Embedded Software 1

General 8051 features (excluding I/O)

General 8051 features (excluding I/O)

CPU8 bit microcontrollerThe basic registers include (more discussed later)

The 8-bit A (accumulator) and B registers, 16-bit program counter (PC) 8-bit program status register (PSW) 8-bit stack pointer (SP)

MemorySeparate program and data memory

Internal ROM used to store the program memory.Internal data memory consists of 256 bytes of internal RAM

8051 Program Memory8051 Program MemoryProgram memory is used to store the program code.

This is the machine code that is obtained from the assembly language or C program. The machine code represents instructions and constants.

When the 8051 is powered up, program execution always starts from the location 0X0000.

After the RESET operation, the program counter (PC) contains 0000h, causing the first instruction to be read from the program location 0000h.

The program memory addresses are 16-bits long. Therefore the 8051 can directly address 216 = 64k locations.

Each program memory location stores a single byte. The CPU can only read from program memory.

8051 Data Memory8051 Data Memory Can read/write from/to data memory.

Therefore the contents of data memory can change as the program executes.

Data memory is used to temporary data used by a program. Also used to store the stack.

The total size of the 8051’s internal RAM is small

256 bytes in total ! Data memory map shows 256 locations

Bottom 32 locations are used224 locations availableAddresses are shown in hex

Lower 128 bytes

8 bits 32 bytesused

Upper 128 bytes

00h

20h

80h

FFh

Simplified Programming Model for 8051

Simplified Programming Model for 8051

Note that there are more important registers to add to this model!!! (See these later)

The 8051 is an accumulator based CPU

Operations on data are performed on data in A and result is placed in A

E.g Add A,#23

Accumulator based machines are common for older 8-bit machines

PC

A

B

R7R6R5R4R3R2R1R0

8-bits

16-bits

Instruction SetsInstruction SetsThe instruction set is the set of instructions that the CPU can decode and

execute. It defines the processor with each processor having it's own instruction set.The general groupings of instructions for any uP are

Arithmetic/Logic Data Movement Transfer of Control Test/Compare Input/Output (only on some CPU's ) Others

An instruction stored in memory is represented by a certain number of byte(s).Some CPUs have fixed length instructions, i.e. all the instructions in the instruction set are

of equal size

Instruction EncodingInstruction EncodingThe instruction contains

what the operation is, the operands for the instruction if there are any. the destination for the result if there is one

The part of the instruction that determines the instruction is called the opcodeExamples are

ADDJMP

An operand may be an immediate value, memory address or a CPU register. The location for the result of an operation could be a

memory address or a CPU register. For an 8 bit machine such as the 8051, an instruction can be represented by 1, 2 or 3 bytes

depending on the instruction

Instruction EncodingInstruction Encoding

Details for the 8051 (8 bit machine) are in notesThe 8051 is an example of an accumulator based machine

Accumulator machines have a special purpose register that normally serves as an operand and destination for an instruction

E.g. ADD A,#27

Encoding 00100100 00011011 24h 1Bh

2 byte instruction

Instruction EncodingInstruction Encoding

Multiples of bytes 8 bit opcode gives 28=256 possible opcodes

255 actually used139 are 1 byte instructions92 are 2 byte instructions24 are 3 byte instructions

Most instructions are one byte, some two, a few threeSee back of notes for complete details on instructions

Example of a 1 byte instruction is the followingCLR C

which is represented by machine code C3H or the equivalent bit pattern 11000011

Instruction TimingInstruction Timing

The microcontroller takes a minimum of 12 clock cycles (machine cycle) to execute an instruction

Some instructions are slower to execute, taking 24 clock cyclesThere are no instructions with 36 clock cycles, but there are 2 instructions

that take 48 clock cycles (or 4 machine cycles)MUL and DIV instructions

See back of lab notes for instruction cycle times for all instructionsNote that the instruction size does not directly imply how long the

instruction cycle is!

