10 Bit Binary to Decimal Convertion Help

[Suraj143]

Hi guys I have a 10 bit result from AD converter which are in two 8 bit GP registers.

Now I want to convert this binary value to decimal value to show in the four segments.

If you have any idea (easy way) please share with me.

 

[futz]

Hi guys I have a 10 bit result from AD converter which are in two 8 bit GP registers.

Now I want to convert this binary value to decimal value to show in the four segments.

If you have any idea (easy way) please share with me.

This is for 8 bits and PIC 18F, but it may give you an idea. You want hex digits, I assume?

fhex	db	"0","1","2","3","4","5","6","7","8","9","a","b","c","d","e","f"

;****************************************************************
; fixhex - subroutine takes byte passed in W - splits it into two
; ascii bytes representing hex digits and outputs them to RS232
fixhex	movwf	temp		;store it for later
	movwf	dec1		;and put it in dec1 too
	movlw	0xf0		;mask out low nybble
	andwf	dec1
	swapf	dec1		;swap nybbles
	movlw	fhex		;point W to table start
	addwf	dec1		;add offset to desired digit
	clrf	TBLPTRH		;set up table pointer
	movff	dec1,TBLPTRL
	tblrd	*		;read the byte
	movf	TABLAT,W	;get it in W
	call	rs_send		;transmit it
fixhx2	movlw	0x0f		;mask out high nybble
	andwf	temp
	movlw	fhex		;point W to table start
	addwf	temp		;add offset
	movff	temp,TBLPTRL
	tblrd	*		;read the byte
	movf	TABLAT,W	;get it in W
	call	rs_send		;transmit it
	return

 

[Suraj143]

Hi futz thanks for your idea I'll make a note on that BTW i have never done 18F yet.

I need 0-9 digits format no need hex.

 

[futz]

Hi futz thanks for your idea I'll make a note on that BTW i have never done 18F yet.

I need 0-9 digits format no need hex.

Then this may help give you a clue where to go. Again it's for 8 bit, so you'll have to take it from here.

This doesn't use the 18F table opcodes, so it will work fine on earlier PICs.

;********************************************************* 
;*fix - subroutine takes byte passed in W - splits it into 
;*two ascii bytes in dec1 and dec2, representing decimal digits. 
fix:	movwf	dec2		;put number in dec2
	clrf	count		;count = 0 
	movlw	0x0a		;W = 10
tens:	subwf	dec2,F		;subtract 10 from number
	btfsc	STATUS,C	;check if result <10 
	goto	again		;no, go again 
	movf	count,W		;yes, put count in W 
	addlw	0x30		;add $30 to make it an ASCII number 
	movwf	dec1		;and store it in dec1 
	movf	dec2,W		;get remainder (2nd digit)
	addlw	0x0a		;put the last subtract back 
	addlw	0x30		;add $30 to make it ASCII 
	movwf	dec2		;store it back in dec2 
	return			;and return
again:	incf	count,F		;increment count 
	goto	tens		;go again

If your code doesn't need ascii (only needs the decimal values) then comment out the two "addlw 0x30" lines.

EDIT: Oops! Just noticed that the code was assuming decimal radix (something dumb I did in the old days). I just edited it so it'll work with the default hex radix you're probably using.

 

[Suraj143]

Hi futz your new coding is very good for me easy to understand.Now I'm studying it how to put the value to the SSD from your coding.

After I understood it I'm going to do the 10 bit one.

