1-Bit Full Adder using Multiplexer

[@APC9906]

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

[@siddhantmeshram4092]

I don't know who you are or where you live but thank you thank you !

[@starkendeavours7072]

Tysm.

[@bhouyem2315]

thank u sm

[@rishavyadav9438]

Thank you

[@satyampatel3244]

tqsm

[@malikhamza2285]

sir you are the only one reason behind my DLD interest.thanks alot for your service.stay blessed

[@abhiramanne9649]

I will be frank I feel like my lecturer for digital electronics is literally following neso academy.

[@ranjeet5806]

Probably your teacher is the one who's running neso academy.

[@tacoz4429]

Thank you. You have pretty much saved me from failing the course. :') There is still hope.

[@saimyousuf2108]

Best channel to learn digital logic

[@josuezalazar]

The logic to use the full Adder carry output is: i0 = 0, i1 = Cin, i2 = Cin, i3 = 1.

[@nandinisharma3360]

Can you explain me how i3 come 1 it's very confused me 😶

[@josuezalazar]

@@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.

[@nandinisharma3360]

@@josuezalazar then why we written i2 = Ci, why not 1 ?😶

[@josuezalazar]

@@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

[@nandinisharma3360]

@@josuezalazar sorry bt it's very confusing me😶😶😞

[@reona3713]

your explanations are better than my prof. Good job

[@kilarisreenivasprasad8030]

Sir,you are my true and best inspiration.................

[@alekdanser]

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

[@hotaru6765]

wow! you made my subject DCD interesting.

[@sameerthor11]

sir you are a living legend hats off sir

[@tabassumhebbalkar]

Thanks a ton! your videos are so helpful :)

[@johncena8336]

great stuff man! keep up the good work.

[@ajayjadhav-vc5wg]

great series of lectures!you are doing good job!

[@joshitapapareddy9845]

Hello sir, How should we apply the same using 8:1 multiplexer? Please reply as soon as possible, thank you!

[@mohammadnaseer5413]

Good and easy explanation 👍🏻

[@sandeepanbhattacharyya828]

can we consider S1 and S2 to be B and Cin respectively

[@HaidarTcm]

Thanks legend, really if i have any course i need help i foud this channel to be the support❤

[@sorryfornoname2157]

Fuck me. this guy explains this better than the teacher at college where i pay to study (the current education system works guys)

[@mssahin1572]

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?

[@sukhdeepsaini1108]

we are talking about carry

[@hitengupta5793]

Because when a=1 and when b=1 it gives C & C bar . So there is nothing common so we take it as one.

[@mvssguptajagadesh3269]

@@hitengupta5793 it should be Ci + Complement of Ci = 1. (Ci + Ci' = 1).... That's why it is one .

[@goldenSpearton16]

@@mvssguptajagadesh3269 tqs brother...!!!

[@riseabovehate9476]

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

[@kunalkadian6792]

why we dont use bc as select line and then just use A and Acomp as input

[@ShivaKumar-xq8yk]

Sir how to implemente the same question with 2 by 1 MUX.....?

[@sanjukansal8206]

What if we have to do it by 8×1 MUX? As it contains 3 selectors.

[@hyphenpointhyphen]

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.

[@WELLINGTONSIBANDA-t1t]

what should I consider in choosing a MUX for implementing a particular Boolean function

[@xXAIvaNOosXx]

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

[@NATANSHROCKS]

Sir could you please tell me the answer of the homework question. Thanks

[@deepakkumrawat8312]

Thank u a lot.. these videos are very helpful

[@vishwarajendran1483]

For Carry output, inputs are 0,Ci,Ci,1.

[@kuldeeptiwari3424]

Kaisa Bhai . Mera nhi aaraha h

[@vijayanethala4973]

How did u get that...I got x,ci,ci,x bcz when A=0,B=0 ci changes from 0 to 1

[@sadhanasavkare7662]

@@vijayanethala4973 then ci=1.but in video its ci=0.correction is needed.

[@ShaanThor]

​@kuldeeptiwari3424 bhai easy he batau?

[@rishavkashyap325]

do we need to connect the select lines of one mux to other ???

[@pratyushsingh2809]

when we use b and c as select lines. Den what will be the inputs in terms of A?

[@legolas433]

all videos are great man thx

[@aqibshafique8671]

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

[@ShubhamSingh-bj3ms]

HAFIZ Aqib Shiekh what u want

[@shahzaibishaque]

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

[@tiboosters9668]

why not make it with one 8x1 MUX, instead of two 4x1 MUX ?

[@Caraval]

why the last input for the second mux is 1, can you explain it? thanks

[@vaibhavahuja4001]

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.

[@deepaksahoo9336]

Vedprakash Meena thanks dude

[@akashdubey9132]

@Vedprakash Meenalooking for the answer. This is the correct one.

[@saurabhraj8477]

As Ci+~Ci=1

[@mukulmohal7492]

why we have used only A and B as S0 & S1 can we use Cin in place of either A or B

[@priyajitmondal6661]

Thank you much sir❤

[@sanjanadogga3856]

Implement a multiplexer using 36 input lines , how many selection lines are needed?

[@chiluvurideepthi6822]

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.

[@sainadhrahul4450]

Since Both Minterms are present I3= Ci + Ci ' which is Equal to 1

[@Zuwwar]

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.

[@chandumistimer8406]

Zuwwar Khan Jadoon we can directly give sum as input den

[@samal-satyajit]

Today is my exam after couple of hours and am watching it now

[@VC-kj9yx]

what if there is a dont care in k-map

[@jaswanthsai1917]

For the last two cases A=1,B=1 then why do we get different ie., Ci, 1

[@yugayadav5906]

THANKS YOU FOR SUPORTING

[@madhavsaha962]

please solve the Boolean function using 8*1 mux please.. thank you.

