Microcontroller (1) Lab Manual

UNIVERSITY OF SCIENCE &TECHNOLEGY

FACUALTY OF ENGNEERING

BIOMEDICAL DEPARTMENT

Microcontroller (1) Lab Manual

Prepared By:

Eng: Mohsen Ali AL-awami

Supervisered By:

Dr: Fadel AL-aqawa

2010-2011

LAB Expeirment (2)

• 
  Main Topics:
  

• 
  Jump ,Loop and Call instructions
  
• 
  Assembely Arthimatics and Logic Operations
  

• 
  Learning Objectives/Tasks:
  

Upon Completion this experiment ,you will be able to :

• 
  Code 8051 assembly language instruction using loops
  
• 
  Code 8051 assembly language conditional jump instructions
  
• 
  Explain condition that determine each conditional jump instruction
  
• 
  Code 8051 subroutines
  
• 
  Describe precautions in using the stack in subroutines
  
• 
  Define the range of numbers possible in 8051 unsigned data
  
• 
  Code addition and subtraction instructions for unsigned data
  
• 
  Define the range of numbers possible in 8051 signed data
  
• 
  Code addition and subtraction instructions for signed data
  
• 
  Explain carry and overflow problems and their corrections
  
• 
  Define the truth table for logic function AND,OR and XOR
  
• 
  Code 8051 assembly language logic function instructions
  

Section 1: Loop and jump instructions

Repeating a sequence of instructions a certain number of times are called a LOOP,

The loop is one of the most widely used actions is performed by the instraction

[‘’ DJNZ reg, lable ‘’].

In this instruction, the register is decremented, if not zero, it jump to target address referred to by the label.

Write a program to

(a) Clear ACC

(b) Add 3 to accumulator ten times

Solution:

Mov A,#0 ;a=0 clear ACC

Mov R2,#10 ; load counter r2=10

AGAIN: ADD A,#03 ; add 03 to acc

Djnz R2, AGAIN ;repeat untel r2=0 (10 times )

Mov R5,A ; save A in R5

Example 1:

Example 2:

What is maximum number of times that the loop in last example can be repeated?

Solution:

Since the holds the count and R2 ia an 8-bites register, it can be hold maximum of FFH

(256 in decimal times)

Loop inside a loop:

As shown in Example 2 the maximum number of count is 256 , what happens if we want to repeat an action more times than 256?

To do that, we use loop in side loop which is called a nested loop

Example 3:


Write a program to

1. 
   load the accumulator with the value 55H
   
2. 
   complement the ACC 700 times
   

Solution:

; Since 700 is larger than 255 (the maximum capacity of any register), so we will use two registers to hold account.

Mov A,#55h ; A=55

Mov R3,#10 ; r3=10

NEXT : Mov R2,#70 ; r2=70

AGAIN: CPL A ;compelement A

DJNZ R2 , AGAIN ;repeat it 70 times (inner loop)

DJNZ R3 , NEXT

• 
  Other Conditional Jumps
  

Conditional jumps for the 8051 are summerized in the next table:

Such as JZ (jump if A =0)

JC (jump if carry =1 )

instraction	Action
JZ	Jump if a=0
JNZ	Jump if a not= 0
DJNZ	Decrement and Jump if a not=0
CJNE A,BYTE	Jump if a not= byte
CJNE REG,#DATA	Jump if a not= #data
JC	Jump if carry=1
JNC	Jump if carry=0
JB	Jump if bit=1
JNB	Jump if bit=0
JBC	Jump if bit=1 and clear bit

All conditional jumps are short jumps:

The address of the target must within -128 to +127 bytes of the contents of PC

Example 4:


Write a program to determine if R5 contains the value 0 .if so, put 55H in it.

Solution:

Mov A,R5 ;copy R5 to A

JNZ NEXT ;jump if A is not zero

Mov R5,#55H

NEXT: …………………………

• 
  JNC(jump if no carry, jumps if cy=0):
  
• 
  In executing ‘’JNC’’ ,the processor looks at the carry flag to see if it raised (cy=1).if it is not ,the CPU starts to fetch and execute instructions from the address of the label .if the carry =1 ,it will not it will execute the next instraction below JNC.
  

It need to be noted that there is also ‘’JC lable ’’ instruction .in the jc instruction, if cy=1 it jumps to the target address.

Example 5:


Find the sum of the values 79H,F5H and E2H.put the sum in the registers R0(low byte)and R5(high byte).

