Lect5 Assembler

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Assembler language 8051

Mairtin O Conghaile


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2

8051 Microcontroller

Complete, highly-integrated microcomputer CPU, RAM, ROM, IO

Port 0

8-bit bidirectional I/O port OR multiplexed low-order address and data bus bytes

Port 1 8-bit bidirectional I/O port

Port 2

8-bit bidirectional I/O port OR high-order address byte Port 3

8-bit bidirectional I/O port OR various special-function signals


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3


Interface Signals (1)P O / A D [ 7 - 0 ]

P 1

P 2 / A [ 1 5 - 8 ]

P 3 . 2 / I N T 0 *P 3 . 3 / I N T 1 *

P 3 . 4 / T 0P 3 . 5 / T 1

P 3 . 6 / W R *P 3 . 7 / R D *

W r i t eR e a d

T i m e r 1T i m e r 0

I n t e r r u p t 1I n t e r r u p t 0

P o r t 0 o r L o A d d r / D a t a

P o r t 1

P o r t 2 o r H i A d d r

P o r t 3 / S p e c i a l F u n c t i o n s :

8

8

8

P 3 . 0 / R X DP 3 . 1 / T X D

S e r i a l I n p u t P o r tS e r i a l O u t p u t P o r t

P S E NA L E / P R O G *A d d r L a t c h E n a / P r o g r a m

P r o g r a m S t o r e E n a b l e

[ B u s T i m i n g ]

[ I n t ]

[ I / O ]


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4


Interface Signals (2)E A * / V P P

X T A L 2 *

X T A L 1 *

E x t e r n a l A c c e s s E n a b l e /

P r o g r a m m i n g V o l t a g e

C r y s t a l / o s c i l l a t o r

C r y s t a l

R e s e t R S T

C l o c k / R e s e t

G e n e r a t o r

( E x t e r n a l )


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Introduction to the 8051

microcontroller Type of memory

Code memory (64k max)

Internal RAM (128 bytes) External RAM (64k max) Special Function Registers (SFR) Bit memory


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Internal RAM


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Basic Registers Accumulator R registers B register Data Pointer (DPTR) 16-bit register Program Counter (PC) 16-bit register

Stack Pointer (SP)


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MOV instruction Simply stated, the MOV instruction copies

data from one location to another. It has the

following formatMOV destination, source This instruction tells the CPU to move (copy) the source

operand to the destination operand, without changingthe content of the source operand.

Examples: MOV A,#55h ; load 55h into register A MOV R0,A ; copy contents of A into R0 MOV R3, #95h ; load value 95h into R3 MOV A,R3 ;copy content of R3 into A


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Some useful pointers Values can be loaded directly into any of the registers

A,B or R0-R7. However, to indicate that it is animmediate value it must be preceeded with a # sign. MOV A,#23H;load 23H into A

MOV R6,#12 ;load 12d into R6 MOV R5,#0f9h

If value is small, rest of bits are assumed to be allzeros. E.g. mov a 4-bit value into an 8-bit register

Moving a value that is too large into a register will

cause an error. To load a value into a register it must be preceeded

with a # sign. Otherwise it means to load from amemory location. MOV A,#17h MOV A,17h


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ADD instruction The ADD instruction has the following

format:

ADD A, source ;Add the source operandto A

This tells the CPU to add the source

byte to reg A and put the result inreg A


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Example Calculate the content of the

accumulator after the program is

executed on the 8051. MOV R5,#25h MOV R7,#34H

MOV A,#0 ADD A,R5 ADD A,R7


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Addressing Modes An "addressing mode" refers to how you are

addressing a given memory location. In summary, the addressing modes are as

follows, with an example of each: Immediate Addressing MOV A,#20h Direct Addressing MOV A,30h Indirect Addressing MOV A,@R0

External Direct MOVX A,@DPTR Code Indirect MOVC A,@A+DPTR

Each of these addressing modes providesimportant flexibility.


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Immediate Addressing Immediate addressing is so-named because the value to be

stored in memory immediately follows the operation code inmemory. That is to say, the instruction itself dictates whatvalue will be stored in memory.

MOV A,#20h This instruction uses Immediate Addressing because the

Accumulator will be loaded with the value that immediatelyfollows; in this case 20 (hexidecimal).

Immediate addressing is very fast since the value to be loadedis included in the instruction. However, since the value to be

loaded is fixed at compile-time it is not very flexible.


