You have no items in your shopping cart.

Subtotal: 0.00

The PS-8086 board which demonstrates the capabilities of the 40-pin 8086 (various families) Sample programs are provided to demonstrate the unique features of the supported devices.

The PS-8086 Kit comes with the following:

  • PS-8086 board
  • Sample devices (INTEL 8086/NEC 8086)
  • Cross cable (RS232)
  • CD-ROM, which contains:
       -  Sample programs
    PS-8086 board User manual
  • Keyboard (101 keys)
8086 Board Overview


8086-board-overview
 
 
 
The PS – 86A board is based on Intel 8086 Microprocessor, which operates at 6.144 MHz using the crystal of 18.432. The board can operate using the 101/104 PC keyboard supplied along with the trainer kit and 2 Line by 16-character LCD display or from the PC (using the Terminal Emulation Software).

Microprocessor’s Address, Data and Control bus pins are brought to the 50 pin FRC connector. PS -86A is equipped with powerful software monitor in two-27C256 EPROM.


The monitor supports Video terminal RS232C interface, local 101keyboard and LCD display. The board has 64KB CMOS static RAM (type 62256). PS -86A works on +9V DC.

8086 Specifications
  • 8086 Microprocessor operating at 18.432 MHz
  • 16KB powerful software monitor two 27C256 EPROM
  • Three 16-bit programmable timers from 8253
  • 48 programmable I/O lines from two nos. of 8255
  • Serial interface using 8251
  • 50 pin FRC connector for system bus expansion
  • 20 pin FRC connector for user interface from 8255
  • 9 pin D type connectors for RS 232 interface
  • Six different selectable baud rates from 150 to 9600
  • 101 PC type keyboard for entering user address/data and for commands
  • Built in line-by-line assemble and disassemble
  • User friendly software monitor for loading and executing programs with break point facility
System Description
Hardware
Processor Clock Frequncy

8086 operates at 18.432 MHz clock.

Memory
  • Monitor EPROM: 0000 –FFFF (SEGMENT)
  • System RAM: 0000 –FFFF (SEGMENT)
       1000 – 3FFF (Reserved For Monitor program)
  • User RAM Area: 1100 – 3FFF
Allocation Of Eprom

START ADDRESS

END ADDRESS

SOCKET NO

IC USED

TOTAL CAPACITY

0000

FFFF

U9

U8

27256

27256

   32 K BYTE

   32 K BYTE

 

Allocation Of Ram

START ADDRESS

END ADDRESS

SOCKET NO

IC USED

TOTAL CAPACITY

0000

 

FFFF

 

U10

U11

62256

62256

   32 K BYTE

   32 K BYTE

 
Parallel Interface

8255 - Programmable peripheral interface.

SYSTEM MAPPING: I/O mapped I/O.

The following are the I/O addresses for 8255(GPIO I):



SOCKET.NO

FUNCTION

ADDRESS

CONNECTOR.NO

 

     U22

 CONTL REG

PORT A

PORT B

PORT C

      FF26

      FF20

      FF22

      FF24

         

      J8

   GPIO I

J9(GPIO I&GPIOII)

 

The following are the I/O addresses for 8255(GPIO II):



SOCKET.NO

FUNCTION

ADDRESS

CONNECTOR.NO

 

     U16

 CONTL REG

 PORT A

 PORT B

 PORT C

      FF36

      FF30

      FF32

      FF34

         

      J6

   GPIO II

J9(GPI0 I&GPIOII)

 
Timer Interface

8253 - Programmable Interval Timer:


SYSTEM MAPPING: I/O mapped I/O.


Channel 2

  • Input clock : 3 MHz
  • Output clock: Depends on selection of baud rate.
  • Used for : Baud rate generation for 8521 USART.

I/O Address

SOCKET.NO

FUNCTION

ADDRESS

CONNECTOR.NO

 

     U12

 CONTL REG

 CHENNAL 0

 CHENNAL 1

 CHANNEL 2

      FF06

      FF00

      FF02

      FF04

         

      J2

 

 
Connector Details
20 Pin Expansion Connectors

The 20 Pin FRC connector is used to interconnect with the Interface cards like ADC, DAC, SWITCH/LED, RELAY buzzer Interfaces etc. Pin details are given below




20-pin-expansion-connectors-for-8086

 50 Pin Expansion Connector

The 50 Pin FRC connector is used to interconnect with the Interface cards like 8255, 8279, 8253/8251, 8259, 8257 and the pin details are given below


50-pin-expansion-connector-for-8086
Keyboard Connector


keyboard-connector-for-8085

9Pin ‘d’ Type (Female)


9pin-d-type-female

 

8251 - Universal Synchronous / Asynchronous Receiver / Transmitter.


RS232 Bridge Converter

 
Baud Clock

Baud clock for 8251 is programmable, provided by Channel 2 of 825

Input Clock for 8251

3.072 MHz


Drivers Used

MAX 232 is used for transmitting receiving of characters.

8251 Uart I/O Address

SOCKET.NO

FUNCTION

ADDRESS

CONNECTOR.NO

      U15

 8251  CONTL REG

 8251  DATA

 

 

      FF10

      FF12

     

 

       D2(SKT)

 

 

LCD Interface



lcd-connector-8051-trainer


Device used: 16 × 2 / 20 × 4 LCD module


System Mapping: I/O mapped I/O.


