1-Bit Full Adder using Multiplexer
ฝัง
- เผยแพร่เมื่อ 20 ธ.ค. 2024
- Digital Electronics: 1-Bit Full Adder using Multiplexer
Contribute: www.nesoacademy...
Website ► www.nesoacademy...
Facebook ► goo.gl/Nt0PmB
Twitter ► / nesoacademy
Pinterest ► / nesoacademy
For question at the end: To get 1 for i3, the shaded region that satisfies A=1 & B=1 shows that there are two 1s, if you group them, you can see that Ci changes from 1 to 0, making the answer 1; to help you more, take each 1 alone (generally bad idea), you'll realize you have Ci+Ci' which is equal to 1
I don't know who you are or where you live but thank you thank you !
Tysm.
thank u sm
Thank you
tqsm
sir you are the only one reason behind my DLD interest.thanks alot for your service.stay blessed
Thank you. You have pretty much saved me from failing the course. :') There is still hope.
Best channel to learn digital logic
I will be frank I feel like my lecturer for digital electronics is literally following neso academy.
Probably your teacher is the one who's running neso academy.
your explanations are better than my prof. Good job
The logic to use the full Adder carry output is:
i0 = 0, i1 = Cin, i2 = Cin, i3 = 1.
Can you explain me how i3 come 1 it's very confused me 😶
@@nandinisharma3360 hey, it's been a while since I studied this. I've just checked it out and I think that i3 = 1 because in the Kmap when A = 1 & B = 1 there're two posible answers. And those answers would be written as Ci or Ci' which equals to 1.
@@josuezalazar then why we written i2 = Ci, why not 1 ?😶
@@nandinisharma3360 because in i2 is when A=1 & B=0 and as you can see we have two options. One of them is 0 and the other is 1. We use the option that's 1, and that option occurs when Ci is 1, or in other words, i2 = Ci
@@josuezalazar sorry bt it's very confusing me😶😶😞
Sir,you are my true and best inspiration.................
wow!
you made my subject DCD interesting.
sir you are a living legend hats off sir
Good and easy explanation 👍🏻
great series of lectures!you are doing good job!
Thanks a ton!
your videos are so helpful :)
Thanks legend, really if i have any course i need help i foud this channel to be the support❤
Thanks a lot for the videos they're really helpful!
Would You mind making one on static hazards and iterative circuits? I find it hard to find a good explanatation regarding those 2 topics anywhere.Thanks again
In an xor gates IC there are four exor gates so output of a xor b can be fed to input of 2nd xor gate with c in the same IC, so in this case also only one IC is required
Hello sir,
How should we apply the same using 8:1 multiplexer? Please reply as soon as possible, thank you!
you can compare cin with both s and Co , if they cin equals to S or Co then you right the value of I = cin if Cin different with S or Co then you give the value of I C'in.
thanks you so much my hero
great stuff man! keep up the good work.
For Carry output,
inputs are 0,Ci,Ci,1.
Kaisa Bhai .
Mera nhi aaraha h
How did u get that...I got x,ci,ci,x bcz when A=0,B=0 ci changes from 0 to 1
@@vijayanethala4973 then ci=1.but in video its ci=0.correction is needed.
@kuldeeptiwari3424 bhai easy he batau?
do we need to connect the select lines of one mux to other ???
Sir how to implemente the same question with 2 by 1 MUX.....?
what should I consider in choosing a MUX for implementing a particular Boolean function
why the last input for the second mux is 1, can you explain it? thanks
see the k map, when A and B both are 1, the shaded area has both of the squares as 1, hence I3 is 1.
Vedprakash Meena thanks dude
@Vedprakash Meenalooking for the answer. This is the correct one.
As Ci+~Ci=1
What if we have to do it by 8×1 MUX? As it contains 3 selectors.
when we use b and c as select lines. Den what will be the inputs in terms of A?
I wondered till your solution of the hw problem how to derive
the value of Ci, so input (last variable) decides what function is used, 00-> 0,11-> 1, Ci Ci'.
Thanks for clearing that up.
Thanks for uploading these videos. They helped me a lot. I have a question. When A = 1 and B =1, how can Cout equal to 1?
we are talking about carry
Because when a=1 and when b=1 it gives C & C bar . So there is nothing common so we take it as one.
@@hitengupta5793 it should be Ci + Complement of Ci = 1. (Ci + Ci' = 1).... That's why it is one .
@@mvssguptajagadesh3269 tqs brother...!!!
Fuck me. this guy explains this better than the teacher at college where i pay to study (the current education system works guys)
can we consider S1 and S2 to be B and Cin respectively
Thank u a lot.. these videos are very helpful
Today is my exam after couple of hours and am watching it now
Sir could you please tell me the answer of the homework question.
Thanks
Can we also add Enable in Full adder using Multiplexer ?
And enable in in 2:1 Multiplexer ?
Why we use Enable in 4:1 Mux special case ?
Kindly solve my problem
Thanks
HAFIZ Aqib Shiekh what u want
Enable is somrthing we use when we want our MUX to give 0 output or 1 by our choice. It is helpful in certain cases e.g in case of 8*1 MUX using 4*1 MUX we use enable to off our one of the two 4*1 MUX
why we have used only A and B as S0 & S1 can we use Cin in place of either A or B
For the 2nd MUX(For carry) , when A=1,B=1 ,then how the output is 1,by k-map...?
And thank you for the video.
Since Both Minterms are present I3= Ci + Ci ' which is Equal to 1
Dear Nesco, i have a question which is that u said that in case of (4:1 multiplexer) your input would b the Cin and A,B would be your select lines. Now in case of (8:1 multiplexer), i would have 3 select lines that is A,B,Cin. so what would be my input then in case of (8:1 MUX)? Because in this case Cin is also considered as a select line.
