Implement a function with 2-input NAND gates

  • Thread starter alfonwob
  • Start date
  • Tags Tags
    Function
7 replies · 4K views · last reply
alfonwob
Messages
4
Reaction score
0
Hello people!

I've just done a K-Map and gotten both the SOP and POS, but have a problem with this function:

f = acd + cab + dab + cdab + abcd ,
f = (a+b+c+d) * (b+c+d) * (b+c+d) * (a+b) * (a+c) * (a+d)

and got to implement it only using NAND gates with 2 inputs. Notice the underline is the negated variable.

Thank you so much in advance.
 
alfonwob said:
Hello people!

I've just done a K-Map and gotten both the SOP and POS, but have a problem with this function:

f = acd + cab + dab + cdab + abcd ,
f = (a+b+c+d) * (b+c+d) * (b+c+d) * (a+b) * (a+c) * (a+d)

and got to implement it only using NAND gates with 2 inputs. Notice the underline is the negated variable.

Thank you so much in advance.

Welcome to the PF.

What did you get from your K-map? Do you know how to make an OR gate from NAND gates to implement your SOP solution?
 
Hi, thanks for your answer. I got (1 / 1 / 0 / 0 ... 0 / 1 / 0 / 0 ... 1 / 0 / 1 / 0 ... 0 / 1 / 0 / 0 ) I know how yo do an OR gate from 2 variables, not for more.

Image last number in right bottom is 0, not 1.
 
Attachments
  • Spreadsheet grid of 0s and 1s in four rows and four columns: row 1 is 1 1 0 0, row 2 is 0 1 0 0, row 3 is 1 0 1 0, row 4 is 0 1 0 1.
    asdfa.png
    439 bytes · Views: 591
alfonwob said:
Hi, thanks for your answer. I got (1 / 1 / 0 / 0 ... 0 / 1 / 0 / 0 ... 1 / 0 / 1 / 0 ... 0 / 1 / 0 / 0 ) I know how yo do an OR gate from 2 variables, not for more.

Image last number in right bottom is 0, not 1.

To build a wider OR gate, you just make a tree of 2-input OR gates...

http://www.electronics-tutorials.ws/logic/log11.gif
Circuit diagram of a 6-input OR function built as a tree: three 2-input OR gates with inputs A and B, C and D, E and F feed outputs A+B, C+D and E+F into a final OR gate producing output Q, labelled 6-input OR Function.
 
  • Like
Likes   Reactions: alfonwob
berkeman said:
To build a wider OR gate, you just make a tree of 2-input OR gates...

http://www.electronics-tutorials.ws/logic/log11.gif
Logic circuit diagram of a 6-input OR function: inputs A and B feed an OR gate giving A+B, C and D feed a second OR gate giving C+D, E and F feed a third giving E+F, and these three outputs feed a final OR gate producing output Q, captioned 6-input OR Function.

Thanks again for helping me.
Ok I understand the image you've attached but I have just to use NAND gates with 2 inputs.

I upload a picture where I show you the step I'm stuck in, just in the the part when I have to add both terms.
 
Attachments
  • Handwritten circuit diagram of NAND gate networks: inputs A and B feed NAND gates producing NOT A and NOT B, then a NAND gate labelled NOT(A+B)... with a further NAND gate on inputs C and D, plus scribbled Boolean expressions on the right.
    20150305_001323[1].jpg
    15.5 KB · Views: 543
You need to include the inverters as part of each OR stage. You show the final stage without those inversions...
 
But I only have NAND gates with 2 inputs. It's a rule from the exercise.
 
alfonwob said:
But I only have NAND gates with 2 inputs. It's a rule from the exercise.

You already show in that drawing how to use a 2-input NAND to make a NOT gate... That's how you got your first 2 NOR gate equivalents...