PIC Beginners Guide

8/3/2019 PIC Beginners Guide

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Understanding & Programming thePIC16C84

A beginner's tutorial

Jim BrownBSc(Eng), HDipEdAd, GDE (Wits)RTF --> HTML Translation by: Dan Creagan

Minor amendments by: Joe da Silva

Contents at a Glance


Contents

List of Programs

Introduction

Introducing the PIC'84

The Instruction Set

05-10-2011 PIC beginners guide

linuxfromscratch.dk//pic-guide.html 1/31


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Interrupts on the '84

Timers on the '84

Using the EEPROM data memory

Appendix: Program listings

Contents


Contents

List of Programs

Introduction

Background

What you need

What you should know


Architecture

Instruction Set

A simple PIC'84 program

Using MPLAB to debug the program

Pause to reflect

The Instruction Set

Instruction format

The STATUS register

1: Move instructions

MOVF f,d (Move f, PIC p58)

MOVWF f (Move W to f, PIC p58)

MOVLW k (Move literal to W, PIC p58)

2: Clear instructions

CLRF f (Clear f, PIC p55)




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CLRW (Clear W, PIC p55)

3: Arithmetic ins tructions

ADDWF f,d (Add W & f, PIC p53)

SUBWF f,d (Subtract W from f, PIC p61)

ADDLW k (Add literal & W, PIC p53)

SUBLW k (Subtract W from literal, PIC p61)

4: Logical functions

ANDWF f,d (AND W with f, PIC p53)

IORWF f,d (Inclusive OR W with f, PIC p58)

XORWF f,d (Exclusive OR W with f, PIC p62)

ANDLW k (AND literal with W, PIC p53)

IORLW k (Inclusive OR literal with W, PIC p57)

XORLW k (Exclusive OR literal with W, PIC p62)

COMF f,d (Complement f, PIC p56)

5: Decrementing & Incrementing

DEC f,d (Decrement f, PIC p56)

INC f,d (Increment f, PIC 57)

6: Bit setting & clearing

BCF f,b (Bit clear f, PIC p54)

BSF f,b (Bit set f, PIC p54)

7: Program control

GOTO k (Go to address, PIC p57)

CALL k (Call subroutine, PIC p55)

RETURN (Return from subroutine, PIC p60)

RETLW k (Return with literal in W, PIC p59)

RETFIE (Return from Interrupt, PIC p59)

8: Skipping instructions

DECFSZ f,d (Decrement f, skip if 0, PIC p56)

INCFSZ f,d (Increment f, skip if 0, PIC p57)

BTFSC f,b (Bit test f, skip if clear, PIC p54)




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BTFSS f,b (Bit test f, skip if set, PIC p55)

9: Rotations & Swap

RRF f,d (Rotate right f thru carry, PIC p60)

RLF f,d (Rotate left f thru carry, PIC p60)

SWAPF f,d (Swap nybbles in f, PIC p62)

10: Sleep & Watchdog Timer

SLEEP (Sleep, PIC p60)

CLRWDT (Clear watchdog timer, PIC p56)

11: Miscellaneous

NOP (No operation, PIC p59)

OPTION (Not recommended)

TRIS (Not recommended)

Pause to reflect


What's an interrupt?

Types of Interrupt and the INTCON register

Servicing an Interrupt

Timers on the '84

The basic idea

The TIMER0 module

Using the timer's overflow



List of Programs

Program 1: simple.asm

Program 2: moves.asm

Program 3: clears.asm

Program 4: arith.as m




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Program 5: inter1.asm



Program 8: time0.asm

IntroductionBackground

When I first started looking into programming the PIC'84, I was daunted by what I saw as the complexity of the thing. After

all, I was only get ting into this for a hobby - not as a control engineer. I had had some experience years before with Zilog's

Z80, so I knew what registers, op-codes and so on were, but was still a little perplexed. The PIC'84 data sheet from

Microchip is quite a technical document, and the MPASM manual is a reference. Neither of these documents served me as a

tutorial, which is what I needed. So this manual is quite s imply my idea of a tutorial for programming the PIC'84, covering

both the '84's instruction set and some MPASM directives. Along the way, it covers the '84 itself, in terms of registers, pins

and so on.

By the end of working through this material you will notbe a fully-fledged '84 expert. But youwill be quite comfortable with

the device, even though you won't necessarily have touched one!

What you need

First, you need a copy of this guide. Preferably your own copy, so you can scribble in it to your

heart's content. Feel free to make as many copies of this as you wish - it's 'freeware'.

Second, you need to have some of Microchip's documentation. As a minimum, you'll need the

PIC'84 data sheet (preferably DS30445B, file '16c84_96.pdf'). This contains all the real info you

will need on the chip itself. I refer to this from time to time: there's no point my trying to duplicate

the detail of the info, or the diagrams. You should also have the MPASM User Guide, which is

available from Microchip's website, at http://www.microchip.com, or from their annual CD

'Microchip Technical Library'. I refer to these thus: [PIC] or [MPASM] with a page, figure or table

number; for instance [PIC figure 4.1] or [MPASM table 6].

Youcouldwork through the material with just the above, but you will get a lot more out of it by

being able to experiment. So, thirdly, load a copy of MPLAB (for Windows) onto your PC. This is

their Interactive Development Environment (IDE), and it contains an editor, assembler and a

simulator. Here, you can create your source code (editor), produce the executable code (assembler)and then run it on your PC (simulator). In the simulator, you can watch the program running,

keeping an eye on the registers and even simulate events like externally caused changes to the I/O

pins. You really should get this, and it's also available as above. Work through its tutorial before

you start this book, at least so you can find your way around it.

That's all youneed, but there's lots more you mightlike. Not least, is a PIC'84 chip and some kind

of PC based programmer. I haven't covered the physical use of it in here, but might in a future

volume. But there are many sources of info for that. I also found it handy to have a calculator that

works in decimal, hex and binary for conversions: there's one in Windows which you just need to

switch to scientific mode.

Of extreme help - perhaps in the longer run - are Microchip's Application Notes (ANs). These are

available from them, and a handy source is the CD-ROM mentioned above. These notes cover all




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sorts of uses of the PICs, with many helpful insights into real-world use. The ANs are in Adobe's

.pdf format: using Adobe Acrobat you can search through the ANs looking for terms like 'motor' or

'serial' and read the ANs appropriate to your needs.

One of the notes, AN585 on real-time operating systems for the PIC, refers toReal time

programming - Neglected topics.I urge you to get your hands on a copy - it is a fascinating tutorial

on the whole subject of interrupts, closed-loop control, and the like.

What you should know

It's not really for me to say what you should know: I don't know who you are or what you do . But, I guess you'll find it

easier going if you do unders tand (or can find out about) basic computer terminology like bits, bytes & EEPROM and

concepts like binary and hex. Of course, if you're using the simulator on your PC, you must feel comfortable with your PC

and Windows, and have played with the IDE tutorial.

