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Line Follower, Help Needed Badly

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a2zmoviebuff

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I have the checked out the following link

**broken link removed**

These are the problems i am facing.

1. This is a robot for following black line. I need my robot to follow a white line.
2. we are very new to microcontrollers, and rom writing

Hence when he says, "MY Code" and then there is a follow up in it. Could someone explain how this program works and how one can modify it to follow a white line on a black background.

3. In the link given as robot_asm what does the initial paragraph mean.
Help needed urgently.
 
The 4 sensors output a digital '1' when it sees white and a '0' when it sees black. You can modify the code to invert the input which will switch the black and white. You could also switch the inputs to the opamps to invert the sensor outputs.

For rom writing you need to look at the website of the manufacturer of your microcontroler.
 
thanks for the reply.
could you just tell me where exactly, i mean which part of the code, you see this digital '1' or digital '0' coming out.

It would be very helpful if you can just explain me the program under "My Code"

"You could also switch the inputs to the opamps to invert the sensor outputs. "

hope you have seen the circuit, but switching inputs just puts the sensor cutdown right.
 
this was written for an 8051 .. it could probably be adapted to your processor..
what controller are you using?


Code:
*#cpu 8051 Tiny
*
* DDS MICRO-C 8031/51 Startup Code & Runtime library for TINY model
*
* Copyright 1991-1999 Dave Dunfield
* All rights reserved.
*
        ORG    $0000         $0800  CODE Starts here (Normally in ROM)
        LJMP   START

        ORG    $0003
        LJMP   SERVICE_EX0

        ORG    $000B
        LJMP   SERVICE_TIMER0_INTERRUPT


* Fixed memory locations for alternate access to the R0-R7 register bank.
* If you are NOT useing BANK 0, these equates must be adjusted.
?R0	EQU	0		Used for "POP" from stack
?R1	EQU	?R0+1		Used to load index indirectly
?R2	EQU	?R0+2		""		""		""		""
?R3	EQU	?R0+3		Used by some runtime lib functions
?R4	EQU	?R0+4
?R5	EQU	?R0+5
?R6	EQU	?R0+6
?R7	EQU	?R0+7
*
* Startup code entry point
*
* If you are NOT using interrupts, you can reclaim 50 bytes
* of code space by removing the following TWO lines.
*        AJMP    *+$0032         Skip interrupt vectors
*        DS      $0032-2         Reserve space for interrupt vectors
*
START   EQU     *
	MOV	SP,#?stk-1	Set up initial stack
        ORL  TMOD,#%00000001    set timer 0 to be counter 16 bit
        SETB    IE.7            $AF  EA
        SETB    IE.1            $A9  ET0 Enable timer 0 interrupt
        SETB    TCON.4          start timer 0


	LCALL	main		Execute program
        SJMP    *               JUMP HERE

* EXIT to MON51 by calling the 'timer1' interrupt vector ($001B).
* This causes MON51 to think that a single-step operation has just
* completed, and therefore it saves the user registers, and performs
* a context switch back to the monitor.
*
* When using 2K addressing (CC51: -Z option, ASM51: -A option) this LCALL
* may fail "Out of range" because it gets translated to ACALL, and $001B
* may not be in the same 2K block as your program. Since 2K devices cannot
* support a debugger, change the ORG to $0000, and ...<continue below>...
*
* If you are NOT using MON51 (or MONICA which works the same), you will
* need to change this to whatever action you desire when main() returns.
* Suggestions: 1:freeze (SJMP *) 2:Restart (SJMP *&$FF00)
exit	LCALL	$001B		Call Timer-1 interrupt
	SJMP	exit		Incase he go's again

**************************** My code *********************************

SERVICE_TIMER0_INTERRUPT   EQU *
   PUSH ACC
   PUSH PSW
   MOV  TH0,#$FF      reload timer 0 for ms
   MOV  TL0,#$00
   INC  tick

   MOV  A,tick
   CJNE A,#100,RIGHT
   MOV  tick,#0

RIGHT
   CLR  C
   SUBB A,speedright
   JC   ON_RIGHT
   CLR  P1.0
   SJMP LEFT

ON_RIGHT
   SETB P1.0

LEFT
   MOV  A,tick
   CLR  C
   SUBB A,speedleft
   JC   ON_LEFT
   CLR  P1.1
   SJMP EXIT_I

ON_LEFT
   SETB P1.1

EXIT_I
   POP  PSW
   POP  ACC
   RETI

SERVICE_EX0 EQU *
   INC  cputick
   RETI


$SE:1
*#map1 Segment 1, initialized variables
$SE:2
*#map2 Segment 2, internal "register" variables
	ORG	$0008		Internal ram ALWAYS starts here

tick         DS  1
speedright   DS  1
speedleft    DS  1   
cputick      DS  1
 
yes i am using 8051 micorprocessor only and hte controller is the same mentioned in hte website.

could you make us understand in plain terms what the program is doing.

Like for example

PUSH ACC
PUSH PSW
MOV TH0,#$FF reload timer 0 for ms
MOV TL0,#$00
INC tick

MOV A,tick
CJNE A,#100,RIGHT
MOV tick,#0

What is happening once the programme starts from the first statement. With that we will know where exactly one can modify the code and get the desired output. As you say it is only the alteration of logic '0' and logic '1' that is to be done.



Secondly, when you say "You could also switch the inputs to the opamps to invert the sensor outputs. "

do you mean to say that this is another alternative, the first being altering the code. The latter one requires the hex code to be altered, could you tell me where i should be doing that.

Thanks for the reply.
 
aquinn_21 said:
Secondly, when you say "You could also switch the inputs to the opamps to invert the sensor outputs. "

do you mean to say that this is another alternative, the first being altering the code. The latter one requires the hex code to be altered, could you tell me where i should be doing that.

You would modify the hardware (the circuit) instead of the code.
You would swap the + and - (inputs) on the op amps. so for sensor one, instead of the + (pin 3) being connected to the 20k pot, it would be connected to the photo transistor (where pin 2 of the op amp is connected) and pin 2 would go to the 20k pot.
you should be able to use the code as-is then...
 
You can change the code without too much trouble

replace :
sensors = (P3 &=0x0f);
with :
sensors = ((~P3) & 0x0f);


All this does is take the bits in from the sensor and change all the ones to zeros and and zeros to ones. The ~ is the bitwise NOT operation.
 
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