Intel 8086 and 8088

Suppose we give a 8-bit instruction ADD B to the microprocessor

then this instruction is not at all understood by microprocessor as it only accepts binary inputs

so first of all it stores the instruction in the INSTRUCTION REGISTOR

then it decodes this instruction ADD B to its suitable binary code 80H in the INSTRUCTION DECODER.. after converting to 80H then the microprocessor understands that ..

yes i have to add the content of the resistor B with that of A(accumulator) and store the result in the accumulator A

this is a small example how microprocessor operates facing the instructions

A permanent address for a mobile node is it's IP address when it is at it's home network. A care-of-address is the one its gets when it is visiting a foreign network.

The answer is at best a moving target; the average six months from now will almost certainly be different. Also, "average" how, and over what population? All laptops still in use? All laptops currently being sold? All laptops currently being manufactured? Do underpowered "netbooks" count? The question is so vague as to be effectively meaningless.

Yes.

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

For the formation of physical address we need Segment address and offset address

Consider an example

Segment Address : 1005H

Offset Address : 5555H

Segment address : 1005H 0001 0000 0000 0101

Shifted by 4 bit positions : 0001 0000 0000 0101 0000

Offset Address : + 0101 0101 0101 0101

Physical Address : 0001 0101 0101 1010 0101

1 5 5 A 5 H

Physical Address of given Segment Address : 155A5H

TEST This input is examined by a 'WAIT' instruction. If the TEST input goes low,

execution will continue, else, the processor remains in an idle state. The input is

synchronized internally during each clock cycle on leading edge of clock.

All of the 8086/8088 registers, AX, BX, CX, DX, SP, BP, SI, DI, CS, DS, SS, ES, IP, and FLAGS, are 16 bit registers. The AX, BX, CX, and DX registers may also be viewed as 8 eight bit registers AH/AL, BH/BL, CH/CL, and DH/DL.

It takes 23 address lines to address 8 mb of memory.

A bus is a collection of conducting wires which connect the processor and other devices in parallel scheme.

The function of an address bus is to carry the address of the memory locations from the processor to the memory device, the address bus is unidirectional(only in one direction) in this processor so the flow of information on this bus is from the microprocessor to the attached device(memory module).

ALE is a signal that means that the data bus contains the lower order address bus values. External hardware should strobe the data bus during ALE time, and lock it on the falling edge of ALE.

220 or 1,048,576 locations, otherwise known as 1 meg. If its an 8 bit bus, we are talking about 1 megabyte. That happens to be the size of the address bus of the original 8086/8088 microprocessor.

A microprocessor is a chip that serves as the central processing unit (CPU) in personal computer(PCs). it is often called a logic chip or master chip. it controls all programs and performs arithmetic calculations with great accuracy. it acts as the 'brain' of the computer.

example:-

Intel 4004 , Intel 8080 , Intel Pentium Pro , Pentium Microprocessor.

You can address 2^20 (or 1,048,576) locations. Each location will consist of 16 bits (1 or 0) which is commonly called a Word (though that isn't always the case).

I think what you're really looking for is the answer in bytes. Each data bus line holds a bit, 8 bits is a byte, so you're data line is 2 bytes. Since each location holds two bytes, this would be 2,097,152 bytes, or 2MB

To provide personal computing power to individual users.

The major difference between the 8085 and the 8086/8088 is that the 8085 is an 8 bit computer, and the 8086/8088 is a 16 bit computer.

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.

We have only one flag register of 8 bits.

Bits description is as follows (Assuming D0=LSB & D7=MSB)

D7=Sign Bit.

D6= Zero Flag

D4= Auxiliary Carry Flag

D2 = Parity Flag

D0= Carry Flag.

The 8086/8088 processor is called a 16 bit processor because its basic architecture is 16 bits wide. Its registers and accumulator are 16 bits wide, and the primary data it manipulates without extra work is 16 bits wide.

When an interrupt is requested, the currently running process is suspended and the handler is

invoked. When the handler exits, control is handed back to the running process. Since interrupts are

practically random and unplanned, when the handler passes back control, it must ensure that all

registers and stacks that it used are restored back to the exact state they were in when the process

got interrupted, so that the process can resume like as though nothing happened.

Access to these registers and stacks can only be done in assembly.

Other reasons include the fact that interrupt handlers must be very fast, and since the interrupt

routines are very short, they can be hand‐crafted to be much faster than anything a compiler

generates.

The 8086/8088 has a clock oscillator circuit. You provide a crystal, and it will generate a clock signal that controls the speed of the processor. In that respect, it has a clock.

The 8086/8088, however, does not have a time of day or date clock. You can build a software entity that keeps day/date time using interrupts from a divider off of the clock oscillator but, that is not the same thing as a non-volatile clock chip such as provided in the PC, but which is not part of the 8086/8088.

literal and absolute direct are the registers