design 3 bit adder from full adders
- Thread starter eldadoh
- Start date
Hello,
The two pictures do not show.
They appear as broken links in my browser.
Can you post the pictures direct to the forum?
Here is how to do this:
http://forum.allaboutcircuits.com/faq.php?faq=vb3_reading_posting#faq_vb3_attachments
Have a look at the following page:
http://www.cs.umd.edu/class/spring2003/cmsc311/Notes/Comb/adder.html
This is one of the pages on arithmetic ciruits on the EDUCYPEDIA:
http://educypedia.karadimov.info/electronics/digitaarithmetic.htm
Bertus
The two pictures do not show.
They appear as broken links in my browser.
Can you post the pictures direct to the forum?
Here is how to do this:
http://forum.allaboutcircuits.com/faq.php?faq=vb3_reading_posting#faq_vb3_attachments
Have a look at the following page:
http://www.cs.umd.edu/class/spring2003/cmsc311/Notes/Comb/adder.html
This is one of the pages on arithmetic ciruits on the EDUCYPEDIA:
http://educypedia.karadimov.info/electronics/digitaarithmetic.htm
Bertus
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Hello, and welcome to the forums.
You will need to put some effort into the solution before we can help you.
What do you think you may have to do?
Also, when you add in decimal, at what point do you carry over to the next digit? Does that occur in binary? If so, where?
When posting in Homework Help, be sure to include your best guess, or at least something you know about the problem so we can have a jumping off point for discussion.
You will need to put some effort into the solution before we can help you.
What do you think you may have to do?
Also, when you add in decimal, at what point do you carry over to the next digit? Does that occur in binary? If so, where?
When posting in Homework Help, be sure to include your best guess, or at least something you know about the problem so we can have a jumping off point for discussion.
well digital design is new for me , i am at the end of my first semester.the soultion is only from the subject of Adders and logic gates.
about my question , I understand the truth table of HA and FA and the logic meaning of those components, but in FA there is sum of 2 binary digits and a carry and in my question there are 3 binary entries and a carry and I don't understand how I can build the correct truth table .
about my question , I understand the truth table of HA and FA and the logic meaning of those components, but in FA there is sum of 2 binary digits and a carry and in my question there are 3 binary entries and a carry and I don't understand how I can build the correct truth table .
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Okay, you know that when you add (in decimal), you start from the lowest significant digit, add the two values and add any digits that exceed the possible values for a single digit (above 0-9) to the next higher digit.
Can you make any connections with doing this in binary?
Can you make any connections with doing this in binary?
in binary I know that when I add :
0+0=the sum is 0 and the carry is 0
0+1=the sum is 1 and the carry is 0
1+0=sum 1 carry 0
1+1=sum 0 carry 1
and when there is a carry :
0+0 +carry 0=sum 0 carry 0
0+0 +carry 1=sum 1 carry 0
0+1 +carry 0=sum 1 carry 0
0+1 +carry 1=sum 0 carry 1
1+0 +carry 0=sum 1 carry 0
1+0 +carry 1=sum 0 carry 1
1+1 +carry 0=sum 0 carry 1
1+1 +carry 1=sum 1 carry 1
but in 3 entries and a carry
0+0+0 +carry 0=sum 0 carry 0
0+0+0 +carry 1=sum 1 carry 0
0+0+1 +carry 0=sum 1 carry 0
0+0+1 +carry 1=sum 0 carry 1
0+1+0 +carry 0=sum 1 carry 0
0+1+0 +carry 1=sum 0 carry 1
0+1+1 +carry 0=sum 0 carry 1 ??
0+1+1 +carry 1=sum 1 carry 1 ??
1+0+0 +carry 0=sum 1 carry 0 ??
1+0+0 +carry 1=sum 0 carry 1 ??
1+0+1 +carry 0=sum 0 carry 1 ??
1+0+1 +carry 1=sum 1 carry 1 ??
1+1+0 +carry 0=sum 0 carry 1 ??
1+1+0 +carry 1=sum 1 carry 1 ??
1+1+1 +carry 0=sum 1 carry 1 ??
1+1+1 +carry 1=sum 1 carry 1 second carry 1 ????? OR is it s1 (s1>sum=s0) =1 sum =0 carry =0 ??
This is how my point of view, and I think i have a mistake.
0+0=the sum is 0 and the carry is 0
0+1=the sum is 1 and the carry is 0
1+0=sum 1 carry 0
1+1=sum 0 carry 1
and when there is a carry :
0+0 +carry 0=sum 0 carry 0
0+0 +carry 1=sum 1 carry 0
0+1 +carry 0=sum 1 carry 0
0+1 +carry 1=sum 0 carry 1
1+0 +carry 0=sum 1 carry 0
1+0 +carry 1=sum 0 carry 1
1+1 +carry 0=sum 0 carry 1
1+1 +carry 1=sum 1 carry 1
but in 3 entries and a carry
0+0+0 +carry 0=sum 0 carry 0
0+0+0 +carry 1=sum 1 carry 0
0+0+1 +carry 0=sum 1 carry 0
0+0+1 +carry 1=sum 0 carry 1
0+1+0 +carry 0=sum 1 carry 0
0+1+0 +carry 1=sum 0 carry 1
0+1+1 +carry 0=sum 0 carry 1 ??
