Microcontroller 8051 - assembly language ppt

Relative, Absolute, & Long Addressing Used only with jump and call instructions: SJMP ACALL,AJMP LCALL,LJMP... Indexed Addressing Mode • This mode is widely used in accessing data eleme

Numerical Bases Used in Programming

• Hexadecimal

• Binary

• BCD

Decimal, Binary, BCD, & Hexadecimal Numbers

(43) 10 =

( 0010 1011 ) 2 =

( 2 B ) 16

Register s

Some 8051 16-bit Register

Some 8-bitt Registers of

the 8051

SP

AT89C51 8752

AT89C52

• RAM memory space allocation in the 8051

7FH

30H 2FH

20H 1FH

17H 10H 0FH 07H 08H 18H

) Stack) Register Bank 1

Register Bank 2 Register Bank 3 Bit-Addressable RAM Scratch pad RAM

Register Addressing Mode

Direct Addressing Mode

Although the entire of 128 bytes of RAM can be

accessed using direct addressing mode, it is most often used to access RAM loc 30 – 7FH.

Register Indirect Addressing Mode

• In this mode, register is used as a pointer to the data.

MOV A,@Ri ; move content of RAM loc

where address is held by Ri into A ( i=0 or 1 )

In other word, the content of register R0 or R1 is

sources or target in MOV, ADD and SUBB insructions.

Immediate Addressing Mode

MOV A,#65H

MOV R6,#65H

MOV DPTR,#2343H

MOV P1,#65H

Relative, Absolute, & Long Addressing

Used only with jump and call instructions:

SJMP ACALL,AJMP LCALL,LJMP

Indexed Addressing Mode

• This mode is widely used in accessing data

elements of look-up table entries located in the program (code) space ROM at the 8051

(A,@A+PC) A= content of address A +DPTR from ROM Note:

Because the data elements are stored in the

program (code ) space ROM of the 8051, it uses the instruction MOVC instead of MOV The “C” means code.

Some Simple Instructions

MOV A,#72H ;A=72H

MOV A,#72H ≠ MOV A,72H

After instruction “MOV A,72H ” the content of 72’th byte of RAM will replace in Accumulator

Note 2:

MOV A,R3 ≡ MOV A,3

ADD A, Source ;A=A+SOURCE

ADD A,#6 ;A=A+6

ADD A,R6 ;A=A+R6

ADD A,6 ;A=A+[6] or A=A+R6 ADD A,0F3H ;A=A+[0F3H]

MUL & DIV

• MUL AB ;B|A = A*B

SETB bit ; bit=1

SETB P0.0 ;bit 0 from port 0 =1

SETB P3.7 ;bit 7 from port 3 =1

SETB ACC.2 ;bit 2 from ACCUMULATOR =1 SETB 05 ;set high D5 of RAM loc 20h Note:

CLR instruction is as same as SETB

i.e.:

But following instruction is only for CLR:

DEC byte ;byte=byte-1

ANL - ORL – XRL

Bitwise Logical Operations:

AND, OR, XOR

Stack in the 8051

• The register used to access

the stack is called SP

(stack pointer) register.

• The stack pointer in the

8051 is only 8 bits wide,

which means that it can

take value 00 to FFH

When 8051 powered up,

the SP register contains

value 07.

7FH

30H 2FH

20H 1FH

17H 10H 0FH 07H 08H 18H

00H Register Bank 0

) Stack) Register Bank 1

Register Bank 2 Register Bank 3 Bit-Addressable RAM Scratch pad RAM

MOV R6,#25H MOV R1,#12H MOV R4,#0F3H PUSH 6

PUSH 1 PUSH 4

SP=08H

F3 12 25

0BH 0AH 09H 08H

SP=08H

12 25

0BH 0AH 09H 08H

SP=09H

LOOP and JUMP Instructions

JNZ Jump if A/=0

DJNZ Decrement and jump if A/=0

CJNE A,byte Jump if A/=byte

CJNE reg,#data Jump if byte/=#data

Write a program to clear ACC, then

add 3 to the accumulator ten time

Solution:

MOV A,#0 MOV R2,#10

DJNZ R2,AGAIN ;repeat until R2=0 (10 times) MOV R5,A

LJMP(long jump)

LJMP is an unconditional jump It is a 3-byte instruction It allows a jump to any memory location from 0000 to FFFFH.

AJMP(absolute jump)

In this 2-byte instruction, It allows a jump to any memory

location within the 2k block of program memory.

SJMP(short jump)

In this 2-byte instruction The relative address range of

00-FFH is divided into forward and backward jumps, that is ,

within -128 to +127 bytes of memory relative to the address of the current PC.

