AND,NOT,OR,XOR,XNOR,NAND,INVERTER using MUX

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vlsitechnology

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not gate using mux

Can anyone send me a material on NOT ,AND ,XOR,XNOR,NAND,INVERTER using MUX ???
Plz its urgent send me a material i will give u 20 points
 

nand using mux

Realization of GATE USING MUX

If any one finds wrong in this reply me

If u other design share with us

with regards
khalandar basha

basha_vlsi@yahoo.co.in[/img]
 
mux logic gates

hii basha u missed Nor gate
 
Reactions: soumendu89

    V

    Points: 2
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    soumendu89

    Points: 2
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inverter using mux

hi vlsi technolgy

i hope i designed not gate correctly.


hi venkat_kvr

as he didn't mentioned abt nor gate i didn't draw
and ur's design for nor gate is correct yaar
and thanks for helping out
 

nand inverter

Sorry but really i cant realize what is the advantage of using mux to design those logic gate ? With respect to digital, again you have to use gates to build mux, and with respect to analog surely the number of transistor you are going to use to build a logic gate by mux will more than the no you have to spend to directly build a gate.
Those above designs are effective in which aspect ? Resolution ( i mean true replica of output waveform ) ?
 

xnor using nand


You're rigth but this kind of questions are very usefull for people preparing themselves for interviews.
 

instrucciones nor, xor, and,in, out, not


This is not only useful for interviews, but sometimes has real use when doing a metal fix.
Some designers like to combine muxes as spare cells. If in later stage you need to add logic, you gotta be able to generate your logic from mux structures. Therefore understanding these implementations is important.

Actually one can also use muxes to create memory elements like latches and flip flops. One of my blog posts dealt exactly with that issue, check it out here:
https://asicdigitaldesign.wordpress.com/2007/09/15/puzzle-10-mux-logic/

hope this helps,

ND.
 
implement or gate and and gate using mux

I feel we should not mix the advantage of implementation of logic in CMOS comparing the same with implementation of logic using gates.

Each has its own importance. The simple implementation of logic using gates (as we realize mux logic using gates) helps a clearer understanding while the reduced implementation in CMOS provides faster operation (in ns) and smaller area too by reducing the transistor count. Finally it is a logic equation which is being realized.

Just imagine the no. of transistors required to implement XOR gate using individual gates and then using mux ... the difference is clearly visible.

Regards,
Anuj.
 

how to use a mux as an inverter


You're right, but such questions are asked in interviews.
 

and≡
take two cases i.e for a==0,a==1;('a' is a selected line for 2:1mux)
take 0 and b as inputs to mux
if(a==0)
out =0;
elsif(a==1)
out=b;[/img][/code]
 

thanks guys, does anyone know how to use mux4 to generate Y=a(~b)(~c)+(~b)(~C)+(~a)(~c)...
that will be helpful..

Added after 1 minutes:

FYI just only one mux4
 

same thing i can get for CMOS i.e. AND,NOT,OR,XOR,XNOR,NAND,INVERTER using CMOS
 

Hi all,

I am a little bit confused on the OR gate.
The OR gate boolean expression is Y = BA' + AA => B(1-A) + A => A + B - AB. Shouldn't the OR gate boolean expression be A + B ? However, the definition of OR operation is A U B = A + B - AB. Did I miss anything? Correct me if I am wrong.

Thank you.
 

Hi,

Just to avoid confusion:

B(1-A) = B•(1-A)
which means "B is *dependent* on the inverse of input A".
Meaning, we only need to check B input if A=0.

B(1-A) + A should not be written as (B + A - A•B) to avoid confusion since we are not dealing with normal maths but Boolean logic.
Comparatively, (B + A•B) can be reduced to B using Boolean logic, i.e.:
B + A•B = B
but this is wrong in conventional maths.

So, how do we define the statement (B - A•B) in words for Boolean logic?
BTW, (1 - A•B) would define the NAND logic of inputs A & B.

Best regards.
 

Hi,

I guess in this case the boolean multiplication is not like algebra multiplication. Some other cases, like

a--
b--OR--
c-------AND-- Y

will get Y = (a+b)•c = ac + bc.

Is there a rule indicates when the boolean multiplication cannot be used as algebra multiplication? How do we derive the b•(1-a)+ a to a + b as the OR gate boolean expression?

Thanks.
 

I think we just opened a can of worms !^_^ .

What I was trying to say was, don't expand minus signs.
Safer to leave 1-XXXXX as 'inverse of'.

Of course some one out there might be able to make sense of A-C.

Best regards.
 

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