Intel 8086 Architecture
Last updated
Last updated
It lunched in 1979
It was the first 16-bit microprocessor
and it has 40 pins
it uses +5v powersupply
has ~29.000 transistors (semiconductors)
has up to 1MB memory, witch is a lot for what it is used.
Memory address ranges from 00000(16) to FFFFF(16)
little-endian organization
Mem organization
Pins
Two main units of the 8086 processor:
BIU – Bus Interface Unit Fetching instructions from memory and decoding them Manages the data transfer between the microprocessor and the memory or I/O devices (I/O – Input/Output) Provides a full 16-bit bi-directional data bus and 20-bit address bus
EU – Execution Unit Executes the instructions
The BIU is responsible for handling all data transfers between the CPU and memory or I/O devices. It does this by managing the B-bus and C-Bus
BIU Contains:
Bus control Logic
So how the BIU works is: It captures six instruction bytes ahead of its current memory, and the instructions a re stored in a Queue (group of registers). And the Queue has FIFO organization, whitch means if MOV command is written first and than ADD, the MOV will be executed first, bcs it was first written-first queued.
JMP and CALL functions are not used when loaded but when they are called, and other instructions ate loaded extend them
Feature of fetching the next instruction while the current instruction is executing is called pipelining
Segment registers are a set of 16-bit registers in x86 processors that hold the base addresses of memory segments.
Segment registers are essential for memory segmentation in 16-bit processors, allowing the CPU to access a 1 MB memory space with 16-bit addresses
Each segment register contains starting address of a segment (called also: Base address or Segment base)
Memory Segmentation
Memory segmentation is a memory management technique used in computer systems, especially in early processors like the Intel 8086, to divide memory into distinct segments. Each segment represents a logical section of memory with a specific purpose, which helps organize data, code, and the stack, allowing for more efficient memory access and usage. Segmentation enables the CPU to access a larger memory space than it could with a single flat address space.
Memory segmentation allows memory addressing capacity to be 1mb .
Internal architecture of 8086 allows usage of 4 active segments with the help of the following four segment registers:
CS - Code Segment
DS - Data Segment
SS - Stack Segment
ES - Extra Segment
The CS register holds the upper 16-bits of the starting address of the segment from which the BIU is currently fetching the instruction code byte.
The SS register is used for the upper 16-bits of the starting address for the program stack
ES register and DS register are used to hold the upper 16-bits of the starting address of the two memory segments, which are used for data.
The instruction pointer register holds the 16-bit address of the next Code byte within the code segment. The value contained in the IP is referred to as an offset.
This value must be offset from (added to) the segment base address in CS to produce the required 20-bit physical address.
The instruction queue is a buffer that holds a 6-byte queue of instructions. This prefetches instructions from memory to reduce the time the processor waits for instructions, allowing faster execution.
The execution unit tells the BIU from where to fetch instructions or data. Decodes instructions and executes instructions.
It contains:
Instruction Decoder
Arithmetic Logic Unit (ALU)
Pointers and Index Registers
▪ The control circuitry in the EU directs the internal operations. ▪ A decoder in the EU translates the instructions fetched from memory into a series of actions which the EU performs. ▪ ALU is 16-bit. It can add, subtract, AND, OR, XOR, increment, decrements, complement and shift binary numbers.
![[Control Circuit, Instruction Decoder, ALU.png]]
▪ A special purpose 16-bit register. ▪ There are 9 valid flag bits. ▪The flag bits are changed to 0 or 1 depending on the value of result after arithmetic or logical operations. ▪ The flag bits can be divided into two sections – control flags and status flags
CF (Carry Flag)
0
Set if an arithmetic operation generates a carry out or borrow into the most significant bit
PF (Parity Flag)
2
Set if the result of an operation has an even number of 1-bits; used for error checking
AF (Auxiliary Carry Flag)
4
Set if there is a carry or borrow between the lower and upper nibbles (used in BCD arithmetic)
ZF (Zero Flag)
6
Set if the result of an operation is zero
SF (Sign Flag)
7
Set if the result of an operation is negative (most significant bit is 1).
TF (Trap Flag)
8
Enables single-step mode for debugging, causing an interrupt after each instruction
IF (Interrupt Flag)
9
Controls the ability to respond to interrupts; if set, the CPU will respond to maskable interrupts
DF (Direction Flag)
10
Controls the ability to respond to interrupts; if set, the CPU will respond to maskable interruptst
OF (Overflow Flag)
11
Set if an arithmetic operation generates a result too large to fit in the destination operand
8 general purpose registers (8-bits each) labelled: AH, AL, BH, BL, CH, CL, DH, and DL.
Can be used individually for temporary storage of 8-bit data.
The AL register is also called accumulator.
Certain pairs of these general-purpose registers can be used together to store 16-bit data, such as AX, BX, CX and DX.
Intel 8086 microprocessor supports 8 types of instructions:
Data Transfer Instructions
Arithmetic Instructions
Bit Manipulation Instructions
String Instructions
Program Execution Transfer Instructions (Branch & Loop Instructions)
Processor Control Instructions ▪ Iteration Control Instructions
Interrupt Instructions
copies the second operand (source) to the first operand (destination)
Syntax:
MOV operand1, operand2 (destionation) (source)
Important:
Never mix an 8-bit register with 16-bit, it is not allowed.
Code segment register (CS) and Instruction Pointer (IP) are never used as destination.
Segment with segment is not allowed.
Memory with memory is not allowed.
When used immediate with memory you must use (byteptr) of
8-bit or (wordptr) of 16-bit for the instructions.
Also used for the instructions have one operand as memory
Example 1: mov ES, DS (Incorrect) mov AX, DS mov ES, AX (correct)
Example 2: mov [SI],[0xF900] (Incorrect) mov BX, [0xF900] mov [SI], BX (correct)