SOCKET.NO

FUNCTION

ADDRESS

CONNECTOR.NO

 

     ----

 LCD COMMAND

 LCD DATA

 

 

      FF40

      FF42

     

 

         

      -----

 

 


RESET

This key is located in the main 8086 Board. On depressing this key the program starts executing from the beginning or reset address 0000. On power on reset it. Display PS - 86 in local LCD display.




reset-button


Power Supply Details

PS trainer kit will work at 0 – 5v (1 amp) from the PS power supply. Provision is made in PS power supply to bring out on the front panel DC regulated voltage output for interfacing with add-on cards.


   +5V      1 amp


POWER SWITCH

power-supply-off

Supply Turned OFF

power-supply-on

Supply Turned ON

 


Keyboard Details

101 PC type keyboard is interfaced to Microcontroller through its port pin. Communication between keyboard and Microcontroller takes place using 2 wires – one for serial clock and serial data (P1.6 and P1.7).




COMMANDS AND KEYS

Reset

This key is located in the main PS-86A board .On depressing this key the programs. Starts executing from the beginning or reset address 0000. On power on reset it. Displays PS- 86A in local LCD display


PS -86

!!!!

H (Help Menu)

This key is used go PS- 86A help menu and it will display the following commands



KEY

FUNCTIONS

A  <BEG>

ASSEMBLE

B

BAUD RATE

D  <BEG>

DISASSEMBLE

E  <BEG><END>

EXAMINE

G  <BEG>

EXECUTE

H

HELP COMMANDS

I    <INSERT>

INTERNAL RAM

L   <OFFSET>

DOWN LODE

M  <ADDR>

MODIFY

N

NORMAL MODE

Q

QUIT

R   <REG>

REGISTER DISPLAY

S

SERIAL TRANSFER

T   <START><END><DS>

BLOCK TRENSFER

U  <BEG><END>

UP LODE

X

DELETE BLACK MEMORY

?

INSTRUCTIONS

 

OPERATING INSTRUCTIONS
Power On

Connect the PS – 8051 board to the power having the following specifications.

+9V DC 1 Amp

Switch on the power supply after ensuring the correct voltages. Following message will appear on the LCD display.

PS -- 86

!!!!


On power on or after reset the display shows PS – 86 as a sign on message. The prompt character – is displayed in the next line informing the user, that the board is ready to accept the commands.

Instruction
PROGRAM ENTRY USING ASSEMBLER
Entering Mnemonics
Example

Press H for help

A1100

Enter the starting Address


Enter Key ↵

User program starts from address 1100 and displays the following and waits for the user data to be typed in the second line

Example

0000 : 1100:

MOV AX,1212

Enter the mnemonics


Enter Key ↵


0000 : 1103:

MOV BX,1212

Enter the mnemonics



Enter Key ↵


Program end.


Exit Command: Double Enter you get the main menu


PS  _  86

_



Program Entry Using Opcode

Modify Memory


Press H for help

_M1100

Enter the starting Address


Enter Key ↵


0000:1100

18 _


0000:1100

18B8_

Enter the opcode


Enter the Space Bar Key


0000:1101:

34 12_

Enter the opcode


Enter the Space Bar Key


Program end. Exit Command:


Double Enter you get the Main Menu


Entering ‘G‘Executing Command

PS  _  86

_G1100

Enter the starting Address


Enter Key ↵


After executing display


PS  _  86

_G1100

Executing display


To EXIT Execution Mode PRESS ’RESET ‘Switch


Entering Result Command

Press H for help

_M

 

 

Enter the Memory Location


Enter Key ↵


Disassembler

Disassemble converts the hex byte stored in the memory into equivalent mnemonics. To enter into disassemble mode, type D in the command mode followed by the memory address.

Example

Press H for help

_D1100

Enter the starting address


Enter Key ↵


1100:  B8  12  12

MOV AX,1212

Enter the Space Bar Key


1103:  BB  12  12

MOV AX,1212

Enter the space bar key


M (Modify External Memory)

Using this command the user can display/modify any external memory address.


Modify External memory


Press H for help

_M1100

Enter the starting Address


R (Register Display)

Example

Press H for help

_R

Enter the starting Address


Enter Key ↵


AX=1104



Enter the Space Bar Key


BX=1204



Enter the space bar to see the remaining registers

T (Transfer Command)
Example

Press ‘T ‘

The source segment addresses 0000. The above command transfer the memory content starting from source start address 1100 to destination start address 1200 till source end address 1500 is reached.


Src  seg address

0000

Starting address

1100

End address

1200

Destination address

1500

 


Block  Transfer

Src  seg :  0000



Enter Key ↵


start  :  1100

end :  1200

Enter the 1100 address

Enter the 1200address


Enter Key ↵


for exit command


dest  :   0  :  1500



Enter Key ↵


Transfer Complete



Enter Key ↵


for exit command


N (Local Mode)

When this key is depressed on PC keyboard, the PS – 8051 Kit starts working through local 101 keyboard. Serial communication is disabled. Following message will appear in the LCD display.


! NORMAL MODE !



B (baud rate)

Press the ‘B ‘


Cur BAUD : 2400

150

Enter the Space Bar Key


Cur BAUD : 2400

9600

Enter the starting Address


Enter Key ↵


SET the 9600 baud rate


Baud rates : 150, 300, 600, 1200, 2400, 4800, 9600


When using the serial Communication.

S (Serial Mode Key)

When this key is depressed the system start communicating through connector. All keys are disabled except reset.

! Serial Mode!

The system displays the message SERIAL MODE. To come back to LCD mode (Normal Mode) user has to press the ‘N’ key in the computer keyboard otherwise press the Reset button.