Solution:

Mov A,#0 ; clear A (A=0)

Mov R5,A ; clear R5

ADD A,#79H ; A=0+79H=79H

JNC N1 ; if no carry ,add next number

INC R5 ;if cy=1,increment R5

N1: ADD A,#0F5H ; A=79+F5=6E

JNC N2 ; jump if cy=0

INC R5 ; if cy=1 ,then incerement R5

N2 : ADD A,#0E2H ; A=6E+E2=50 and CY=1

JNC over ; jump if cy=0

INC R5 ; if cy=1 ,then incerement R5

OVER: mov R0,A ;Now R0=50h , and R5 =02 SS

• 
  The unconditional jump is a jump in which control is transferred unconditionally to the target location
  

• 
  LJMP (long jump)
  

• 
  3-byte instruction
  
• 
  First byte is the opcode
  
• 
  Second and third bytes represent the 16-bit
  

target address

– Any memory location from 0000 to FFFFH

• 
  SJMP (short jump)
  

• 
  2-byte instruction
  
• 
  First byte is the opcode
  
• 
  Second byte is the relative target address
  

– 00 to FFH (forward +127 and backward

-128 bytes from the current PC).

Call instructions

• 
  Call instruction is used to call subroutine
  

• 
  Subroutines are often used to perform tasks
  

that need to be performed frequently

• 
  This makes a program more structured in
  

addition to saving memory space

• 
  LCALL (long call)
  

• 
  3-byte instruction
  

• 
  First byte is the opcode
  
• 
  Second and third bytes are used for address of target subroutine
  

– Subroutine is located anywhere within 64K byte address space

• 
  ACALL (absolute call)
  

• 
  2-byte instruction
  

• 
  11 bits are used for address within 2K-byte range
  

• 
  When a subroutine is called, control is transferred to that subroutine, the processor
  

• 
  Saves on the stack the the address of the instruction immediately below the LCALL
  
• 
  Begins to fetch instructions form the new location
  

• 
  After finishing execution of the subroutine
  

• 
  The instruction RET transfers control back to the calle
  
• 
  Every subroutine needs RET as the last instruction
  

Example 6:


ORG 0

BACK: MOV A,#55H ;load A with 55H

MOV P1,A ;send 55H to port 1

LCALL DELAY ;time delay

MOV A,#0AAH ;load A with AA (in hex)

MOV P1,A ;send AAH to port 1

LCALL DELAY

SJMP BACK ;keep doing this indefinitely

;---------- this is delay subroutine ------------

ORG 300H ;put DELAY at address 300H

DELAY:MOV R5,#0FFH ;R5=255 (FF in hex), counter

AGAIN:DJNZ R5,AGAIN ;stay here until R5 become 0

RET

END

001 0000 ORG 0

002 0000 7455 BACK: MOV A,#55H ;load A with 55H

003 0002 F590 MOV P1,A ;send 55H to p1

004 0004 120300 LCALL DELAY ;time delay

005 0007 74AA MOV A,#0AAH ;load A with AAH

006 0009 F590 MOV P1,A ;send AAH to p1

007 000B 120300 LCALL DELAY

008 000E 80F0 SJMP BACK ;keep doing this

009 0010

010 0010 ;-------this is the delay subroutine------

011 0300 ORG 300H

012 0300 DELAY:

013 0300 7DFF MOV R5,#0FFH ;R5=255

014 0302 DDFE AGAIN: DJNZ R5,AGAIN ;stay here

015 0304 22 RET ;return to caller

016 0305 END ;end of asm file

0A
09	00
08	07

• 
  Stack fram after the first LCALL
  

Low byte goes first then high byte
08
SP (stack pointer) = 09

• 
  The use of ACALL instead of LCALL
  

can save a number of bytes of program ROM space .

ARITHMETIC & LOGIC

INSTRUCTIONS AND

PROGRAMS

• 
  Assembely Arthimatics Operations:
  

• 
  Addition of unsigned numbers
  

• 
  ADD A,source ;A = A + source
  
• 
  The instruction ADD is used to add two operands
  

• 
  Destination operand is always in register A
  
• 
  Source operand can be a register, immediate data, or in memory
  
• 
  Memory-to-memory arithmetic operations are never .
  

Example 1:


Show how the flag register is affected by the following instruction.

MOV A,#0F5H ;A=F5 hex

ADD A,#0BH ;A=F5+0B=00

Solution:

F5H 1111 0101

+ 0BH + 0000 1011

-------- -----------

100H 0000 0000

• 
  When adding two 16-bit data operands,the propagation of a carry from lower byte to higher byte is concerned.
  

Example 2:

1

3C E7

+ 3B 8D

---------

78 74


Write a program to add two 16-bit numbers. Place the sum in R7 and

R6; R6 should have the lower byte.

