I'm just having some difficulty understanding the full-adder and the how the carry-in/out are obtained and what effect it has on the circuit.

See figure attached.

The original question was to design the simpliest circuit that could determine how many bits in a three-bit unsigned number are equal to 1.

With the circuit posted in mind lets consider the scenario where z2=z1=z0=1.

With this we should obtain Cout = 1 and S = 1, giving us a binary value of 3, thus indicating 3 ones were counted in the input.

The part I don't understand is how the full adder obtains these results.

How does z2 + z1 = s = 1, and why is Cout = 1? Also, what is the significance of the carry in? If I change the values of the carry in how does it change the output?

If anyone could clarify it would be greatly appreciated.

Thanks again!

 

Hello,

Take a look at the following links:
http://www.allaboutcircuits.com/vol_4/chpt_9/3.html
http://isweb.redwoods.cc.ca.us/INSTRUCT/CalderwoodD/diglogic/full.htm

Bertus

 

Well... binary numbers only consist of 1 and 0.

The Full Adder is using the method of binary addition

0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 10

So if your input Z1 = Z2 = 1, the addition result will be 1 + 1 = 10

Since there is a Z0 Carry In which is also 1, the result will be 10 + 1 = 11.

Then the 11 divided as 1 for S and 1 for Carry Out.

Another example is if you give 0 for the carry in, the result will be 10 (1 + 1 + 0 = 10), the S will be 0 and Carry Out will be 1.

If the result is only 1 or 0, then it goes to S, and the value of Carry Out is 0.

Cheers