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How many memory address does this number of address lines allow the 8086 to access directly?
Wiki User
∙ 13y ago
The 8086/8088 has 20 address lines. It can access 220, or 1MB, or 1,048,576 bytes of memory.
Wiki User
∙ 13y ago
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How many memory addresses does this number of address lines allow the 8086 to access directly?
The 8086/8088 has 20 address lines. It can access 220, or 1MB, or 1,048,576 bytes of memory.
How many address lines are needed to access 256KB of main memory?
The number of address lines needed to access N-KB is given by log2N Then the number of address lines needed to access 256KB of main memory will be log2256000=18 address lines.
Minimum number of address lines needed to access a physical memory of 1GB?
You need 30 address lines to access 1G of memory. 230 = 1,073,741,824. log2 (1,073,741,824) = 30.
A name or number used to identify a storage location is called?
It refers to Memory Address since variable will never be a NUMBER it is Memory Address like an HexaDecimal Address of FFFFH. Eds
How many nuMBer of address lines required for 8 MB of memory?
It takes 23 address lines to address 8 mb of memory.
How many address lines are required to access 1MB RAM using microprocessor 8086?
The 8086/8088 microprocessor has a 20 bit address bus, so the number of memory locations it can address is 220 or 1,048,576.
What is the mean of Real memory addressing mode?
In real mode addressing for x86 cpus, memory is addressed with pairs of a segment and offset. The offset is added to the segment address multiplied by 16 to yield a 20-bit (20 binary digits, in other words, from the number 0 to (2^20)-1=1,048,575) address that points to a specific byte (8-bit number) in memory. Real mode is different than protected mode (which is used by Windows 95+, Linux on x86, etc) in that there is no segment protection, no inherent multitasking support, and it is possible to directly access the BIOS interrupts. Note also that the 20-bit address number prevents more than one (1) megabyte of memory from being addressed at a time.
What is the address of the last byte in a 512 mega byte memory expressed as a decimal number?
The address of the last byte in a 512 mega byte memory, expressed as a decimal number, is 536,870,911.
What is the memory's addressibility?
HI I am Ahtarva,The addressibility is how many bits does that particular processor or micro-controller's architecture use to specify the address of a memory location in the memory. For example if someone say that addressibility is 8 bit then your memory address contains 8 bits and at maximum you have 2^8 different memory locations (or say memory addresses in your device). Here 2^8 is called Address space.
What number system is used to display a memory address?
it is decimal unsigned number system...
The number of memory locations that a CPU with 16 bit program counter can address is?
The number of memory locations that a CPU with 16 bit program counter can address is 65,536. However, the 8086/8088 has a segmented architecture giving a total addressibility of 20 bits or 1,048,576 locations. Without changing the code segment register, though, you can only access 65,536 locations.
The temporary storage available inside the computer is called?
There are a number of people who have asked this same question. The Answer is still Random Access Memory