Solution:

CLR C ;make CY=0

MOV A, #0E7H ;load the low byte now A=E7H

ADD A, #8DH ;add the low byte

MOV R6, A ;save the low byte sum in R6

MOV A, #3CH ;load the high byte

ADDC A, #3BH ;add with the carry

MOV R7, A ;save the high byte sum

• 
  The binary representation of the digits 0 to 9 is called BCD (Binary Coded Decimal)
  

• 
  Unpacked BCD
  

In unpacked BCD, the lower 4 bits of the number represent the BCD number, and the rest of the bits are 0 .

Ex. 00001001 and 00000101 are unpacked BCD for 9 and 5.

• 
  Packed BCD
  

In packed BCD, a single byte has two BCD number in it, one in the lower 4 bits, and one in the upper 4 bits .

Ex. 0101 1001 is packed BCD for 59H.

• 
  Adding two BCD numbers must give a BCD result.
  

Example 2:


MOV A, #17H

ADD A, #28H

The result above should have been 17 + 28 = 45 (0100 0101).

To correct this problem, the programmer must add 6 (0110) to the

low digit: 3F + 06 = 45H.

• 
  DA A ;decimal adjust for addition
  

• 
  The DA instruction is provided to correct the aforementioned problem associated with BCD addition
  
• 
  The DA instruction will add 6 to the lowe nibble or higher nibble if need .
  

Example 3 :


MOV A,#47H ;A=47H first BCD operand

MOV B,#25H ;B=25H second BCD operand

ADD A,B ;hex(binary) addition(A=6CH)

DA A ;adjust for BCD addition

(A=72H)

The “DA” instruction works only on A. In other word, while the source

can be an operand of any addressing mode, the destination must be in

register A in order for DA to work.

• 
  Subtraction of unsigned numbers :
  
• 
  In many microprocessor there are two different instructions for subtraction: SUB and SUBB (subtract with borrow)
  

• 
  In the 8051 we have only SUBB
  
• 
  The 8051 uses adder circuitry to perform the subtraction
  

SUBB A,source ;A = A – source – CY
To make SUB out of SUBB, we have to make CY=0 prior to the execution of the instruction

• 
  Notice that we use the CY flag for the borrow
  
• 
  SUBB when CY = 0
  

• 
  1. Take the 2’s complement of the subtrahend (source operand)
  
• 
  2. Add it to the minuend (A)
  
• 
  3. Invert the carry
  

CLR C

MOV A,#4C ;load A with value 4CH

SUBB A,#6EH ;subtract 6E from A

JNC NEXT ;if CY=0 jump to NEXT

CPL A ;if CY=1, take 1’s complement

INC A ;and increment to get 2’s comp

NEXT: MOV R1,A ;save A in R1

Solution:

4C 0100 1100 0100 1100

- 6E 0110 1110 1001 0010

---- ----- -----

-22 01101 1110
Example 4:

• 
  SUBB when CY = 1
  

• 
  This instruction is used for multi-byte numbers and will take care of the borrow of the lower operand .
  

Example 5:


CLR C

MOV A,#62H ;A=62H

SUBB A,#96H ;62H-96H=CCH with CY=1

MOV R7,A ;save the result

MOV A,#27H ;A=27H

SUBB A,#12H ;27H-12H-1=14H

MOV R6,A ;save the result

Solution:

We have 2762H - 1296H = 14CCH.

SIGNED ARITHMETIC INSTRUCTIONS

(Signed 8-bit Operands )

• 
  D7 (MSB) is the sign and D0 to D6 are the magnitude of the number
  

• 
  If D7=0, the operand is positive, and if D7=1, it is negative
  
• 
  Positive numbers are 0 to +127
  
• 
  Negative number representation (2’s complement)
  

1. Write the magnitude of the number in 8-bit binary (no sign)

2. Invert each bit

3. Add 1 to it.

D7

D6

D5

D4

D3

D2

D1

D0

----

Sign Magnitude

Show how the 8051 would represent -34H

Solution:

1. 0011 0100 34H given in binary

2. 1100 1011 invert each bit

3. 1100 1100 add 1 (which is CC in hex)

Signed number representation of -34 in 2’s complement is CCH


SIGNED ARITHMETIC INSTRUCTIONS

(Overflow Problem)

• 
  If the result of an operation on signed numbers is too large for the register.
  

• 
  An overflow has occurred and the programmer must be noticed.
  

Examine the following code and analyze the result.

MOV A,#+96 ;A=0110 0000 (A=60H)

MOV R1,#+70 ;R1=0100 0110(R1=46H)

ADD A,R1 ;A=1010 0110

;A=A6H=-90,INVALID

Solution:

[phone]

+ [phone]

----- -------------

+ 166 1010 0110 and OV =1
According to the CPU, the result is -90, which is wrong. The CPU

sets OV=1 to indicate the overflow

SIGNED ARITHMETIC INSTRUCTIONS

(2's Complement)

• 
  To make the 2’s complement of a number
  

CPL A ;1’s complement (invert)

ADD A,#1 ;add 1 to make 2’s comp.