0+1+1 +carry 1=sum 1 carry 1 ??
1+0+0 +carry 0=sum 1 carry 0 ??
1+0+0 +carry 1=sum 0 carry 1 ??
1+0+1 +carry 0=sum 0 carry 1 ??
1+0+1 +carry 1=sum 1 carry 1 ??
1+1+0 +carry 0=sum 0 carry 1 ??
1+1+0 +carry 1=sum 1 carry 1 ??
1+1+1 +carry 0=sum 1 carry 1 ??
1+1+1 +carry 1=sum 1 carry 1 second carry 1 ????? OR is it s1 (s1>sum=s0) =1 sum =0 carry =0 ??
This is how my point of view, and I think i have a mistake.
You are looking at the big picture, while I am taking about a single stage whose outputs become the next stage's inputs - the principle behind a ripple-carry adder.
So, if you have \(a + b, where a ={a_{2},a_{1},a_{0}} and b ={b_{2},b_{1},b_{0}} \), you add them in stages.
So, let's add this is decimal and let a=133 and b=496
Stage 0:
\(a_{0}+b_{0} = 3+6 =9\)
The sum(\(s_{0}\)) is 9 with no carry(\(c_{0}\)).
Stage 1:
\(a_{1}+b_{1} +carry_{0} = 3+9+0 =12\)
The sum(\(s_{1}\)) is 2 with a carry(\(c_{1}\)) of 1.
Stage 2:
\(a_{2}+b_{2} +carry_{1} = 1+4+1 =6\)
The sum(\(s_{2}\)) is 6 with a carry(\(c_{2}\)) of 0.
The result is comprised of the sum of each stage s and the final carry, where \(s={s_{2},s_{1},s_{0}}\), so the result is \( result={carry_{2},s_{2},s_{1},s_{0}} = {0,6,2,9}\), which tells us the result of 133 + 496 = 629.
Looking at it like this, can you extend this method to a binary adder using full adders?
So, if you have \(a + b, where a ={a_{2},a_{1},a_{0}} and b ={b_{2},b_{1},b_{0}} \), you add them in stages.
So, let's add this is decimal and let a=133 and b=496
Stage 0:
\(a_{0}+b_{0} = 3+6 =9\)
The sum(\(s_{0}\)) is 9 with no carry(\(c_{0}\)).
Stage 1:
\(a_{1}+b_{1} +carry_{0} = 3+9+0 =12\)
The sum(\(s_{1}\)) is 2 with a carry(\(c_{1}\)) of 1.
Stage 2:
\(a_{2}+b_{2} +carry_{1} = 1+4+1 =6\)
The sum(\(s_{2}\)) is 6 with a carry(\(c_{2}\)) of 0.
The result is comprised of the sum of each stage s and the final carry, where \(s={s_{2},s_{1},s_{0}}\), so the result is \( result={carry_{2},s_{2},s_{1},s_{0}} = {0,6,2,9}\), which tells us the result of 133 + 496 = 629.
Looking at it like this, can you extend this method to a binary adder using full adders?
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This is an intuitive approach - it is what you do everytime you add things, but you must look at it as a series of steps in order to do this.
If you truly want a truth table, fill this out understanding that your sum is now two bits and the carry is still just one bit:
A|B|C|Cin|S1|S2|Cout
0|0|0|0
0|0|0|1
0|0|1|0
0|0|1|1
0|1|0|0
0|1|0|1
0|1|1|0
0|1|1|1
1|0|0|0
1|0|0|1
1|0|1|0
1|0|1|1
1|1|0|0
1|1|0|1
1|1|1|0
1|1|1|1
A truth table covers any and all possible inputs and shows the resulting outputs.
If you truly want a truth table, fill this out understanding that your sum is now two bits and the carry is still just one bit:
0|0|0|0
0|0|0|1
0|0|1|0
0|0|1|1
0|1|0|0
0|1|0|1
0|1|1|0
0|1|1|1
1|0|0|0
1|0|0|1
1|0|1|0
1|0|1|1
1|1|0|0
1|1|0|1
1|1|1|0
1|1|1|1
A truth table covers any and all possible inputs and shows the resulting outputs.
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Consider how you add two decimal numbers. Say
When you are adding the hundreds column, what information do you need to know what the hundreds digit is in the answer and whether or not there is a carry to the thousands digit?
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