The Intel 8085 microprocessor uses an 8-bit data bus and a 16-bit address bus. It can address 64 KB of memory, with each memory location capable of storing a byte. Therefore, the maximum positive number that can be represented in an 8-bit register is 255 (2^8 - 1), while in a 16-bit address space, the maximum is 65,535 (2^16 - 1) when considering the entire memory range.
The memory address of an integer variable a can be obtained using the address-of operator & in languages like C or C++. For example, using the expression &a will yield the memory address where the variable a is stored. This address is typically represented in hexadecimal format.
To write the value 0xaa55 to the physical memory address 0x67a9, you would typically use a direct memory access method in a programming language that supports low-level memory manipulation, such as C or assembly. For example, in C, you might use a pointer: *(volatile unsigned short *)0x67a9 = 0xaa55;. This ensures that the value is written directly to the specified address in memory. Remember to ensure that you have the necessary permissions and that the memory address is valid for your system.
Pointer.
A variable is a named memory address in which a value may be stored and mutated.
Different microprocessor can address different amounts of memory. The motherboard design should allow for maximising the physical memory to what the microprocessor can address
There is no microprocessor with !t of address memory, only virtual memory. the firt one was the 80486.
microprocessor can access 2^8 points which is 256 then we have 8 bit memory = 1 bytes then 1*256 =256 bytes
256
Even though the 8085 is an 8 bit microprocessor, it can address 64K memory, because it has a 16 bit address bus.
If you assume that it has a 16-bit data bus, then it would be 128k so the microprocessor can access 2^16 points, which is 64k (from it being a 16bit address) 16bits = 2 bytes (memory) so through a 16 bit memory, it can access 2*64k, which is 128k alternatively, if its 8bit memory, 8bits=1byte 1*64k = 64k I'm no expert, and i was searching for the answer myself, hope this helped
16KB
The maximum memory capacity of the 8086 microprocessor is 1 megabyte (MB). This limitation arises from its 20-bit address bus, which allows it to address up to 2^20 memory locations, equating to 1,048,576 bytes or 1 MB. The memory is organized into segments, with the segmentation scheme allowing for more efficient memory management within this address space.
The Intel 8088 microprocessor has 20 address lines. This allows it to address up to 1 MB (2^20 bytes) of memory. The address lines are labeled A0 to A19, enabling the microprocessor to access a wide range of memory locations.
The 8085 microprocessor has a 16-bit address bus, allowing it to address a maximum of (2^{16}) memory locations, which equals 65,536 bytes or 64 KB of memory. This limitation is due to the architecture of the 8085, where each address corresponds to a unique byte in memory. Therefore, while the term "64-bit" may be misleading in this context, it actually refers to the maximum addressable memory space rather than a true 64-bit capability.
Usually memory banks made up of SRAMs or DRAMs or EPROMs consist of the storage area provided on a microprocessor. For understanding how the address space of a 20 bit address line microprocessor is organised, read about address decoding for even and odd memory addressing through SRAMs and EPROMs.
You can address 214 or 16384 different locations with 14 address lines.