Programming The 8086 Trainer Kit

Procedure 1: TO ENTER THE MNEMONICS


1) Initially connect the 9V adaptor to J10 connector


2) Switch ON the PS-8086 kit using slide Switch SW1


3) “PS - 86” will be displayed on the LCD


4) Connect the Keyboard in PS/2 connector


5) Depress “A” starting address of the program for Ex: A1100


For ex: A1100 enter key

Type the mnemonics MOV AX, 1212press Enter key

Type the mnemonics MOV BX, 1212 press Enter key and continue the same procedure till the end of the program



ADDRESS

OPCODES

MNEMONICS

1100

B8 12 12

MOV AX,1212

1103

BB 12 12

MOV BX,1212

1106

01 D8

ADD AX,BX

1108

BE 00 12

MOV SI,1200

110B

89 04

MOV [SI],AX

110D

F4

HLT

 


6) To verify the code depress D starting address and depress space bar to see next memory location


For Ex: D1100 and press spacebar till the end of the program


7) To execute the program Depress “G staring address for Ex: G1100.


8) To see the result depress “M result address” for Ex: M1200.


9) To view the output in the Register depress ‘R’ and press enter key in keyboard.


Procedure 2: TO ENTER THE OPCODE

Follow the same procedure till step 4


1) Depress “M” starting address of the program for Ex: M1100


For ex: M1100 press enter


Type the opcode B8 space bar


Type the opcode 12 space bar and continue the same till the end of the program



ADDRESS

OPCODES

MNEMONICS

1100

B8 12 12

MOV AX,1212

1103

BB 12 12

MOV BX,1212

1106

01 D8

ADD AX,BX

1108

BE 00 12

MOV SI,1200

110B

89 04

MOV [SI],AX

110D

F4

HLT

 


2) To view the code depress D starting address and depress space bar to see next memory


For Ex: D1100 and press spacebar till the end of the program


3) To execute the program Depress “G staring address for Ex: G1100.


4) To see the result depress “M result address” for Ex: M1200.


5) To view the output in the Register depress ‘R’ and press enter key in keyboard


Note:


  • “M” is used for displaying the result, for Ex: M8500
  • “M” is used to entering the Opcode.
  • “M” is used for entering the data.

Note:


There are two ways to enter the program


  • Mnemonics method
  • Opcode method

Sample program is given to enter the program in both the methods




PROGRAMMING DETAILS

Programming 8086 Overview

  • The 8086 Microprocessor uses a multiplexed 16 bit address and address bus
  • During the first clock of machine cycle the 16 bit address s sent out on address/data bus
  • These 16 bit addresses may be latched externally by the address latch enable signals(ALE)
  • 8086 Microprocessor can access 1024kb of external memory using its 20 bit address and memory read/write signals
  • The 8086 provide s0, s1 and s2 signals for bus control.
  • The 8086 Microprocessor has a 16 bit program counter (IP) and 16 bit stack pointer (sp)

It has following set of 16 bit Registers:


AX –Accumulator


BX, CX, DX (These four register can be used as two 8 bit register individually)


Index Register


SI → Source index


DI → Destination index


BP → Base pointer index


Segment Register


CS → Code segment register


DS → Data segment register


ES → Extra segment register


SS → Stack segment register


FL → Flag register


Interrupts

The 8086 have two interrupt


  • External mask able interrupt (INTR)
  • Non mask able interrupt (NMI)

Break Point Display In Local Mode

When break point is encountered, all the register values are saved and the Acc. “AX=XXXX “Value is displayed in the LCD display. Now use SPACE key to check register values one by one

EXAMPLE PROGRAMS
Addition of Two Bytes of Data
Flow Chart



addition-two-bytes-data-for-8086


Algorithm
  • Initialize the pointer to the memory for data and result.
  • Load the data into AX, BX.
  • Add the two data of AX and BX registers.
  • Store the result into Memory from AX registers.
Input
  • Input data’s (2 byte) are loaded into Memory address 1500.
  • LSB in 1500, MSB in 1501 – 1st data.
  • LSB in 1502, MSB in 1503 – 2nd data.
Output
  • Result stored in Memory address 1520.
  • LSB in 1520, MSB in 1521.
Program

ADDRESS

OPCODE

MNEMONICS

COMMENTS

1100

BE 00 15

MOV      SI, 1500

Move 1500 into SI pointer

1103

AD

LODSW

Load the first data into AX

1104

89 C3

MOV      BX, AX

Move AX value  into BX

1106

AD

LODSW

Load the second data into AX

1107

01 C3

ADD       BX, AX

Add BX and AX registers

1109

BF 20 15

MOV      DI, 1520

Load 1520 address location  into DI

110C

89 1D

MOV       [DI], BX

Store BX value into memory

110E

74

HLT

HALT

 


Subtraction of Two Bytes of Data

Flow Chart



subtraction-two-bytes-data-for-8086


Algorithm

  • Initialize the pointer to the memory for data and result.
  • Load the two data’s into AX, BX.
  • Subtraction of these two bytes of data.
  • Store the result into Memory address 1520.
Input
  • Input data’s (2 byte) are loaded into Memory address 1500.
  • LSB in 1500, MSB in 1501 – 1st data.
  • LSB in 1502, MSB in 1503 – 2nd data.
Output

  • Result stored in Memory address 1520.
  • LSB in 1520, MSB in 1521.