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Direct Addressing Direct addressing is so-named because the value to be stored in

memory is obtained by directly retrieving it from another memorylocation. For example:MOV A,30h

This instruction will read the data out of Internal RAM address 30(hexidecimal) and store it in the Accumulator.

Direct addressing is generally fast since, although the value to beloaded isnt included in the instruction, it is quickly accessablesince it is stored in the 8051s Internal RAM.

It is also much more flexible than Immediate Addressing since thevalue to be loaded is whatever is found at the given address--which may be variable.

Also, it is important to note that when using direct addressing anyinstruction which refers to an address between 00h and 7Fh isreferring to Internal Memory. Any instruction which refers to anaddress between 80h and FFh is referring to the SFR controlregisters that control the 8051 microcontroller itself.


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Indirect Addressing Indirect addressing is a very powerful addressing mode which

in many cases provides an exceptional level of flexibility.MOV A,@R0

This instruction causes the 8051 to analyze the value of the

R0 register. The 8051 will then load the accumulator with thevalue from Internal RAM which is found at the addressindicated by R0.

For example, lets say R0 holds the value 40h and InternalRAM address 40h holds the value 67h. When the aboveinstruction is executed the 8051 will check the value of R0.Since R0 holds 40h the 8051 will get the value out of Internal

RAM address 40h (which holds 67h) and store it in theAccumulator. Thus, the Accumulator ends up holding 67h.


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Indirect

Addressing(contd) Indirect addressing always refers to Internal RAM; it never

refers to an SFR. Thus, in a prior example we mentionedthat SFR 99h can be used to write a value to the serial port.Thus one may think that the following would be a validsolution to write the value 1 to the serial port:

MOV R0,#99h ;Load the address of the serial portMOV @R0,#01h ;Send 01 to the serial port -WRONG!!

On an 8051 these two instructions would produce anundefined result since the 8051 only has 128 bytes ofInternal RAM.


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External Direct

Addressing External Memory is accessed using "External Direct"

addressing. There are only two commands that use External Direct

addressing mode:

MOVX A,@DPTRMOVX @DPTR,A

As you can see, both commands utilize DPTR. In these instructions, DPTR must first be loaded with the

address of external memory that you wish to read orwrite. Once DPTR holds the correct external memory

address, the first command will move the contents ofthat external memory address into the Accumulator.

The second command will do the opposite: it will allowyou to write the value of the Accumulator to theexternal memory address pointed to by DPTR.


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External Indirect

Addressing External memory can also be accessed using a

form of indirect addressing. This form of addressing is usually only used in

relatively small projects that have a very smallamount of external RAM. An example of thisaddressing mode is:MOVX @R0,A

Once again, the value of R0 is first read and

the value of the Accumulator is written to thataddress in External RAM. Since the value of@R0 can only be 00h through FFh the projectwould effectively be limited to 256 bytes ofExternal RAM.


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Question 1 Write a short program for the 8051

microcontroller that carries out the followinginstructions:

Loads the accumulator with the value 40h Loads R7 with 12d Copies the content of R7 to address 30h directly Loads the register R0 with 30h Indirectly moves the contents of 30h to reg B

Indicate clearly the contents, at each stage,of all memory locations and registersinvolved.


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Structure of Assembly

language Consists of a series of assembly

language instructions. Instruction consists of four fields:[label1:] mnemonic [operands] [;comment] Label field allows the program to refer to a

line by name. Comment field must begin with a

semicolon


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Example of Assembly

ProgramORG 0H ;start at mem loc 0MOV R5,#25h ;load 25h in r5MOV R7,#34H ;load 34h into r7

MOV A,#0 ;clear AADD A,R5 ;A=A+R5ADD A,R7 ;A=A+R7ADD A,#12h ;A=A+12

HERE: SJMP HERE ;stay in this loopEND ;end of asm source ;file


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8051 Data Types and

Directives Only one type of data type 8-bit

Its the job of the programmer to

break down data larger than 8-bits.


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DB The DB directive is the most widely used data directive in

assembler. Used to define the 8-bit data. When DB is used, the numbers can be decimal, binary,

hex or ASCII formats. The only directive that can be used to define ASCII strings

larger than two characters.

ORG 500HDATA1: DB 39HDATA2: DB 2591 ;ASCII NUMBERS

ORG 518HDATA3: DB Computer Engineering


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Assembler Directives ORG: Used to indicate the

beginning of the address. EQU: Used to define a constant

without occupying amemory location.

e.g. Count EQU 25

END: Indicates the end of thesource file.