I certainly don't expect that you have any PIC84 knowledge - this tutorial is aimed at the absolute neophyte.


Architecture

A microcontroller such as the '84 is - by its nature - a complete computer on a chip. It has 1: a processor and 2: registers , as

well as 3: program and 4: data memory. This makes it different from 'mere' CPUs which have only the process or and

registers.

1: The PIC'84 has an 8-bit processor, and that 's all we need to know about it!

2: The 51 registers are the processor's internal working storage. Referring to [PIC figure 4-2], you'll see that some of

these have names (STATUS, PORTB etc): these have special functions and we'll examine them later. PORTA, for

instance, contains the contents of one of the I/O ports, namely port A. There are 36 general purpose registers, and

we can consider these as ours (as opposed to the processor's special ones) and use them as our 'scratch pad'.

3: The '84 has 1K x 14 program memory: that means there are 1024 program locations, and each is 14 bits wide.

You'll see from [PIC fig 4-1] that the memory goes from 0000h to 03FFh: 03FFhex is 1023 in decimal so there are 1024

locations including 0000. The '14 bits wide' part comes from the length of each instruction: look at any one in [PIC

table 9.2] and you'll see that each is 14 bits long. For instance, ADDLW is 11 111X kkkk kkkk : don't worry about

what this means, just count it! We'll be using this memory extensively later on.

4: It has 64 bytes data memory, which is used for storage of, for ins tance, calibration values. We'll look at the use of

this EEPROM memory las t in this booklet.

In addition, the device includes 5: four types of interrupt and 6: 13 I/O pins.

5:Interrupts are means of something preempting the program. A button press by an operator, could interrupt the

program's normal flow and cause some special treatment. We'll examine the '84's interrupt techniques later.

6: The use of theI/O pins is key to the us e of a device like the '84, since any process consists of 3 parts - the input

to the process , the process itself, and the output. The '84 has 13 pins configured as 2 ports: port A comprises 5 pins

and port B has 8; and we'll be us ing them later.

Instruction Set

There are 35 instructions in the PIC84 instruction set; all are 14 bits long, and cons ist of an opcode and operand/s.

Basically, the opcode specifies what to do and the operand/s specify how or where. These instructions are split into 3




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groups: byte-oriented, b it-oriented and literal & control. Later on we will use each and every one of thes e operations , firstly

individually (usually trivially) and secondly with others in more meaningful scenarios.

For now, let's look at one of the instructions. Concentrate more on theformatthereof, rather than thefunction which we'll

see closely later. I've chosen the instruction ADDWF - see [PIC table 9.2]. In this tutorial, I have adopted the Courier font

below for all coding examples. The syntax of this instruction is:

ADDWF f,d

and you'll see this sort of look in many other instructions.

This is an appropriate time to introduce the working register: this is a s pecial register (over and above those already

mentioned) known as the W register, and it is where the arithmetic and logic unit (ALU) does the maths. Any instruction

with the 'W' in the name acts on the working register. Generally, some other register is known as 'F' in these commands. So -

almost intuitively - we can understand that the command ADDWF adds the working register to register F. But where is

register F?

Have a look at [PIC fig 4-2]: each register (other than W, that is, which isn't in the figure since it is a special one) is given a

hexadecimal address. For instance, PORTA is 05h, TRISA is 85h and we could use one of the 36 GP registers between 0Ch

and 2Fh like 3Ch. The 'f' in the command is a place holder for the actual address of the register F; remember there is no

register F as such - it is a generic name for some register or other (stands forfile register). So, we would code theinstruction as:

ADDWF 3Ch,d

Well, not quite - what's the 'd' for? It is the des tination of the result, and you'll see this in many operations . Depending on

our needs at the time, we need to choose to put the result either back into the working register or register F itself. [PIC page

51] explains that 'd' may be 1 or 0: a 0 causes the result to go to W, and a 1 puts it in F. Finally, therefore, this command

could be either of the following, depending on our needs:

ADDWF 3Ch,0 ; adds W to reg 3C, result to W

ADDWF 3Ch,1 ; adds W to reg 3C, result to 3C

Let's get on with actually writing a simple program . . . .

A simple PIC'84 program

This sample program serves a number of purposes . Apart from showing how to use a few instructions, it also introduces a

couple of the assembler concepts and will also show some simple simulator techniques. The program, simple.asm is

presented below; I'll walk you through it line by line.

Program 1: simple . asm

;simple.asm

;to demonstrate the look & feel of a program

; and introduce some instructions & directives

;***************** setup *******************************

processor 16C84 ;processor type

org 0010h ;origin of program in memory

w equ 0 ;in byte instructions, use w & f

f equ 1 ; instead of 0 & 1, it's much clearer




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my_reg_1 equ H'10' ;position my 2 registers in file

my_reg_2 equ H'15' ; map at h'10' & H'15'

;***************** program ******************************

;all we're going to do, is load the accum (reg w) with

; some values and play around with some arithmetic in w

; and my_reg_1&2

movlw H'08' ;put value H'08' into w register

movwf my_reg_1 ;move contents of w to my_reg_1

; note - same as movwf 10h, since

; my_reg_1 and 10h are same thing

movlw 32h ;put 32h into w reg

addwf my_reg_1,f ;add contents of w to that of my_reg_1

; answer goes into my_reg_1, due to f

movlw 92h ;put value 92h into w register

movwf my_reg_2 ;move contents of w to my_reg_2

; note - same as movwf 15h, since

; my_reg_2 and 15h are same thing

movlw 26h ;put 26h into w reg

subwf my_reg_2,w ;subtr w from my_reg_2

; answer goes into w

end

Let's examine this sample program. You should verify the following by finding each directive or instruction in the

appropriate Microchip book:

Anything after a ';' is a comment. Every programming book in the world recommends comments to explain what's

going on (at least to others , if not to yourself!).

In the part called 'setup', we meet three assembler directives:

processortells the assembler which model of Microchip the program is for. If this is incorrect, then

you might use an instruction in the code, for a processor which doesn 't support that instruction and an

assembly error will result.

orgtells the assembler where to s tart putting the code in program memory. From the memory map [PIC

figure 4.1], you can see that you s hould at least put your program beyond 0004h.equis an equivalence. It s imply means that the items on either side ofequmean the same thing. For

instance, the ADDWFinstruction expects a 0 or 1 in the 'd' place: by equating a 'w' with a '0', it means we can

use the 'w' in the instruction rather than '0'. This is eas ier to remember during coding, and to read later.

Similarly, by equatingmy_reg_1with 10h, every time I want to refer to register 10h, I can more easily refer




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to it by its meaningful, more eas ily remembered, name.

