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Microcontroller 8051 Assembly Language

Microcontroller 8051 Assembly Language. Numerical Bases Used in Programming. Hexadecimal Binary BCD. Hexadecimal Basis. Hexadecimal Digits: 1 2 3 4 5 6 7 8 9 A B C D E F A=10 B=11 C=12 D=13 E=14 F=15.

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Microcontroller 8051 Assembly Language

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  1. Microcontroller 8051 Assembly Language

  2. Numerical Bases Used in Programming • Hexadecimal • Binary • BCD

  3. Hexadecimal Basis • Hexadecimal Digits: 1 2 3 4 5 6 7 8 9 A B C D E F A=10 B=11 C=12 D=13 E=14 F=15

  4. Decimal, Binary, BCD, & Hexadecimal Numbers (43)10= (0100 0011)BCD= ( 0010 1011 )2 = ( 2 B )16

  5. A B R0 DPTR DPH DPL R1 R2 PC PC R3 Some 8051 16-bit Register R4 R5 R6 R7 Some 8-bitt Registers of the 8051 Registers SP

  6. Memory mapping in 8051 ROM memory map in 8051 family 0000H 0000H 0FFFH 1FFFH 8751 AT89C51 8752 AT89C52 4k 8k

  7. 7FH Scratch pad RAM 30H 2FH Bit-Addressable RAM 20H 1FH Register Bank 3 18H 17H Register Bank 2 10H 0FH (Stack) Register Bank 1 08H 07H Register Bank 0 00H • RAM memory space allocation in the 8051

  8. Addressing Modes • Register • Direct • Register Indirect • Immediate • Relative • Absolute • Long • Indexed

  9. Register Addressing Mode MOV Rn, A ;n=0,..,7 ADD A, Rn MOV DPL, R6 MOV DPTR, A MOV Rm, Rn

  10. 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. MOV R0, 40H MOV 56H,A MOV A, 4 ; ≡ MOV A, R4 MOV 6, 2 ; copy R2 to R6 ; MOV R6,R2 is invalid !

  11. 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 ) MOV @R1,B In other word, the content of register R0 or R1 is sources or target in MOV, ADD and SUBB insructions.

  12. Immediate Addressing Mode MOV A,#65H MOV R6,#65H MOV DPTR,#2343H MOV P1,#65H

  13. Relative, Absolute, & Long Addressing Used only with jump and call instructions: SJMP ACALL,AJMP LCALL,LJMP

  14. 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 MOVC A,@A+DPTR (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.

  15. Some Simple Instructions MOV dest,source ; dest = source MOV A,#72H ;A=72H MOV R4,#62H ;R4=62H MOV B,0F9H ;B=the content of F9’th byte of RAM MOV DPTR,#7634H MOV DPL,#34H MOV DPH,#76H MOV P1,A ;mov A to port 1 Note 1: 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

  16. 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] SUBB A, Source ;A=A-SOURCE-C SUBB A,#6 ;A=A-6 SUBB A,R6 ;A=A+R6

  17. MUL & DIV • MUL AB ;B|A = A*B MOV A,#25H MOV B,#65H MUL AB ;25H*65H=0E99 ;B=0EH, A=99H • DIV AB ;A = A/B, B = A mod B MOV A,#25 MOV B,#10 DIV AB ;A=2, B=5

  18. SETB bit ; bit=1 CLR bit ; bit=0 SETB C ; CY=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.: CLR C ;CY=0 But following instruction is only for CLR: CLR A ;A=0

  19. DEC byte ;byte=byte-1 INC byte ;byte=byte+1 INC R7 DEC A DEC 40H ; [40]=[40]-1

  20. RR – RL – RRC – RLC A EXAMPLE: RR A RR: RRC: RL: RLC:

  21. ANL - ORL – XRL Bitwise Logical Operations: AND, OR, XOR EXAMPLE: MOV R5,#89H ANL R5,#08H CPL A ;1’s complement Example: MOV A,#55H ;A=01010101 B L01: CPL A MOV P1,A ACALL DELAY SJMP L01

  22. Stack in the 8051 7FH Scratch pad RAM 30H 2FH Bit-Addressable RAM 20H 1FH Register Bank 3 18H 17H Register Bank 2 10H 0FH (Stack) Register Bank 1 08H 07H Register Bank 0 00H • 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.

  23. 0BH 0BH 0BH 0BH 12 25 F3 12 25 25 0AH 0AH 0AH 0AH 09H 09H 09H 09H 08H 08H 08H 08H SP=08H SP=09H SP=08H Start SP=07H Example: MOV R6,#25H MOV R1,#12H MOV R4,#0F3H PUSH 6 PUSH 1 PUSH 4

  24. LOOP and JUMP Instructions Conditional Jumps :

  25. DJNZ: Write a program toclear ACC, then add 3 to the accumulator ten time Solution: MOV A,#0 MOV R2,#10 AGAIN: ADD A,#03 DJNZ R2,AGAIN ;repeat until R2=0 (10 times) MOV R5,A

  26. 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.

  27. 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 2Kbyte range.

  28. Example: Write a program to copy a block of 10 bytes from RAM location starting at 37h to RAM location starting at 59h. Solution: MOV R0,#37h ; source pointer MOV R1,#59h ; dest pointer MOV R2,#10 ; counter L1: MOV A,@R0 MOV @R1,A INC R0 INC R1 DJNZ R2,L1

  29. Decimal Addition 16 Bit Addition • 1A44 + 22DB = 3D1F 156 + 248

  30. Performing the Addition with 8051 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.

  31. 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

  32. The Whole Program ;Load the first value into R6 and R7 MOV R6,#1Ah MOV R7,#44h ;Load the first value into R4 and R5 MOV R4,#22h MOV R5,#0DBh ;Call the 16-bit addition routine LCALL ADD16_16 ADD16_16: ;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

  33. Timer & Port Operations • Example: Write a program using Timer0 to create a 10khz square wave on P1.0 MOV TMOD,#02H ;8-bit auto-reload mode MOV TH0,#-50 ;-50 reload value in TH0 SETB TR0 ;start timer0 LOOP: JNB TF0, LOOP ;wait for overflow CLR TF0 ;clear timer0 overflow flag CPL P1.0 ;toggle port bit SJMP LOOP ;repeat END

  34. Interrupts • Enabling and Disabling Interrupts • Interrupt Priority • Writing the ISR (Interrupt Service Routine)

  35. 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.

  36. Interrupt Vectors

  37. Writing the ISR Example: Writing the ISR for Timer0 interrupt ORG 0000H ;reset LJMP MAIN ORG 000BH ;Timer0 entry point T0ISR: . ;Timer0 ISR begins . RETI ;return to main program MAIN: . ;main program . . END

  38. Structure of Assembly language and Running an 8051 program EDITOR PROGRAM Myfile.asm ASSEMBLER PROGRAM Myfile.lst Other obj file Myfile.obj LINKER PROGRAM OH PROGRAM Myfile.hex

  39. Examples of Our Program Instructions • MOV C,P1.4 JC LINE1 • SETB P1.0 CLR P1.2

  40. 8051 Instruction Set 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 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

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