Is this possible?? Digital Logic question

[brunswickite]

Is this possible?

my DLD is a little rusty

but i wanted to know if its possible to make an OR gate with only AND gates and INVERTERS.

 

[pillage2001]

Yes.

Compliment it.

A+B = A'.B'

 

[brunswickite]

Originally posted by: pillage2001
Yes.

Compliment it.

A+B = A'.B'

the compliemnt of an AND gate is a NAND gate

AND gate truth table
0 0 0
0 1 0
1 0 0
1 1 1


OR gate truth table
0 0 0
0 1 1
1 0 1
1 1 1

 

[pillage2001]

Originally posted by: brunswickite

Originally posted by: pillage2001
Yes.

Compliment it.

A+B = A'.B'

the compliemnt of an AND gate is a NAND gate

AND gate truth table
0 0 0
0 1 0
1 0 0
1 1 1


OR gate truth table
0 0 0
0 1 1
1 0 1
1 1 1

LOL, OH NOES........Been out of college for too long..hang on. 😀

wait a minute, did I get it wrong? A'.B'= C'

Try it. 😀

 

[Ameesh]

demorgans law

~(p ^ q) = ~p V ~q

 

[Templeton]

Originally posted by: pillage2001
Yes.

Compliment it.

A+B = A'.B'

Close... A + B = A'' + B'' then by demorgans, you have (A'B')' so you need a nand gate with both inputs inverted.

 

[pillage2001]

Originally posted by: Templeton

Originally posted by: pillage2001
Yes.

Compliment it.

A+B = A'.B'

Close... A + B = A'' + B'' then by demorgans, you have (A'B')' so you need a nand gate with both inputs inverted.

Ah...the demorgan fella... LOL Totally forgotten about him but I was close. 😀

 

[brunswickite]

Originally posted by: Templeton

Originally posted by: pillage2001
Yes.

Compliment it.

A+B = A'.B'

Close... A + B = A'' + B'' then by demorgans, you have (A'B')' so you need a nand gate with both inputs inverted.

ahh nice thanks