Microcontroller Fundamentals & Programming

1. Microcontroller Fundamentals & Programming Arithmetic Instructions

2. Arithmetic Instructions The 68HC11 Arithmetic Instructions are : • ADD, ADD with Carry , • SUBTRACT, SUBTRACT with Carry, • DECIMAL ADJUST Instruction, • INCREMENT, DECREMENT, • COMPARE, • MULTIPLICATION, and • DIVISION

3. ADD Instructions • Example: • ADDA , ADDB , ADDD • Add contents in accumulator with contents in memory. • The result after addition will be placed in accumulator. • The flag bits in CCR will change according to the result of the addition. Memory 11 23 35 + Result Goes Back xx ACC

4. Example: ADD Instruction LDAA #$29 ACCA = 0 0 1 0 1 0 0 1 ADDA #$3A 3A = 0 0 1 1 1 0 1 0 + Result: ACCA = $63 Binary= 0 1 1 0 0 0 1 1 Hex = 6 3 The flag bits in the CCR change as follow: H = 1 A “1” is carried over from bit-3 to bit-4 N = 0 Answer or result is positive Z = 0 The result is not zero V = 0 No overflow. Answer is correct. C = 0 No carry.

5. Memory 11 23 35 + ADD with Carry Instructions Example: ADCA , ADCB • Add the contents in accumulator with contents in memory and with the C-bit in the CCR. CCR C-bit xx ACC Result

6. Example: ADD with CARRY instruction SEC ; Set Carry to “1” LDAA #$29 ; ACCA = 0 0 1 0 1 0 0 1 ADCA #$3A ; $3A = 0 0 1 1 1 0 1 0 + C-bit = 0 0 0 0 0 0 0 1 + Result:ACCA = $64 Binary = 0 1 1 0 0 1 0 0 Hex. = 6 4 The flag bits in the CCR change as follow: H = 1 A “1” is carried over from bit-3 to bit-4 N = 0 Answer or result is positive Z = 0 The result is not zero V = 0 No overflow. Answer is correct. C = 0 No carry.

7. Example: ADDD (Add 16-bit values) LDD #$2299 ACCD = 0010 0010 1001 1001 ADDD #$0800 ;$0800 = 0000 1000 0000 0000 + Result:ACCD = $2A99 ;Binary = 0010 1010 1001 1001 Hex. = 2 A 9 9 The flag bits in the CCR change as follow: H = X Not use. N = 0 Answer or result is positive. Z = 0 The result is not zero V = 0Answer is correct. C = 0 No carry.

8. 11 23 35 Memory _ SUBTRACT Instructions Example: SUBA , SUBB , SUBD • Subtract memory contents from the accumulator. • The result after subtraction placed in accumulator. • Flag bits in CCR will change according to the result of subtraction. ACC xx Result

9. Example: SUBTRACT instruction C = 1 LDAA #$29 ; ACCA = 0 0 1 0 1 0 0 1 SUBA #$3A ; $3A = 0 0 1 1 1 0 1 0 - Result:ACCA = $EF Binary = 1 1 1 0 1 1 1 1 Hex. = E F The flag bits in the CCR change as follow: H = X Not use. N = 1 Answer or result is negative Z = 0 The result is not zero V = 0 Answer is correct. C = 1 $3A larger than $29, so a borrow is required.

10. _ Subtract with Carry Instructions • Example: • SBCA , SBCB • Subtract memory contents and the C-bit from accumulator. • The result after the subtraction will be placed in accumulator. • The flag bits in CCR will change according to the result of subtraction. Memory ACC xx 11 23 35 Result C-bit CCR

11. Example: Subtract with Carry Instruction SEC ; Set Carry to “1” LDAA #$3A ; ACCA = 0 0 1 1 1 0 1 0 SBCA #$23 ; $23 = 0 0 1 0 0 0 1 1 - C-bit = 0 0 0 0 0 0 0 1 -Result:ACCA= $16Binary = 0 0 0 1 0 1 1 0 Hex. = 1 6 The flag bits in the CCR change as follow: H = X Not use N = 0 Answer or result is positive Z = 0 The result is not zero V = 0 No overflow. Answer is correct. C = 0 No borrow. .

12. Decimal Adjusted Instruction Example: DAA • DAA instruction adjusts accumulator A contents immediately following an ADDA or ADCA only. • DAA will adjust the result to BCD format. LDAA #$29 ; ACCA = $2 9 ADDA #$3A ; $3A = $3 A + DAA ; before DAA = $6 3 (6) + Result: ACCA = $69 ; after DAA = $6 9

13. Decimal Adjusted Instruction (Rule) • Example: DAA • Rule for BCD addition • If result is equal orless than 9 and a carry is not produced, then answer is correct. (No adjustment needed) • If result is greater than 9 or a carry is produced, a correction of +6 must be added to the sum.

14. Examples of DAA (1) Example 1: Equal or less than 9 and NO carry LDAA #$25 ; ACCA = $2 5 ADDA #$33 ; $33 = $3 3 + DAA ; before DAA = $5 8 Result:ACCA = $58 ; after DAA = $5 8

15. Examples of DAA (2) Example 2: Greater than 9 and NO carry LDAA #$25 ; ACCA = $2 5 ADDA #$36 ; $36 = $3 6 + DAA ; before DAA = $5 B 6 + Result:ACCA = $61 ; after DAA = $6 1

16. Examples of DAA (3) Example 3: Greater than 9 and a carry LDAA #$66 ; ACCA = $6 6 ADDA #$3A ; $3A = $3 A + DAA ; before DAA = $A 0 6 6 + Result:ACCA = $06 ; after DAA = $10 6 C-bit =1

17. Increment Instructions • Example: • INC , INCA , INCB , INX , INY • After executing the increment instruction a value of one is added to the memory or registers. • Flag bits N, Z, V will be affected.

18. Examples: INCREMENT instructions LDX #$2000 ; IX = $2000 LDAA #$20 ; ACCA = $20 LDAB #$35 ; ACCB = $35 STAA $1200 ; Memory ($1200) = $20 INCA ; ACCA = $20 + $01 = $21 INCB ; ACCB = $35 + $01 = $36 INC $1200 ; Memory ($1200) = $20+$01=$21 INX ; IX = $2000+$01 = $2001 WAI

19. Decrement Instructions • Example: • DEC , DECA , DECB , DEX , DEY • After executing the decrement instruction a value of one is subtracted from the memory or registers. • Flag bits N, Z, V will be affected.

20. Examples: DECREMENT instructions LDX #$205A ; IX = $205A LDAA #$2E ; ACCA = $2E LDAB #$89 ; ACCB = $89 STAB $1500 ; Memory ($1500) = $89 DECA ; ACCA = $2E - $01 = $2D DECB ; ACCB = $89 - $01 = $88 DEC $1500 ; Memory ($1500)= $89 - $01= $88 DEX ; IX = $205A - $01 = $2059 WAI

21. Compare Instructions Example: CMPA , CMPB, CMPD , CPX , CPY • are used to compare the contents of registers and memory data. • after instruction has been executed: • the flag bits are updated according to the result. • contents of register and memory data will notchange. • Compare and conditional branch instructions are usually use together.

22. Example: Compare LDAA #$09 ; ACCA = $09 CMPA #$09 ; ACCA subtract $09 BEQ DISP ; if result equal = zero - - - ; branch to DISPLAY - - - - ; content of ACCA = $09 - - - DISP LDX $1000

23. Thank You