The 8085 is not pipelined.
the 8085 microprocessor is a 8-bit microprocessor and these are bidirectional but the address lines are unidirectional.these address lines are used to address the location of the instruction in memory .these data lines are used to transfer data between processor and peripheral devices. when the address of the instruction will be recognized by the address lines the data will be send to the processor therefore the 16 address lines are not act as a data lines in 8085
8 data lines
timing
There are eight datalines, D0 through D7, in the 8085 microprocessor. They are shared, or multiplexed with the eight low order address lines, A0 through A7, and are called AD0 through AD7 on the pinout drawing.
A Bus is a set of data/control lines which are basically just a group of wires. e.g if 8085 wants to send 0xFF (255) to an Output device, it would make all the lines of the data bus high, bcoz 0xFF is 11111111 in binary, that is to say all the wires of the bus are 1. (1 means at +5V). -munish
there are mainly 3 buses are there in 8085. They are: Address bus :-Used to carry address Data bus :- Used to carry data Control bus :-Used to carry signals such as control and timing signals
It provides timing signals.
Because that's how Intel designed it. The 8085 is an 8-bit computer operating on a 16-bit address space.
The pins used to control interrupts in the 8085 are INTR/INTA, RST5.5, RST6.5, RST7.5, and TRAP.
It works on 8 bit lower address lines has to be multiplexed with data lines Appropriate control signals have to be generated to interface with memory and I/O devices 8 bit processors are slow Memory is just 64 kb By Anmol Bhardwaj SUSCET Tangori Mohali ( India)
It has to be calculated, you can not generate the control word it's an hex value for controlling the ports of 8255A