Install Steam
login
|
language
简体中文 (Simplified Chinese)
繁體中文 (Traditional Chinese)
日本語 (Japanese)
한국어 (Korean)
ไทย (Thai)
Български (Bulgarian)
Čeština (Czech)
Dansk (Danish)
Deutsch (German)
Español - España (Spanish - Spain)
Español - Latinoamérica (Spanish - Latin America)
Ελληνικά (Greek)
Français (French)
Italiano (Italian)
Bahasa Indonesia (Indonesian)
Magyar (Hungarian)
Nederlands (Dutch)
Norsk (Norwegian)
Polski (Polish)
Português (Portuguese - Portugal)
Português - Brasil (Portuguese - Brazil)
Română (Romanian)
Русский (Russian)
Suomi (Finnish)
Svenska (Swedish)
Türkçe (Turkish)
Tiếng Việt (Vietnamese)
Українська (Ukrainian)
Report a translation problem
Notice how the first addition adds two digits and produces one sum digit and one carry digit. The second addition adds three digits, one of which is the carry from the previous addition, and produces one sum digit and one carry digit. And so on.
Here's a simple illustration of that: https://imgur.com/NVRzgt4
We all know that 7+3 cannot be 6, so what went wrong? In the third digit (the "4's place"), the 7 has its bit set while the 3 has its bit clear. When you add them, you don't get a carry, so the fourth digit (the "8's place") of the final result does not get set.
The sum outputs all go to the output pins. Where, then, can you connect the carry outputs to ensure that they get involved in the next digit's addition?
Your image and your spoiler tip provided me with what I needed to solve the problem. My rather ugly looking solution is linked here. What I needed to do was instead of carrying the output to the same output pin I needed to put it into the next adder.
https://imgur.com/a/oHh6qPL
Who thought circuits were this much fun.
Happy to help!
What you have here is a "ripple carry" adder, where the carries "ripple" across the sum as it's being generated. There are other adders that are faster at the expense of extra circuitry, but this is a great introduction.
Thanks!