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## PowerPoint Slideshow about 'Microcontroller Fundamentals & Programming' - jola

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### Microcontroller Fundamentals & Programming

Arithmetic Instructions

Arithmetic Instructions

The 68HC11 Arithmetic Instructions are :

- ADD, ADD with Carry ,
- SUBTRACT, SUBTRACT with Carry,
- DECIMAL ADJUST Instruction,
- INCREMENT, DECREMENT,
- COMPARE,
- MULTIPLICATION, and
- DIVISION

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

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.

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

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.

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.

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

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.

_

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

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.

.

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

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.

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

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

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

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.

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

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.

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

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.

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

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