
#1
Oct2410, 05:12 PM

P: 122

I'm trying to create a full adder using one 3to8 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?




#2
Oct2410, 05:54 PM

P: 54

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. 



#3
Oct2410, 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....




#4
Oct2410, 06:11 PM

P: 54

Creating a full adder using a 3to8 decoderanyway 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 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 



#5
Oct2410, 06:15 PM

P: 122





#6
Oct2410, 06:17 PM

P: 54

So L3 being on represents X=0 Y=1 C=1 



#7
Oct2410, 06:22 PM

P: 4,513

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. 



#8
Oct2410, 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?




#9
Oct2410, 06:26 PM

P: 54

L0L7 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. 



#10
Oct2410, 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 Kmap with s and c_out with the Li's since each output of the decoder L has three components. Eh.




#11
Oct2410, 06:42 PM

P: 54

You know the basic design of an adder right? edit "quote" this to make it easier to read the spacing is off 



#12
Oct2410, 06:47 PM

P: 122

Yes, I know how to make one using two XCORs, two ANDs, and an OR gate.




#13
Oct2410, 06:54 PM

P: 54

You can make any function using NAND using boolean algerbra. so, if you want to make OR




#14
Oct2410, 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. 



#15
Oct2410, 07:05 PM

P: 54

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? 



#16
Oct2410, 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.




#17
Oct2410, 07:35 PM

P: 122

Also, should this kmap have the first inputs X and Y Cin also? 



#18
Oct2410, 07:37 PM

P: 54

yes sums of products does add to 1 


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