16777216 bits
2097152 bytes
2147483648 bytes
2 bytes or 16 bits
The 16 bytes (128 bits) at internal RAM locations 0x20-0x2F are bit-addressable.
See the related link. According to that info, you could say about 2048 bytes RAM (it had 1K of 16 bit words, but a byte is 8 bits). And 12K ROM.
a) To provide a memory capacity of 4096 bytes using 256x8 RAM chips, you need 4096 bytes / 256 bytes per chip = 16 chips. b) Each memory address for the 256 locations in a chip will require 8 bits (since 2^8 = 256). Therefore, each chip will require 8 address lines to select one of the 256 locations.
Of the 128-byte internal RAM of the 8051, only 16 bytes are bit-addressable. The rest must be accessed in byte format. The bit-addressable RAM locations are 20H to 2FH.
The bit addressable memory in 8051 is compose from 210 bits: - bit address space: 20H - 2FH bytes RAM = 00H - 7FH bits address; - SFR registers; The following addresses are NOT bit addressable, only 1-byte addressable: - 32 bytes RAM from 00H to 1FH (R0 - R7 registers in all four banks); - 80 bytes RAM general user from 30H to 7FH.
3 Gigabytes
2.00000000.000000
A hypothetical 32x1 RAM chip provides storage for 32 bits or 4 bytes. 256k bytes require 256 * 1000 * 8 = 2048000 bits (or 256 * 1024 * 8 = 2097152 bits, if the k is interpreted to mean kibibyte rather than kilobyte, using the IEC nomenclature).Because 2048000 / 32 = 64000, you'd need 64000 chips.
~ 23 068 672
A stick of 512 megabytes of RAM can hold about 512 megabytes of data. It cannot, however, store it for long because it is volatile and is not designed to store data.