# NOR Gate Question

Discussion in 'Electronic Basics' started by Bart, Aug 17, 2006.

1. ### BartGuest

My thinking:
+5volt logic works usually with a "high" of near 5 volts and "low" near zero
volts.
If I'm thinking right, the "inversion bubble" on a NOR gate means that if
both inputs are "low", the output logic becomes true but is sinking. If
either input is "high" then the output logic is false and is.....what?
I just can't get this straight in my head. Any help is appreciated as I need
a gate where if both inputs are "low" the output goes "high" (+5volts)

Bart

2. ### EeyoreGuest

You're presumably thinking of TTL ? No-one uses it any more. CMOS 74HC etc
happily sinks and sources current.

Graham

3. ### John PopelishGuest

You are making this more complicated than it needs to be.

An or gate produces a 1 output, if either OR both inputs are 1. It
produces a zero output only if both inputs are zeros.

Most logic is positive logic (1 state is represented by the more
positive state)

The ball on the output of the gate is an inverter that reverses the
state of the output upstream of the bubble.

So a NOR gate produces a zero if either or both inputs are 1.
It produces a 1 output only if both inputs are zeros. Just the
inverted output of an OR gate.

Yes, you need a NOR gate.

4. ### Alan BGuest

Nope. Grab yer TTL data book, and look up the 74LS02. Both inputs low is
a high output and a source current.
Low, and sinking. And only false if your application is true=high.
Reckon you want a NOR gate.

5. ### DJ DelorieGuest

Well, it's TTL and CMOS logic, not "+5 logic". But, yes.
For a NOR gate, the inversion is on the output. So, first consider a
plain OR gate:

Either input high, output high. Neither input high, output low.

Now invert the OUTPUT (which is where the bubble is).

Either input high, output LOW. Neither input high, output HIGH.

As an additional example, consider an inverted-input OR (equivalent to
a nand gate):

--O\--,--O/--'

Plain OR: Either input high, output high. Neither input high, output low.
The above: Either input LOW, output high. Neither input LOW, output low.

A truth table helps too:

A B OR NOR AND NAND XOR NXOR
--------------------------------
L L L H L H L H
L H H L L H H L
H L H L L H H L
H H H L H L L H

Just find the column that does what you want, and see which gate that
is.

6. ### BartGuest

Thanks John,
For heaven's sake, I read many different explanations of the "inversion
bubble" and logic gates that I read way too much into it. Now I understand
the shape of an OR gate and NOR gate are the same except for the ball,
meaning "invert" or reverse the truth table. Seems so simple now. Thanks
again!
Bart
PS: You scare me when you say "most" logic is positive logic. When will I
know if "negative" logic is inferred?

7. ### Guest

Bubbles on the inputs. Example - a 7400 gate (NAND) drawn as an OR
with bubbles on the inputs. Ampex used to do this is the '70s. I never
got comfortable with it though it makes sense.

GG

8. ### DJ DelorieGuest

You'll know. One common example is RS-232.

Plus, the specs say what Vih, Vil, Voh, and Vil must be, usually
spec'd as maximum or minimum voltages relative to the various
supply/ground pins.

9. ### John PopelishGuest

Beats me, but I have come across negative logic a few times. Made my

10. ### Guest

i think u r some what confused

high and low are two states.since in digital we deal with two states-
high or low ;1or 0 like that

for repersenting a high we use +5 volt and for a low we use 0 v and
we can use
for repersenting a high we use 0 volt and for a low we use -5 v
these r all depand upon the logic which we use. dont get confused with
that

NOR gate is a gate where both inputs are "low" the output goes "high"
(+5volts)

11. ### Alan BGuest

Damn good question. Like "stratus" said, many times you'll see the bubble
on the input, which is an indication that the "true" state of that input is
low. Another example is a "data selector," which, for instance, might have
three inputs and eight outputs. One of the outputs will go low according
to the bit pattern of the three inputs - the low output is "true" in that
it is the line that is selected. This gets really common when creating
microprocessor controls and such.

Part of the reason to use negative logic is that the low state is the
current hog (see the sink and source numbers for your chips), so it pays
power-wise to have most of the nodes in the circuit high at any one time,
and only the active (or true) nodes low. Part of knowing whether negative
logic is used just comes from experience with a circuit.

12. ### Guest

Bart wrote:
NAND

if you draw a truth table. A,B, What you Want.
0 0 1
....... .......

Then you'd see AND is a similar pattern, you want NAND. And
logically, you sohuld guess NAND.

Also, googling, maybe you can build a NAND with NORs. I seem to recall
you can build anything from NORs, and probably anything from NAND.

To put NAND into NORs. Know how to put AND into NORs, and how to put
NOT into NORs. Feed your AND into a NOT.

