I need to build a circuit that adds two n-bit 2's complement numbers. the two numbers have to be stored in shift registers. The circuit should start with the least significant bits of each number and add each pair of bits using an adder. The sum bit from the adder will then be stored into one of the registers. WE need to use 2 shift registers, an adder and one dflipflop.
Help designing a serial adder
adderflipflopshift-register
Related Solutions
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.
Note that when adding 2 to a binary number, the low bit doesn't change. You therefore simplify the problem to pass the low bit, and a 2-bit adder you add one to for the upper bits. Yes, it really is that easy.
By the way, if you're only ever going to add 1 with a adder, then it's generally called a "counter" instead of a "adder".
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Best Answer
As it seems to be a homework in which you don't show that you tried anything before posting I will just lead you to the answer, although you could have found it with a simple Google search.
Everything you need can be found in those two links:
Registers and Counters
4-Bit Binary Adder / EE 584 – Introduction to VLSI