The 8086/8088 microprocessor family is a 16 bit microprocessor. The 8086 implementation also has a 16 bit data bus, but the 8088 implementation has an 8 bit data bus, comparable to the 8085. The 8088 implementation was intended as a logical upgrade from the 8085, while keeping the complexity of the system on an equal footing as the 8085.
so they fit correctly into an 8bit or 16bit slot
Rimm
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
If you assume that it has a 16-bit data bus, then it would be 128k so the microprocessor can access 2^16 points, which is 64k (from it being a 16bit address) 16bits = 2 bytes (memory) so through a 16 bit memory, it can access 2*64k, which is 128k alternatively, if its 8bit memory, 8bits=1byte 1*64k = 64k I'm no expert, and i was searching for the answer myself, hope this helped
16bit refers to the number of bits that can be used in a single data format element. 16bit graphics are found in games that can be played on the original Nintendo Entertainment System console.
The h-L register in computing refers to a pair of registers used in some processors to store a 16-bit memory address. It is often used for memory operations and calculations involving memory addresses. The h register stores the high-order byte while the l register stores the low-order byte of the address.
Standard VGA is 16bit.
48bit does not belong in the list. In relation to desktop computers, it goes: 8bit, 16bit, 32bit, 64bit, 128bit... It refers to the internal design of the microprocessor. 8bit was the original register width in the first generation of computers etched into a single chip. The register size doubles from previous generations to their successors. It's not that a 48bit computer cannot be designed and manufactured, it's more of a decision to expand the capacity of the machine by copying the current design next to the existing hardware and weaving the original and copy together.
give address according to size of memory like 128x8 FOR SINGLE CHIP it means 27=128, 7no off line and 16bit add bus &data bus 8bit, add start 0000H to 007F A15 A14 A13 A12 A11 A10 A9 A8 AD7 AD6 AD5 AD4 AD3 AD2 AD1 AD0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1
Its 16bit microprocessor,and-> the 8086 has a 16bit databus 20bit address bus-> the intel 8086,is designed to operate in two modes namely(1) minimum mode(2) maximum mode
use two of them for 8 bits.
He does Let's Plays of Old and Recent 8bit and 32bit games.