8051 Instruction set8051 Instruction set

ACALL addr11 DIV AB LJMP addr16 RETIADD A,<src> DJNZ <byte>,<rel8> MOV <dest>,<src> RL AADDC A,<src> INC <byte> MOV DPTR,#data16 RLC AAJMP addr11 INC DPTR MOV bit,bit RR AANL <dest,<src> JB bit,rel8 MOVC A,@A+<base> RRC AANL C,<bit> JBC bit,rel8 MOVX <dest>,<src> SETB bitCJNE <dest>,<src>,rel8 JC rel8 MUL AB SJMP rel8CLR A JMP @A+DPTR NOP SUBB A,<src>CLR bit JNB bit,rel8 ORL <dest>,<src> SWAP ACPL A JNC rel8 ORL C,bit XCH A,<byte>CPL bit JNZ rel8 POP direct XCHD A,@RiDA A JZ rel8 PUSH direct XRL <dest>,<src>DEC <byte> LCALL addr16 RET

How many instructions??

Lecture 2

Getting started with the

instruction set

Getting started with the

instruction set

A key requirement is the ability to move data

Between registersBetween memory and

registersBetween memory

The MOV instruction can be used

Different modes of addressing available

Lower 128 bytes

8 bits 32 bytesused

Upper 128 bytes

00h

20h

80h

FFh

PC

A

B

R7R6R5R4R3R2R1R0

8-bits

16-bits

Introducing the MOV instructionIntroducing the MOV instruction

The MOV instruction is of the formMOV <dest>,<src>

And has the effect of moving a byte from <src> to the <dest> location.

There are 15 variations!!!Example MOV A,#12

Lower 128 bytes

8 bits 32 bytesused

Upper 128 bytes

00h

20h

80h

FFh

PC

A

B

R7R6R5R4R3R2R1R0

8-bits

16-bits

12

Aside: Representing hexadecimal valuesAside: Representing hexadecimal values

On paper, a hexadecimal value is represented in one of the following ways

0xA4

A4hHowever, for an assember, a hexadecimal value starting

with the digit A to F must be preceded by 0x.This is so as to distinguish the number from an identifier.Therefore, the representation 0xA4 must be used with an

assembler.

Introducing the MOV instructionand immediate

values

Introducing the MOV instructionand immediate

values

Example MOV A,#12

Here the destination is the A register and the source is an immediate value

All immediate values are preceded by the hash symbol

Using an immediate value means that the source byte is constant

The value can be decimal or hex

Lower 128 bytes

8 bits 32 bytesused

Upper 128 bytes

00h

20h

80h

FFh

PC

A

B

R7R6R5R4R3R2R1R0

8-bits

16-bits

12

Introducing the MOV instruction

and register addressing

Introducing the MOV instruction

and register addressing

Register addressing is another method for accessing data

Example MOV R3,#32

MOV A,R3 Firstly the decimal value 32 is

copied into R3, while the second instruction copies the value in R3 into A.

Any one of the 8 registers R0-R7 can be used

Use Rn to indicate any register when describing register addressing

Lower 128 bytes

8 bits 32 bytesused

Upper 128 bytes

00h

20h

80h

FFh

PC

A

B

R7R6R5R4R3R2R1R0

8-bits

16-bits

32

32

Introducing the MOV instruction with register addressing

and immediate values

Introducing the MOV instruction with register addressing

and immediate values

MOV R3,#32

MOV A,R3

Destination operand

Uses register addressing

R3

Source operand

Uses immediate value

32Destination operand

A

Source operand

Uses register addressing

R3

MOV R3,#32

MOV A,R3

Introducing the MOV instruction

showing instruction encoding

Introducing the MOV instruction

showing instruction encoding

01111011 00100000

7B 20

11101011

EB

Aside, what is the benefit of short instructions?

Introducing the MOV instruction

and direct addressing

Introducing the MOV instruction

and direct addressing

To access data in lower RAM, direct addressing is used

Example MOV 0x20,A

Here, the destination is hex address 20 and the source value, which is in A, is copied into RAM

Direct addressing can only be used to access the lower 128 bytes and NOT the upper 128 bytes

Lower 128 bytes

8 bits 32 bytesused

Upper 128 bytes

00h

20h

80h

FFh

PC

A

B

R7R6R5R4R3R2R1R0

8-bits

16-bits

15

Note the difference between immediate

values and direct addressing: Be careful!!!