Zuwwar Khan Jadoon we can directly give sum as input den
why we dont use bc as select line and then just use A and Acomp as input
How do u determine the select line variables y did u take a and b why not b and cin
Vedprakash Meena my xams over!!
Can you please tell why I3=1 for "carry" in Kmap ?
Explain??
Thank you much sir❤
Implement a multiplexer using 36 input lines , how many selection lines are needed?
plz explain about 2 to 1 multiplexer by using 74LS150
Thanks Sir😀😀
how 1 will be there in the input of carry MUX?
Because its Ci + Ci(bar) = 1
all videos are great man
thx
Hey how do you design a 1 bit full adder using exnor gates and a mux ? Please respond... I need help
please solve the Boolean function using 8*1 mux
please..
thank you.
B and ci are the selector variable??
Suppose s1 = 0 and s0 = 0 for sum then how will Ci's value be taken ... As 1 or as 0 as TT has both at m0 and m1
please solve implement of full adder using using 4:1 mux
awesome explaination
sir can u plz explain how to design 32*1 mux using 8*1 and 4*1 mux
i did home task of carry but i didnt inderstand how you write 1 as last input of carry circuit can you elaborate it plz?
you can see from the k-map of carry out, the values of lat two blocks is 1,i.e., when A and B both are 1's the value of i4=c'+c which is equal to 1.
hope this helps
ctrl C, for copy
ctrl V, for paste,
Did you just assume him to be using a keyboard? He may be using a touchpad and mouse.
@@studentcommenter5858 he uses wacom tablet and wacom digital pen.
Sir...if we take select lines different is there any change in answer...?
Pls make a video that shows how to implement a 4 variable Sop or pos expression using a 8:1 Mux. Suppose the expression is Y(A, B, C, D) = Summation of m(0,2,3,6,8,9,12,14)
I did that bro
Sir, how the value of I4=1 in the case of Carry ???
In case of I3 C(i ) is changing from 1 to 0 i.e. the last two cells
How is it 1?
Exactly. Did you figure it out yet?
My question too.
Yep so u have to take both the cells as we do in normal K map.
So Ci +C ̅i=1 by Boolean Algebra Laws
i mean in 8:1 MUX... A,B,Cin these all are your select lines, so what would be your input then?
***** So when would it be "0" and when would it be "1" ?
***** thanks :)
Can you explane me...
Can we make 4-bit adder with this structure for 1-bit adder?
How? Are we connect some outputs or make another true tables?
***** Thank you a lot! I understand now :)
Thank you sir
THANKS YOU FOR SUPORTING
How do you get 1 and 0 as the answer?
+Jason Ramia sir, as for i3 c iz changed from 1 to zero then how could we can take it as 1 ...if smthng changes we dnt take it in account. please explain.
in carry combinational circuit you represent i0=0 and i3=1,on the other hand in sum circuit you represent them by c and c complement why?
because on analyzing the truth table for carry Mux selector inputs (A, B in this case), we found that :
When S1= 0 , S0 =0 ; we have 0 in K map, hence i0 = 0
&
when S1= 0, S0 = 1 ; we have Cin = 1 in K map, hence i1 = Cin (Same is the case when S1 =1 & S0 =0 & i2 = Cin)
&
When S1 = 1, S = 1 ; we have Both A' & A = 1 in K map, hence i3 = 1.
Hope it helps. :)
I think I3 is 0
By grouping we get 0
How to determine I3 for carry
why not make it with one 8x1 MUX, instead of two 4x1 MUX ?
How should we know which mux should we use like here he used 4 x 1 mux
For the last two cases A=1,B=1 then why do we get different ie., Ci, 1
what about the i0.....there is no 1 in shaded region when a and b both are zero
sir can we use one 8x1 to make 1 bit full adder
Can we select selector variables as per our wish
I need to know for last c whose value is 1
Thank you
sir how is I3=1 in 4*1 mux of carry output
plz explain it for carry
how to implement 4 variable function of POS using 4x1 muxF(A,B,C,D)=POS (0,1,4,5,7,8,11,13) using 4x1 multiplexer
this playlist does not include videos of multiplexer...please upload those videos...
Could you show how to do combinational circuit with multiplier of 4 bits
For i3,how can we represent 1 in the input by using logic gates?
If A=1 and B=1, Then carry output =B AND in that case we have B=1
i could not get the Cout as you described it there is something am missing?
Can you explain it?
Thanks a lot man.
Sir I'm a big fan of your videos, but could you please make one on "the implementation of a full adder using 2 4x1 mux". That'd be very helpful.
+Neso Academy Thanks!!
but there are already 2 4×1 mux in it 😐
What am i missing? This has 2 4*1 mux and this is a full adder? +thesorcerer111 mate did you find the answer?
Can u again explain i0=ci ,i1=c1(bar)....why ???
Down part is not visible
How to implement this by 8x1 MuX
Can you do a video on ALU
god bless you sir
if we take BC as selectors it will be more easy I think
what if there is a dont care in k-map
can we not take b and ci as the selector lines? then get the values of I0-I3 in terms of A?
Why can't you use implementation table??
how carry calculate
To implement the inputs of last mux is not suitable for understanding....still have some doubts
thank you so much..
how to design a 2-bit binary control adder/subtractor using 2 full adders , 2 NOR gates , and two 2x1 mux
Dat kmap method is obsolete use design table method its much easier n quicker
Thank you!
For carry
Sir for I0 Ci=no minterms
for I3. Ci=0 or 1
Then what is the I0 I3 value in this case