 

[pdfruth]

Found this algorithm out on piclist;

  ;======= Bin2Dec10G.asm ============= Sep 04 =========
  ;       Test of 10b binary to 4 digit conversion as 
  ;       based on division by powers of 10. 
  ;
  ;       E. A., Grens
  ;
  ;       For most PICs - here for 16F628 with 4 MHz
  ;       internal oscillator.
  ;
  ;       Average processor cycles for execution of
  ;       subroutine over the input range 0x000 to 0x3FF
  ;       is 88.  No variables required except the
  ;       input blh, blo and the out BCD digits.
  ;       Subroutine length 38.
  ;======================================================

        list p=16f628
        radix   dec
        #include 
        __config 0x3D30

  
  ;-------Set Up RAM Variables---------------------------
  blo   equ     0x20            ;lsbyte of 10b binary
  bhi   equ     0x21            ;msb
  d4    equ     0x23            ;msdigit
  d3    equ     0x24
  d2    equ     0x25
  d1    equ     0x26            ;lsd
  ;-------Program Code--------------------------------
                                         
        org     0x00            ;Effective Reset Vector
  ;
        nop
        goto    Start           ;Jump by interrupt
  ;
  Start
        movlw   0x07
        movwf   CMCON           ;Digital I/O
        movlw   0x03            ;Test input
        movwf   bhi
        movlw   0xFF
        movwf   blo
        movlw   0x04
        call    Bin2decg
  Hold
        goto    Hold            ;Finished

  Bin2decg                      ;10b binaey to BCD
        clrf    d1
        clrf    d2
        clrf    d3
        clrf    d4
        clrc
        rrf     bhi,f           ;N = 4*Q + R
        rrf     blo,f           ;blo = Q
        rrf     d1,f            ;d1 = R as temp, R25 
        goto    B2d2            ;repeat
  B2d3
        addwf   blo,f           ;get remainder
                                  ;    after /100, C = 0
        rlf     d1,f            ;*4 and add R back in
        rlf     blo,f
        rlf     d1,f
        rlf     blo,f           ;4*blo + R = N - 
                                  ;    1000*d4 - 100*d3
                movlw   10
  B2d4
        subwf   blo,f           ;blo = N - 1000*d4 -
                                  ;     100*d3 - 10*d2
        skpc
        goto    B2d5            ;blo10
        goto    B2d4
  B2d5
        addwf   blo,f           ;put last 10 back, C=0
        movf    blo,w            
        movwf   d1
        return
   
        end

 

[Suraj143]

Hi pdfruth thanks for your excellent coding.For sure I'll work with that after futz coding.

@ futz
I'm dreaming why its subtracting 10.

 

[futz]

Hi pdfruth thanks for your excellent coding.For sure I'll work with that after futz coding.

@ futz
I'm dreaming why its subtracting 10.

It counts the number of times it can successfully subtract 10 from the value (the value is less than 10 at that point). The count (number of successful subtracts of 10) becomes the 10s digit.

Then the remainder gets the last subtract (the one that didn't work) added back. That value becomes the 1s digit.

You may need to add a 100s count as well, and possibly even a thousands.

 

[Suraj143]

Hi futz this is the place I got stucked.

Let say the AD result is 19
Now subtract 10 from 19 will be = 9

Now the carry bit is clear (a successful subtract) so the new values will be

D1 =0
D2= 19

 

[futz]

The count variable is the one to watch. It keeps track of the number of successful subtractions. Count is the value that becomes your tens digit, not dec1 or dec2.

The remainder, after the last unsuccessful subtract, with the last 10 added back to make it non-negative (and correct), becomes your ones digit.

With 10 bits, you will have to also add a thousands count (only has to count to 1, so a simpler check would be easiest) and a hundreds count, with corresponding multiple subtract routine.

Hmm... I should actually write one. Might be more to it than that, but shouldn't be too difficult.

 

[Suraj143]

Hi futz I have small problem.

The CARRY bit will set when subtracting from smaller value.
Ex: 5-10 will set carry bit.

In my example 19-10 = 9 so the carry bit will CLEAR.

It will never goto increment count.

So the new result is

D1= 0
D2= 19

Please tell me am I right or wrong?

 

[futz]

Hi futz I have small problem.

The CARRY bit will set when subtracting from smaller value.
Ex: 5-10 will set carry bit.

In my example 19-10 = 9 so the carry bit will CLEAR.

It will never goto increment count.

So the new result is

D1= 0
D2= 19

Please tell me am I right or wrong?

