Register to reply

Creating a full adder using a 3-to-8 decoder

by dmatador
Tags: 3to8, adder, creating, decoder
Share this thread:
dmatador
#1
Oct24-10, 05:12 PM
P: 122
I'm trying to create a full adder using one 3-to-8 decoder and some nand gates. As of now I know I will have X, Y, and C_in as my inputs. I am having trouble with figuring out what the 8 outputs of the decoder should be, so I am unsure about where and how to use the nand gates. Anyone able to give me a nudge in the right direction?
Phys.Org News Partner Science news on Phys.org
'Smart material' chin strap harvests energy from chewing
King Richard III died painfully on battlefield
Capturing ancient Maya sites from both a rat's and a 'bat's eye view'
niceboar
#2
Oct24-10, 05:54 PM
P: 54
Quote Quote by dmatador View Post
I'm trying to create a full adder using one 3-to-8 decoder and some nand gates. As of now I know I will have X, Y, and C_in as my inputs. I am having trouble with figuring out what the 8 outputs of the decoder should be, so I am unsure about where and how to use the nand gates. Anyone able to give me a nudge in the right direction?
so you want to create a full adder for the three bits? Or is one of the bits a carry?

edit: guessing C is carry.


Draw out the map for X, Y C, and L0 through L7 first

Then draw out the map for X, Y and C.

Then you can correspond the values e.g. L1 would correspond to an output of 1 for the sum and 0 for the carry. L7 corresponds to bit x added to bit y added to the carry which the output should be
sum of one carry one

So you feed sum into a NAND and get the COMPLEMENT of what you want since you will eventually feed it into another NAND gate
ditto for c

should get you started might be different from what I say just off the top of my head.
dmatador
#3
Oct24-10, 05:59 PM
P: 122
Right. C_in is the carry in and X and Y are the two bits I want to add together....

niceboar
#4
Oct24-10, 06:11 PM
P: 54
Creating a full adder using a 3-to-8 decoder

Quote Quote by dmatador View Post
Right. C_in is the carry in and X and Y are the two bits I want to add together....
Scratch that last thing about getting the compliment

anyway just realize that
for input L0 you want an output of s=0 c=0
L1 s=1 c=0
L2 s=1 c=0
L3 s=0 c=1
etc
then minimize the function

I am having trouble with figuring out what the 8 outputs of the decoder should be
I missed this

So a 3 - 8 decoder has 3 inputs and 8 outputs. each output corresponds to a combination of the input. so there are 2^3 combinations of x,y,c there will be one and only one output for each combination.

input of
000 turns on the L0 line
001 turns on the L1 line
010 L2
011 L3
100 L4
etc
dmatador
#5
Oct24-10, 06:15 PM
P: 122
Quote Quote by niceboar View Post


Draw out the map for X, Y C, and L0 through L7 first

Then draw out the map for X, Y and C.

Then you can correspond the values e.g. L1 would correspond to an output of 1 for the sum and 0 for the carry. L7 corresponds to bit x added to bit y added to the carry which the output should be
sum of one carry one



should get you started might be different from what I say just off the top of my head.
Could you be a little clearer with the first part about the maps? Should I have a K-map for inputs X Y Cin and output Cout, and then one for X Y Cin and output Sum? What exactly will my 8 outputs from the decoder be?
niceboar
#6
Oct24-10, 06:17 PM
P: 54
Quote Quote by dmatador View Post
Could you be a little clearer with the first part about the maps? Should I have a K-map for inputs X Y Cin and output Cout, and then one for X Y Cin and output Sum? What exactly will my 8 outputs from the decoder be?
Edited my last post.

So L3 being on represents X=0 Y=1 C=1
Phrak
#7
Oct24-10, 06:22 PM
P: 4,512
If you are selecting decoder outputs with a nand gate, I have to assume the selected output of your decoder is active low.

Adding X, Y and carry_in is really just adding three bits. X_out of your full adder will be high whenever you have an odd number of input ones. So for this you need a 4 input nand gate fed with the decoder outputs generated from these inputs.
dmatador
#8
Oct24-10, 06:22 PM
P: 122
Okay, that makes sense. I guess my big problem is the gates. Do I need to add these 8 outputs a certain way to yield one sum and one carry out?
niceboar
#9
Oct24-10, 06:26 PM
P: 54
Quote Quote by dmatador View Post
Okay, that makes sense. I guess my big problem is the gates. Do I need to add these 8 outputs a certain way to yield one sum and one carry out?
Make an input output chart of
L0-L7 and Sum Carry

Remember that one and only one output will be active at a time no matter what combination of X Y C you choose.

