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.
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.
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.
3 Gigabytes
2.00000000.000000
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.
~ 23 068 672
I do not understand the question. You can buy parts that are 8 bits by 512 which is 512 bytes. Static RAM by it's nature is fairly self contained with little in additional circuits since it does not require refresh.