What happens if you double the width of the address bus?

By increasing the width of the address bus, more memory locations can be directly addressed. Each time the width is increased by one wire, the address capacity is doubled. A 32-bit ARM processor could address up to 232 = 4,294,967,296 memory locations! In older computers, the data bus was a mere 8 bits wide (d=8).

How many address lines are needed for memory?

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. Thus, n=10.

How many lines are there in address bus?

The width of the address bus (that is, the number of wires) determines how many unique memory locations can be addressed. Modern personal computers and Macintoshes have as many as 36 address lines. This theoretically allows them to access 64 gigabytes of main memory.

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

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

How is address bus and memory size related?

Address Bus stores the location of a byte of memory. If an address bus is of size 32 bits, that means it can hold upto 232 numbers and it hence can refer upto 232 bytes of memory = 4GB of memory and any memory greater than that is useless. Data bus is used to send the value to be written to/read off the memory.

How many memory locations can a 16 bit address bus address?

65,536
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.

How do you calculate address lines needed?

Just keep count of the number of times you have doubled to get the address lines needed. 16 address lines would give 64kbits (65536) if I remember correctly.

How many address lines are required to represent 32K memory?

4K bytes) has 12 address lines. A JEDEC 27256 EPROM (32K bytes) has 15 address lines. A JEDEC 27512 EPROM (32K words, aka. 64K bytes) also has 15 address lines, but holds twice as much as a 27256.

What is memory address bus?

An address bus is a computer bus architecture used to transfer data between devices that are identified by the hardware address of the physical memory (the physical address), which is stored in the form of binary numbers to enable the data bus to access memory storage.

What is the number of address bus lines and data bus lines used with a memory of 64 Kword?

For example, 16 address lines would be 64k bytes, assuming 8 data lined. Also, in the past, DRAM chips were 1 bit. A 64k bit DRAM was using 8 address lines. The number of Address lines gives you the memory capacity.

How many address and data lines will be there for a 16M 32 memory system Mcq?

How many address and data lines will be there for a 16M x 32 memory system? d. 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 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 addresses can a 32-bit address bus represent?

a system with a 32-bit address bus can address 2 32 (4,294,967,296) memory locations. Total 8 values. So using ABC, you can access any of those eight values, i.e., you can reach any of those memory addresses. So, TL;DR, the simple relationship is, with n number of lines, we can represent 2 n number of addresses.

How many address lines does it take to fill 64K of memory?

If your machine always loaded say 64B cache lines, and your RAM was set up to deliver 64B bursts from a requested address, you’d only need 10 address lines to cover the same 64k of memory. The CPU would sort out which byte the load actually wanted internally, without needing to put the .

How many bits are in an 8-line data bus?

Every time you read a location (by loading its address on the address bus), the 8 bits that are stored at that location are loaded (by the memory chip) on the 8-line data bus. As obvious, your memory has 1024*8 bits (8192 bits). Or simply, 1024 bytes 🙂

How many unique values can you put on the address bus?

Since there are 2^16 unique values you can put on the address bus, and you want to be able to address your memory with byte granularity, each value maps to a single byte.