In the part called 'program', we meet a number of '84 instructions . Check the full descriptions in [PIC] if you like; we'll be

looking at them later, anyway.

MOVLW k[PIC page 58] causes the value k (eg 08h) to be put into the working register.

MOVWF f[PIC page 58] copies the contents of W into register f. Note thatmovwfis s trictly a misnomer,

since this is really a copy (W is not cleared), not a move (in which W would be emptied). Note too,

Microchip's convention in [PIC] by comparingmovlwand movwf: in describing the operation, the

parenthes es () mean 'the contents of'. Thus k(W) means the value k becomes the contents of W; (W)

(f) means that the contents of W become the contents of f. Lastly, be sure you understand the use of the

equconcept with respect to registers. The 'f' inMOVWF frefers to a register; [PIC figure 4-2] shows that

registers have hex address es and we would expect the instruction to readMOVWF 10hfor instance.

Equating my_reg_1 with 10h, means I can writeMOVWF my_reg_1with the same result.

ADDWF f,dand SUBWF f,d[PIC pages 53 & 61] respectively perform the arithmetic addition &

subtraction on the contents of W and f. Note the use of the equivalencies here too; we can refer to

my_reg_1 as before, and also replace the allowed values of d (ie, 0 &1) with w & f respectively. Hence we

can writeADDWF my_reg_1,fin place ofADDWF 10h,1.

Let's assemble and run this program . . .

Using MPLAB to debug the program

This is a three s tep process: edit the source code, assemble it, and then run the simulator to see what happens. If you

worked through the MPLAB tutorial, you will already know about the editing and assembling. I'll ass ume you have and do,

and so we can move to the s imulation.

With sample.asm successfully assembled, and the editor being the only window open in MPLAB, single-step through the

program (the foots tep icon). Not too helpful - you see each line of the program being highlighted as it is activated, but that's

it.

So, let's start to use a few of the simulator's other facilities. I sugges t you open the following other windows in MPLAB.

They all serve similar purposes - namely to see what's happening in the registers - but all implement differently. I found it

useful to have them all open at once, to compare the implementations :

Window > File Register

This window lists the contents of all file registers from 00 to 2F - ie, it does not show those in the region 80 to 8B. Asyou s tep through the program, you will be able to see the hex contents of registers 10h and 15h change. Note that

any register whose contents have just changed is shown in red - this goes back to blue on the next step (unless

there's another change of course).

Window > Special Function Registers

Here you can see the SFRs by name, rather than location which is what the previous window shows. The contents

are shown in decimal, hex and b inary.

Window > New Watch Window

This is your custom window into the registers, and it allows you to select which registers you want to monitor. Allregisters are available to choos e, both SFRs and GPRs. When you choose New Watch Window, you are prompted

with an input box and a down arrow. Click the arrow, and a list of all available registers appears: choose the one you

want (eg, my_reg_1) and then OK. You'll see the contents of the register displayed in the window. To add other

registers you can either click in the watch window's top-left-corner icon, and then choose Add Watch; or just press




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the Insert key (while the watch window is selected). You can save oft used combinations as .wat files, and then load

them back later with Window > Load Watch Window.

A word ofwarning . . . Where did the list of register names to select from, come from? The SFRs are obvious: the

MPLAB knows about them anyway. The others, like my_reg_1&2, come from the equdirectives in your code,

which is great. However, in our example, we have also usedw equ 0andf equ 1although these are notintended as register names. However, they s till appear in the list of poss ibilities. So what's the problem - just don't

choose them you say. But don't forget that w is a register: so now there are 2 w entries in the list, one was there all

along (the working register), and we caused the other with theequ. Using either of them, causes a symbol notfounderror in your watch table, meaning that you cannot monitor the working register if you have declared aw

equin your code. Try it yourself, then comment the offending line out with a ;, re-assemble and try again. Now youcan use the working register in the watch.

With these three windows, s tep through your program: you've now got some insight into what's going on. Satisfy yourself

that all is working to plan.

Pause to reflect

Where s hall we go from here? Well, where are we now? We've looked at the architecture and instruction set briefly; we've

created a simple program and examined the code to get the feel of it; we've edited & assembled this program; and we'veseen some of the simulator's facilities.

I think we should move on to look more fully at the instruction set. I know that [PIC] covers all the registers, flags,

interrupts , timers and s tuff before going on to the instructions, but I found this the better way.

The Instruction Set

Microchip present full details of the instruction set in [PIC, section 9, page 51 on]. They group these into 3 categories which

I don't find particularly useful. These are Byte- and Bit-oriented and Literal / Control operations. I have taken the view of

grouping them into what we use each command for, such as performing arithmetic. Page 52 in [PIC] gives a one-page

summary of the commands ; I made a copy of this page loose from the data sheet, for easy reference. A windows help file is

available on the Internet from Trisys Inc: I found this very useful. It contains details - similar to those in [PIC] - on the

instruction set and the special function registers. It's a normal windows help file, with the expected hypertext links &

searches.

For each instruction, I have explained the operation as described in [PIC], and have expanded on [PIC]'s explanations where

appropriate. (For instance, I have explained what an inclusive oractually is in IORWF.) Perhaps more importantly, I have

suggested exercises to us e the instructions as presented, with a poss ible solution given in the Appendix. The exercises

suggested are quite trivial, but will afford you the opportunity to use each instruction, and also practice using the s imulator.

Remember to use things like the watch window to see what's going on as you step your program: keep an eye on registers

you decide to use, as well as W and STATUS.

Instruction format

Microchip's grouping does keep instructions of s imilar format together. The instructions look like th is:

byte_command f,d

where f is the file register designation and d is the des tination; if d=0, the result goes to the working register, w; if

d=1, the result goes to register f

bit_command f,b

where f is the file register, and b is the bit therein; bits are numbered from 0 on the right, to 7 on the left. In the text, a

bit is written as FILE_REG, for instance INTCON means bit no 4 in the intcon register at 0B

other_command k




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where k is an 8-bit constant or literal.

The STATUS register

The '84 implements a STATUS register at 03h, which contains the arithmetic status of the arithmetic & logic unit; see [PIC

fig 4-3]. Have a glance down the 'Status Affected' column of [PIC table 9-2], and you will see that many '84 instructions

affect certain parts of STATUS. Many instructions affect STATUS, Z - the Zero flag, which gets set if the result of an

operation was zero. Certain operations affect the carry bits: STATUS, C - the Carry bit, is set if a carry out occurs from

the left-mos t bit in the result; STATUS, DC - the Digit Carry bit, is set if there is a carry between the hex digits (ie, from

the right hand nybble (bit 3) to the left hand nybble (bit 4). Two commands affect STATUS, the Power Down bit PD, and

STATUS, the Time Out bit TO.

1: Move instructions

We have met some of these instructions, which do nothing other than put things in registers.