Program

ADDRESS

OPCODE

MNEMONICS

COMMENTS

1100

BE 00 15

MOV SI,1500

Load 1500 into SI

1103

AD

LODSW

Load  the first data

1104

89 C3

MOV BX, AX

Move AX value  into BX

1106

AD

LODSW

Load the second data

1107

01 C3

SUB BX, AX

subtract AX from BX

1109

BF 20 15

MOV DI, 1520

Load 1520 address into DI

110C

89 1D

MOV [DI],BX

Load BX value into DI

110E

CC

INT 3

Break point

 


Multiplication Of Two Byte Data

Flow Chart



multiplication-two-byte-data-for-8086


Algorithm

  • Initialize the pointer to the memory for data and result.
  • Load the multiplier value into AX register.
  • Load multiplicand value in BX register.
  • Multiply of these two data’s.
  • Store the result into Memory address 1520.

Input

  • Input data’s (2 byte) are loaded into Memory address 1500.
  • Load the multiplier value in 1500.
  • Load the multiplicand value in 1502.

Output

  • Result stored in Memory address 1520.

Program


ADDRESS

OPCODE

MNEMONICS

COMMENTS

1100

BE 00 15

MOV SI,1500

Load 1500 into SI

1103

AD

LODSW

Load the multiplicand value

1104

89 C3

MOV BX, AX

Load AX value into BX

1106

AD

LODSW

Load the multiplier value

1107

F7 E3

MUL BX

Multiply two data

1109

BF 0 5 15

MOV DI, 1520

Load 1520 address into DI

110C

89 05

MOV [DI], AX

Store AX value into DI

110E

47

INC DI

Increment the DI

110F

47

INC DI

Increment the DI

1110

89 15

MOV [DI], BX

Store BX value into DI

1112

CC

INT 3

Break point

 


Division (2 Byte/ 1 Byte)

Flow Chart



division-2-byte-or-1-byte


Algorithm

  • Initialize the pointer to the memory for result.
  • Load the dividend value into AX register.
  • Load the divisor value into BX register.
  • Divide these two data’s.
  • Store the result into Memory address 1520.

Input

  • Dividend value loaded into AX register.
  • Divisor value loaded into BX register.

Output

  • Result stored into 1520 address.
  • Quotient stored into 1522 address.
  • Remainder stored into 1523 address.

Program


ADDRESS

OPCODE

MNEMONICS

COMMENTS

1100

BA 00 00

MOV DX, 0000

Clear DX registers

1103

B8 FD FF

MOV AX, FFFD

Load the dividend in AX

1106

B9 0F 00

MOV BX, 0F

Load the divisor value in BX

1109

F7 F1

DIV BX

Divide the two data’s

110B

BF 00 15

MOV DI, 1520

Load 1520 address into DI

110E

88 05

MOV [DI], AL

Load AL value into DI

1110

47

INC DI

Increment DI

1111

88 25

MOV [DI], AH

Load AH value into DI

1113

47

INC DI

Increment DI

1114

89 15

MOV [DI], DX

Load DX value into DI

1116

CC

INT3

Break point

 


Block Move From One Location to Another

Flow Chart



block-move-from-one-location-to-another


Algorithm

  • Initialize the pointer to the memory where data to be transformed.
  • Load the AL register with the data from memory.
  • Initialize destination pointer to the memory where data to stored.
  • Store data from AL register.

Input

  • Input data from address 1500 which is pointed SI, transferred to the desired Location.
  • Number of byte in CL.

Output

  • Output – data in address 1550 is the moved data.

Program


ADDRESS

OPCODE

MNEMONICS

COMMENTS

1100

B1 08

MOV CL, 08

Load 08 value into CL

1102

 

BE 00 14

 

MOV SI, 1500

Load 1500 into SI

1105

BF 50 14

MOV DI, 1550

Load 1550 into DI

1108

AC

LODSB

Load the data in AL  Register

1109

88 05

MOV [DI], AL

Store the result in specified  Location

110B

47

INC DI

Increment the pointer

110C

FE C9

DEC CL

Decrement the pointer

110E

75 F8

JNZ 1108

Loop continues until the counter is zero

1110

CC

INT 3

Break point

 


Searching a Byte

Flow Chart



searching-a-byte


Algorithm

  • Initialize the pointer to the memory for storing data and result.
  • Load DL with search byte.
  • Load CL with count.
  • Load AL with data from memory. Compare AL with DL if its equal store the result else decrement counts go to step2.
  • Store the result.

Input

  • (Search the byte) A in 50 locations from 1500.

Output

  • Store the result byte in 1600.

Program


ADDRESS

OPCODE

MNEMONICS

COMMENTS

1100

BF 00 13

MOV DI, 1600

Load 1600 into DI

1103

BE 00 12

MOV SI, 1500

Load 1500 into SI

1106

B1 50

MOV CL, 50

Load 50 into CL

1108

B2 0A

MOV DL, 0A

Load 10 into DL

110A

AC

LODSW

Load CL register with the count

110B

38 C2

CMP DL, AL

Compare DL and AL register values

110D

FE C9

DEC CL

Decrement CL register

110F

75 05

JZ 1114

If count is zero then jump into 1114

1111

75 F7

JNZ 110A

If count is not zero then jump into 110A

1113

F4

HLT

 

1114

88 05

MOV [DI], AL

Load AL value into DI

1116

4E

DEC SI

Decrement SI  register

1117

89 F3

MOV BX, SI

Load SI value into BX

1119

47

INC DI

Increment DI

111A

88 1D

MOV [DI], BL

Store BL value into DI

111C

47

INC DI

Increment DI

111D

88 3D

MOV [DI], BH

Store BH value into DI

111F

CC

INT 3

Break point

 


Gray Code Conversion (Look Up Table)

Flow Chart



gray-code-conversion


AlgorithmFlow Chart

  • Load the memory with truth table of gray codes.
  • Initialize the pointer to the memory for data and result.
  • Load AL with the data from memory.
  • Convert gray code for that data.
  • Store the result into Memory.

