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How many address bits are required to address a 4096 bytes of memory?

How many address bits are required to address a 4096 bytes of memory?

Using 12 address lines you can access whole 4096 bytes of memory location.

How many memory locations can be referenced in 16 bits?

A 16-bit integer can store 216 (or 65,536) distinct values. In an unsigned representation, these values are the integers between 0 and 65,535; using two’s complement, possible values range from −32,768 to 32,767. Hence, a processor with 16-bit memory addresses can directly access 64 KB of byte-addressable memory.

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How many data lines will be required to address a 16 bit RAM memory of 4k size?

12 address lines are require for 4k memory.

How many address bits are required to represent 4k memory?

12 bits
So, 12 bits are needed to address 4k memory locations.

How many address bits are needed to represent 4096 locations?

Answer is “9”

How many address bits are required for a 4096 KB memory which has a data width capacity of 16 bits?

This is simple math. 4096 x 16 = 65536 bits, or 16 bits 4096 times.

What is the maximum supported memory size in a 16-bit architecture with 16-bit memory address if each memory address holds 1 byte data?

The microprocessor can access 216=65536 memory addresses, regardless of the size of data held at each memory address. If each memory address holds one byte (8 bits) of data then the total memory size that can be addressed directly is 216 bytes, which is 26 KiB=64 KiB.

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How many address lines are needed to address each machine location in a 4096 4 memory chip?

ANSWER: 11 It means that a memory of 2048 words, where each word is 4 bits.

How many address and data lines are in 1m 16 ROM system?

None of the above Correct solution is (c). Since there are 16M words, the number of address lines will be 24, since 224 = 16M.

How many address bits are required for a 4096 KB memory which has a data with capacity of 16 bits?

How many address bits are required to represent 8K memory?

So, therefore we need an additional 3 address bits to increase this by a factor of 8 (8 = 2 cubed), giving a total of 13 address bits.