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

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

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

How many address bits are required for a 1024 x 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 address lines are available in a 16k x 8 Eprom chip?

A JEDEC 27256 EPROM (32K bytes) has 15 address lines.

How many address bits are needed for a memory of size 2kb x 8?

There are k address lines and therefore 2k addresses. Each of these addresses contains an n-bit word. In this instance, 2k = 2048 = 211. You need 11 address lines.”

What is 16K memory?

16K means 16,384 ‘storage locations’ of 8 bits, thus 16KB. 64 x 16 means 64 storage locations of 16 bits, thus 128 (8 bit) bytes.

What is the size of the main memory?

Main-memories generally store and recall rows, which are multi-byte in length (e.g. 16-bit word = 2 bytes, 32-bit word = 4 bytes). Most architectures however, make Main Memory byte-addressable rather than word addressable.

How many bits are transferred in one memory cycle of the MBR is 64 bit?

64 bits

What is the difference between MBR and Mar?

To access data in memory, CPU makes use of two internal registers: MAR (memory address register) and MBR (memory buffer register). MAR specifies the address in memory for the next read/write; MBR otherwise contains the data to be written into memory, or data to be read from memory.

What is the difference between IBR and IR register?

Instruction Buffer Register IBR is holds the full instruction from a word memory. IBR holds the instruction of a word… while MBR stores the word itself. instruction register In computing, an instruction register (IR) is the part of a CPU’s control unit that stores the instruction currently being executed or decoded.

What allows you to specify exactly how many bits of the address are fixed?

Most modern computers are byte-addressable. Each address identifies a single byte (eight bits) of storage. Data larger than a single byte may be stored in a sequence of consecutive addresses.

Are 4 bits enough to store 8 directions?

Four bits are not enough to store the eight directions. Eight bits are needed for the new version of the game.