MOVF f,d (Move f, PIC p58)

(f)(dest)

MOVWF f (Move W to f, PIC p58)

(w)(f)

MOVLW k (Move literal to W, PIC p58)

k(W)

Exercise: Write a program to us e these commands to (for ins tance) put something into W, move it from there to anotherregister, put something else into W, and then move the original thing back to W. See moves.asm.

2: Clear instructions

These 2 commands clear a register.

CLRF f (Clear f, PIC p55)

00h(f)

CLRW (Clear W, PIC p55)

00h(W)

Exercise: Expand the above program to clear registers at the end. See clears.asm .

3: Arithmetic instructions

Performing arithmetic is pretty important: that's why computers are called computers after all. The '84 can only add and

subtract though.

Arithmetic occurs either between W and an f register:

ADDWF f,d (Add W & f, PIC p53)

(W)+(f)(dest)




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SUBWF f,d (Subtract W from f, PIC p61)

(f)-(W)(dest)

or between W and a literal:

ADDLW k (Add literal & W, PIC p53)

(W)+k(W)

SUBLW k (Subtract W from literal, PIC p61)

k-(W)(W), [PIC p61]

Exercise: Use thes e and previous commands to load some registers and W, and do some adding and subtracting. Keep a

close eye on the carry bit and on the zero bit in the status register. See arith.asm.

Technical aside:You may see in [PIC] that subtraction is performed using the 2's complement method. Que? I hear you

ask. This is the normal way that subtraction occurs in binary. I'll explain, then look at an example. Express both numbers in

binary. Leave the number from which you are taking away, unchanged. Form the 2's complement of the one which is being

taken away thus : change all 0s to 1s and all 1s to 0s (this is the complement), add 1 to the right hand digit, carrying to the

left as necess ary. Now add the result to the unchanged other number. Discard the carry at the left, if there is one. That's the

answer. Let's check . .

We want to subtract 20 from 27 - we should get 7. Proceed as follows:

Convert to binary:2711011 . . . . x

2010100 . . . . y

Make 2's complement of y: complement:01011

add 1:01100 . . . . z

Add x and z: +11011 . . . . x

(1) 00111 = 7 QED.

The 1 in brackets is the carry, which gets discarded.

4: Logical functions

At this stage, before we examine the logical functions provided by the '84, we'll discuss what logical functions are. Cons ider

for the time being an electronic device of some description, with 3 wires attached. Let's say that 2 wires are inputs , and that

what gets output on the 3rd

wire depends on the inputs. Further, we'll say that this is a digital device, so that the 3 wires can

only have binary values of 1 & 0.

What relationships can exist, then, between the 2 inputs and the output? The input combinations are easy: if wire/pin 1 can

be 0 or 1, and so can pin 2, then the combinations are 00, 01, 10 & 11. What about pin 3: perhaps it must be a 1 if and only if

the inputs are both 1 (ie 11); or perhaps a 1 if either or both inputs is 1 (ie 01, 10, 11). Or whatever!

The basic relationships are known as andand or; andmeans both inputs while ormeans either or both. Most explanations

of this resort to the s o called truth table, which I have drawn below for the following logical operations: and, or, xor, nand,nor. I'll explain them once you've had a look at the table.

Inputs A B A AND B A OR B A XOR B A NAND B A NOR B




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both either, both either, not both not both neither, notboth

00 0 0 0 1 1

01 0 1 1 1 0

10 0 1 1 1 0

11 1 1 0 0 0

The function andmeans that the result is 1 only when both inputs are 1. The ormeans that either input may be a 1 for

output to be 1, but so may both. The functionxormeans exclusive or, and means that either input as 1 will cause a 1 to be

output, and specifically excludes the s ituation where both inputs are 1. Lastly, the nandand norare the negations ofand

and orrespectively: compare the columns and you'll see what this means.

By the way, the orfunction (as opposed to thexorfunction) is sometimes known as the inclusive or (ior). The '84 in fact

uses th is term.

The '84 provides a number of logical operations which act on 2, 8-bit values; these values are compared bit for bit. Forexample - without yet looking at an '84 instruction - consider the anding (is there such a word?) of the numbers H'5F'

(equivalent to D'95' and B'01011111') and H'A3' (which is D'163' and B'10100011'); resulting in H'03' (which is D'3' or

B'00000011').

5F: 01011111

A3: 10100011

and: 00000011

Clearly, only in the rightmost 2 pos itions is the andsatisfied, resulting in 1s there and 0s elsewhere. You should check this

on the calculator in Windows, which has logical functions built in.

So, let's move to the instructions as provided by the '84.

Comparison occurs either between W and an f register:

ANDWF f,d (AND W with f, PIC p53)

W. .and. (f)(dest)

IORWF f,d (Inclusive OR W with f, PIC p58)

W. .or. (f)(dest)

XORWF f,d (Exclusive OR W with f, PIC p62)

W. .xor. (f)(dest)

or between W and a literal:

ANDLW k (AND literal with W, PIC p53)

W. .and. k(W)

IORLW k (Inclusive OR literal with W, PIC p57)




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W. .or. k(W)

XORLW k (Exclusive OR literal with W, PIC p62)

W. .xor. k(W)

Lastly, you may have noted that the '84 doesn't provide nandor norfunctions. Rather, it provides the means of

complementing (negating) a register; this means to nandyou must first andand then com:

COMF f,d (Complement f, PIC p56)

complement of (f)(dest)

Exercise: I suggest 2 things here. First , using Windows' calculator, verify the results of the examples in [PIC] with each o f

the above: this will ensure that you understand the concepts. Then, write an as sembler program to verify the '84

instructions, by loading the appropriate registers, performing the actions , then checking the results. See logic.asm.

5: Decrementing & Incrementing

Two s imple instructions can decrement or increment the contents of a register, thus:

DEC f,d (Decrement f, PIC p56)

(f)-1(dest)

INC f,d (Increment f, PIC 57)

(f)+1(dest)

Exercise: In a program, perhaps by adding to one of the previous ones in which you do some arithmetic or some logic, check

out that these commands do work. Keep an eye on the Zero flag, which is set if either command causes the register inquestion to go to zero: we'll rely on this fact in 2 more commands later. See dec_int.asm

6: Bit setting & clearing

Using the following 2 commands , you can set or clear any bit b in register f. But why would we want to? Two reasons come

to mind:

First, as an example, the register in question might be a port controlling some external equipment. Each bit could

be switching a different device: a motor, a light or whatever. Setting and clearing each bit switches each device on or

off.

Second, as we'll see in 8. below, we can use the fact that a bit is set or clear to skip instructions in our program.

Being able to set or clear any bit gives us the ability to control this process.

