Microcontroller Fundamentals &

Programming

Arithmetic Instructions

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Arithmetic InstructionsThe 68HC11 Arithmetic Instructions are :

ADD, ADD with Carry ,

SUBTRACT, SUBTRACT with Carry,

DECIMAL ADJUST Instruction,

INCREMENT, DECREMENT,

COMPARE,

MULTIPLICATION, and

DIVISION

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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.

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Memory

ACC

+

Result

Goes Back

ADD Instructions

xx

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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 3The 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.

Example: ADD Instruction

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Example:

ADCA , ADCB

Add the contents in accumulator with contents in memory and with the C-bit in the CCR.

ACC

Result

Memory

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+

C-bit

ADD with Carry Instructions

CCR

xx

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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.

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

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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 Answer is correct.

C = 0 No 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 A 9 9

Example: ADDD (Add 16-bit values)

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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.

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Memory

ACC

Result

_

SUBTRACT Instructions

xx

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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.

C = 1LDAA #$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

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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.

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MemoryACC

_

C-bit

Result

Subtract with Carry Instructions

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CCR

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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..

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

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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

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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)

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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

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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

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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

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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

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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

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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

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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

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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

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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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