Note the difference between immediate

values and direct addressing: Be careful!!! It is important to understand the difference between an

immediate value and direct addressing

MOV A,#0x64 MOV A,0x64

Here, the value 64h is stored in A

Here, the value at address 64h is stored in A.

Find out in lab the machine code for both of these?

PC

A

B

R7R6R5R4R3R2R1R0

8-bits

16-bits

Introducing the MOV instruction

and indirect addressing

Introducing the MOV instruction

and indirect addressing

To access data ìn any one of the 256 RAM locations, indirect addressing can be used

Example MOV @R0,A

Indirect addressing can be specified using either @R0 or @R1

when discussing indirect addressing, it is common to refer to @Ri, where i could be 1 or 0.

How does it work??Read on!!!

Lower 128 bytes

8 bits 32 bytesused

Upper 128 bytes

00h

20h

80h

FFh

20h

13

PC

A

B

R7R6R5R4R3R2R1R0

8-bits

16-bits

Introducing the MOV instruction

and indirect addressing

Introducing the MOV instruction

and indirect addressing

Example MOV @R0,A

The contents of R0 are interpreted as an address in RAM. If R0 contains the value 20h, then the above instruction will copy the value in A into address 20h in RAM.

In effect, it is now possible to indirectly access a memory location

Have you seen this idea before, C programmers?

Lower 128 bytes

8 bits 32 bytesused

Upper 128 bytes

00h

20h

80h

FFh

20h

13

Summary of addressing modes

Summary of addressing modes

Immediate value#data8e.g. MOV A,#56h

Regsister addressingRie.g. MOV R6,R3

Direct Addressingdirecte.g. MOV R6,4Eh

Indirect addressing@Rie.g. MOV @R1,R6

Good news for us this semester

The first 3 are the addressing modes that will be mostly used and that indirect addressing will be rarely used

Full details for MOV instructions

Full details for MOV instructions

MOV A,sourceMOV A,#dataMOV dest,sourceMOV dest,#data

Where dest and source can be any of Rn, direct or @Ri

AppendixAppendix

8051 Block diagram 8051 Block diagram

ROM RAMTimer

Timer 1

Instruction encodingInstruction encoding

Another example is

ADDC A,R1

This instruction adds the contents of A and the contents of R1 using the carry and stores the result in A

The machine code is 39H. In general for this instruction, it will be encoded as follows

0011 1

5 bits 3 bits

Rn

ExercisesExercises

1. Draw a block diagram of the 8051.2. What is the purpose of the program memory on the

8051?3. At what address does the 8051 start executing code

from upon powerup?4. How many memory locations in the 8051 program

memory?5. What register contains the address of the next program

instruction?a) How many bits in this register?

ExercisesExercises6. Can program memory be written to during program execution?7. What is the purpose of the 8051 data memory?8. What size is the data memory?9. Draw a simplified programming model for the 8051, making sure to

label all registers.10. What does it mean to say that a processor is an accumulator based

machine11. What is an instruction set?12. List the general groupings of instructions that make up an

instruction set?

ExercisesExercises

13. What information does an instruction contain?14. On the 8051, what size are the instructions?15. What is machine code for the following instruction and

how many clock cycles does it take?ADD A,R5

16. Do we need to remember the machine code for each instruction?

17. How many 8051 instructions are there?18. What is the purpose of the MOV instruction?

ExercisesExercises

19. What is an immediate value?20. How is it represented?21. What instruction is used to move the immediate

decimal value 12 into the accumulator?22. What is register addressing?23. Move the value in register R4 into the accumulator.

What is the resulting machine code?24. What is direct addressing? Give an example.

ExercisesExercises

24. What is indirect addressing? Give an example25. Write out the code to move the decimal values 0,1,15

and 8 into the regsiters R0 to R3 respectively?26. Write out the code to mode the hexadecimal values

0,a5,15 and c into the registers R0 to R4 respectively.27. Write out the code to move the initialise the memory

locations 20h to 24h with zero.28. Write out the code to swap the values in R0 and R1.