pic programming.................. what next?

[Nigel Goodwin]

You're thinking too deeply! - the controller on the LCD module has it's own memory, and stores the first nibble internally 'somewhere', there's no reason to know where, and probably no way to find out.

As far as I'm concerned (and you!) the LCD module is a black box, there's no need to know what happens inside it, as long as you know it's response to external actions.

 

[tama182]

No im not thinking too deeply !!, i dont understand the program you've written,

youve sent a load of letters to the LCD, where each letter takes two passes, as a letter is 8 bit and it set up to do only 4 bit,

heres how im understanding it, and i dont see how it works

------------------------

Send H to LCD (01001000)

send 00000100 to Porta (first half of H)

Porta=00000100

send 00001000 to Porta (second half of H)

Porta now becomes = 00001000

----------------------

so if porta is now 00001000 (second half of H), how does it remember the first half because that has been overwritten

 

[Nigel Goodwin]

so if porta is now 00001000 (second half of H), how does it remember the first half because that has been overwritten

As I said above, it's stored internally in the LCD module.

 

[tama182]

so then once it starts E, is H also stored? so that in the end the LCD stores the boths words?

 

[Nigel Goodwin]

so then once it starts E, is H also stored? so that in the end the LCD stores the boths words?

They don't provide you all the internal details of the LCD controller, only the details you need to use it - certainly all text you send is stored in the display buffer, until overwritten or cleared. Command instructions are not stored in the display buffer though, but may be stored elsewhere?.

 

[tama182]

i just noticed this, i understand that you have given porta the name LCD_PORT, and that you have also given some of the bits in porta names as well, to make it easy to read, ie LCD_RS and LCD_RW,

LCD_PORT	Equ	PORTA
LCD_TRIS	Equ	TRISA
LCD_RS		Equ	0x04		
LCD_RW		Equ	0x06
LCD_E		Equ	0x07

but when you write

LCD_RS.......... Equ........ 0x04

, how come it doesn't try to give register 4 (fsr) the name of lcd_rs, how does it know that you mean the bits of register 5 (porta)

 

[Nigel Goodwin]

The equates and includes are simply text substitution items - that line "LCD_RS Equ 0x04" simply instructs the assembler to replace every occurance of 'LCD_RS' with '0x04' - so it in no way affects anything else.

 

[tama182]

The equates and includes are simply text substitution items - that line "LCD_RS Equ 0x04" simply instructs the assembler to replace every occurance of 'LCD_RS' with '0x04' - so it in no way affects anything else.

yes i understand that, but is'nt the hex number 0x04 the location of register fsr,

INDF                         EQU     H'0000'
TMR0                         EQU     H'0001'
PCL                          EQU     H'0002'
STATUS                       EQU     H'0003'
FSR                          EQU     H'0004'
PORTA                        EQU     H'0005'
PORTB                        EQU     H'0006'
PCLATH                       EQU     H'000A'
INTCON                       EQU     H'000B'
PIR1                         EQU     H'000C'
TMR1L                        EQU     H'000E'
TMR1H                        EQU     H'000F'
T1CON                        EQU     H'0010'
TMR2                         EQU     H'0011'
T2CON                        EQU     H'0012'
CCPR1L                       EQU     H'0015'
CCPR1H                       EQU     H'0016'
CCP1CON                      EQU     H'0017'
RCSTA                        EQU     H'0018'
TXREG                        EQU     H'0019'
RCREG                        EQU     H'001A'
CMCON                        EQU     H'001F'

 

[eblc1388]

yes i understand that, but is'nt the hex number 0x04 the location of register fsr,

But "4" is simply a value which is not to mean just a register address. You asked the question, "What is the address of FSR?", the answer is four.

Depending on the circumstances, many other things can have the value of four.

Sides_of_a_square EQU 4
Legs_of_a_cow EQU 4
....

 

[tama182]

ok, let me just ask this first, going from the 628 text,

INDF EQU H'0000'
TMR0 EQU H'0001'
PCL EQU H'0002'
STATUS EQU H'0003'
FSR EQU H'0004'
PORTA EQU H'0005'
PORTB EQU H'0006'

when the pic master wrote his code, he put at the top

LCD_PORT Equ PORTA
LCD_TRIS Equ TRISA
LCD_RS Equ 0x04
LCD_RW Equ 0x06
LCD_E Equ 0x07

now going back to the 628 test its says that PORTA EQU H'0005',

so Nigel could have witten this instead, which is the same thing

LCD_PORT Equ 0x05 (=H'0005')
LCD_TRIS Equ TRISA
LCD_RS Equ 0x04
LCD_RW Equ 0x06
LCD_E Equ 0x07

is that correct?

