeeks.org/digital-electronics-bcd-adder/
I thought the same thing
yes. you can use case two only. but to understand it better use condition 3.
Yes I agreed
Best teaching!!
Great Job!!
When will you complete your videos?
Very GOOD Keep it up! man
very nicely explained.
Helped alot. Thanks
Really helpful tomorrow is my exam and I got the logic from this
lol same here!
Thank u Its was a good help
Thank you, this helped quite a lot!
+Mohammad Sultan Khaja I know, watching it Porota!
Farhan Waseehahahhahahahhahahahhah
Thanks yaar bhai Maza aagya💙
awesome sir.u are great
Is there any video in ur series for serial adder?
Why were 0&3 bits were called min term and maxterm ? Those terms are defined entirely different in some other video! Anyways excellent explanation. Thanks for that 👍
what's the need of case 3?
10 and 11 can be differentiated from 8 and 9 with help of case 2 only..then why go for case 3.
true..even i thought that
He did eliminate that -_-
Exactly! Even I was thinking about that.
He could have done that but wanted everyone to understand so simplified the equation later.
Yea thinking the same thing
super lecture salute u
Thanks you are awesome :D
Sir last carry will be from II 4 bit adder, or directly from c*
We can use "Add-3 method" also to convert binary to bcd sum ,is it?
it was easy to understand.....ty
Thnx for easy explanation
lecture is awesome
thank you so much
thank you nice explanation
How to implement bcd to excess 3 code converter using full adder?
May I know why we are adding 0110->6 to 2 digital Decimal while converting to bcd?
What happens to the input carry of 2nd adder that you made?
thank you so much !!!
please make lecture on digital signal processing also.it is the toughest subject.
superb lecture sir
Nice explanation
Do we only give 0-9 as input, since it is BCD? Or any decimal numbers like 21, 38 etc can be given input?
Very well explain sir. But how do you write this is verilog code?
attractive style
very nice lecture sir .can u please tell me which app do u use to capture your screen???
excuse me Professor, unfortunately i could not understand those two options thay you had done, i mean multiplications between S3(S2+S1) and S3(S1), what is the reason for doing that? and how did you get those multiplications? thanks before
hii neso academy plz do a tutorial on how to design 4 bit bcd adder using IC 7483 using Nand gates only
Would you please lecture on XS-3 Adder ?
Why should we give separate conditions for 10-11 and 12-15 doesn't the second condition apply to all of them.?
Thank You So Much
whether if we take k map for all output s3,s2,s0.s1 and or the sop ,is that the ans you got will be same?
yes
Thank you sir :)