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

  1. May 13, 2012 #1
    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

    Code (Text):
    Overflow = C[SUB]n[/SUB] - C[SUB]n-1[/SUB]

    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

    Code (Text):
    O=C[SUB]7[/SUB] XOR C[SUB]6[/SUB]
    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
     
  2. jcsd
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