digital-logiclogic-gatespower electronics
A full adder is a combinational circuit that performs the arithmetic sum of three bits: A, B
and a carry in, C, from a previous addition
Q) What is the meaning of this bold text? (i.e from a previous addition)
I know what a carry bit is, but when it comes to Full Adder I get confused by this statement…
This circuit is more complicated than necessary, but it looks like C2 does give the correct output. The simplified expression for C2 is: $$ C_2 = \overline{\overline{(A_1+B_1)}+(\overline{C_1} \cdot \overline{A_1B_1})}\\ =(A_1+B_1)\overline{(\overline{C_1}\cdot\overline{A_1B_1})}\\ =(A_1+B_1)(C_1+A_1B_1)\\ =C_1A_1+C_1B_1+A_1A_1B_1+A_1B_1B_1\\ = C_1A_1 + C_1B_1 + A_1B_1 $$
The standard adder / subtract circuit is given by the figure below.For an explanation of how it works please see my previous answer here : Subtraction using adder circuit
Now notice that if you were to change the input that is tied to '0' to Cin, then the circuit would now perform the operation: A + ~B + (Cin ^ SUB), meaning that:
1) it would be a normal Adder/Subtract circuit when Cin is 0.
2) if Cin = 1 and Sub = 1 it would evaluate to:
A + ~B + (1 ^ 1) = A + ~B = A + ~B + 1 - 1 = A + (~B + 1) - 1 = A - B - 1 ,
This means that the normal Add/Sub circuit can achieve the operation you require, all you need to do is change the input tied to '0' your Cin.
You could also design the desired circuit by using the usual method of writing all your possible inputs into a truth table, then deriving the Boolean equations for them, then simplifying them to get your final circuit equation.
Best Answer
For example if you want to add 2 numbers 2 bits each. $$AB+CD$$ First you preform B+D+0 (because you have no carry from a previous addition) and then you preform A+C+carry from a previous addition, i.e. from adding B+C. You do this the same way you do a long addition you did in school.