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

  • Thread starter hime
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  • #1
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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
 

Answers and Replies

  • #2
berkeman
Mentor
61,278
11,780
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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