AH AL
BH BL
CH CL
DH DL
SP
BP
DI
SI
CS
DS
SS
ES
IP
Internal
communication
register
Σ
EU
Control
System
Temporary Register
Flags
ALU
BUS
Control
Logic
ALU Data BUS (16 bit)
Instruction Queue
Queue
BUS
(8bit)
Data BUS
(16 bit)
8086
BUSES
Add BUS (20 bit)
General Registers
4
3
2
1 6
5
1. BIU O/Ps IP
2. IP=IP+1
3. Instruction  Queue
4. EU draws Instr. &
Execute it.
5. BIU Fetches another
Inst. While in the
same time EU
executing the 1st
Inst.
PIPE LINEING
FIFO
FETCH
overlaps with
Execute

1MB 0
7
S4
64KB
S1
64KB
S3
64KB
S2
64KB
00000H
FFFFFH
S4
64KB
S1
64KB
S3
64KB
S2
64KB
S4
64KB
S1
64KB
S3
64KB
S2
64KB
S4
64KB
S1
64KB
S3
64KB
S2
64KB
XXXX:0000
XXXX:FFFF
XXXX:0000
XXXX:FFFF
XXXX:0000
XXXX:FFFF
XXXX:0000
XXXX:FFFF
X100:0000
X100:FFFF

OFFSET VALUE
15 0
Effective Address
EA
SEGMENT REGISTER
15 0
Segment Base Register
SBA
0000
20-BIT
19 0
Physical Memory Address
PA
Adder

0
7
SEGMENT
64KB
00000h
FFFFFh
0022h
15 0
1234h
15 0
0h
SBA+EA=12340+0022=12362
19 0
Adder
EA
SBA PA
12340h
12362h
=
22
+

PA = SBR or (SR) : EA
= : + +
The Operand resides
in an internal Register.
Examples:
AND AL,DH
AND BX,CX
It is a method for specifying an Operand
The Operand is a part
of the Instruction.
Examples:
AND AL,80h
AND 0800h,BX
Direct Add.
Register
indirect Add.
Based Add.
Indexed Add.
Based
Indexed Add.
CS
DS
ES
SS
BX
BP
SI
DI
8-BIT or
16-BIT Displacement

(1)

SHIFT
&
ROTATION
Group
BRANCH
Group
CHARRECTER
STRING
Group
DATA
Transfer
Group
LOGICAL
Group
ARITHMATIC
Group
8086
CONTROL
Group
















1MB 0
7
S4
64KB
S1
64KB
S3
64KB
S2
64KB
00000H
FFFFFH
XXXX:0000
XXXX:FFFF
XXXX:0000
XXXX:FFFF
XXXX:0000
XXXX:FFFF
XXXX:0000
XXXX:FFFF
SS
64KB
SS:0000
SS:FFFF
[SP]= TOS

;This program is used to test PPI1 Mode2 and . . . . . . . .
;Date: 28/7/2008
code segment
assume cs:code,ds:code
org 0
start:
mov ax,cs ;Comments
mov ds,ax ;ds=cs
hlt
DSeg DB 00,10h,0F2h,
code ends
end start
Assembly
Program
Structure
LABLES
COMMENTS
DIRECTIVES
INSTRUCTIONS

The DB & DW directives:
Examples:
Variable Name DB Aah ;VARIABLE NAME(BYTE) = AAh
DATA1 DW Aah ;DATA1(WORD) = AAh
X DB 50DUP(?) ;50 BYTES OF MEM. ARE RESERVED
;FOR X WITH NO INITIAL VALUE
STUDENTS DW 38DUP(?) ;38 WORDS OF MEM. ARE RESERVED FOR
; STUDENTS WITH NO INITIAL VALUE
TABLE DB 11h,0Abh,29h,50h,33h,0FDh,0B0h
;THESE VALUES ARE ASSIGNED FOR
;TABLE VARIABLE
NAMES DB 100DUP(‘U’)
;FILL 100-BYTES OF MEMORY STARTING
;AT EA=NAMES WITH ASCII (U)

INPUTS & OUTPUTS FOR MASM & LINK
MASM LINK
prog.obj
prog.exe
prog.asm
prog.lst

Examples: Convert the following Instruction mnemonics
to there machine codes.
(A) MOV BL,AL
1- OPCODE = 100010 2- W=0 (8-BIT)
3- Let REG be the source i.e: D=0 4- then REG = 000 for AL
5- MOD =11 6- Then R/M = 011
100010 0 0 11 000 011 = 88C3h

D
0
1
REG = code for Source reg
REG = code for Dest. reg
MODE

(B) ADD AX,[SI]
1- OPCODE = 000000 2- W=1 (16-BIT)
3- The REG is dest. i.e: D=1 4- then REG = 000 for AX
5- MOD =00 6- Then R/M = 100
000000 1 1 00 000 100 = 0304h

D
0
1
REG = code for Source reg
REG = code for Dest. reg
MODE

(C) XOR CL,[1234h]
1- OPCODE = 001100 2- W=0 (8-BIT)
3- The REG is dest. i.e: D=1 4- then REG = 001 for CL
5- MOD =00 6- Then R/M = 110
001100 1 0 00 001 110 = 320E3412h
0011 0100 0001 0010

D
0
1
REG = code for Source reg
REG = code for Dest. reg
MODE

(C) ADD [BX][DI][1234],AX
1- OPCODE = 000000 2- W=1 (8-BIT)
3- The REG is source i.e: D=0 4- then REG = 000 for AL
5- MOD =10 6- Then R/M = 001
000000 0 1 10 000 001 = 01813412h
0011 0100 0001 0010

D
0
1
REG = code for Source reg
REG = code for Dest. reg
MODE

(C) MOV WORD PTR [BP][DI][1234],0ABCDh
1- OPCODE = 110001 2- W=1 (16-BIT)
3- D=1 from table 4- then REG = 000 from table
5- MOD =10 6- Then R/M = 011
110001 1 1 10 000 011
= 01813412CDABh
0011 0100 0001 0010 1100 1101 1010 1011

D
0
1
REG = code for Source reg
REG = code for Dest. reg
MODE

(C) MOV [BP][DI][1234],DS
1- OPCODE = 10001100
2- then REG = 011 from table
3- MOD =10
4- Then R/M = 011
100011 0 0 10 011 011 = 8C9B3412h
0011 0100 0001 0010