BCF f,b (Bit clear f, PIC p54)

0(f)

BSF f,b (Bit set f, PIC p54)

1(f)

We read thes e two operations as "0 (or 1) becomes the content of bit 'b' in register 'f'".

Exercise: Sprinkle these commands into any of the programs above, and watch the changes to individual bits in your watch

window. See bits.asm




15/31

7: Program control

For many reasons, we need to control the flow through our program. Normally, flow proceeds linearly from the top; often

this is not suitable.

Firstly, we often need to loop through certain instructions: we might have a system in which one process is a continuous

one. This might be the control of a bucket conveyor which continuous ly adds ingredients to a vat. Here, when a bucketful

is added, we take it from the top and go through the same steps again, adding more on each pass .

Second, there may be part of our program which is useful in many other parts. Rather than repeating this code in manyplaces, we separate the piece we like from the rest of the code; then we merely call it in as often as we like, from wherever.

The reusab le piece is called a subroutine. The subroutine returns control to the point from which it was called when it's

finished.

Let's look at looping first:

GOTO k (Go to address, PIC p57)

k(PC)

This instruction literally 'goes to k', which it does this by loading the address of k into the program counter, PC. In order touse goto you s hould s tart the line to which you want to go, with a label. Then you goto label. See [MPASM p15] for rules

on labels.

Exercise: Modify one of the programs you have already written. You could put a label near the top, and a goto later on. As

you step, you'll see from the highlighted line that your program loops through the s ame part again and again. Look at the

program counter (pcl) in a watch window and you'll verify this. See goto.asm

Now we shall examine subroutines . We need to understand the concept of the stackfirst . Look at [PIC figure 4-1] and you'll

see a connection between the program counter and the s tack. The stack - which has 8 levels in the '84 - is the repository of

where the PC was before the subroutine was invoked. This is so that the program can go back to take up where it left off

when the subroutine is finished.Strictly speaking, the stack is loaded with the address of the instruction immediately afterthe one doing the sending; otherwise it would keep on going back and performing the instruction it had done earlier - we

really want the next one. We refer to loading the stack aspushing and taking a value off later aspopping. The stack is only

accessible at the top: it's like a pile of plates in one of those hoppers in a restaurant. The top of the stack is abbreviated as

TOS.

There are 2 instructions ass ociated with any subroutine - one to s end the program off to it, the other to bring it back:

CALL k (Call subroutine, PIC p55)

(PC)+1TOS, k(PC)

The call to a subroutine pushes the current PC+1 onto the stack, then changes the PC to the address k. This results in

program flow jumping to the s ubroutine, but the address to come back to later is safely ensconced on the stack for later

retrieval.

RETURN (Return from subroutine, PIC p60)

TOS(PC)

Return is the last instruction in the subroutine itself. The return instruction pops the stack, and so the program resumes in

the right place. Think about the depth of the stack: it means that calls can be nested, and flow will be correct as long as the

calls are matched by returns. Consider the following, which is OK:

CALL . . . . . push 1st

address

CALL. . . . push 2nd

address




16/31

RETURN. . . pop 2nd

address

RETURN. . . . . pop 1st

address.

But we don't want this, which is not OK:

CALL . . . . push 1st

address

CALL . . . push 2nd address

RETURN . . . . pop 1st

address

RETURN . . . pop 2nd

address.

As an alternative to the vanilla-flavoured return, you can bring a value back with you, thus :

RETLWk (Return with literal in W, PIC p59)

k(W), TOS(PC)

This is the normal return, with the k(W) added.

Exercise: Write a program to execute s ome code - anything you like - and then branch to a subroutine to do something else.

Check that you return correctly. Use both types of return.

Problems? You probably had a little difficulty wondering where to put the subroutine in the source. How do you s top the

subroutine being run by mistake? Because wherever you put it - either above or below the main part of the program, the

program's natural flow will go through the subroutine, which is called subby here. If the stack is empty , this causes a stack

underflow error, as when the subroutine's return is encountered before a call has been made. If the s tack has indeed been

pushed by some other event - like an interrupt - then the return would be honoured but incorrectly.

;example 1 ;example 2

. .

. .

label subby call subby

. subby code here .

. and here .

return label subby

. . subby code here

. . and here

call subby return

. .

. .

The way around this seems to be to proceed more or less as example 1, with the subroutines at the top. But, down below

where the main part of the program is, have a label like start; above the subroutines, have a goto startto hop over them in

the normal course of things. When a subroutine is called, the program shoots to the top: one subroutine will not flow into




17/31

another, because the return will take care of this. Thus:

goto start

label subby1

.

return

label subby2

.

return

start

.

.

call subby2

.

call subby1

.

. etc

See subby.asm

Associated with these last 2 commands is another type of return, which is to do with interrupts. We

will look at interrupts later, but for completeness I've put the return from interrupt here:

RETFIE (Return from Interrupt, PIC p59)

TOS(PC), 1GIE

When an interrupt occurs, the PC is pushed to the stack like with a subroutine call, so retfie pops it back out. Also, though,

the occurrence of an interrupt disable further interrupts: one doesn 't want interrupts interrupted after all. What happened

was, that the interrupt caused the global interrupt enable flag, GIE (INTCON) to be set to 0. Returning from an interrupt

means that further interrupts s hould be allowed, hence retfie sets GIE back to 1.

8: Skipping instructions

There are 4 ins tructions which allow you to s kip over the next instruction. I'll explain this before we look at the commands

themselves.

In the conveyor example above, we don't want the vat to overflow, so let's have a level sensor in the vat (just like the ball-

valve in the loo) which switches on when the vat is full. The conveyor is running, adding buckets ful; the program needs to

monitor the switch, and when the vat is full (the switch closes) the program switches off the conveyor and starts some

other part of the process , perhaps the mixer. This process is shown below: the dots show some or other process happeningbefore and after the part of interest. I've used line numbers for ease of reference. This is notin '84 assembler!

.




18/31

line # 40: start conveyor

.

.

line # 63: is the level sensor switch set?

line # 64: no - go back to line # 63 and look again

line # 65: yes - stop the conveyor

line # 66: start the mixer and carry on

.

.

In this example, we need to skip line 64 when the vat is full. When the vat still has room, this code will cause more ingredientto be added; the conveyor will stop and the mixer will start when the vat full sensor has switched on.

Now we can look at the commands which do this skipping in the '84:

DECFSZ f,d (Decrement f, skip if 0, PIC p56)

(f)-1(dest), skip if result = 0

INCFSZ f,d (Increment f, skip if 0, PIC p57)

(f)+1(dest), skip if result = 0

The above commands are based on the decfand incfcommands seen earlier: the sz part means 'skip if zero'.

Exercise: Verify these 2 instructions by loading a start value into a register called 'count ' or something. Then use either

instruction to change it, looping through this block of code. Check that the instruction following the decfsz or incfsz

(probably a goto, to caus e the looping) is skipped or not, as appropriate. (This approach would work if the conveyor had to

add say 10 bucketsful, rather than wait for a sensor.) See skip.asm.