Input

  • Data in 1500.

Output

  • Result in 1501.

Lookup Table

  • Start from 1600.

The look up table is provided by hex or of two bits in a byte the value ranges from 00 to 0f. 1600 - 00 01 03 02 06 07 05 04 0c 0d 0f 0e 0a 0b 09 08.


Program


ADDRESS

OPCODE

MNEMONICS

COMMENTS

1100

BB 00 12

 

MOV BX, 1600

 

Load 1200 into BX

1103

BE 50 11

MOV SI, 1500

Load 1500 into SI

1106

AC

LODSB

Load the accumulator with the data

1107

D7

XLAT

Check gray code for that data

1108

BF 51 11

MOV DI, 1501

Load 1501 address into DI

110B

88 05

MOV [DI], AL

Store the gray code of the given data

110D

CC

INT3

Break point

 


Sum of N Consecutive Numbers

Flow Chart



sum-of-n-consecutive-numbers


Algorithm

  • Load the value of n.
  • t (n) = t (n - 1) + t (n - 2).
  • t (n - 1) = t (n - 2) + 1.
  • n = n - 1.
  • if n > 0 continue else go to step2.
  • Initialize the pointer to memory for storing the result.
  • Store result.

Input

  • Load the value of n into CL.

Output

  • Result is stored in 1600.

Program


ADDRESS

OPCODE

MNEMONICS

COMMENTS

1100

B1 O4

MOV CL,04

Load CL with value 04

1102

B0 00

MOV AL,00

Initialize 00 value into AL

1104

B3 01

MOV BL,01

Initialize 01 value into BL

1106

00 D8

ADD AL,BL

Add previous and next value

1108

FE C3

INC BL

Increment BL

110A

FE C9

DEC CL

Decrement CL

110C

75 F8

JNZ 1106

Loop executes until the desired value of n is reached

110E

BF 00 20

MOV DI,1600

Store the result in 1600

1111

89 05

MOV [DI],AX

Load AX value into DI

1113

CC

INT3

Break point

 


Ascii to Hex Code Conversion

Flow Chart



ascii-to-hex-code-conversion


Algorithm

  • Load the input data in AL register.
  • Subtract 30 from AL register value.
  • If data is less than or equal to 16 terminate the program.
  • Else subtract 7 from AL register value.
  • Result stored in AL register.

Input

  • Data input in AL register.

Output

  • Data output in AL register.

Program


ADDRESS

OPCODE

MNEMONICS

COMMENTS

1100

B0 31

MOV AL,31

Get data 31 into AL

1102

2C 30

SUB AL,30

Subtract 30 with the AL

1104

3C 10

CMP AL,10

If data is less than or equal to 16 go to 110C

1106

72 04

JB 110C

If 1st operand is below the 2nd operand then short jump into 110C

1108

74 02

JZ 110C

If count zero then jump into to 110C

110A

2C 07

SUB AL,07

Else subtract 7 from AL register value

110C

CC

INT 3

Break point

 


Bcd to Hexa Decimal Conversion

Flow Chart



bcd-to-hexa-decimal-conversion


Algorithm

  • Load the data in AL register.
  • Separate higher nibbles and (in) lower nibbles.
  • Move the higher nibbles (in) to lower nibbles position.
  • Multiply AL by 10.
  • Add lower nibbles.
  • Store the result into Memory.

Input

  • Data in AL register.

Output

  • Result in AL register.

Program


ADDRESS

OPCODE

MNEMONICS

COMMENTS

1100

B0 10

MOV AL,10

Load register AL with the data 10

1102

88 C4

MOV AH,AL

Load AL value into AH

1104

80 E4 OF

AND AH,0F

Mask higher bits

1107

88 E3

MOV BL,AH

Load AH value into BL

1109

24 F0

AND AL,F0

Mask lower bits

110B

B1 O4

MOV CL,04

Load 04 value into CL

110D

D2 C8

ROR AL,CL

Rotate the data from last 4bits to  first 4 bits

110F

B7 0A

MOV BH,0A

Load 10 value into BH

1111

F6 E7

MUL BH

Multiply by 10

1113

00 D8

ADD AL,BL

Add lower nibble to the multiplied   data

1115

CC

INT3

Break point

 


Hexa Decimal to Ascii Code

Flow Chart



hexa-decimal-to-ascii-code


Algorithm

  • Load AL with the input data.
  • Check If (AL<=9) then add 30 with AL register.
  • Else add 7 with AL register.
  • Result stored into AL register.

Input

  • Data in AL register.

Output

  • Result in AL register.

Program


ADDRESS

OPCODE

MNEMONICS

COMMENTS

1100

B0 0A

MOV AL,0A

Load register AL with the data 10

1102

3C 09

CMP AL,09

If data less than 9 add 30 to the data

1104

74 04

JZ 110A

If count is zero then go to 110A

1106

72 02

JB 110A

If 1st operand is below than 2nd operand then short jump into 110A

1108

04 07

ADD AL,07

Else Add AL with 07

110A

04 30

ADD AL,30

add 30 with AL

110C

CC

INT3

Break point

 


Matrix Addition

Flow Chart



matrix-addition-for-8086


Algorithm

  • Initialize the pointer to memory for data and result.
  • Load CL with count.
  • Add two matrices by each element.
  • Process continues until CL is 0.
  • Store the result into Memory.

Input

  • Data in 2000 consecutive location as rows and columns for first matrix.
  • Data in 3000 consecutive location as rows and columns for second matrix.

Output

  • Data in 3000 with 9 entries.

