1. Memory Read - (PC) and then increment PC - Op Code C1 - Four clocks 2. Memory Read - (SP) and then increment SP - Transfer to C - Three clocks 3. Memory Read - (SP) and then increment SP - Transfer to B - Three clocks
The 8085 and 8086/8088 operate on 5 volts.
The delay is calculated in 8085 by making the use of formulas. Because of the complexity of the mathematical operations a calculator may also be required.
80bit adderss register
Because that's how Intel designed it. Even though the 8085 is an 8-bit computer, the program counter and stack pointer are 16 bits wide in order to support the address bus, which is also 16 bits wide. In the case of the 8086/8088, the program counter and stack pointer are still 16 bits wide, even though the address bus is 20 bits wide, because the 8086/8088 adds segmentation through the 16 bit segment register which is left shifted by 4.
The 8085 is called an 8 bit microprocessor because the accumulator is 8 bits in size. This means that the primary data size is 8 bits. This is true even though the address bus, certain registers, and certain register operations are 16 bits in size.
There is an example of a LHLD 5000H diagram on this website: atelier-drachenhaus.de/timing-diagram-8085. This will provide an idea of how to draw the diagram.
Stack/explicit.
Refer http://wiki.answers.com/Q/How_to_draw_timing_diagram_for_8085_microprocessor_instruction
The OUT instruction on the 8085 uses 10 T cycles, 3 for opcode fetch, 1 for opcode decode, 3 for port address fetch, and 3 for port data store. Any wait states encountered are above and beyond that.
evaluation of microprocesser 8085 upto this year
8086 has two blocks Bus Interfacing Unit(BIU) and Execution Unit(EU).The BIU performs all bus operations such as instruction fetching, reading and writing operands for memory and calculating the addresses of the memory operands. The instruction bytes are transferred to the instruction queue.EU executes instructions from the instruction system byte queue.Both units operate asynchronously to give the 8086 an overlapping instruction fetch and execution mechanism which is called as Pipelining. This results in efficient use of the system bus and system performance.BIU contains Instruction queue, Segment registers, Instruction pointer, Address adder.EU contains Control circuitry, Instruction decoder, ALU, Pointer and Index register,Flag register.
There is no exit instruction in the 8085. Do you mean return, as in from a function or interrupt? If so, the instruction is RET.
Not in the 8085.
In the 8085 microprocessor, the opcode fetch machine cycle is not sufficient on its own for executing the MOV MA instruction. While the opcode fetch cycle is responsible for retrieving the instruction from memory, additional machine cycles are required to perform the data transfer or execution of the instruction. Specifically, the MOV MA instruction involves both an opcode fetch cycle and a memory access cycle to complete the operation. Therefore, multiple machine cycles are necessary for executing this instruction effectively.
There are 74 instructions in the 8085 microprocessor.
The timing diagram for the LDA (Load Accumulator Direct) instruction in the 8085 microprocessor involves several key phases. Initially, the opcode is fetched from memory, which takes 4 clock cycles. Next, the address of the data to be loaded into the accumulator is specified in the next two cycles, followed by another two cycles to read the data from the specified memory location into the accumulator. The entire process typically takes 7 machine cycles, including the necessary memory access time.
MUL is not an 8085 instruction.