answersLogoWhite

0


Best Answer

8 bits

User Avatar

Wiki User

โˆ™ 2011-10-11 18:20:55
This answer is:
User Avatar
Study guides

Add your answer:

Earn +20 pts
Q: Memory word size of 8085 microprocessor?
Write your answer...
Submit
Still have questions?
magnify glass
imp
Related questions

What is memory word size required in an 8085 system?

64KB


What is memory word addressing capability of 8085 ยตp?

64K


What determines whether a microprocessor is considered an 8-bit a 16-bit or a 32-bit microprocessor?

The word size (8 bits,16 bits, or 32 bits) of a microprocessor is the size of the data path in the execution unit. Typically, this is the size of the accumulator.Note: This is the execution unit size, not the bus interface unit size. An example where this matters is the 8088, which is a 16 bit computer running on an 8 bit bus.The 8085 is 8 bits. The 8086/8088 is 16 bits. The 80386 is 32 bits. Modern Intel processors are 64 bits.


Refers to the number of bits that a microprocessor can manipulate at one time?

word size


What determines microprocessor is 8 or 32 bit?

The word size of a computer is generally determined by the size of its accumulator. The 8085 was 8 bits. The 8086/8088 was 16 bits. Later incarnations of the Intel x-86 line were 32 bits. Current incarnations of the Intel X-64 line are 64 bits.


What is the instruction format of 8085?

An instruction is a command to the microprocessor to perform a given task on specified data. Each instruction has two parts: One is task to be performed,called the operation code (opcode). Second is the data to be operated on, called the operand. It can be specified in various ways,it may include 8bit/16bit data, an internal register, a memory location , or 8bit/16bit address. In some instructions, the operand is implicit. The 8085 instruction set is classified into three groups according to Word size. They are- 1. One word / 1 byte instructions 2. Ttwo word / 2byte instructions 3. Three word / 3byte instructions


What is the need for address and data alignment for DMA transfer?

Data alignment pertains to the starting address of some block of memory being a multiple of the basic data size, such as a 16 or 32 bit word, or being related to the block size of an IO device, such as a disc drive with a sector size of 512 bytes. The 8085 microprocessor, itself, does not care about alignment, as it is an 8 bit computer running on an 8 bit bus. (The 8086 microprocessor does care, because misalignment can cause performance penalties or corruption, but this article deals with the 8085, not the 8086.) External devices, however, such as the 8237 DMA Controller, may well impose limitations on the starting address of a block transfer, because of their internal design. Each case is specific, so you need to consult the documentation of your specific device.


Types of instruction format of 8085?

The 8085 instruction set is classified into three groups according to its Word size. They are 1. One word /1 byte instructions 2. Two word / 2 byte instructions 3. Three word / 3 byte instructions


What refers to the number of bits that a microprocessor can manipulate at one time?

As quoted from Google Books, "Word size refers to the number of bits that a microprocessor can manipulate at one time."


How do the instruction of 8085 is classified based on their function and word length?

The 8085 instruction set is classified into the following three groups according to word size: 1. One-word or 1-byte instructions 2. Two-word or 2-byte instructions 3. Three-word or 3-byte instructions


What is the wordlength of 8085?

The 8085 is an 8 bit processor, so its word length is 8 bits.


What is the minimum and recommend memory requirement for the OS your system is running?

It depends on a type of microprocessor. The simplest microprocessors had memory word only four bits wide and they could have 15 bites of memory. I suppose that it was enough for some simple calculators.

People also asked