Program


ADDRESS

OPCODE

MNEMONICS

COMMENTS

1100

B1 09

MOV CL, 09

Initialize 09 into CL register

1102

BE 00 20

MOV SI, 2000

Load 2000 into SI for 1st matrix

1105

BF 00 30

MOV DI, 3000

Load 3000 into DI for 2nd matrix

1108

8A 04

MOV AL, [SI]

Load AL with data of first matrix

110A

8A 1D

MOV BL, [DI]

Load BL with data of second matrix

110C

00 D8

ADD AL, BL

Add two data of AL and BL

110E

88 05

MOV [DI], AL

Store AL with data into DI

1110

47

INC DI

Increment DI

1111

46

INC SI

Increment SI

1112

FE C9

DEC CL

Decrement CL

1114

75 F2

JNZ 1108

Loop continues until all elements of

Matrix to added

1116

CC

INT3

Break point 

 


Seperating Odd and Even

Flow Chart



seperating-odd-and-even


Algorithm

  • Initialize the pointer to memory for data and result.
  • Loaded the data in AL register from memory.
  • Rotate the AL register by one bit.
  • If carry flag is set then go to step2.
  • Store the even number as a result into the Memory.

Input

  • Data in 2000 (mixer of odd and even numbers).
  • Count: number of bytes in CL.

utput

  • Even numbers stored in 3000.

Program


ADDRESS

OPCODE

MNEMONICS

COMMENTS

1100

B1 08

MOV CL, 08

Initialize 08 into CL

1102

BE 00 20

MOV SI, 2000

Load 2000 address into SI

1105

BF 00 30

MOV DI, 3000

Load 3000 address into DI

1108

AC

LODSB

Load the counter value

1109

D0 C8

ROR AL,1

Rotate AL in one time

110B

72 FB

JB 1108

If carry occurs go to L1 (odd

  Data)

110D

D0 C0

ROL AL, 1

Else rotate by left to get original data

110F

88 05

MOV [D1], AL

Store the even data

1111

47

INC DI

Increment DI

1112

FE C9

DEC CL

Decrement CL

1114

75 F2

JNZ 1108

Loop executes until counter is zero

1116

CC

INT3

Break point

 


Fibonacci Series

Floew Chart



fibonacci-series-for-8086


Algorithm

  • Initialize the pointer to memory for storing result.
  • Number of the counts loaded into CL register.
  • T (n + 1) = t (n) + t (n - 1).
  • Repeat the above process until count is 0.

Input

  • Load number of terms in CL.

Output

  • Result in 2000 (clear the memory from 2000 by 00 before executing the program).

Program


ADDRESS

OPCODE

MNEMONICS

COMMENTS

1100

B1 10

MOV CL, 10

Initialize 10 into CL register

1102

B3 00

MOV BL, 00

Initialize 00 into BL register

1104

B2 01

MOV DL, 01

Initialize 01 into DL register

1106

BF 00 20

MOV DI, 2000

Load 2000 into DI

1109

88 D0

MOV AL, DL

Move DL value into AL

110B

00 D8

ADD AL, BL

Add BL value with AL register

110D

88 05

MOV [DI],AL

Store AL value into DI.

110F

47

INC DI

Increment DI

1110

88 D3

MOV BL, DL

Move DL value BL register

1112

88 C2

MOV DL, AL

Move AL value DL register

1114

FE C9

DEC CL

Decrement CL

1116

75 F3

JNZ110B

If count is zero then go to 110B

1118

CC

INT3

Breakpoint

 


Factorial of a Number

Flow Chart



factorial-of-a-number-for-8086


Algorithm

  • Load the counter with value of n into CL register.
  • T (n) = t (n - 1) * t (n - 2).
  • Repeat the process until n becomes to store result.
  • Initialize the pointer to memory to store result.
  • Store the result into Memory address 2000.

Input

  • Load the value of n into CL register.

Output

  • Result stored in Memory address 2000.

Program


ADDRESS

OPCODE

MNEMONICS

COMMENTS

1100

B1 04

MOV CL, 04

Load the value of 04 in CL

1102

B0 01

MOV AL, 01

Initialize 01 into AL

1104

B3 01

MOV BL, 01

Initialize 01 into BL

1106

F6 E3

MUL BL

Multiply previous value by next Value

1108

FE C3

INC BL

Increment BL

110A

FE C9

DEC CL

Decrement CL

110C

75 F8

JNZ 1106

Loop continues until count is Zero

110E

BF 00 20

MOV DI, 2000

Load 2000 address into DI

1111

89 05

MOV [DI], AX

Store AX value into DI

1113

CC

INT3

Break point

 


Find the Largest Number in an Array

Flow Chart



find-largest-number-in-an-array-for-8086


Algorithm

  • Take the first number of the array.
  • Compare with next number.
  • Take the bigger one of the them.
  • Decrement the count in CL register.
  • If the count is not zero then continue from step 2.
  • Store the result into Memory address 9500.

Input

Enter the size of array (count) in 9000.

Enter the data starting from 9001.


Output

Result is stored in 9500.


Program


ADDRESS

OPCODE

MNEMONICS

COMMENTS

1100

BE 00 90

MOV SI,9000

Load 9000 address into SI

1103

8A 0C

MOV CL,[SI]

Load SI value into  CL

1105

46

INC SI

Increment SI

1106

8A 04

MOV AL,[SI]

Move the first data in AL

1108

FE C9

DEC CL

Reduce the count

110A

46

INC SI

Increment SI

110B

3A 04

CMP AL,[SI]

if AL> [SI] then go to jump1 (no swap)

110D

73  02

JNB 1111

If count is zero then jump into 1111

110F

8A  04

MOV AL,[SI]

Else store large no in to AL

1111

FE C9

DEC CL

Decrement the count

1113

75 F5

JNZ 110A

If count is not zero then jump into 110A

1115

BF 00 95

MOV DI,9500

Else store the biggest number at  9500

1118

88 05

MOV [DI],AL

Store the AL value into DI

111A

CC

INT3

Break point

 


Average Of An Array

Flow chart



average-of-an-array-for-8086


Algorithm

  • Add the bytes one by one up to the count (CL).
  • Then divide the total with the count.

