Hi,
Kindly guide me

How many full and half adders are required to add 16 bit numbers?

Kindly guide me with this question.

Zulfi.

 

How many half/full adders are require to add 1 bit numbers?

How about 2 bit numbers?

How about n bit numbers?

 

Guidance:

1. Half Adders take two operand bits and produce outputs: sum and carry out
2. Full Adders take two operand bits and a carry in and produce outputs: sum and carry out

Are you seriously telling me that you could not divine this distinction by looking at the definition?!!?
Does that help you with the question?

 

Hi,
Kindly guide me

How many full and half adders are required to add 16 bit numbers?

Kindly guide me with this question.

Zulfi.

Come on, Zulfi!

You have been to the rodeo enough times by now to know that you need to show some kind of attempt at answering the problems you are asking for help on.

This is one where you should be able to at least take a shot at it by looking at what it takes to add 1-bit numbers and then to add 2-bit numbers, and then 3-bit numbers, and make an attempt to figure out the pattern involved.

 

Hi,
Thanks all for trying to help me with this question.
I know the folllowing definitions and i know their truth tables also but i cant understand the pattern.

Half Adders take two operand bits and produce outputs: sum and carry out
Full Adders take two operand bits and a carry in and produce outputs: sum and carry out

How many half/full adders are require to add 1 bit numbers?
H.A=1, F.A=0
How about 2 bit numbers?
I dont know.
How about 3 bit numbers?
F.A=1

Sorry, I cant create the pattern.

Zulfi.

 

Hi,
Thanks all for trying to help me with this question.
I know the folllowing definitions and i know their truth tables also but i cant understand the pattern.

How many half/full adders are require to add 1 bit numbers?
H.A=1, F.A=0
How about 2 bit numbers?
I dont know.
How about 3 bit numbers?
F.A=1

Sorry, I cant create the pattern.

Zulfi.

A full adder adds two 1-bit numbers along with a single 1-bit number (the carry-in) producing a 1-bit sum and a 1-bit carry-out.

A 3-bit adder adds two 3-bit numbers along with a single 1-bit number (the carry-in) producing a 3-bit sum and a 1-bin carry out.

Now, Zulfi, there are millions of webpages out there, not to mention the E-book here, that talk about how to construct multibit adders from full adders. Surely you can do better research than this.

 

Hint: (you could drive a truck through)

Fact: A half adder cannot process a CARRY IN!!!!!

Q: In a multibit adder how many stages, in the range [0,..,n] DO NOT REQUIRE a CARRY IN?

Does that help?

 

The question is a bit ambiguous because we don't know if the adder they have in mind has to process a carry in or not. But that is easily addressed either by stating your assumption or by answering it for both possibilities.

 

The sense I got from the original question seemed to imply a non-zero value for both numbers. In a stand alone adder this might be a reasonable implementation to not process a CARRY IN. In a multifunction ALU the CARRY IN provides a number of useful functions at very little cost.

It might also be worth pointing out that high speed adders don't even use a logic gate implementation. See Mead & Conway (1972)

 

Hi,
Thanks for responding me.
I have seen a web site related to 3 - bit adders. It shows 9 inputs. So we can use 3 FA?
For 16-bit we have to use 16 FA?

Kindly guide me.

Zulfi.

 

You appear to be on the right track, but I have no idea what the 9 inputs for your 3-bit adder would be.

Could you PLEASE provide a drawing of some kind instead of assuming that our crystal balls are good enough to let us peer through your eyes to see what you were looking at?

 

Hi,
Thanks for your quick response. I got information about 3-bit adders from the following site:

http://www.hobbyprojects.com/combination_logic/3_bit_adder.html

Is my answer correct for 16- bit numbers?

Zulfi.

 

Thanks for posting the reference. I still don't see how you are claiming 9 inputs. Please list the nine inputs you are refering to (or perhaps take a screen capture and use paint to circle them).

Look at the block for the least significant bits. Is the carry-in used? If not, then consider what Papabravo said in Posts #3 and #7.

 

Hi,
I have attached the file with 9 inputs encircled. The figure shows that the cin in the first fa is not used. Even then we 8 inputs.

Kindly guide me.

Zulfi.

 

The Cin ports are NOT inputs to the 3-bit adder. If you put this into a box and labeled it "Zulfi's 3-bit adder in a box", which wires would you pass into the box for your input and which wires would you bring out of the box for your output? You would only pass in the two 3-bit values A[2:0] and B[2:0] (so six inputs, total) and you would only bring out the three sum bits, sum[2:0] and the carry out of the final stage (which you might call Cout or you might just call it sum[3]), so you would have four outputs.

 

Hi,
Thanks for your response.

The Cin ports are NOT inputs to the 3-bit adder. If you put this into a box and labeled it "Zulfi's 3-bit adder in a box", which wires would you pass into the box for your input and which wires would you bring out of the box for your output? You would only pass in the two 3-bit values A[2:0] and B[2:0] (so six inputs, total) and you would only bring out the three sum bits, sum[2:0] and the carry out of the final stage (which you might call Cout or you might just call it sum[3]), so you would have four outputs.

This clears the confusion of Cin and Cout. But even then we have to do internal connections. So I think we would need 3 F.A? Is this correct?

Zulfi.

 

If the Cin of an FA is grounded (meaning that it is not used), how is this different, functionally, from a HA?

 

Hi,
Thanks for your reply. This means in case of 3 bit adder we need two F.A and one H.A. Right? I would check for four bit adder and let you know.

Zulfi.

 

Yes. At this point you should be able to generalize this to an N-bit adder. Take a look at what Papabravo said in Post #3 again.

 

Hi,
I have found the diagram of 4-bit adder.
It has 3- F.A and one half adder. So 16- bit adder would have 15 F.A and one HA. Is it right?

Zulfi.