BTFSC f,b (Bit test f, skip if clear, PIC p54)

skip if (f)=0

BTFSS f,b (Bit test f, skip if set, PIC p55)

skip if (f)=1

Read thes e instructions as 'bit test f, skip if clear/set".

These commands could be used to check the vat full sensor mentioned above. The bit to be tes ted would be the bit of a

port on the '84: it would be set or cleared by the sensor from time to time.

Exercise: Write a program with a block of code to loop through. These can be just abou t any instructions, and will simulate

the conveyor as nonsense. Have a btfss at the bottom, followed by a goto back to the top. Have some more code then,which simulates the mixer coming on. The btfss will need to refer to one of the '84's I/O ports as its f (H'05' or H'06' for porta

& b respectively), and you can use any bit 0-4 on porta, 0-7 on portb you wish. You should have another goto right at the

bottom, to make a sort of outer loop back to the top - this might not be too real-life, but you'll want the whole program to

loop to tes t what happens when the pin changes. See bits.asm




19/31

But how do we get the pin to change its state, simulating the vat full sensor? There's an eas y way of doing this in MPLAB's

simulator. Go Debug > Simulator stimulus > Asynchronous stimulus; you'll get a table of Stim0 to Stim12 buttons. Right

click on any one, perhaps Stim7, and click Assign pin. Double click on the pin you've elected to use as the s ensor, perhaps

it was RA3, say. Stim7 then changes to RA3 . Now right click on the button again: you can choose to pulse, make low, make

high or toggle the pin. Here, we would want to toggle the pin, since the vat sensor will be changing from time to time

probably, as it empties and fills.

Now we're all set to tes t. Step through your program. Depending on the initial state of the pin, which [PIC table 4-1] says is

unknown at power up, the program will loop or not. (At the inner loop, that is; it will always go back to the top from the very

bottom, for testing purposes.) Any time you like, left click on your chosen Stim button, now known as RA3 or whatever. (Ifyou have the port showing in a watch window, you'll see the pin change state at the next instruction boundary.) In any

case, next time the program gets to the btfss step, it should behave differently, since we've toggled the pin and so the btfss

should get a different answer.

9: Rotations & Swap

Three instructions allow you to manipulate the bits within a register. Of these, two slide the bits to the right or left (through

the carry bit), the third switches the register's two nybbles over.

RRF f,d (Rotate right f thru carry, PIC p60)

Each bit in the f register is moved 1 to the right; the one that falls o ff the right is circled round to the carry, and the

carry moves in from the left.

RLF f,d (Rotate left f thru carry, PIC p60)

Each bit in the f register is moved 1 to the left; the one that falls off the left moved into the carry, and the carry is

circled round to the right.

SWAPF f,d (Swap nybbles in f, PIC p62)

(f


20/31

OPTION (Not recommended)

TRIS (Not recommended)

Pause to reflect

We have now met each '84 ins truction, with the necessary background if appropriate, and us ed them in simple programs.

Now, we'll move on to explore two important areas of the PIC'84: interrupts and timers. It is these two areas which move the

'84 into the real world.


What's an interrupt?

Put simply, it's exactly what it says: a means of getting the computer's attention. Refer to the conveyor/mixer scenario

mentioned earlier (p17), where we saw the following snippet of 'pseudocode':

.

line # 40: start conveyor

.

.

line # 63: is the level sensor switch set yet?

line # 64: no - go back to line # 63 and look again

line # 65: yes - stop the conveyor, then

line # 66: start the mixer and carry on

.

.

Here, the program will loop around line 63 and 64 for ever if necess ary, waiting for the vat to fill up. That's OK, the program's

job is to run the process! Well, true, but it's wasteful. It might take a few seconds, minutes or hours to fill the vat; we couldhave the program doing something useful while it's waiting. But if the program's off somewhere else - doing the payroll? -

and the switch closes, how will the program know? It's not looking.

Same way as you get someone's attent ion when you come to visit. They're not standing looking out the front door all day,

they're going about their daily bus iness ; your knock on the door interrupts this, and they come to the door.

Our program would look something like this:

.

line # 40: switch on the conveyor

.

line # 50: do some payroll work




21/31

line # 51: and some more, and stay up here until interrupted

.

.

.

line # 60: only come here when interrupted

line # 61: stop the conveyor

line # 62: switch on the mixer

line # 63: go back to where you were when you got sent here

Now, the program spends most of its time in the payroll section, not looking at the s witch in the vat. When the switch

closes , the hardware takes care of things: the switch is connected to the computer by the interrupt request line (IRQ), and

program control is passed to a predetermined point. That point is line # 60 in the example above, and what happens at line

60 is up to the program/mer.

Notice something crucial here: once the interrupt has been taken care of, control must go back to where it was earlier. Not

only that, but the state of the sys tem must be put back: if any registers were changed while handling the interrupt, they

must be restored. We'll look into this later.

Now we know what an interrupt is in concept, we can look at how the '84 allows for this notion.

Types of Interrupt and the INTCON register

The '84 has 4 different types of interrupt [PIC p44]:

an external interrupt on pin 6, aka RB0

a change on any of pins 10-13, aka RB4-7

a timer overflow

an EEPROM write complete.

(Of these 4 types, the first one is the kind we'd use in the above conveyor/payroll program.)

In order to allow interrupts, and to determine the source thereof, the processor uses the INTCON register (0Bh) whosecontents are detailed in [PIC table 4-5], and which I'll explain here. Each bit serves a specific purpose, and interrupts work

according to the value (set or clear) of the various bits.

Firstly, interrupts as a whole may be enabled or disabled by INTCON, the GIE (Global Interrupt Enable) bit. If this bit is

clear, then no interrupts may occur: this is the power-up value by the way. Also, if an interrupt occurs, then GIE is cleared

to prevent the interrupt being interrupted; returning from the interrupt with the retfie instruction re-enables interrupts .

Second, each of the interrupt types must be enabled with its own bit in INTCON, before it can be used:

external interrupt on pin 6: INTCON, the INTE bit

change on any of pins 10-13: INTCON, the RBIE bit

timer overflow: INTCON, the T0IE bit

EEPROM write complete: INTCON, the EEIE bit.




22/31

Third, when interrupts occur, certain bits (known as interrupt flags)are set so that we may determine the source of the

interrupt:

external interrupt on pin 6: INTCON, the INTF bit

change on any of pins 10-13: INTCON, the RBIF bit

timer overflow: INTCON, the T0IF bit.

Why do we need to know the s ource? Well, depending on the type of interrupt, we will take different action: checking theappropriate bit tells us which interrupt it is. Note that the interrupt flags always get set when there's an interrupt, regardless

of the state of the corresponding enable bit.

