Basic logic problem

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I want a truth table as follows, where A and B are inputs, X and Y are
Outputs:

ABXY
0001
0110
1011
1111

I would like to use one chip to decode, such as a 74x02 quad two-input nor
gate. I've found two solutions. One is five gates, as shown at
http://www.miketurco.com/cae/gate1.GIF , which is one gate too many. The
other solution is to use a 74x138. I consider the 74x138 to be a cheat!

In going through this, I'm starting to realize how rusty I'm getting at
doing this kind of stuff.

I imagine there's an easy to see, simple solution to this problem. Please
enlighten me, I think my brain needs a kick.

Thanks,

Mike




Re: Basic logic problem
oN 07/10/04, Mike Turco said:

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There used to be an CD4XXX chip that had an odd mix of gates inside. I
can't recall the number, have no idea if ot's still made, nor if the
mix of gates was right for this. I suppose you have no leftover gates
or inverters anywhere? Half a 7432 and a single inverter or transistor
would get it done.

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Why? It is a decode function, after all...

--
Bill
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Re: Basic logic problem
oN 07/10/04, William Meyer said:

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Sorry, I meant two inverters...

--
Bill
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Re: Basic logic problem

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I think I can get it done with a single 7402 or 7404 and one transistor as
an inverter. Since the 74138 works & is only one chip, I guess I'll go with
that.

What concerns me, I think, is my having to think this through for hours and
still not remember the definitive way, e.g. using Boolean algebra, to solve
this problem. Maybe a five gate solution is the only answer, but how do I
know that for sure?



Re: Basic logic problem
oN 07/10/04, Mike Turco said:

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With Boolean algebra, and an equation for each output term. For things
as simple as that one, it can be frustrating, when we're so accustomed
to high density components, but on the other hand, there's the cost of
the time it takes to be more clever about it. <g>

Of course, you could use diodes and resistors, too ;)

--
Bill
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Re: Basic logic problem
oN 07/10/04, Mike Turco said:

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Wow, I am rustier than I thought! I just worked it through and realized
(and proved) that my first answer was correct.

  X = (A OR B)
  Y = (A or (NOT B))

Therefore, one inversion and two sections of a 74C32. So in the absence
of an unused inverter, I'd grap a transistor, a couple of resistors,
and the 'C32.

--
Bill
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Re: Basic logic problem

news:aw%Hc.8854

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Or a single inverter in a SOT23 package like the Fairchild NC7S14 (about 7
cents US).

Cheers,
Alf



Re: Basic logic problem
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ABX
000
011
101
111

That's X = (A OR B)


ABY
001
010
101
111

That's Y = (A OR NOT B)

X = A OR B = NOT ( NOT A AND NOT B )
Y = A OR NOT B = NOT ( NOT A AND B )

Which is 4 NAND.

Re: Basic logic problem

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nor
Please

OK, I more-or-less came to all of the above on my own, except that I used
NOR gates. (Boolean/logic is another area in which a custom font would come
in handy because I could throw in some negated letters, but that's another
thread.)

I haven't used Boolean it since George Boole & I used to drink poisonous ale
and make logic gates out of cat hairs, so please bare with me....

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OK, I get that with three nand gates:
http://www.miketurco.com/cae/nandor.GIF .

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I get two gates for that because you have to invert one of the inputs.

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I get 5 gates again, except that you use NAND gates and I use NOR gates.

Mike



Re: Basic logic problem

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The
cheat!
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at
come
ale

You are missing the obvious.  Both equations have a term in common which
is NOT A.  So Y output can be realized using the NOT A term from X, and
one more gate, for a total of 4.  Simply adding up the number of gates for
each equation will not yield the lowest number possible.

Hope this helps,

Mike Anton



Re: Basic logic problem
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X = A OR B ==> NOT X = A NOR B
    Use one NOR with A, B as inputs
    Use another NOR to invert NOT X
Y = A OR (NOT X) ==> NOT Y = A NOR (NOT X)
    Use one NOR with A, (NOT X) as inputs
    Use another NOR to invert NOT Y
4 NOR gates, right?  (Assuming 3 gate delays is okay)

Tom Taylor

Re: Basic logic problem
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X = A OR B ==> NOT X = A NOR B
    Use one NOR with A, B as inputs
    Use another NOR to invert NOT X
Y = A OR (NOT X) ==> NOT Y = A NOR (NOT X)
    Use one NOR with A, (NOT X) as inputs
    Use another NOR to invert NOT Y
4 NOR gates, right?  (Assuming 3 gate delays is okay)

Tom Taylor

Re: Basic logic problem
On Sat, 10 Jul 2004 14:12:29 -0700, "Mike Turco"

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You could always use one inverter and four diodes!

Feed A through two diodes to X and Y

Feed B through one diode to X

Feed B through one inverter and a diode to Y

(with pull downs on X and Y of course)

Uses only 1/4 of  a chip!

Alan

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Re: Basic logic problem

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Two simple solutions:

7400:
Two gates for outputs
x = !(!A*!B)
y = !(!A*B)

Two other gates used as inverters for !A and !B

1/2 74x139 - use out 0 for X and out 1 Y



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