Numerical Bases Used in Programming

  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 Decimal, Binary, BCD, & Hexadecimal Numbers (43) =

  10 (0100 0011) = BCD ( 0010 1011 ) =

  2 ( 2 B )

  16

  Registers SP

  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 Memory mapping in 8051

• ROM memory map in 8051 family

  0000H

  0FFFH 0000H

  1FFFH 8751

  AT89C51 8752

  AT89C52

  4k 8k

• RAM memory space allocation 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

  Addressing Modes

• Register • Direct • Register Indirect • Immediate • Relative • Absolute • Long • Indexed

  Register Addressing Mode MOV Rn, A ;n=0,..,7 ADD

A, Rn MOV DPL, R6 MOV DPTR, A MOV Rm, Rn

  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 !

  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

  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

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

  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. ADDA, Source ;A=A+SOURCE ADDA,#6 ;A=A+6 ADDA,R6 ;A=A+R6 ADD A,6 ;A=A+[6] or A=A+R6 ADDA,0F3H ;A=A+[0F3H]

  SUBB

A, Source ;A=A-SOURCE-C

  SUBB A,#6 ;A=A-6 SUBB A,R6 ;A=A+R6

  

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 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

  

DEC byte ;byte=byte-1

  

INC byte ;byte=byte+1

  INC R7 DEC A DEC

  40H ; [40]=[40]-1 RR – RL – RRC – RLC A EXAMPLE: RR A RR: RRC:

  C RL: RLC: ANL - ORL – XRL Bitwise Logical Operations:

AND, OR, XOR

  

CPL A ;1’s complement

Example: MOV A,#55H ;A=01010101 B L01: CPL A MOV P1,A ACALL DELAY SJMP L01

  

Stack in the 8051

• The register used to

  access the stack is called SP (stack pointer) register.

  7FH Scratch pad RAM

• The stack pointer in the

  30H

  8051 is only 8 bits wide,

  2FH

  which means that it can

  Bit-Addressable RAM take value 00 to FFH.

  20H

  When 8051 powered up,

  1FH Register Bank 3

  18H

  the SP register contains

  17H Register Bank 2 value 07.

  10H

  0FH Register ) Stack (

  08H Bank 1

  07H Register Bank 0

  00H Example: MOV R6,#25H MOV R1,#12H MOV R4,#0F3H PUSH

  6 PUSH

  12

  09H

  0AH

  0BH

  25

  12

  08H SP=08H

  09H

  0AH

  0BH

  25

  F3

  1 PUSH

  08H SP=08H

  09H

  0AH

  0BH

  25

  08H Start SP=07H

  09H

  0AH

  0BH

  4

  08H SP=09H

  LOOP and JUMP Instructions JZ Jump if A=0 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 JC Jump if CY=1 JNC Jump if CY=0 JB Jump if bit=1 JNB Jump if bit=0 JBC Jump if bit=1 and clear bit Conditional Jumps :

  DJNZ: Write a program to clear 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

  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

  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

  . 100's 10's 1's .

  1

  5

  6

• 2

  4

  8 =

  4

  16 Bit Addition

4 Decimal Addition 156 + 248

  1A44 + 22DB . 256's 16’s 1's

.

  1 A 4 4

  = 3D1F

• 2 2 D B

  

Performing the Addition with 8051

. 65536's 256's 1's .

R6 R7

• R4 R5

  = R1 R2 R3 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.

  

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.

  

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

  Interrupts

  1. Enabling and Disabling Interrupts

  2. Interrupt Priority

  

3. Writing the ISR (Interrupt Service

Routine)

  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.

  Interrupt Vectors Interrupt Vector Address System Reset 0000H External 0 0003H Timer 0 000BH External 1 0013H Timer 1 001BH Serial Port 0023H Timer 2 002BH

  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 . 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 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 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 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