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PSW (Program StatusWord)

Addresses D0h, Bit-Addressable The Program Status Word is used to store a

number of important bits that are set and clearedby 8051 instructions. The PSW SFR contains the carry flag, the auxiliary

carry flag, the overflow flag, and the parity flag. Additionally, the PSW register contains the

register bank select flags which are used to selectwhich of the "R" register banks are currently

selected.


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Example

State the contents of the RAMlocations after the following program:

SETB PSW.4MOV R0,#99H

MOV R1,#85H

MOV R2,#99H

MOV R7,#85H

MOV R5,#99H


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Solution

By default PSW.3 = 0 and PSW.4 =0; thereforeline 1 sets RS1=1 and RS0 = 0, therebyselecting register bank 2.

Register Bank 2 uses RAM locations 10H 17H.After execution of this program we have thefollowing: RAM location 10H has value 99H RAM location 11H has value 85H RAM location 12H has value 3FH RAM location 17H has value 63H RAM location 15H has value 12H


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Stack in the 8051

Section of RAM used by the CPU tostore information temporarily.

Information can be data or anaddress.

The CPU needs this storage area

since there are only a limitednumber of registers.


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How stacks are accessedin the 8051

The register used to access the stack is called theSP (stack pointer) and is 8-bits wide (00h-ffh).

When the 8051 is powered up the SP contains thevalue 07. This means that RAM location 08 is the first location

used for the stack. Final location is 1F (20h -> used for bit- addressable

memory) Storing of a CPU register in the stack is called a

PUSH. Loading the contents of the stack back into a CPUregister is called a POP.


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Pushing onto the stack

The SP points to the last used location of thestack.

As we push data onto the stack, the stackpointer is incremented by one.

When you pop a value off the stack, the8051 returns the value from the memory

location indicated by SP, and thendecrements the value of SP.


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Example

Show the stack and stack pointer for the following:

MOV R6,#25H

MOV R1,#12H

MOV R4,#0F3HPUSH 6 ;push onto stack from R6

PUSH 1 ;push onto stack from R1

PUSH 4 ;push onto stack from R4

POP 3 ;pop stack into R3POP 5 ;pop stack into R5

POP 2 ;pop stack into R2


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Upper Limit Ram locations 08 1F used for the

stack. If more than 24bytes of stack required,

then the SP must be changed to pointto RAM locations 30h-7Fh using theinstruction

MOV SP,#xx Also may need to shift SP if a given

progam needs register bank1,2 or 3.


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Example

Show the stack and stack pointer for thefollowing instructions:MOV SP,#5FH

MOV R2,#25HMOV R1,#12HMOV R4,#0F3HPUSH 2PUSH 1PUSH 4


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Questions Write a simple program in which the value 55h is added five times.

Show the stack and the stack pointer for each line of the following:Org 0MOV SP,#70HMOV R5,#66HMOV R2,#7FH

MOV R7,#5DHPUSH 5PUSH 2PUSH 7CLR AMOV R2,AMOV R7,APOP 7POP 2POP 5


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Program Flow

When an 8051 is first initialized, it resets thePC to 0000h.

The 8051 then begins to execute instructions

sequentially in memory unless a programinstruction causes the PC to be otherwisealtered.

There are various instructions that can modifythe value of the PC; specifically, conditionalbranching instructions, direct jumps and calls,and "returns" from subroutines.

Additionally, interrupts, when enabled, cancause the program flow to deviate from itsotherwise sequential scheme.


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Loop and JumpInstructions

Repeating a sequence of instructions a certainnumber of times is called a loop.

The loop action is performed by the instruction

DJNZ reg,label In this instruction, the register is decremented;

if it is not zero, it jumps to the target addressreferred to by the label.

Prior to the start of the loop the register isloaded with the counter for the number ofrepetitions.


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Example

Write a program to clear the Acc, thenadd 3 to the accumulator ten times.

MOV A,#0

MOV R2,#10

AGAIN: ADD A,#03DJNZ R2,AGAIN

MOV R5,A


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Question

Write a program to load theaccumulator with the value 10h

and then complement the Acc700times. (Hint: Try two separateloops to achieve the overall of 700)


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Solution

MOV A,#10H

MOV R3,#10

NEXT: MOV R2,#70

AGAIN: CPL A

DJNZ R2,AGAINDJNZ R3,NEXT


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Other Conditional Jumps

Instruction Action

JZ Jump if A = 0

JNZ Jump if A 0

DJNZ Decrement and jump if A 0

CJNE A,byte Jump if A byte

CJNE reg,#data Jump if byte #data

JC Jump if CY = 1

JNC Jump if CY = 0

JB Jump if bit = 1

JNB Jump if bit = 0


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Example

Find the sum of the values 79H,F5H and E2H. Put the sum of the

registers in R0 (low byte) and R5(high byte)