Also the answer is yes you do. You need to turn 1 signal into 2 depending on the input and desired output.
dmatador
#10
Oct24-10, 06:36 PM
P: 122
I'm sorry you are helping a lot but I am very slow with this stuff. I am not sure how to make a K-map with s and c_out with the Li's since each output of the decoder L has three components. Eh.
niceboar
#11
Oct24-10, 06:42 PM
P: 54
Quote Quote by dmatador View Post
I'm sorry you are helping a lot but I am very slow with this stuff. I am not sure how to make a K-map with s and c_out with the Li's since each output of the decoder L has three components. Eh.
no doesn't need to be a kmap no need to reduce it yet

Input        intermediate                                   output
x | y | c | L0 | L1 | L2 | L3 | L4 | L5 | L6 | L7 | Sum | Carry
0   0    0    1     0     0    0     0     0     0     0     0        0 
0   0    1    0     1     0    0      0    0     0     0     1        0
0   1    0    0     0     1    0      0    0     0     0     1        0
etc

You know the basic design of an adder right?

edit "quote" this to make it easier to read the spacing is off
dmatador
#12
Oct24-10, 06:47 PM
P: 122
Yes, I know how to make one using two XCORs, two ANDs, and an OR gate.
niceboar
#13
Oct24-10, 06:54 PM
P: 54
Quote Quote by dmatador View Post
Yes, I know how to make one using two XCORs, two ANDs, and an OR gate.
oh so its just the NANDs throwing you off?

You can make any function using NAND using boolean algerbra.

so, if you want to make OR

a ---          a`
        NAND ---- --------------         !(a`b`) apply DeMorgans  a+b
a-----                                 NAND------------------------------
                                       |
b`----        b`                   |
         NAND----------------|
b-----
quote this as well
dmatador
#14
Oct24-10, 06:56 PM
P: 122
Ok so now I have the graph and the decoder with the outputs that yield a sum of 1 hooked up with a nand gate.

I understand how to use the gates to make other gates. What is confusing me is just taking the particular outputs of the decoder that have sums or whatever and translating that using nand gates. The whole idea of using a decoder for this confuses me since I know how to do it with just gates... the function and relationship between the decoder and the gates is what is throwing me off.
niceboar
#15
Oct24-10, 07:05 PM
P: 54
Quote Quote by dmatador View Post
Ok so now I have the graph and the decoder with the outputs that yield a sum of 1 hooked up with a nand gate.

I understand how to use the gates to make other gates. What is confusing me is just taking the particular outputs of the decoder that have sums or whatever and translating that using nand gates. The whole idea of using a decoder for this confuses me since I know how to do it with just gates... the function and relationship between the decoder and the gates is what is throwing me off.
Alright so what you want to do is

L -> Sum, Carry
L0 -> 0 0
but since you can only use NAND

L0 -> 0 0
L1 -> 1 0
L2 -> 1 0
etc


Okay I don't know if you can actually do this but I don't see why you couldn't (someone confirm?)

write out the equation for this function in sums of products form

and partially apply DeMorgan's
i.e.
ABC + CDF

applying DeMorgan's we get
!(ABC) * !(CDF)

look familiar?
dmatador
#16
Oct24-10, 07:20 PM
P: 122
No you are helping a lot. But I am most clueless on how to use these gates after the decoder to complete the circuit.
dmatador
#17
Oct24-10, 07:35 PM
P: 122
Quote Quote by niceboar View Post
Alright so what you want to do is

L -> Sum, Carry
L0 -> 0 0
but since you can only use NAND

L0 -> 0 0
L1 -> 1 0
L2 -> 1 0
etc


Okay I don't know if you can actually do this but I don't see why you couldn't (someone confirm?)

write out the equation for this function in sums of products form

and partially apply DeMorgan's
i.e.
ABC + CDF

applying DeMorgan's we get
!(ABC) * !(CDF)

look familiar?
OK OK this is looking familiar. I will need to formulate an equations that relates L to S and C, and then implement this with the outputs from the decoder? Will I treat L1 as 001, or just as L1? Is this just notation at this point? And also, am I just adding up the sums that go to 1?

Also, should this k-map have the first inputs X and Y Cin also?
niceboar
#18
Oct24-10, 07:37 PM
P: 54
Quote Quote by dmatador View Post
OK OK this is looking familiar. I will need to formulate an equations that relates L to S and C, and then implement this with the outputs from the decoder? Will I treat L1 as 001, or just as L1? Is this just notation at this point? And also, am I just adding up the sums that go to 1?

Also, should this k-map have the first inputs X and Y Cin also?
Since you are using the decoder outputs you want to relate the Ls to Cout and Sum

yes sums of products does add to 1


Register to reply

Related Discussions
Creating a 4 bit full adder logic circuit Engineering, Comp Sci, & Technology Homework 2
Full Adder using a 3-to-8 Decoder in VHDL Electrical Engineering 0
Need help badly..about full adder circuit Engineering, Comp Sci, & Technology Homework 1
Full adder/seven segment Engineering, Comp Sci, & Technology Homework 6
Full adder using decoder? Electrical Engineering 1