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t.man · 14th January 2009, 06:43 AM

how can one go about in representing the following Boolean function using NAND gates with a minimum number of gates?

z = A' BC + AB'C + ABC' + ABC

MrAl · 14th January 2009, 09:31 AM

Hi,

One way is to reduce the equation using Boolean Algebra.

Another way is to generate a truth table and look at the
pattern of 1's and 0's to look for a simpler logic representation.

Example using Boolean Algebra:
z = ABC' + ABC
reduces to:
z = AB + AB
because it doesnt matter whether C is high or low, so it gets eliminated,
and of course that reduces to:
z = AB
so
z = ABC' + ABC = AB
and that would be implemented using one two-input 'AND' gate.

moghli_amateur · 28th January 2009, 02:15 PM

Quote:

Originally Posted by moghli_amateur

originally posted byt.man
how can one go about in representing the following Boolean function using NAND gates with a minimum number of gates?

z = A' BC + AB'C + ABC' + ABC

Typically logic designers use K-map but in this particular case the boolean laws can be easily used.
z= A' BC + AB'C + AB
z= A' BC +A(B'C+B)
z=A'BC+A(B+C)
z=A'BC+AB+AC
z=B(A'C+A)+AC
z=AB+BC+AC
As you can see this is just a minimal expression in Sum of Products form.
Since "AND-OR" combination is equivalent to "NAND-NAND" combination, you can relaise the above expression by three 2 input nanad gate and one three input nand gate.

Pessisim never won a war

Chaerl · 13th February 2009, 04:41 PM

With this boolean equation (which is correctly derived), we can do 6 NAND gates to get the right circuit. I hope I'm right.

Krumlink · 14th February 2009, 05:36 AM

AND each logic input and OR each group together.
z = A' BC + AB'C + ABC' + ABC

blueroomelectronics · 14th February 2009, 05:43 AM

Whoa long time no see Krumlink

Chaerl · 14th February 2009, 12:04 PM

Quote:

Originally Posted by Krumlink

AND each logic input and OR each group together.
z = A' BC + AB'C + ABC' + ABC

In this case, you got it wrong. It should be the minimum number of NAND. In this case, we must simplify first the equation with the result of z=AB+BC+AC.
Afterwards, we can represent each the AND and OR with its equivalent NAND circuit. To further lessen the circuit, we must simplify more the equation by grouping. Thus, z=AB+(BC+AC). You can choose which term you want to group. I'll try to attach the circuit next time.

Hero999 · 14th February 2009, 12:55 PM

Quote:

Originally Posted by Krumlink

AND each logic input and OR each group together.
z = A' BC + AB'C + ABC' + ABC

Welcome back.

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14th January 2009, 06:43 AM	#1
t.man	logic diagrams! how can one go about in representing the following Boolean function using NAND gates with a minimum number of gates? z = A' BC + AB'C + ABC' + ABC Last edited by t.man; 14th January 2009 at 06:48 AM.

14th January 2009, 09:31 AM	#2
MrAl	Hi, One way is to reduce the equation using Boolean Algebra. Another way is to generate a truth table and look at the pattern of 1's and 0's to look for a simpler logic representation. Example using Boolean Algebra: z = ABC' + ABC reduces to: z = AB + AB because it doesnt matter whether C is high or low, so it gets eliminated, and of course that reduces to: z = AB so z = ABC' + ABC = AB and that would be implemented using one two-input 'AND' gate.

13th February 2009, 04:41 PM	#4
Chaerl	With this boolean equation (which is correctly derived), we can do 6 NAND gates to get the right circuit. I hope I'm right. Last edited by Chaerl; 13th February 2009 at 04:42 PM.

14th February 2009, 05:36 AM	#5
Krumlink	AND each logic input and OR each group together. z = A' BC + AB'C + ABC' + ABC __________________ IF YOU WANT TO HELP FINE IF NOT PLEASE DON"T PEE IN MY CHEERO'S

14th February 2009, 05:43 AM	#6
blueroomelectronics	Whoa long time no see Krumlink __________________ Bill Smart Kits build Smart People http://www.blueroomelectronics.com/