Input

  • Size of array (count) in CL = 6 (see the program).
  • Enter the data starting from 9000h.

Output

  • Average is stored in AX register.
  • Quotient in AL and the reminder in AH.

Program


ADDRESS

OPCODE

MNEMONICS

COMMENTS

 

1100

BB 00 00

MOV BX,0000

Load 0000 into BX

1103

BE 00 90

MOV SI,9000

 Array start address

1106

B8 00 00

MOV AX,0000

Load 0000 into AX

1109

B1 06

MOV CL,06

Initialize 06 into CL register

110B

88 CD

MOV CH,CL

Load the count value into CH

110D

8A 1C

MOV BL,[SI]

Get the data byte

110F

00 D8

ADD AL,BL

Add the data byte

1111

46

INC SI

Increment the SI pointer

1112

FE C9

DEC CL

Check the count

1114

75 F7

JNZ 110D

If count is not zero then go to 110D

1116

F6 F5

DIV CH

Find the average by sum/count

1118

CC

INT3

 Break point

 


Generate Square Wave

I/O ADDRESS FOR 8253 /8254:


Counter 0 → FF00


Counter 1 → FF02


Counter 2 → FF04


Counter reg → FF06


Flow chart



generate-square-wave-for-8086


Program


ADDRESS

OPCODE

MNEMONICS

COMMENTS

1100

B0 B7

MOV AL,36

Load 36 into AL for generating SQUARE

1102

BA 06 FF

MOV DX,FF06

Load FF06 into DX

1105

EE

OUT DX,AL

Send the data to the timer

1106

B0 02

MOV AL,FF

Load LSB count in the AL

1108

BA 04 FF

MOV DX,FF04

Port address in DX

110B

EE

OUT DX,AL

Output the AL contents to CLK 2

110C

B0 00

MOV AL,00

Load MSB count in the AL

110E

BA 04 FF

MOV DX,FF04

Load FF04 into DX

1111

EE

OUT DX,AL

Output the AL content to CLK 2

1112

CC

INT3

Break point

 


Descending Order

Algorithm

  • Get the first data and compare with the second data.
  • If the two data are in descending order then no swap.
  • Else swap the data byte by descending order and then again compare the other data bytes up to the count.
  • Do the above the array is a ranged in descending order.
  • Finally the array is arranged in ascending order.

Input

Enter the count in location 9000.

Enter the data location starting from 9001.


Output

Result in descending order in the location 9001.


Flow chart



descending-order-for-8086


Program


ADDRESS

OPCODE

MNEMONICS

COMMENTS

1100

BE 00 90

MOV SI, 9001

Load 9000 into SI

1103

8A 0C

MOV CL, [SI]

Load SI value into CL

1105

BE 00 90

MOV SI, 9000

get the count

1108

8A 14

MOV DL, [SI]

Load SI count value into DL

110A

46

INC SI

Increment the pointer

110B

8A 04

MOV AL, [SI]

first data in AL

110D

FE CA

DEC DL

Decrement DL

110F

74 16

JZ 1127

If count is zero then jump into 1127

1111

46

INC SI

Increment SI

1112

8A 1C

MOV BL, [SI]

Load SI count value into BL

1114

3A C3

CMP AL, BL

if al > bl go to (jump1)

1116

72 07

JB 111F

 

1118

4E

DEC SI

Decrement SI

1119

88 04

MOV [SI],AL

Load ACC value in SI

111B

88 D8

MOV AL, BL

Store the greatest data

111D

EB 03

JMP 1122

Jump into 1122

111F

4E

DEC SI

Decrement SI

1120

88 1C

MOV [SI], BL

Store the smallest data in memory

1122

46

INC SI

Increment SI

1123

FE CA

DEC DL

Decrement DL

1125

75 EA

JNZ 1111

If count is not zero then jump into 1111

1127

88 04

MOV [SI], AL

Load AL value into SI

1129

FE C9

DEC CL

Decrement CL

112B

75 D8

JNZ 1105

If count is not zero then jump into 1105

112D

CC

INT3

Break point

 


Ascending Order

Algorithm

  • Get the first data and compare with the second data.
  • If the two data are in ascending order then no swap.
  • Else swap the data byte by ascending order and then again compare the other data bytes up to the count.
  • Do the above the array is arranged in ascending order.
  • Finally the array is arranged in ascending order.

Input

Enter the count in location 9000.

Enter the data location starting from 9001.


Output

Result in ascending order in the location 9001.


