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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.
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How many bits of address bus are required to address 1mb memory?
You need 20 bits of address bus to address 1 Mb of memory.
How many address bits are required in an 8K X 8 memory?
15 bits
How many address lines are required to address 64 kbytes of memory?
You need 12 address lines to access 4K of memory. 212 = 4096. log2 (4096) = 12.
How many lines of the address must be used to access 2048 bytes. How many of these lines are connected to the address input of all chips?
That depends on the memory architecture of the system.if the memory chips are byte wide and not used to create a multibyte bus, 11 address bits are needed.if the memory chips are 32 bits wide, 9 address bits are needed (with the CPU internally selecting which of the 4 bytes it will use).it the memory chips are 64 bits wide, 8 address bits are needed (with the CPU internally selecting which of the 8 bytes it will use.if the memory chips are 4 bits wide, 12 address bits will be needed and the CPU must perform 2 memory cycles per byte that it needs. (yes, I have seen a computer that worked this way!)etc.
Why is it if MARIE has 4k words of main memory and addresses must have 12 bits?
4K (4096) of addressable space is defined by 12 bits of address space, because 212 = 4096.
Consider a logical address space with 4 pages of 1024 bytes per pageeach mapped onto a physical memory of 64 frames. how many bits are there in logical address?
4 pages -> 2^2 bits 1024 bytes -> 2^10 bits 64 frames -> 2^6 bits Therefore: Logical memory = 2+10=12 bits Physical memory = 10 +6 =16 bits
When the low order bits of the address are used to select the memory bank it is?
When the low order bits of the address are used to select the memory bank it is interleaved.
What is memory address register of Intel 8085?
8 bits
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 memory locations can 14 address bits access?
16384
How many address bits are required for accessing 1GB RAM?
It requires 30 address bits to address 1GB of RAM.230 = 1,073,741,824
Consider a logical address space of eight pages of 1024words each mapped onto a physical memory of 32 frames how many bits are there in the logical address?
As was given for a 4 Page, 1024 words & 64 frames (shown below) 4 pages -> 2^2 bits 1024 bytes -> 2^10 bits 64 frames -> 2^6 bits Therefore: Logical memory = 2+10=12 bits Physical memory = 10 +6 =16 bits The answer for this problem is 13. 8 pages -> 2^3 bits 1024 bytes -> 2^10 bits 32 frames -> 2^5 bits Therefore: Logical memory = 3+10=13 bits (Page + Word) Physical memory = 10 + 5 =15 bits (Word + Frame)
