8 bit adder/sub with 4 bit ripple adder and muxes only

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SUMMARY

The discussion centers on designing an 8-bit adder/subtractor using a single 4-bit ripple carry adder and multiplexers (muxes). The proposed solution involves calculating the sum of the first 4 bits with the ripple carry adder and using full adders constructed from muxes for the next 4 bits, requiring 8 muxes. Participants seek advice on simplifying this design while ensuring it accommodates subtraction functionality. A request for a block diagram is made to facilitate further suggestions.

PREREQUISITES
  • Understanding of 4-bit ripple carry adders
  • Knowledge of full adder design using multiplexers
  • Familiarity with binary addition and subtraction concepts
  • Basic digital circuit design principles
NEXT STEPS
  • Research optimization techniques for ripple carry adders
  • Learn about alternative adder designs, such as carry-lookahead adders
  • Explore the implementation of full adders using fewer muxes
  • Study block diagram creation for digital circuit designs
USEFUL FOR

Digital circuit designers, electrical engineering students, and anyone interested in optimizing arithmetic logic unit (ALU) designs.

hime
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Hi
I'm having trouble with coming up with a minimal solution to this problem: create a 8 bit adder/sub using only one 4 bit ripple carry adder and muxes. This is what I'm thinking but I don't know how to make it any bit simpler:

We can calculate the sum of first 4 bits using the ripple carry adder and then calculate the sum of the next 4 bits(bits 4-7) using full adders built with mux's. A full adder needs 2 muxes(one for sum output and another for carry out) and this means I need to use 8 mux's to calculate the sum of bits 4-7...any hints on how to improve on my idea or how to make this design more simpler? Thanks
 
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hime said:
Hi
I'm having trouble with coming up with a minimal solution to this problem: create a 8 bit adder/sub using only one 4 bit ripple carry adder and muxes. This is what I'm thinking but I don't know how to make it any bit simpler:

We can calculate the sum of first 4 bits using the ripple carry adder and then calculate the sum of the next 4 bits(bits 4-7) using full adders built with mux's. A full adder needs 2 muxes(one for sum output and another for carry out) and this means I need to use 8 mux's to calculate the sum of bits 4-7...any hints on how to improve on my idea or how to make this design more simpler? Thanks

Are you saying you need to implement the subtractor too?

Can you show a block diagram of your initial solution? That may help us to suggest possible simplifications.
 

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