Flow chart



ascending-order-for-8086


Program


ADDRESS

OPCODE

MNEMONICS

COMMENTS

1100

BE 00 90

MOV SI, 9000

Load 9000 into SI

1103

8A 0C

MOV CL, [SI]

Load SI value into CL

1105

BE 00 90

MOV SI, 9000

Get second data

1108

8A 14

MOV DL, [SI]

Load SI second data into DL

110A

46

INC SI

Increment SI

110B

8A 04

MOV AL, [SI]

Load SI value into AL

110D

FE CA

DEC DL

Decrement DL

110F

74 16

JZ 1127

If count is zero then go to 1127

1111

46

INC SI

Increment SI

1112

8A 1C

MOV BL, [SI]

Load SI value into BL

1114

38 D8

CMP AL, BL

if AL > BL go to (jump1)

1116

72 07

JNB 111F

 

1118

4E

DEC SI

Decrement SI

1119

88 04

MOV [SI],AL

Load AL value into SI

111B

88 D8

MOV AL, BL

Load BL value into AL

111D

EB 03

JMP 1122

 

111F

4E

DEC SI

Decrement SI

1120

88 1C

MOV [SI], BL

Load BL value into SI

1122

46

INC SI

Increment SI

1123

FE CA

DEC DL

Decrement DL

1125

75 EA

JNZ 1111

If count is not zero then go to 1111

1127

88 04

MOV [SI], AL

Load AL value into SI

1129

FE C9

DEC CL

Decrement CL

112B

75 D8

JNZ 1105

If count is not zero then go to 1105

112D

CC

INT3

Breakpoint

 


ADDITIONAL PROGRAMS ON 8086

Compare string


ADDRESS

MNEMONICS

1100      

LEA SI, [1200]

1104      

LEA DI, [1300]

1108      

MOV CX, 0003H

110b      

CLD

110c      

REPE CMPSB

110e      

JNZ NOTEQUAL

1110      

MOV AL, 01

1112      

MOV [1400], AL

1115      

HLT

1116      

NOTEQUAL:           MOV AL, 00

1118      

MOV [1400], AL

111b      

HLT

 


Condition 1: (Same String In Data1 And Data2)


1ST INPUT

2ND INPUT

1200

11

1300

 11

1201 

22

1301

 22

1202 

33

1302

 33

 


Output

1400 : 01


Condition 2: (Different String In Data1 And Data2)


1ST INPUT

2ND INPUT

1200

11

1300

 44

1201 

22

1301

 55

1202 

33

1302

 66

 


Output

1400


MOV string program


ADDRESS

OPCODE

1100      

MOV       CX,[1500]

1104      

LEA         SI,[1600]

1108      

LEA         DI,[1700]

110c      

CLD

110d      

REP         MOVSB

110f       

HLT

 


Input location


COUNT INPUT

DATA INPUT

1500

03

1601

 22

1601 

11

1602

 33

 


Out location


OUTPUT

1700

11

1701 

22

1703

33

 


ONE'S COMPLEMENT OF A 16-BIT NUMBER

Objective

To find the one's complement of the data in register pair AX and store the result at 1400.


Theory

In the one's complement of a binary number the ones are changed to zeros and vice versa. It is one way of representing negative numbers. All negative numbers start with a 1 at the MSBit. For instance considering the hex number 5600 For ex: 5600 = 0101 0110 0000 0000 One's complement = 1010 1001 1111 1111


= A9FF


Example

The example given is to find the one's complement of 1234 and store it in memory location 1400.


Input   :  


Data  :   (AX) = 0001 0010 0011 0100 = 1234

Result  :   [1400] = 1110 1101 1100 1011 = EDCB



MEMORY ADDRESS

OPCODE

MNEMONICS

1100

C7 C0 34 12

MOVAX, 1234

1103

F7 D0

NOT AX

1106

89 06 00 14

MOV [1400],AX

110A

F4

HLT

 


Procedure

  • Enter the above mnemonics into RAM memory from 1100 using the assembler command.
  • Using GO command execute the program and enter 1100. This is the address from where execution of your program starts.
  • Press ENTER key to start execution.
  • Reset the kit using RESET key.

Masking Off Bits Selectively

Objective

To clear 8 selected bits, the 2nd HN and the HN in a 16 bit number.


Theory

The logical AND instruction is used for masking off bits. The bits which have to be cleared are to be AND with a logical zero and the other bits are to be high. Hence to achieve the above objective, AND with 0F0F.


Example

The 16 bit number is at location 1200 and the result is at location 1400.


Input: [1200] = FF


[1201] = FF


Result: [1400] = 0F


[1401] = 0F



MEMORY ADDRESS

OPCODE

MNEMONICS

1100

8B 06 00 12

MOV BX,1200

1104

81 E0 0F0F

AND AX,0F0F

1108

89 06 00 14

MOV [1400],AX

110C

F4

HLT

 


Procedure

The procedure outlined for previous exercises is to be followed for this program also.


Computing a Boolean Expression

Objective

To obtain a Boolean expression F which has 4 terms and 8 variables A,B,C,D,E,F,G,H. F = {(AB'CDE' + A'BCD(BCD+EFGH)}


Theory

Evaluation of Boolean expressions through minimization procedures is customary. But this example seeks to do the same using the 8086 registers. The 4 minterms are in FOUR 8 bit registers. Use of logical instructions to perform this is consequential. Don't care variables are represented by set bits. The correspondence is, ABCDEFGH)))) D7 D6 D5 D4 D3 D2 D1 D0


Example

Input: AL = 10110111B ------- B7


AH = 01111111B ------ 7F


BL = 11111111B ------ FF


BH = 11111111B ------ FF


Result: [1100] = 11111111B ------ FF



MEMORY ADDRESS

OPCODE

MNEMONICS

1100

C6 C0 B7

MOV AL, B7

1103

C6 C4 7F

MOV AH, 7F

1106

C6 C3 FF

MOV BL, FF

1108

C6 C7 FF

MOV BH, FF

110C

08 FB

OR BL, BH

110E

20 DC

AND AH, BL

1110

08 E0

OR AL, AH

1112

88 06 00 12

MOV [1200], AL

1116

F4

HLT