A nand gate becomes ___ gate when used with negative logic ?

NAND gate becomes AND gate when used with negative logic. 
 
NAND gate turth table(with -ve logic) 
0 0 0 
0 1 0 
1 0 0 
1 1 1 -> this is truth table of AND gate 

C this website. Ans is Nor gate. that is convert one tozeroes and vice versa. 
 
http://www2.ele.ufes.br/~ailson/digital2/cld/chapter2/Ex02.doc.html

It becomes an and gate

my lord.it becomes a AND Gate ....such a simple question.

When used with negative logic means that the inputs are inverted.

If the inputs of a nand gate are inverted and given, the output will be an OR of the inputs. De morgans law (A'.B')' = A + B

OR gate is the right answer

ANS is nor gate

negative logic means a high is interpreted as low and vice-versa

poornima is right

it is so simple it becomes an AND gate when we use negative logic becuase NAND means that Negaive AND

The NAND gate becomes a "OR" gate because
(a'.b')' = ((a+b)')' = a+b (by De-Morgan's)

Nor gate

it is NOR ofcourse

it is NOR gate. Lets analyse how.

In positive logic 0s are considered low and 1s as high voltages. In negative logic 0s are high and 1s are low. Now the truth table for Nand is

A B Y         A(V) B(V) Y(V)-> positive logic

0 0 1         low    low   high

0 1 1         low   high   high

1 0 1         high   low   high

1 1 0         high   high  low

Now when used as negative logic then the truth table will be

A B Y

1 1 0

1 0 0

0 1 0

0 0 1

hence it is NOR gate.

an and gate with negative logic becomes nor gatetruth table for and gate is0 0 - 01 0 00 1 01 1 1we get 1 1 1 0 it is for nor gate

OR gate

OR gate

OR gate

hi , it will become AND gate at there.

First of all, a NAND gate is not (a'.b'), it is (a.b)' -- that is, (a'+b'). When you negate this, you get (a'+b')' which you can then de-morgan to get (a.b). So a NAND gate become AND

it becomes 'OR' gate..so simple apply Demorgan's laws..u will get 'OR'.

The ans is OR GATE.

ediots go to page no 64 fig 2.11 NAND with negative logic is nor

ya sudhansu has explained it correctly...
actually from de-morgan's
(a'b')'=((a+b)')=(a+b)

he dude it will b or gate so simple logic----
(a'b')'=a''+b''=a+b

Yes i think it becomes nor gate because using a negative logic doesn't mean doing any mathematically. it means just suposing zero to be one and viceversa

OR GATENAND gate in negative logic (convert 1s to 0s & vice versa of positive logic)
1 1 1
1 0 1
0 1 1
0 0 0
This is truth table of OR gate

It becomes a NOR gate

L L H
L H H
H L H
H H L


Now assume
H=1 and L=0 we get NAND
H=0 and L=1 we get NOR

Similarly for AND gate we get OR gate

NAND gate with negative logic becomes a NOR gate with positive logic.

It is OR gate.

NOR gate