Servicing an Interrupt

In the conveyor/payroll example above, there must be a means of sending the program to the right place to sort out the

interrupt; line 60 in the example. How does the processor know where to go? In the PIC84, all interrupts are sent to 0004h:

in the parlance of microprocessors, this point is the interrupt vector, and the program is said to vectorthere. At the vector,

we must make sure that we provide whatever code is necess ary to take care of the problem.

A crucial point is that we might need to save the status of the machine as it was before interruption, before we service the

interrupt. Clearly, the activities undertaken during servicing could change such things as the W & STATUS registers; we

would need these restored after handling the interrupt, in order that our other work may resume. The '84 only saves the PC

to the stack (PC+1, actually), and it's up to us to ensure we save and restore anything else we need!

Let's walk through a scenario in pseudocode, without yet looking at the '84 code. Foster sugges ts 3 parts to a program

which must handle simple interrupts: an initialize section, where in our case we'll enable the interrupts; a main section,

where the bulk of the time is s pent (doing the payroll, or whatever) and the handlerpart where the interrupt is taken care of.

Initialize:

set GIE, INTCON, to enable interrupts

set INTE, INTCON, to enable pin 6 interrupt

Handler:

save the status of the machine; probably W & STATUS

check interrupt flags for source; INTF, RBIF, T0IF

branch to the correct 'sub-handler' for the interrupt type (we've only

set INTE, though)do whatever is required for the interrupt

restore the status

return

Main:

do the payroll calculations

Exercise:Start slowly with interrupts - this can get quite hairy. Write a program to provide the

absolute minimum interrupt ability: enable them, have an interrupt service routine that does little ifanything other than re-enable interrupt and then return, and a main part that merely loops through

some meaningless code waiting for an interrupt. You need not save the status of the machine, nor

check for the type of interrupt at this stage. In order to see this program working properly in the

simulator, I suggest using the interrupt on change in port b, using the asynchronous stimulus




23/31

technique discussed before. Use a toggle on the pin to effect the change: while the program is

looping in the main section, toggle the pin and check that the interrupt occurs when you next step

the program. While the program is then in the isr, toggle the pin again: if it's all working, the pin

change willnotcause an interrupt, because GIE will have been cleared. See inter1.asm.

Exercise: Now add to the program, to save the state of the machine at the interrupt. To test this,

make sure you load W (say) in the main part, then change it in the ISR, and check that it gets

restored properly. Hint: see [PIC p46, example 8-1]. See inter2.asm

Exercise: Finally, allow for more than 1 kind of interrupt - change on one of portb pin 4:7 as well as

an interrupt on the rb0/int pin. Now, you'll need to determine the kind of interrupt, and handle it

accordingly. See inter3.asm.

Timers on the '84

The basic idea

The basic idea, is that we will want to use our '84 to time something. Time, that is, in the way we would use a wall clock: we

might switch on our fish-tank lights for ins tance, or close the curtains at 18h00 each night. But there's no clock in the '84: no

clock in the wall clock sense anyway.

What there is though, is a simple relationship between the carrying out of instructions and the s peed with which the

process or ticks away. The book tells us [PIC p1] that an instruction completes in 1 cycle, namely 400ns when the process or

is being clocked at 10MHz.

Let's verify this, and introduce a handy feature of MPLAB at the same time: the Stopwatch. In MPLAB prepare to run any

of the programs you have written up to now. Open the s topwatch by going Window > Stopwatch. You'll see a simple

window, containing 3 important pieces of information: the number ofsteps completed in what time, and at whatfrequency.

Check that the frequency is set to 10MHz for now, clickzero and s tep your program once. You should see that you

progress to 1 cycle in 400ns. (Unless this was coincidentally a program branch, which is a 2-cycle instruction.) Experiment

with various frequencies; for instance 400ns becomes 40ns at 100MHz, or 800ns at 5MHz.

So, we have the glimmerings of a wall clock type of timer here - we know how much time has pas sed if we know how many

steps have occurred. Enter Timer0 . . .

The TIMER0 module

The PIC'84 timer, TIMER0, works on the bas is that each time an ins truction cycle occurs , a certain interval of time has

elapsed. When TIMER0 is running, its register (TMR0) is incremented each cycle; each 400ns if the clock is a 10MHz one.

Therefore, the value of TMR0 represents the 'time' in 400ns steps. But TMR0, like all registers, is an 8-bit one and can thus

only count as high as FF: that means time only goes as far as 255 x 400ns = 102000ns or about 100s which is not too long.Cast your mind back to the discus sion of interrupts earlier, where we said that one of the interrupt types was a timer

overflow interrupt: that means that the TMR0 register has gone over the top at FF and back to 00. That means the interrupt

occurs every 100s of real time. It's up to us to do s omething with this interrupt, which probably means incrementing aregister of our own, clockedsay: every time clockedgets to a certain value, that means 1 second of real time, and we'd

probably then go off and increment another register, perhaps seconds

Let's walk through the process , then, of simply getting TIMER0 going, using the initialize, handler, main interrupttechnique used earlier.

Initialize: set GIE, INTCON, to enable interrupts




24/31

set T0IE, INTCON, to enable timer interrupt

clear T0CS, OPTION, to enable timer - see [PIC 6.0]

set PSA, OPTION, to keep prescaler out of this

Handler: increment 'clocked', our overflow counter

Main: have some trivial loop to keep the program going

Exercise:Write a program to implement the above simple timer, bearing in mind we won't be doing

anything with the timer overflows other than clicking them inclocked.Check the program in the

simulator, having visible any registers you think you should monitor, also the stopwatch. Before

you start, you'll need to figure out how to access theoption register, which is not quite as simple as

accessing saystatus. I've explained it below. See time0.asm.

Accessing registers in Bank1: You'll have noticed in [PIC fig 4-2, p12] that some registers like

optionare in what they refer to as Bank1, as opposed to Bank 0. Others, like statusare in both

Banks. What this means in practical terms, is that these Bank 1 registers are not normally

available, because our default sphere of operations is Bank 0. We use thestatusregister to effect a

switch between banks, with RP0 & RP1 (STATUS & ) doing the trick. This simply means

that we must set RP1 to go to Bank1 and clear it to return. For those registers which are in both

banks, you can of course access them no matter where you are, but remember the register has 2

different addresses:statusis known as h'03' and h'83'.

Using the timer's overflow

Now we can probably see the wood through the trees: we have a register clocked which is updating whenever the timer

overflows. That's roughly every 100s. Being an 8-bit register itself, clocked can of course only count to 255, and so it will

overflow at 255 x 100s which is 2.55 x 10-2 seconds. We'll need to count thes e overflows too, into say clocked2. Now,we're getting closer to the actual second, which is what we're heading for: how high do we need to count in clocked2 to

reach 1 second? Incrementing every .0255 seconds there are roughly 39 increments in a second , so this means, finally, we

can count real-time seconds by incrementing another register, perhaps SECONDS, every 39 steps . To get to minutes and

hours, we can simply overflow SECONDS into MINUTES at 60 seconds and MINUTES into HOURS every 60 minutes .