[@tarunmarisetty2839]

awesome explaination

[@utkarshshukla231]

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

[@rohitsayandip6741]

Thanks a lot man.

[@DoubleL_Leo]

Thank you sir

[@ANIKETKUMAR-su1kf]

please solve implement of full adder using using 4:1 mux

[@jackdanksterdawson112]

How do u determine the select line variables y did u take a and b why not b and cin

[@jackdanksterdawson112]

Vedprakash Meena my xams over!!

[@yugeshvemula8044]

plz explain about 2 to 1 multiplexer by using 74LS150

[@SomiMm-q1g]

Thank you

[@sureshdudyala2589]

Down part is not visible

[@gouravmanchanda9301]

this playlist does not include videos of multiplexer...please upload those videos...

[@ashibindal6644]

sir can u plz explain how to design 32*1 mux using 8*1 and 4*1 mux

[@roosterstudio]

Can you explane me... Can we make 4-bit adder with this structure for 1-bit adder?

[@roosterstudio]

How? Are we connect some outputs or make another true tables?

[@roosterstudio]

***** Thank you a lot! I understand now :)

[@gunduboinadileep9523]

if we take BC as selectors it will be more easy I think

[@RadixElektra]

ctrl C, for copy ctrl V, for paste,

[@studentcommenter5858]

Did you just assume him to be using a keyboard? He may be using a touchpad and mouse.

[@NaveenSword]

@@studentcommenter5858 he uses wacom tablet and wacom digital pen.

[@SahaabMansha]

sir can we use one 8x1 to make 1 bit full adder

[@_zerox_X]

How should we know which mux should we use like here he used 4 x 1 mux

[@AshranBaig]

how 1 will be there in the input of carry MUX?

[@ravimaurya9674]

Because its Ci + Ci(bar) = 1

[@Zuwwar]

i mean in 8:1 MUX... A,B,Cin these all are your select lines, so what would be your input then?

[@Zuwwar]

***** So when would it be "0" and when would it be "1" ?

[@Zuwwar]

***** thanks :)

[@alagarhari9410]

Can we select selector variables as per our wish

[@nitishkumarprajapati6096]

Thanks Sir😀😀

[@ankit4376]

How to determine I3 for carry

[@ajaysrivastav5412]

Can you please tell why I3=1 for "carry" in Kmap ?

[@explorewithvikas3019]

Explain??

[@baldeepsingh1113]

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

[@anastariq8464]

what about the i0.....there is no 1 in shaded region when a and b both are zero

[@jamestoh8853]

How do you get 1 and 0 as the answer?

[@Vishtory]

+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.

[@bjornnct8809]

Hey how do you design a 1 bit full adder using exnor gates and a mux ? Please respond... I need help

[@abdulrahmanhamdan4355]

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?

[@akashbisht22]

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. :)

[@rprvk]

I think I3 is 0 By grouping we get 0

[@khaldounal-nuaimi3594]

Thank you!

[@animeshbanerjee4379]

Can u again explain i0=ci ,i1=c1(bar)....why ???

[@Naveenkumar-qu9lk]

Sir...if we take select lines different is there any change in answer...?

[@aravindosarkar341]

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)

[@santoshodeyar9736]

I did that bro

[@nurulfarhana823]

Could you show how to do combinational circuit with multiplier of 4 bits

[@thesorcerer111]

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.

[@thesorcerer111]

+Neso Academy Thanks!!

[@abudriaz9678]

but there are already 2 4×1 mux in it 😐

[@utsavprabhakar5072]

What am i missing? This has 2 4*1 mux and this is a full adder? +thesorcerer111 mate did you find the answer?

[@adhishravi20]

B and ci are the selector variable??

[@gunar3939]

I need to know for last c whose value is 1

[@tharindukrishan6789]

Thanks a lot!

[@AKJOSHI28]

Sir, how the value of I4=1 in the case of Carry ???

[@swarnavanath461]

In case of I3 C(i ) is changing from 1 to 0 i.e. the last two cells How is it 1?

[@abhinavbhat6629]

Exactly. Did you figure it out yet?

[@mdnahidislam1908]

My question too.

[@aswinp888]

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

[@zainadiab2699]

how to design a 2-bit binary control adder/subtractor using 2 full adders , 2 NOR gates , and two 2x1 mux

[@HabibUrRahman-mu8zf]

Why can't you use implementation table??

[@renuka8372]

[@aradhanakaushal5422]

I did this way .. but out of 5 marks only 3 was given stating u need to draw reduced truth table for this...

[@arslanmalik6151]

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?

[@nikhilkoditala369]

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

[@souravkarmakar2497]

To implement the inputs of last mux is not suitable for understanding....still have some doubts

[@letsbehelpinghands246]

thank you so much..

[@ishikamakkar6622]

I was doing the homework so i found that for carry..when A=0,B=0 there is no value for c..what are we supposed to write???

[@shirsenduchandra2430]

0 is the answer

[@_cxc_8198]

thank you so much :D

[@TheShahzebrizvi]

Can you do a video on ALU

[@justinrosario3053]

god bless you sir

[@siriparuchuri9722]

how to implement 12:1 mux using only 4:1 mux

[@TheAmigodaniel]

i could not get the Cout as you described it there is something am missing? Can you explain it?

[@nagasairamtaluri8933]

For carry Sir for I0 Ci=no minterms for I3. Ci=0 or 1 Then what is the I0 I3 value in this case

[@NANDANKUMAR-ps3nw]

plz explain it for carry

[@anishpaul6300]

For i3,how can we represent 1 in the input by using logic gates?

[@ankit_02_nce44]

If A=1 and B=1, Then carry output =B AND in that case we have B=1