What is the number of bits required for addressing 4kb memory?

Discussion Forum

Que.	How many address bits are required to represent 4K memory
b.	12 bits
c.	8 bits
d.	10 bits
	Answer:12 bits

How many bits would you need to address a 4kb memory if the memory is word addressable with a word size of 16 bits?

Memory is word addressable and a cache block size is 16 words meaning 4 offset bits. 128 sets mean 7 set bits are required. = 320 bytes. So, option B.

How many bits would be in the memory of a computer with 4kb?

4kb means 4000 bits.

How many address bits are needed for all memory?

Likewise, you need 20 bits to address every byte in a megabyte, and 30 bits to address every byte in a gigabyte. 232 = 4294967296, which is the number of bytes in 4 gigabytes, so you need a 32 bit address for 4 GB of memory.

How do I calculate an address size?

To put it another way, Step 1: calculate the length of the address in bits (n bits) Step 2: calculate the number of memory locations 2^n(bits) Step 3: take the number of memory locations and multiply it by the Byte size of the memory cells.

How many bits would you need to address a 2M 32 memory if?

(2M x 32) memory module. 1. Individual Byte Addressing – 2 bits required. required to resolve all (2^2 = 4) bytes in each 32-bit word.

How many bits are needed for opcode?

The instruction consists of opcode and operands. Given the instruction set of size 12, 4 bits are required for opcode (2^4 = 16). As there are total 64 registers, 6 bits are required for identifying a register.

How many bits would you need to address a memory if the memory is byte addressable?

Therefore 34 bits are required to uniquely address each byte. . Therefore 28 address bits are needed. For (b) & (c) we need to divide the address by the number of bytes in a memory word since we have byte addressing, e.g. in this case divide by 4 since there are 4 bytes in a 32-bit memory- word.

How many RAM chips are there per memory word?

The design calls for two chips per 16–bit word. One chip holds the high–order byte for the word. The other chip holds the low–order byte for the word. Suppose that a 2M x 16 main memory is built using 256K x 8 RAM chips and memory is word addressable.

How many 256×8 RAM chips are needed to provide a memory capacity of 2048 words?

Each RAM chip has 64 x 1 byte = 64 bytes. Thus the number of chips to address a memory capacity of 2048 bytes will be, 2048/64 = 32 chips.

What is high order interleaving?

1. High Order Interleaving – In high-order interleaving, the most significant bits of the address select the memory chip. The least significant bits are sent as addresses to each chip. One problem is that consecutive addresses tend to be in the same chip.

How are RAM chips calculated?

Each chip contains 8 bits. Because it takes 8 bits to make 1 byte, the capacity of the module can be calculated by grouping the memory chips on the module into groups of eight. If each chip contains 512Mb, a group of eight means that the module has a size of 512MB (512Mb x 8 = 512MB).

How many 128×8 RAM chips are needed to provide?

Each RAM chip has 64 x 1 byte = 64 bytes. Thus the number of chips to address a memory capacity of 2048 bytes will be, 2048/64 = 32 chips….GO Book for GATECSE 2022.

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What is the memory capacity if we have 8 modules of 8 bytes each?

You may have encountered examples of chip densities, such as “64Mbit SDRAM” or “8M by 8”. A 64Mbit chip has 64 million cells and is capable of holding 64 million bits of data. The expression “8M by 8” describes one kind of 64Mbit chip in more detail.

How many 1024 * 1 RAM chips are required to construct a 1024 * 8 Memory System 2 points?

RAM chip size = 1k ×8[1024 words of 8 bits each] RAM to construct =16k ×16 Number of chips required = (16k x 16)/ ( 1k x 8) = (16 x 2) [16 chips vertically with each having 2 chips horizontally] So to select one chip out of 16 vertical chips, we need 4 x 16 decoder.

How many address bits are required for a 1024 * 8 memory?

Or 1024*8 gives 8192, or 2^13. so 13 bits. If it ain’t a 1024 bits multiplied by 8 bits, or there are 1024 arrays of each size of 8 bit, then 1024 arrays, so 10 bits of addresses will do it.

How many chips are needed to construct a 16K 16 Ram?

So, to select one chip out of 16 vertical chips, we need 4 × 16 decoder. Available decoder is 2 × 4 decoder….Correct Option: B.

Number of chips required =	16K × 16	= 16 × 2
Number of chips required =	1K × 8	= 16 × 2

What will be Decoder size if we want to design 512 * 8 RAM from given 128 * 8 RAM?

In a 512 x 8 RAM, there is a 9-bit address, we will divide this 9-bit address into 2 and 7-bit addresses. The 2-bit address will be input as a 2 x 4 decoder, this decoder will have 4 output and the input will have 2 bits. Also, we can write 128 x 8 RAM chip as 27 x 8, every RAM chip will need a 7-bit address.

Is ROM main memory?

Memory Basics Random Access Memory (RAM) is primary-volatile memory and Read Only Memory (ROM) is primary-non-volatile memory. It is also called as read write memory or the main memory or the primary memory. It is a volatile memory as the data loses when the power is turned off.