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Solution

MOV A,#0 ;clear A

MOV R5,A ;clear R5

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

JNCN_1 ;if no carry, add nextINCR5 ;if CY=1, increment R5

N_1: ADD A,#0F5H ;A=79H+F5H=6EH and CY1=1

JNCN_2 ;jump if CY=0

INCR5 ;if CY=1, increment R5

N_2: ADD A,#0E2H ;A=6E+E2=50H and CY=1JNCOVER ;jump if CY=0

INCR5 ;if CY=1, increment R5

OVER: MOV R0,A ;now R0=50H and R5=02


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Unconditional JumpInstructions

All conditional jumps are short jumps,meaning that the address of the target

must be within -128 and +127 bytes ofthe contents of the program counter(PC).

Unconditional jump instructions are: LJMP (Long jump) 3 byte instruction SJMP (Short jump) 2 byte instruction


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CALL instructions

CALL instruction is used to call a subroutine

LCALL (long call) 3 byte instruction

ACALL (absolute call) 2 byte instruction When a subroutine is called, control is

transferred to that subroutine. After finishing execution of the subroutine, the

instruction RET (return) transfers control backto the caller.


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Time Delay Generationand Calculation

For the CPU to execute an instruction takes acertain number of clock cycles.

In the 8051 family, these clock cycles are

referred to as machine cycles. We can calculate a time delay using the

available list of instructions and theirmachine cycles.

In the 8051, the length of the machine cycledepends on the frequency of the crystaloscillator connected to the 8051 system.


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Time Delay Generationand Calculation (contd)

The frequency of the crystal connectedto the 8051 family can vary from 4MHz

to 30MHz. In the 8051, one machine cycle lasts 12

oscillator periods.

Therefore, to calculate the machine

cycle, we take 1/12 of the crystalfrequency and then take the inverse.


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Example

The following shows crystalfrequency for three different 8051-

based systems. Find the period ofthe machine cycle in each case.

(a)11.0592MHz

(b) 16MHz(c)20MHz


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Solution

1/11.0592MHz = period per oscillation

Machine cycle = 12x

= 1.085s 1/16MHz = period per oscillation

Machine cycle = 12x

= 0.75s 1/20MHz = period per oscillation

Machine cycle = 12x

= 0.6s


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Delay Calculation

A delay subroutine consists of two parts:

(a) setting a counter

(b) a loop Most of the time delay is performed by the body of the loop. Very often we calculate the time delay based on the

instructions inside the loop and ignore the clock cyclesassociated with the instructions outside the loop.

Largest value a register can hold is 255; therefore, one wayto increase the delay is to use the NOP command.

NOP, which stands for No Operation simply wastes time.


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Example

Find the size of the delay in the following program, ifthe crystal frequency is 12MHz.

MOV A,#55HAGAIN: MOV P1,A

ACALL DELAY

CPLA

SJMP AGAIN

DELAY: MOV R3,#200HERE: DJNZ R3,HERE

RET What does the above program do?


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Solution

Crystal Cycle

DELAY: MOV R3,#200 12

HERE: DJNZ R3,HERE 24

RET 12

Therefore, we have a delay of[(200X24)+12+24]x0.083s = 402s


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Loop inside a loop delay

Another way to get a large delay is to use a loop insidea loop, which is also called a nested loop. E.g.

Crystal Cycle

DELAY: MOV R3,#250 12HERE: NOP 12

NOP 12

NOP 12

NOP 12

DJNZ R3,HERE 24RET 24

Time Delay=[250.(12+12+12+12+24)]x0.083s +(12+24)x0.083s = s


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Question

For a machine cycle of 1s, find the time delay ofthe following subroutine.

DELAY:MOV R2,#200

AGAIN: MOV R3,#250

HERE: NOP

NOPDJNZ R3,HERE

DJNZ R2,AGAIN

RET


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Solution

HERE Loop: (4x250)x1s = 1000 s AGAIN Loop: Repeats the HERE loop 200

times i.e. 200x1000s = 200ms The instructions MOV R3,#250 and DJNZ

R2,AGAIN at the beginning at end of theAGAIN loop will add (3x200x1s) = 600s tothe delay time.

Total execution time 200.6ms (anapproximation since we have ignored thefirst and last instructions in the subroutine)

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