How many address bits are required to access 128 bytes?

How many address bits are required to access 128 bytes?

The memory address space is 128 MB, which means 227. However, each word is 8 (23) bytes, which means that you have 224 words. This means you need log2 224 or 24 bits, to address each word.

How are memory locations addressed?

Each memory location has a physical address which is a code. The memory controllers’ bus consists of a number of parallel lines, each represented by a binary digit (bit). The width of the bus, and thus the number of addressable storage units, and the number of bits in each unit, varies among computers.

How many bytes is a memory address?

Most commonly, modern systems are what you call “byte-accessible”. This means: One memory location stores 1 byte (8 bits).

How do you find the starting and ending address of memory?

In this example, the starting addresses of all memory chips are integer multiple of the sizes of the corresponding memory chips. ending address = starting address + memory chip size – 1. For example, if the starting address of EEPROM is 2000H and the EEPROM size is 1K (400H), the ending address of EEPROM is 23FFH.

How many bits can a 2K memory unit store?

In general when discussing RAM, EEPROM and FLASH memories the capacity is given in 2^X format. Therefore 2K represents 2^11 = 2048 Bytes. Assuming the starting address is 0x00, and the memory is organized as 8 bit words, the last address would be 0x7FF (2047d).

How many bits does an address line have?

3 Answers. One address addresses one byte. Using 16 bits, you can write 65536 addresses (from 0 to 65535, that’s 65536 different addresses), and address 65536 bytes.

What is the size of physical address of any memory location?

Physical memory is 32MB, byte-addressable, and words are 4 bytes each. Virtual addresses are 32 bits, and pages are 16kB.

What is a starting address?

Starting address can mean two different things. It may refer to the first address of a specific Modbus slave table e.g. holding registers start at address 40001. Or it may refer to the first address when reading for example multiple holding registers.

How many bits can a 4K memory store?

The “4K x 8” notation indicates memory organization: it means there are 4096 memory locations, each containing 8 bits.

How many address bits are required for a 512 * 4 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. OK so for 512 Mbyte RAM the equation is 512*1024²=2^x (?)

How big is the last byte of memory?

To access the last byte, CPU skips 1073741823 bytes, so its address is 1073741823. Now you have to know what 32-bit actually means. As I mentioned before, it’s the size of a machine word.

Is the address of memory a byte or a byte?

With possibly rare exceptions, current products address memory in units of bytes. so an “address” is a “byte address”. Now saying that as you dig into the logic, there are MANY cases where buses will have non-byte addressing.

Which is the last address for a 16 bit transfer?

But 0x1FFE is the last address for a 16 bit transfer and 0x1FFC the last address for a 32 bit transfer. Side note: memory in general when you are working as a chip engineer or board designer is in units of bits.

How to calculate the ending address of memory?

If a chip is said to have 8KBytes of flash in its documentation then yes that means 0x8 * 0x400 = 0x2000 so there are 0x2000 individual bytes that can be accessed in that flash in the address space of BASE+0x0000 to BASE+0x2000-1 or BASE+0x0000 to BASE+0x1FFF. If BASE is 0x00000000 then 0x0000 to 0x1FFF, that is correct.