# Question: How Do You Calculate Address And Data Line?

## How many address lines are required for 1mb memory?

20Data are identified by the physical memory address and the addresses are stored in the form of binary format to allow the data bus to access the memory.

Because 1MB is equal to bytes..

## How is memory size calculated?

In your example for Range 1, you are correct. That is the size of the memory, stated in hexidecimal, in bytes. You may gain the most insight by first converting 00FF FFFF to a decimal number, then converting that number of bytes into megabytes. 1 MB = 1 Megabyte = 1024 * 1 KB = 1,048,576 bytes.

## What is address bus width?

Microprocessor and Memory Basics a is the width of the address bus, while d is the width of the data bus. In many older computers, the address bus was 16 bits wide (a = 16). … ARM processors normally have 32-bit wide address buses. A 32-bit ARM processor could address up to 232 = 4,294,967,296 memory locations!

## How many different addresses does a memory containing 20k words required?

There are 1024 bytes in one k byte. A 16k ram would therefore need 1024*16 = 16384 addresses. Add to that 16*2^4 for the peripherals and you get 16640. You then require at least 15 bits to represent those addresses.

## How many address lines are needed for memory?

If n=1, you can only address 2 locations (0 and 1). If n=2, you can address 2 locations (0, 1, 2, and 3). As you can see, number of addressable locations = n^2. This means that n=log(1024) to the base 2.

## How do you find the memory address?

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 much data can be saved in each memory location?

One memory location stores 1 Byte (8 bits). The basic storage unit for memory is 1 byte. If you need to store 4 bytes, and place the first byte at 0001, the last byte will be at 0004.

## How many bits are stored by a 256 * 4 memory chip?

Please log in or register to add a comment. nd we have ram chips of 256*4. which means there are 256 rows in the ram with 4 bits in each row that is one nibble of data. well total number will be 256.

## What is address line in memory?

An address line usually refers to a physical connection between a CPU/chipset and memory. They specify which address to access in the memory. So the task is to find out how many bits are required to pass the input number as an address. … If your input is 64 kilobytes, the answer is 16 (65536 = 2^16).

## How many separate address and data lines are needed for a memory of 8k 16?

1 comment. 8K∗16=213∗16, thus 13 address lines and 16 data lines.

## How many address lines are required to represent 32k memory?

Ans. 32K = 25 x 210 = 215, Hence 15 address bits are needed; Only 16 bits can address this.

## How many address lines are required for a 2k memory?

11 address linesIt is 2k in size. 11 address lines are needed to address all the addresses inside the EPROM. A similar calculation reveals that the 2K RAM also needs 11 address lines. The PIO chip only has 4 bytes inside, so it only needs 2 address lines.

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

12 address12 address lines are require for 4k memory. How does a 32-bit address represent 1 byte of memory?

## What is the size of a memory address?

32 bitsMemory addresses are numbers too, so on a 32-bit CPU the memory address consists of 32 bits. It works exactly like usual addition, but the maximum digit is 1, not 9. Decimal 0 is 0000 , then you add 1 and get 0001 , add one once again and you have 0010 .

## What is data bus width?

Bus width refers to the number of bits that can be sent to the CPU simultaneously, and bus speed refers to the number of times a group of bits can be sent each second. A bus cycle occurs every time data travels from memory to the CPU.

## What is the highest address in a 48k memory?

2 AnswersA) In a 48K memory Number of bits required is =log(48 * 1024)=16 bit.Now this 16 bit can represent highest address as (2^16)-1 =65535.But since the memory is only 48K it will represent highest address as (48*1024)-1=49151 (since one of the address will be 0)

## What is K in memory?

Technically, therefore, a kilobyte is 1,024 bytes, but it is often used loosely as a synonym for 1,000 bytes. For example, a computer that has 256K main memory can store approximately 256,000 bytes (or characters) in memoryat one time. … In computer literature, kilobyte is usually abbreviated as K or KB.

## How do I find the address bus size?

So to work out the amount of addressable memory, we must multiply the number of addresses by their size.Total Addressable Memory = (2^address bus width) * Data bus width.IE a machine with a 16 bit Data Bus and 32 bit address bus would have.(2^32)*16 bits of accessible storage.or 8GB – Do the math yourself to prove it.