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16-bit addition/subtraction with an 8-bit adder

by cyenko
Tags: 16bit, 8bit, adder
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cyenko
#1
May13-12, 05:37 PM
P: 4
1. The problem statement, all variables and given/known data

Suppose we have only one 8-bit ripple carry adder but need to do 16-bit addition and subtraction.
Design a sequential circuit (FSM) with only one 8-bit ripple carry adder to implement a 16-bit adder-subtractor.
You are allowed to use MUXs and need to generate the overflow signal.

2. Relevant equations

Overflow = Cn - Cn-1

3. The attempt at a solution

I've attempted to find similar circuit schematics, boolean expressions, diagrams, etc. online to no avail. I am familiar with the implementation with an 8-bit ripple carry adder, and can obtain the final overflow signal through

O=C7 XOR C6
Since I am prohibited from using two 8-bit adders, I was thinking of encoding the given 16 bit numbers using a 16-4 encoder, and then doing the operations, but this would not be valid when multiple bits of the given 16-bit numbers are set high.

I'm looking for guidance on where to begin. I think once I have the general idea on how 16 bits can be stuck in the 8 bit adder, I'll be able to do this problem. Specifically, where should I route the overflow to, and what combination of bits from the two original numbers should I put into the adder?

Thanks for your time,
Chris
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