CALL Instructions

Another control transfer instruction is the CALL

instruction, which is used to call a subroutine.

• LCALL(long call)

This 3-byte instruction can be used to call

subroutines located anywhere within the 64K byte address space of the 8051.

• ACALL (absolute call)

ACALL is 2-byte instruction the target

address of the subroutine must be within 2K

byte range.

Write a program to copy a block of 10 bytes from RAM location starting at 37h to RAM location starting at 59h.

Solution:

Performing the Addition with 8051

. 65536's 256's 1's

1.Add the low bytes R7 and R5, leave the answer in R3

2.Add the high bytes R6 and R4, adding any carry from step 1, and leave the answer in R2 3.Put any carry from step 2 in the final byte, R1

Steps 1, 2, 3

MOV A,R7 ;Move the low-byte into the accumulator

ADD A,R5 ;Add the second low-byte to the accumulator

MOV R3,A ;Move the answer to the low-byte of the result

MOV A,R6 ;Move the high-byte into the accumulator

ADDC A,R4 ;Add the second high-byte to the accumulator, plus carry MOV R2,A ;Move the answer to the high-byte of the result

MOV A,#00h ;By default, the highest byte will be zero

ADDC A,#00h ;Add zero, plus carry from step 2

MOV R1,A ;Move the answer to the highest byte of the result

The Whole Program

;Load the first value into R6 and R7

;Step 1 of the process

MOV A,R7 ;Move the low-byte into the accumulator

ADD A,R5 ;Add the second low-byte to the accumulator

MOV R3,A ;Move the answer to the low-byte of the result

;Step 2 of the process

MOV A,R6 ;Move the high-byte into the accumulator

ADDC A,R4 ;Add the second high-byte to the accumulator, plus carry MOV R2,A ;Move the answer to the high-byte of the result

;Step 3 of the process

MOV A,#00h ;By default, the highest byte will be zero

ADDC A,#00h ;Add zero, plus carry from step 2

MOV MOV R1,A ;Move the answer to the highest byte of the result

;Return - answer now resides in R1, R2, and R3 RET

Timer & Port Operations

• Example:

Write a program using Timer0 to create a 10khz square wave on P1.0

END

Interrupt Enable (IE) Register :

• EA : Global enable/disable.

• - : Undefined.

• ET2 :Enable Timer 2 interrupt.

• ES :Enable Serial port interrupt.

• ET1 :Enable Timer 1 interrupt.

• EX1 :Enable External 1 interrupt.

• ET0 : Enable Timer 0 interrupt

• EX0 : Enable External 0 interrupt.

Writing the ISR

Example:

Writing the ISR for Timer0 interrupt

LJMP MAIN

.

.

.

END

Structure of Assembly language and Running an 8051 program

EDITOR PROGRAM

ASSEMBLER PROGRAM

LINKER PROGRAM

OH PROGRAM

Examples of Our Program Instructions

• MOV C,P1.4

JC LINE1

• SETB P1.0

CLR P1.2

8051 Instruction Set

ACALL: Absolute Call

ADD, ADDC: Add Acc (With Carry)

AJMP: Absolute Jump

ANL: Bitwise AND

CJNE: Compare & Jump if Not Equal

CLR: Clear Register

CPL: Complement Register

DA: Decimal Adjust

DEC: Decrement Register

DIV: Divide Accumulator by B

DJNZ: Dec Reg & Jump if Not Zero

INC: Increment Register

JB: Jump if Bit Set

JBC: Jump if Bit Set and Clear Bit

JC: Jump if Carry Set JMP: Jump to Address JNB: Jump if Bit Not Set JNC: Jump if Carry Not Set JNZ: Jump if Acc Not Zero JZ: Jump if Accumulator Zero LCALL: Long Call

LJMP: Long Jump MOV: Move Memory MOVC: Move Code Memory MOVX: Move Extended Memory MUL: Multiply Accumulator by B NOP: No Operation

ORL: Bitwise OR POP: Pop Value From Stack

PUSH: Push Value Onto Stack RET: Return From Subroutine RETI: Return From Interrupt RL: Rotate Accumulator Left RLC: Rotate Acc Left Through Carry RR: Rotate Accumulator Right

RRC: Rotate Acc Right Through Carry SETB: Set Bit

SJMP: Short Jump SUBB: Sub From Acc With Borrow SWAP: Swap Accumulator Nibbles XCH: Exchange Bytes

XCHD: Exchange Digits XRL: Bitwise Exclusive OR Undefined: Undefined Instruction