Avr Micro Controller Family Assembly Language Programming 337

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AVR Microcontroller Family

Assembly Language ProgrammingUniversity of Akron

Dr. Tim Margush


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4/30/2012 Dr. Tim Margush - AssemblyLanguage Programming

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AVR

Atmel AVR 8-Bit Processors come in a

variety of configurations and packages

They all share a common core registers,instructions, basic I/O capabilities

Our focus is the ATMega16


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ATMega16 Specs

131 Instructions

32 8-bit GP registers

Throughput up to 16 MIPS

16K programmable flash (instructions)

512Bytes EEPROM 1K internal SRAM

Timers, serial and parallel I/O, ADC


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AVR CPU

PC: address of next

instruction

IR: prefetchedinstruction

ID: current instruction

GPR: R0-R31 ALU: Note internal

data path


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AVR Memory

Flash: Machine

instructions go here

SRAM: For runtime data

Note bus independence for

data and instructions

EEPROM: Secondary

storage

EEPROM and Flashmemories have a limited

lifetime of erase/write cycles


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Flash Memory

Programs reside in word addressable flash storage Word addresses range from 0000-1FFF (PC is 13 bits)

Byte addresses range 0000-3FFF (0x4000=16K)

Harvard Architecture It is possible to use this storage area for constant data as

well as instructions, violating the true spirit of thisarchitecture

Instructions are 16 or 32-bits Most are 16-bits and are executed in a single clock

cycle


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SRAM

The ATMega16 has 1K (1024 bytes) of byte

addressable static RAM

This is used for variable storage and stack spaceduring execution

SRAM addresses start at $0060 and go through

$045F The reason for not starting at zero will be covered

later


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EEPROM

Electrically Erasable Programmable Read

Only Memory

Programs can read or write individual bytes

This memory is preserved when power is

removed

Access is somewhat slow; it serves as a form ofsecondary storage


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Clock

All processors are pushed through their fetchexecute cycle by an alternating 0-1 signal, called aclock

The ATMega16 can use an internal or externalclock signal Clock signals are usually generated by an RC oscillator

or a crystal The internal clock is an RC oscillator programmable to 1, 2, 4,

or 8 MHz

An external clock signal (crystal controlled) can be moreprecise for time critical applications


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AVR Machine Language

AVR instructions are 16 or 32-bits

Each instruction contains an opcode

Opcodes generally are located in the initial bits of aninstruction

Some instructions have operands encoded in the

remaining bits

Opcode and operands are numbers, but their containers

are simply some of the bits in the instruction


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Load Immediate

LDI Rd, K Load a constant into a register (Rd = K)

1110 bbbb rrrr bbbb Limitations: 16


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LDI Examples

LDI R16, $2C 1110 0010 0000 1100 or 0xE20C

LDI R27, $0F 27 is 11011

1110 0000 1011 1111 or 0xE0BF

Note that this instruction always

starts with E, has the two nybbles of K at positions 2 and 0,

and encodes the register in nybble 1


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Add Registers

ADD Rd, Rr

Add the contents of two registers, store result in

Rd (Rd = Rd + Rr) 0000 11rd dddd rrrr

Any registers: 0


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ADD Examples

ADD R16, R3 ddddd is 10000 and rrrrr is 00011

0000 1101 0000 0011 or 0x0D03

ADD R27, R17 ddddd is 11011, rrrrr is 10001

0000 1111 1011 0001 or 0x0FB1

Note that this instruction always starts with 0, followed by C, D, E, or F

and can have any values in the second byte


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Expanding Opcodes

The following are not

legal opcodes 1, 11, 111

No opcode longer than

4 bits starts 1110

Except ser Rd

1110 1111 dddd 1111

Similarly, no prefix of

000011 is an opcodeand no opcode longer

than 6 begins with this

sequence

Except lsl Rd

0000 11dd dddd dddd

LDI: 1110011110110011

ADD: 0000111100101101

How does theprocessor"know" where

the opcodeportion stops?


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A Machine Language Program

Load program into memory At address 0 we place the word $E20C

At address 1 we place the word $E01F At address 2 we place the word $0F01

Execute the three instructions in sequence

Set PC to 0 and perform 3 fetch-execute cycles Observe result

R16 has the sum of $2C and $0F, $3B


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A Machine Language Program

0000: $E20C LDI R16, $2C

0001: $E01F LDI R17, $0F

0002: $0F01 ADD R16, R17

R16 = R16 + R17

= $2C + $0F

= $3B


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AVR Studio

An integrated development environment

Provides a text editor

Supports the AVR assembler

Supports the gnu C compiler

Provides an AVR simulator and debugger

Provides programming support for the AVRprocessors via serial interface


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AVR Studio: New Project

Start AVRStudio

Click New

Project Select type:

Assembler

Choose aproject name

Select createoptions andpick alocation

Location should be a folder to hold allproject folders

Each project should be in its own folder


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AVR Studio: New Project

On the next dialog,

select the Debug

platform: AVRSimulator

Pick the device type:

ATMega16

Finish


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AVR Studio: Interface

Enter the program

in the assembly

source file that isopened for you.

Click the

Assemble button

(F7)

Assemble

Workspace

Output

Editor


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AVR Studio: Assembler Report

Assembler summary indicates success

6 bytes of code, no data, no errors


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AVR Studio: Debugger

Start the debugging session

Click Start Debugging

Next instruction is shown with yellow arrow

Choose I/O View

View registers 16-17

Step through program

F10 is Step Over

Start

Debugging


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AVR Studio: Debugger

The first 2 instructions are completed

R16 and R17 have the expected values from the

LDI instructions

The sum is placed in R16

$3B is the sum


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AVR Studio: Memory

Memory contents may be viewed (and

edited) during debugging

You can view program (flash), data (SRAM),or EEPROM memory

You can also view the general purpose and I/O

registers using this tool The Program


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Using Mnemonics

Rather than code themachine languageprogram as a sequence

of numeric wordvalues expressed inhexadecimal,assembly language

programmers usuallyuse instructionmnemonics

This program willassemble and runidentically to the first ldi R16, $2C

ldi R17, $0F

add R16, R17


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What's Next?

In our sample

program, we executed

three instructions,what comes next?

Undefined! Depends

on what is in flash

How do we terminatea program?

Use a loop!

0000: $E20C LDI R16, $2C

0001: $E01F LDI R17, $0F

0002: $0F01 ADD R16, R17

0003: $???? ???

If a program is to simply

stop, add an instruction

that jumps to its own

address


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Relative Jump

RJMP K 1100 kkkk kkkk kkkk

-2048


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Relative Jump

Here: RJMP Here 1100 kkkk kkkk kkkk

K = -1 ($FFF)

1100 1111 1111 1111 or $CFFF0000: $E20C LDI R16, $2C

0001: $E01F LDI R17, $0F

0002: $0F01 ADD R16, R17

0003: $CFFF RJMP -10004: $???? ???

This distance is -1 word

Remember that the program counter is

incremented before the instruction is executed

Avr Micro Controller Family Assembly Language Programming 337

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