LOGIC AND COMPARE INSTRUCTIONS

1. 
   AND LOGIC)
   

ANL destination,source ;dest = dest AND source

• 
  This instruction will perform a logic AND on the two operands and place the result in the destination
  
• 
  The destination is normally the accumulator
  
• 
  The source operand can be a register, in memory, or immediate
  

Example 1:


Show the results of the following.

MOV A,#35H ;A = 35H

ANL A,#0FH ;A = A AND 0FH

35H 0 0 1 1 0 1 0 1

0FH 0 0 0 0 1 1 1 1

= ------- -------- -----------

05H 0 0 0 0 0 1 0 1

2. 
   OR LOGIC)
   

• 
  ORL destination,source ;dest = dest OR source
  

The destination and source operands are ORed and the result is placed in the destination .

• 
  The destination is normally the accumulator
  
• 
  The source operand can be a register, in memory, or immediate .
  

Example 2:


Show the results of the following.

MOV A,#04H ;A = 04

ORL A,#68H ;A = 6C
04H 0 0 0 0 0 1 0 0

68H 0 1 1 0 1 0 0 0

=----- --------- ---------

6CH 0 1 1 0 1 1 0 0

(3- XOR)

XRL destination, source ;dest = dest XOR source

• 
  This instruction will perform XOR operation on the two operands and
  

place the result in the destination

• 
  The destination is normally the accumulator
  
• 
  The source operand can be a register, in memory, or immediate .
  

Show the results of the following.

MOV A,#54H

XRL A,#78H

54H 0 1 0 1 0 1 0 0

78H 0 1 1 1 1 0 0 0

= ------ -------- ----------

2CH 0 0 1 0 1 1 0 0
Example 3:

4- Compare Instruction

CJNE destination,source,rel. addr.

• 
  The actions of comparing and jumping are combined into a single instruction called CJNE (compare and jump if not equal)
  
• 
  The CJNE instruction compares two operands, and jumps if they are not equal.
  
• 
  The destination operand can be in the accumulator or in one of the Rn registers The source operand can be in a register, in memory, or immediate The operands themselves remain unchanged.
  
• 
  It changes the CY flag to indicate if the destination operand is larger or smaller.
  

Example 4:


CJNE R5,#80,NOT_EQUAL ;check R5 for 80

... ;R5 = 80

NOT_EQUAL:

JNC NEXT ;jump if R5 > 80

... ;R5 < 80

NEXT: ...

• 
  Compare Carry Flag
  

• 
  destination ≥ source CY = 0
  
• 
  destination < source CY = 1
  

• 
  The compare instruction is really a Subtraction.
  

Rotating Right and Left

• 
  RR A ;rotate right A
  
• 
  In rotate right
  

• 
  The 8 bits of the accumulator are rotated right one bit, and
  
• 
  Bit D0 exits from the LSB and enters into MSB, D7
  

MSB LSB



MOV A,#36H ;A = 0011 0110

RR A ;A = 0001 1011

RR A ;A = 1000 1101

RR A ;A = 1100 0110

RR A ;A = 0110 0011

t’)

• 
  RL A ;rotate left A
  

• 
  In rotate left
  

• 
  The 8 bits of the accumulator are rotate left one bit, and
  
• 
  Bit D7 exits from the MSB and enters into LSB, D0
  

MSB LSB



MOV A,#72H ;A = 0111 0010

RL A ;A = 1110 0100

RL A ;A = 1100 1001


Rotating through Carry
RRC A ;rotate right through carry

• 
  In RRC A
  

• 
  Bits are rotated from left to right
  
• 
  They exit the LSB to the carry flag, and the carry flag enters the MSB.
  

MSB LSB

CY


CLR C ;make CY = 0

MOV A,#26H ;A = 0010 0110

RRC A ;A = 0001 0011 CY = 0

RRC A ;A = 0000 1001 CY = 1

RRC A ;A = 1000 0100 CY = 1


RLC A ;rotate left through carry

• 
  In RLC A
  

• 
  Bits are shifted from right to left.
  
• 
  They exit the MSB and enter the carry flag,
  

and the carry flag enters the LSB.


MSB LSB


CY


Write a program that finds the number of 1s in a given byte.

MOV R1,#0

MOV R7,#8 ;count=08

MOV A,#97H

AGAIN: RLC A

JNC NEXT ;check for CY

INC R1 ;if CY=1 add to count

NEXT: DJNZ R7,AGAIN