Practice question · Put in order
Order what happens when a four-bit ripple-carry adder computes a sum.
- The carry-out of the leftmost stage becomes the adder's final carry
- Its carry-out arrives at the carry-in of the stage to its left
- The carry travels left through each remaining stage in the same way
- The rightmost stage adds the two least significant bits together with the incoming carry
- That stage can now settle and produce its own sum bit and carry-out
Hints
- Ask what each stage is waiting for before it can produce a settled answer.
- No stage except the rightmost has a valid carry-in at the start.
Show the answer
- The rightmost stage adds the two least significant bits together with the incoming carry
- Its carry-out arrives at the carry-in of the stage to its left
- That stage can now settle and produce its own sum bit and carry-out
- The carry travels left through each remaining stage in the same way
- The carry-out of the leftmost stage becomes the adder's final carry
Why
Each stage needs its carry-in before it can settle, so the stages finish one after another from right to left rather than all at once. Assuming the stages work in parallel is exactly the misconception that makes ripple-carry delay surprising, the sum bits appear in sequence, not simultaneously.
Practise Adders, Multiplexers, and Decoders
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