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Total Addressable Memory
65,536 bytes
Total Addressable Memory vs Address Bus Width (bits)
Formula
## Memory Address Space The address bus width determines how many unique memory locations a processor can access. ### Formula **Locations = 2^Address_Bits** **Total Memory = Locations x (Data_Bus_Width / 8) bytes** ### Common Examples - 8-bit address: 256 bytes - 16-bit address: 64 KB - 20-bit address: 1 MB (original PC) - 32-bit address: 4 GB - 64-bit address: 16 EB (theoretical)
Esempio Risolto
16-bit address bus with 8-bit data bus.
- 01Locations: 2^16 = 65,536
- 02Bytes per location: 8/8 = 1
- 03Total: 65,536 x 1 = 65,536 bytes
- 04In KB: 65,536 / 1024 = 64 KB
Domande Frequenti
Why can 32-bit systems only address 4 GB?
2^32 = 4,294,967,296 bytes = 4 GB. Physical Address Extension (PAE) can extend this on some architectures.
Does data bus width affect address space?
Not the number of locations, but it affects total bytes. A 16-bit data bus accesses 2 bytes per location.
What is memory-mapped I/O?
Hardware registers are placed in the address space, reducing available RAM but simplifying peripheral access.
Impara