Intel 8086 and 8088

http://hamradio.lakki.iki.fi/new/Datasheets/transistor_pinouts/case-14.gif

Picture in context can be found here:

http:/hamradio.lakki.iki.fi/new/Datasheets/transistor_pinouts

4K (4096) of addressable space is defined by 12 bits of address space, because 212 = 4096.

• LXI H, 4150 ; Point to data
• LXI B, 0000 ; Initialize hundreds= 0, Tens=0
• MOV A, M ; Get hex data to A
• LOOP: SUI 64
• JC LOOP 1
• INR B ; hundreds= hundreds+1
• JMP LOOP
• LOOP 1: ADI 64 ; if subtracted extra, add it clear carry flag
• LOOP 2: SUI 0A
• JC LOOP 3
• INR C ; Tens=tens+1
• JMP LOOP 2
• LOOP 3: ADI 0A ; If subtracted extra, add it again
• INX H ; A = Units
• MOV M, B ; store hundreds
• MOV B, A ; Combine Tens in C &
• MOV A, C ; Units in A to form a
• RLC ; Single 8-bit number
• RLC
• RLC
• RLC
• ADD B
• INX H
• MOV M, A ; Store tens & Units
• HLT

SP is the stack pointer and can be used in the stack operations. It cannot be used for fetching the instruction

The highest memory address in the 8086/8088 is FFFFFH.

For any segment base address, segment first physical address will have 0 in the least

significant position in hexadecimal format.

Let say, our Segment base =0x1234, and we calculate segment first physical address as

0x12340

The CPU comprised of three main parts: AlU(ARITHMETIC LOGIC UNIT):- Does the actual Logic comparisons that need to be processed. Control Unit:-Can execute and store the results coming out of ALU. Registers:- store the data that is to be executed next.

I am not quite sure how significant this is. The reason, however, is that accessing a register, which is part of the CPU, is faster than accessing RAM memory, which is located on a separate chip.

I am not quite sure how significant this is. The reason, however, is that accessing a register, which is part of the CPU, is faster than accessing RAM memory, which is located on a separate chip.

I am not quite sure how significant this is. The reason, however, is that accessing a register, which is part of the CPU, is faster than accessing RAM memory, which is located on a separate chip.

I am not quite sure how significant this is. The reason, however, is that accessing a register, which is part of the CPU, is faster than accessing RAM memory, which is located on a separate chip.

Do you have a virtual copy of game plan sharpshooter,coney island or pinball lizard .How much would it cost ?

The address bus in the 8085 is 16 bits wide.

The return address.

Address Resolution Protocol (ARP)

ARP is primarily used to translate IP Addresses to Ethernet MAC Addresses.

Minimum mode in the 8086/8088 provides the functionality of the 8288 bus controller, which means that functions like HOLD, HLDA, WR-, M/IO-, DT/R-, DEN-, ALE, and INTA- come off of the 8086/8088.

Maximum mode means that these functions come off of the 8288, in somewhat enhanced form, and the pins are replaced with new meanings like RQ-/GT0-, RG-/GT1-, LOCK-, S2-, S1-, S0-, QS2, and S21, giving the processor extra capabilities.

There are four segment registers in the 8086/8088 processor, CS, DS, ES, and SS, also known as Code Segment, Data Segment, Extra Segment, and Stack Segment. Any time an address is generated by the processor, it is added to the value of one of the segment registers, after that segment register is effectively multiplied by 16, or left shifted four bits, in order to generate the physical address that accesses memory. This gives an effective address range of 20 bits, or 1mb, but note that only 64kb is addressable through any segment register at one time, unless you stop to change the contents of that segment register. This is known as a segmented architecture. By default, the CS register is used when fetching instructions, the DS register is used when accessing data, the SS register is used when accessing the stack, and the ES register is used during certain string type instructions. If desired, an instruction prefix can be used to override, such as forcing use of CS instead of DS when using a table contained within opcode space.

Minimum bus cycle duration = 4 clock cycles Bus clock = 8 MHz Maximum bus cycle rate = 8 M / 4 = 2 M /s Data transferred per bus cycle = 16 bit = 2 bytes Data transfer rate (per second) = Bus cycle rate * data per cycle = 2 M * 2 = 4 M bytes per second

Interrupt is the signal generated by the input/output devices in order to take the attention of the processor.

When the processor receives the interrupt signals it checks the priority status and finish the current fetch and execute cycle and (if the priority status is high) allow the input/output device to process their tasks. Then previous fetch and execute cycle is continued. This is called interrupt service routine.

The programs are broken down in to small sub programs which are called subroutines. Then the program reusability, readability, maintainability ...etc will be increased.

Code segment (CS) is a 16-bit register containing address of 64 KB segment with processor instructions. The processor uses CS segment for all accesses to instructions referenced by instruction pointer (IP) register. CS register cannot be changed directly. The CS register is automatically updated during far jump, far call and far return instructions. Data segment (DS) is a 16-bit register containing address of 64KB segment with program data. By default, the processor assumes that all data referenced by general registers (AX, BX, CX, DX) and index register (SI, DI) is located in the data segment. DS register can be changed directly using POP and LDS instructions. Code segment (CS) is a 16-bit register containing address of 64 KB segment with processor instructions. The processor uses CS segment for all accesses to instructions referenced by instruction pointer (IP) register. CS register cannot be changed directly. The CS register is automatically updated during far jump, far call and far return instructions. Data segment (DS) is a 16-bit register containing address of 64KB segment with program data. By default, the processor assumes that all data referenced by general registers (AX, BX, CX, DX) and index register (SI, DI) is located in the data segment. DS register can be changed directly using POP and LDS instructions.

The offset address in an 8086/8088 is the logical address that the program "thinks about" when it addresses a location in memory. The Execution Unit (EU or CPU) is responsible for generating the offset address. The Bus Interface Unit (BIU), on the other hand, takes the offset address and adds it to four times the selected segment register value in order to determine a real address, which is now 20-bits in length.

Some programs do deal with segment addresses as well - these are called far pointers instead of near pointers - but the program has to do more than one step to load both the offset and the segment address - a complexity created by running in a 16-bit environment.

8 bits

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

1st address for operand.

2nd address for another operand.

3rd address for store the result.

4th address for next instruction.

in 8086 there is 20 bit address bus,so it can address 1,048,576 address.

At each address we can store 8 bit address (1-byte)but if want to write a word(16-bit)into a memory segment to store data in byte form then we write the data in two consecutive memory address which are even(low) and odd(high) memory.

Mov ax,[2020h]

mov bx,[2022h]

mul bl

mov [2024h],ax

hlt