1. Microcontroller Fundamentals & Programming Arithmetic Instructions

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

3. 3 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. 11 23 35 Memory ACC + Result Goes Back ADD Instructions xx

4. 4 LDAA #$29ACCA = 0 0 1 0 1 0 0 1 ADDA #$3A3A = 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 = 1A “1” is carried over from bit-3 to bit-4 N = 0Answer or result is positive Z = 0The result is not zero V = 0No overflow. Answer is correct. C = 0No carry. Example: ADD Instruction

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

6. 6 The flag bits in the CCR change as follow: H = 1A “1” is carried over from bit-3 to bit-4 N = 0Answer or result is positive Z = 0The result is not zero V = 0No overflow. Answer is correct. C = 0No carry. 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 Example: ADD with CARRY instruction

7. 7 The flag bits in the CCR change as follow: H = XNot use. N = 0Answer or result is positive. Z = 0The result is not zero V = 0Answer is correct. C = 0No carry. LDD #$2299 ACCD = 0010 0010 1001 1001 ADDD #$0800 ;$0800 = 0000 1000 0000 0000 + Result: ACCD = $2A99 ;Binary = 0010 1010 1001 1001 Hex. = 2 A9 9 Example: ADDD (Add 16-bit values)

8. 8 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. 11 23 35 Memory ACC Result _ SUBTRACT Instructions xx

9. 9 The flag bits in the CCR change as follow: H = XNot use. N = 1Answer or result is negative Z = 0The result is not zero V = 0Answer is correct. C = 1$3A larger than $29, so a borrow is required. 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 Example: SUBTRACT instruction

10. 10 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. 11 23 35 Memory ACC _ C-bit Result Subtract with Carry Instructions xx CCR

11. 11 The flag bits in the CCR change as follow: H = XNot use N = 0Answer or result is positive Z = 0The result is not zero V = 0No overflow. Answer is correct. C = 0No borrow.. 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 = $16 Binary = 0 0 0 1 0 1 1 0 Hex. = 1 6 Example: Subtract with Carry Instruction

12. 12 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 Decimal Adjusted Instruction

13. 13 Example: DAA Rule for BCD addition  If result is equal or less 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. Decimal Adjusted Instruction (Rule)

14. 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. 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. 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. 17 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. Increment Instructions

18. 18 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 Examples: INCREMENT instructions

19. 19 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. Decrement Instructions

20. 20 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 Examples: DECREMENT instructions

21. 21 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 not change.  Compare and conditional branch instructions are usually use together. Compare Instructions

22. 22 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 Example: Compare

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