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 16 bytes (128 bits) at internal RAM locations 0x20-0x2F are bit-addressable.
a byte is abasic storage unit in memory. when application program instructions and data are transferd to memory from storage devices. byte addressable memory refers to memory address that is accessed one byte (8 bits) at a time as opposed to 2 byte(16 bits), 4 byte(32 bits) or 8 byte(64 bits) addressable memory.
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.
4K (4096) of addressable space is defined by 12 bits of address space, because 212 = 4096.
The number of bits in a logical address depends on the architecture of the system. For example, a 32-bit architecture typically uses 32 bits for logical addresses, allowing for 4 GB of addressable memory. In contrast, a 64-bit architecture uses 64 bits, enabling a theoretical maximum of 16 exabytes of addressable space, though practical limits are usually much lower. The actual number of bits can vary based on the specific system design and memory management.
Having 12 bits for a memory address allows for addressing 2^12, or 4,096 unique memory locations. This is commonly sufficient for smaller systems, as it provides a total addressable memory space of 4 kilobytes (KB). The choice of 12 bits balances the need for a compact addressing scheme while still accommodating reasonable amounts of RAM for various applications. In larger systems, more bits may be used to increase the addressable memory capacity.
A subnet mask of 255.255.255.248 corresponds to a /29 prefix, which means there are 3 bits available for host addresses (since 32 bits total - 29 bits for the network = 3 bits for hosts). The formula for calculating the number of addressable hosts is 2^n - 2, where n is the number of host bits. Thus, 2^3 - 2 equals 6 addressable hosts on this network.
There are 16 bits in a port address, which gives a port range of 0 - 65535
There is no such thing as 14-bit byte addressable memory. There IS, however, 16-bit byte addressable memory, which I hope is what you meant. In that case, you simply divide 16 by 4 to get 16/4 = 4 nibbles. If you did in fact mean 14-bit address, then it would be 14/4 = 3.5 nibbles
Hey ! in bit addressable , one can reach the bits (PORT 1.0,PORT1.1,PORT1.2,..etc..) of the ports in the programming while byte addressable only allows the bytes(PORT 1) to be addressed. therefore,there is difference in programming instructions too.
The 8086/8088 has an internal 20-bit address bus and 16-bit data bus. Externally, the address bus is 20-bits, and the data bus is 16-bits for the 8086 and 8-bits for the 8088.The data bus in the 8086 is 16 bits in size, while the address bus is 20.
Bit addressability refers to the capability of a computer system to access and manipulate individual bits within a byte of memory. Unlike byte addressable systems, where memory is accessed in units of one byte (8 bits), bit addressable systems allow for operations on specific bits, enabling finer control and efficient data manipulation. This feature is particularly useful in applications requiring precise control over data storage, such as in embedded systems or digital signal processing.