Wrong! (I think) This is a working algorithm, in working code, which I have used on many different cpu's since the early 80's.

C: Carry/borrow bit
For ADDWF, ADDLW, SUBLW, and SUBWF instructions
1 = A carry-out from the Most Significant bit of the result occurred
0 = No carry-out from the Most Significant bit of the result occurred
Note: For borrow, the polarity is reversed. A subtraction is executed by adding the two’s
complement of the second operand.

There it is. When borrowing it works the other way.

 

[Suraj143]

The count variable is the one to watch. It keeps track of the number of successful subtractions. Count is the value that becomes your tens digit, not dec1 or dec2.

Oh I see I thought other way.Count is the one that I put to SSD (tens digit).

But I still thinking what is successful subtract is?

In the coding when a successful subtract occurred all the time count variable is zero.Because carry bit is clear when successful subtract occurred.

any idea.

 

[Pommie]

Have a look at , about half a page down is 10 bit to BCD conversion.

Mike.

 

[simrantogether]

simple...

simple...

10 bit binary...

e.g.

10 1100 1001

Hence , the answer is:

1 * 2^0 +0 * 2^1 +0 * 2^2 +1 * 2^3 +0 * 2^4 +0 * 2^5 +1 * 2^6 +1 * 2^7 +0 * 2^8 +1 * 2^9

The answer is to be read from right to left of the given binary value..

Regards,

Simran..

 

[Nigel Goodwin]

Check my analogue PIC tutorial, it includes excellent routines to convert the analogue readings to decimal and display them - I've used it for years (it came off the PICList originally) and it's always worked flawlessly.

 

[futz]

Check my analogue PIC tutorial, it includes excellent routines to convert the analogue readings to decimal and display them - I've used it for years (it came off the PICList originally) and it's always worked flawlessly.

Ah, excellent! I started writing a 16 bit converter last night. Ran out of time. Hadda sleep. I'm gonna finish mine tonight without cheating, and then look at what's probably a better algorithm at Nigel's site.

 

[blueroomelectronics]

I would think the DAW instruction might come in handy, not used it yet myself...

 

[eng1]

I would think the DAW instruction might come in handy, not used it yet myself...

I bookmarked this page some time ago, but did not study the algorithm in detail. It's based on the DAW instruction. The author says that the routine takes the same number of cycles, no matter what its input is. That would be interesting with 16+ bit variables.

 

[Suraj143]

8 Bit BCD to 10 Bit BCD coding help

Hi guys I studied the 8 bit BCD conversion & I added another digit for that so I can count up to 1023.

Unfortunately in 8 bit the maximum value is 255. But my AD is giving 10bit result. I want to use this method for 10 bit conversion. I used ADRESL only I don’t know how to combine with ADRESH.

I'm not familiar with rrf/rlf BCD version its too high for me.I like this method.
So please help me to combine these two for 10 bit pattern.

Start		movf	ADRESL,W	;move lower byte of AD to W
		call	BCD10
		movf	Digit2,W
		call	BCD100
		movf	Digit3,W
		call	BCD1000
                goto   Start

BCD10	    	clrf   	Digit2
	    	movwf  	Digit1	
Tens	    	movlw  	.10
	    	subwf  	Digit1,W
            	btfss  	STATAUS,c
	    	goto   	DoneDigit2
	    	movwf  	Digit1
            	incf   	Digit2,F
	    	goto   	Tens
DoneDigit2  	Return


BCD100		clrf	Digit3
		movwf	Digit2
Hundreds	movlw	.10
		subwf	Digit2,W
		btfss	STATUS,C
		goto	DoneDigit3
		movwf	Digit2
		incf	Digit3,F
		goto	Hundreds
DoneDigit3  	Return


BCD1000		clrf	Digit4
		movwf	Digit3
Thousands	movlw	.10
		subwf	Digit3,W
		btfss	STATUS,C
		goto	DoneDigit4
		movwf	Digit3
		incf	Digit4,F
		goto	Thousands
DoneDigit4  	Return