13. ### Jerry RGuest

Nope, that'd be a NOR gate (positive locic):

A B OR NOR NAND
----------------
0 0 0 1 1
0 1 1 0 1
1 0 1 0 1
1 1 1 0 0

....jerry

14. ### Alan BGuest

I don't know about NOR, but it is correct that NAND logic can be used as a
building block for any logic circuit, no matter how complex.

15. ### Guest

You can use NOR to build a NAND, and vice versa.

We can convert between OR and AND.
NOTs getting in the way aren't a problem, becuause we can build NOTs
from NANDs, and we can build NOTs from NORs.

I suppose that the "product of sums equation" and alternative "sum of
products equation" , you get from a truth table , also come from the
idea that we can use ANDs or ORs. It's the same equation just NOTed
both sides. I'm rusty though, I learnt this boolean algebra for a short
period years ago.

16. ### Guest

yikes, thanks!

17. ### Don KlipsteinGuest

How about "Karnaugh Map"? I remember doing a lot with those when I took
a "computer logic circuits" course (I hope I got the course name right) at
Drexel University around 1983 or so. If I remember correctly, this
"combinational logic optimization method" relied upon "minimized sum of
products" and "minimized product of sums" being similarly optimal, and I
somewhat remember needing freedom to invert some inputs.

Keep in mind that inverting the inputs to a NAND gate makes it an OR,
and inverting the inputs to a NOR gate makes it an AND. (I fairly
remember that one and I hope I figured out confirming this memory of mine
to be true!)

A NOR gate with inputs tied together and a NAND gate with inputs tied
together is an inverter.

Going back to the Karnaugh Map method - it generally proposes whatever
multiple gates to be limited to two "layers". And if the first layer is
inverting (NAND rather than AND or NOR rather than OR) then the result is
that you can make both layers of multi-input (2 or more inputs) gates both
NAND or both NOR, with the second layer being changed from AND to NOR or
from OR to NAND, and with the first layer being changed from OR to NOR or
from AND to NAND.

Hope this helps, even if this is my roughly annual half-answering a
school homework problem!

- Don Klipstein ()

18. ### Guest

funnily enough I had written "Karnaugh map" then I deleted it. Because
I remembered that you look at the truth table and derive the product of
sums or sum of products equation.

Then you use the karnaugh map. (unless you can do it or the
corresponding boolean alebra in your head ). The karnaugh map is just
a tool to help simplify it. It didn't involve any inverting. I think it
just did some kind of boolean factorization and also, simplfying things
like 0 OR A , 1 AND A
I know that AND doesn't change very nicely into NOR.

A AND B
= NOT NOT (A AND B. )
= NOT (NOT A OR NOT B)
= (NOT A) NOR (NOT B)

Oh, I see you're onto something
invert one layer, and you get NOR . But that's not inverting AND into
NAND.
(NOT A) AND (NOT B)
= A NOR B

googled "product of sums" "sum of products" NAND NOR
"A sum of products logic expression can be realized by NAND gates by
replacing ... A product of sums expression can be realized by NOR gates
by replacing all ... "
meseec.ce.rit.edu/eecc341-winter2001/341-12-18-2001.ppt

Looking at that ppt,
I think you are talking about **double inverting**, so as not to change
the thing. The SOP or POS never starts off with a NOT over the whole
thing(the whole outer layer), or over each inner layer. So i'm not
looking for a NAND or NOR in any layer.

That is very interesting, I never did that to a POS or SOP.

And this is for expressions rather than equations. So you're double
inverting to keep it the same.

I think I see what you're saying
Great concept
based on
NOT A OR NOT B = NOT (A AND B) i.e. A NAND B
NOT (A AND B ) = NOT A OR NOT B i.e. A NOR B

For a product of sums, outer layer = AND, invert the whole thing twice
and convert to NOR.

For a sum of products, outer layer = OR, invert the inner layer
twice, and convert to NAND.

nice, thanks.

I don't know if that was what you were saying, but it's very
interesting!

19. ### PeteSGuest

K-maps depend on Sum of products form :

i.e. Y = ABC + DEF + ... and so forth. This is sometimes also known as
a minterm.

product of sums is

Y = (A + B) * (C + D) * and so forth, where * denotes the AND operator
(logical multiplication). This is also sometimes known as maxterm.

For the conversion across base types, use DeMorgan's theorem

if

Y = A * B then
!Y = !A + !B

You can see this from the truth table, incidentally

Only AND / OR / INVERT gates can be treated this way (as they are the
only true priimitives).

To apply DeMorgan's theorem, invert the output and all inputs, and swap
all signs from AND <-> OR.

So a 2 input AND with active high inputs is equivalent to a 2 input NOR
with active low inputs (assuming 1 = high and 0 = low which is not
always necessarily true).

The same can be done for each primitive - a 2 input OR with active high
inputs is equivalent to a 2 input NAND with active low inputs.

As one may convert either NAND or NOR to any gate (by the use of
inverters, which each of those inherently have), they are both known as
universal gates.
2 input NAND

Y = !(A * B) converts to:

!Y = A + B

Cheers

PeteS  