 

[Nigel Goodwin]

ok, let me just ask this first, going from the 628 text,

INDF EQU H'0000'
TMR0 EQU H'0001'
PCL EQU H'0002'
STATUS EQU H'0003'
FSR EQU H'0004'
PORTA EQU H'0005'
PORTB EQU H'0006'

when the pic master wrote his code, he put at the top

LCD_PORT Equ PORTA
LCD_TRIS Equ TRISA
LCD_RS Equ 0x04
LCD_RW Equ 0x06
LCD_E Equ 0x07

now going back to the 628 test its says that PORTA EQU H'0005',

so Nigel could have witten this instead, which is the same thing

LCD_PORT Equ 0x05 (=H'0005')
LCD_TRIS Equ TRISA
LCD_RS Equ 0x04
LCD_RW Equ 0x06
LCD_E Equ 0x07

is that correct?

Yes it is, but the whole point of using the includes is that you don't even need to know what the register addresses are - I'm assuming that you are correct that PortA is at address 0x05?.

The example above is also designed to make it easy to change, you simply change PortA and TRISA to PortB and TRISB, which is easier than remembering obscure numbers. Also, when you read the lines it's obvious what it means - the numbers mean very little!.

 

[eblc1388]

so Nigel could have witten this instead, which is the same thing

LCD_PORT	Equ	0x05 (=H'0005')
LCD_TRIS	Equ	TRISA
LCD_RS		Equ	0x04		
LCD_RW		Equ	0x06
LCD_E		Equ	0x07

The code generated is the same but there is a difference. Say if one wants to use PORTB for the LCD instead, one just change LCD_PORT Equ into PORTB.

While in your case you have to lookup the register address for PORTB and manually input 0x06 into the definition.

 

[tama182]

yep hang on though, thats not the bit that i dont understand, heres the bit i dont understand,

LCD_PORT Equ 0x05 (=H'0005')
LCD_TRIS Equ TRISA
LCD_RS Equ 0x04
LCD_RW Equ 0x06
LCD_E Equ 0x07


you said that the above would work, in that lcd_port equals porta, and

LCD_RS Equ 0x04
LCD_RW Equ 0x06
LCD_E Equ 0x07

all are bits in porta,

but howcome writing LCD_PORT Equ 0x05, signifies a regisiter, and LCD_RS Equ 0x04 signifies a bit in a register,

ok say i wanted to call the register FSR, the name LCD_RS, then i would write LCD_RS Equ 0x04, (FSR is a at h'0x04', look at table in previous posts)

but thats the same code a giving bit 4 in the porta register a name, now can you see where i dont get it?

 

[Pommie]

I'll try and describe it a different way.

   bsf PORTA,legs_on_a_table
   bsf PORTA,FSR
   bsf PORTA,4

The above are all the same instruction.
It is where the number is used that defines what it is.

An Alternative

  movlw FSR
  movlw legs_on_a_table

The above will put 4 into W

Also

  movfw FSR
  movfw legs_on_a_table

The above will get the contents of location 4 and put it in W.

Mike.

 

[eblc1388]

but howcome writing LCD_PORT Equ 0x05, signifies a regisiter, and LCD_RS Equ 0x04 signifies a bit in a register

4 by itself is meanlingless. If your software requires an address, then the 4 becomes an address.

So MOVWF LCD_PORT would expect and require 0x5 to mean an actual port address.

If the software requires an reference to a particular bit, then 4 become the position of the bit.

BSF LCD_PORT, LCD_RS would mean the 4th bit of address 5.

Now if you write:

MOVWF LCD_RS ; <<<<< mistake, LCD_RS is not a port address

You are sending the content of W to address 4, which is a big mistake.

 

[tama182]

ahh get it.......when nigel was writing

LCD_RS Equ 0x04

i was thinking that lcd_rs was only bit 4 of porta, and that no matter where you wrote lcd-rs, it would refer to bit 4 of porta,

but lcd-rs = 4, and just 4


ah so........ ((lcd-rs x lcd-rs)/(lcd-rs x lcd-rs))x(lcd-rs) = lcd- rs :lol:

 

[tama182]

im using a picdem plus 2 and an icd 2, to program my 16f628, i gather that the icd 2 is simply there to just debug, so can i simply plug the pic dem straight into my computer?