Intel 8085

manager

The opcode fetched from the memory is being decoded for the next steps and moved to the appropriate registers.

Fetch operands from memory if necessary: If any operands are memory addresses, initiate memory read cycles to read them into CPU registers.

no it is not

The BIOS function in the 8086 microprocessor is called an interrupt function. It is an interrupt function because it is not called by a function call instruction.

You can find examples of a timing diagram online. Finding an image might make it easier to see exactly what is needed.

We know the formula

G3=B3

G2=B2 XOR B3

G1=B1 XOR B2

G0=B0 XOR B1

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In 8086 the instruction queue is 6 byte long. This is because even the longest 8086 instruction is 6 byte long. Thus it is possible to prefetch even the longest instruction in the instruction set.

In the 8085, the LDA instruction loads the accumulator from memory, while the STA instructionstores the accumulator to memory. LDA is a read, while STA is a write. LDA is opcode 3AH, while STA is opcode 32H.

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peripherals are anything that can plug into the computer to help feed it data to run

as a counter

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The CPU has an automatic program stack. This stack can be located anywhere in the address space and can be any size up to the amount of memory available in the system. Normally the SP is initialized by one of the first instructions in an application program. The stack is configured as a data structure that grows downward from high memory to low memory. At any given time, the SP holds the 16-bit address of the next free location in the stack. The stack acts like any other stack when there is a subroutine call or on an interrupt. ie. pushing the return address on a jump, and retrieving it after the operation is complete to come back to its original location. There are instructions that push and pull the A and B accumulators and the X and Y index registers. These instructions are often used to preserve program context.

latches remember the last state but buffers do not

The Intel 8259A Programmable Interrupt Controller is used to manage interupts from various sources, source as the hard disk, the floppy disk, the communications ports, the keyboard, the real-time clock, etc., and provide interrupt vectors as specified by the program. Originally designed for the 8086, in 8086 mode it works with the 8086/8088 and its functionality is included in most South Bridge chipsets today.

The HOLD pin on the 8085 is an external request for control of the bus. Upon receipt of HOLD, the 8085 will complete its current cycle and assert HLDA (HOLD Acknowledge), and then it will float the address, data, and control bus one half clock cycle later. The external hardware is then free to use the bus. When it is done, it releases HOLD, the 8085 releases HLDA, and the 8085 takes control of the bus and continues with the next cycle. HOLD is used by external DMA controllers, such as the 8257, to transfer data to and from memory on behalf of high speed peripherals, without requiring 8085 attention to that data transfer.

We can implement "Divided by '2' " operation by using RRC.

Embedded processors still use 8 and 16 bit, especially if it's legacy hardware being supported. General purpose desktops have been 32 or 64 bit for awhile.

RET instruction needs 3 machine cycles. One to fetch and decode the instruction(4 T states), and two more machine cycles(i.e. 2*3=6 T states) to read two bytes from the stack(stack is exterior to microprocessor, stack is in R/W memory, so to exchange data with stack needs machine cycles). Thus, RET instruction needs total 3 machine cycles and 10 T-states.

DI = Disable all interrupts (except TRAP)

EI = Enable all interrupts

Also, SIM can selectively mask RST5.5, RST6.5, and RST7.5.

Flags that are allowed in to many many places

In the 8085, trap, intr, and rstx.5 lines are checked on the falling edge of clock, at the beginning of the last T cycle of an instruction, or approximately one clock cycle before ALE.

ANSWER There are 2128 combinations of addresses. This is about 3.4 x 1038 locations. Assuming each address holds a 32-bit word, that's 1.2 x 1039 bytes. That's a LOT of memory.