Exercise: Create a program to implement the above, at least as far as the SECONDS register.



The following programs are intended as illustrative of the use of the commands discussed. They are certainly trivial, and

probably not paragons of programming technique! In each, I have added a section calledsimulator: this section is asuggestion for things you might like to do during the simulation of your program. For instance, you might need to have a

watchwindow open showing the W register and portb, as well as having the stackopen too. Of course, it's up to you how

you design and code these programs, and how you experiment with them in the simulator.

Program 1: simple.asm




25/31

See above (A simple PIC'84 program)

Program 2: moves.asm

;moves.asm to show how MOVF, MOVWF & MOVLW work

;*********************************** simulator ***

;watch window: reg1, w, pcl

;*********************************** setup ***

processor 16c84

reg1 equ h'10'

;*********************************** program ***

start: movlw h'05' ;load w

movwf reg1 ;move (w) to reg1

movlw h'82' ;change w

movf reg1,0 ;restore w

end

Program 3: clears.asm

;clears.asm to show how clrf & clrw work

;based on moves.asm

;*********************************** simulator

;watch window: reg1, w, pcl

;*********************************** setup

processor 16c84

reg1 equ h'10'

;*********************************** program

start: movlw h'05' ;load w

movwf reg1 ;move (w) to reg1

movlw h'82' ;change w

movf reg1,0 ;restore w

clear: clrf reg1 ;clear reg1




26/31

clrw ;clear w

end

Program 4: arith.asm

;arith.asm to show using ADDWF, SUBWF, ADDLW, SUBLW

;************************************* simulator

;watch window: reg1,reg2,status,w,pcl

;************************************* setup

processor 16c84

reg1 equ h'10'

reg2 equ h'12'

;************************************* program

loads: movlw d'20' ;load w

movwf reg1 ;load reg1

movlw d'80' ;load w anew

movwf reg2 ;load reg2

arith: addlw d'05' ;add d'05' to w

sublw d'100' ;sub w from d'100'

addwf reg1,1 ;add w to reg1, into reg1

subwf reg2,1 ;sub w from reg2, into reg2

end


;inter1.asm is a simple interrupt handler -

; it does not save the state of the machine

; nor does it determine the kind of interrupt.

;

;*********************************************simulator

;watch window: intcon, pcl




27/31

;stack: have this open too - see the return address come & go

;asynch stimulus: have a toggle on rb4

;********************************************* setup

processor 16c84

movlw h'0'

movwf h'0b' ;clear intcon

goto main ;hop over the isr

;********************************************** isr start

isr: org h'0004' ;interrupts always vector to here

nop ;here we actually do the stuff

nop ; of the interrupt

bcf h'0b',0 ;clear int before going back

retfie ;this re-sets intcon gie

;*********************************************** isr ends

;

;*********************************************** main start

main: org h'0020' ; leave enough room for the isr!

bsf h'0b',7 ; set global int enable

bsf h'0b',3 ; set change on b int enable

payrol: nop ; loop here until interrupted,

nop ; doing payroll stuff

goto payrol ;

nop

nop

end


;inter2.asm does save the state of the machine

; but does not determine the kind of interrupt.




28/31

;

;****************************************************simulator

;watch window: intcon, pcl, portb, w, w_saved


;asynch stimulus: have a toggle on rb4

;**************************************************** setup

processor 16c84

w_saved equ h'10' ;a place to keep w

movlw h'0'


goto main ;hop over the isr at the beginning

;**************************************************** isr start


movwf w_saved ;save w as it was in main

movlw h'65' ;do something to change w

nop ;do more isr stuff

;now, restore w: there is

; no "movfw" - 2 swapf's seems to be

; the way to do this....

swapf w_saved,1 ; first, swap w_saved into itself

swapf w_saved,0 ; then, swap it into w

bcf h'0b',0 ;clear int on b before going back

retfie this re-sets intcon gie

;***************************************************** isr ends

;

;****************************************************** main start

main: org h'0020' ; leave room for the isr!








29/31

movlw h'0f' ; simulate a calc by loading w

goto payrol ;

nop

nop

end


;inter3.asm saves the state of the machine

; and determines the kind of interrupt.

;

;**************************************************simulator

;watch window: intcon, pcl, portb, w_saved, w


;asynch stimulus: have a toggle on rb4(for rbi) and rb0(int)

;************************************************* setup

processor 16c84

w_saved equ h'10' ;a place to keep w

movlw h'0'


goto main ;hop over the isr at the beginning

;*************************************************** isr start


movwf w_saved ;save w as it was in main

;find out what kind of interrupt

btfsc h'0b',0 ; is it an rbi?

call _rbi ; yes - call the rbi 'sub-handler'

btfsc h'0b',1 ; is it an int?

call _int ; yes - call the int 'sub-handler'

;end up here after sub-handler

;now, restore w

swapf w_saved,1 ; first, swap w_saved into itself




30/31

swapf w_saved,0 ; then, swap it into w

bcf h'0b',0 ;clear rbif before going back

bcf h'0b',1 ;clear intf before going back

retfie ; this re-sets intcon gie

_rbi: movlw h'65' ;do something to change w

return

_int: movlw h'24' ;do something different to w

return

;***************************************************** isr ends

;

;**************************************************** main start

main: org h'0020' ; leave room for the isr!



bsf h'0b',4 ; set ext int on pin6 enable

; we've got 2 types of int!



movlw h'0f' ;simulate a calc by loading w

goto payrol ;

nop

nop

end

Program 8: time0.asm

;time0.asm to see how timer0 works

;

;************************************ setup

processor 16c84

STATUS EQU H'03'




31/31

OPTIO EQU H'81'

tmr0 EQU H'01'

BANK_ EQU H'05'

T0CS EQU H'05'

PSA EQU H'03'

clocked equ h'10'

INTCON equ h'0B'

T0IE equ h'05'

GIE EQU H'07'

goto top

;************************************ isr

isr: ORG h'0004'

incf clocked ;clocked is the 'wall clock'

BCF INTCON,2

retfie

;************************************ program

top: org h'0010'

clrf tmr0 ;clear the tmr0 register

clrf clocked

mode: bsf STATUS,BANK_ ;switch to page 1

bcf OPTIO,T0CS ;go from counter to timer

bsf OPTIO,PSA ;no prescaler yet

bsf INTCON,GIE

bsf INTCON,T0IE

loop: nop

nop

goto loop

END


Documents

PIC Beginners Guide