How to convert 10bit ADC data to BCD?

BCD?

This is for a PIC, but it should be possible to translate it since it's not that complex. It converts a 16 bit unsigned binary value to 5 packed (two digits per byte) BCD digits. binH and binL contain the 16 bit binary value. count and temp are single byte work areas.

binary_to_bcd bcf STATUS, 0 ; clear the carry bit movlw .16 movwf count clrf tenthousands clrf thousands_and_hundreds clrf tens_and_ones loop16 rlf binL, F rlf binH, F rlf tens_and_ones, F rlf thousands_and_hundreds, F rlf tenthousands, F ; decfsz count, F goto adjDEC RETLW 0 ; adjDEC movlw tens_and_ones movwf FSR call adjBCD ; movlw thousands_and_hundreds movwf FSR call adjBCD ; movlw tenthousands movwf FSR call adjBCD ; goto loop16 ; adjBCD movlw 3 addwf INDF,W movwf temp btfsc temp,3 ; test if result > 7 movwf INDF movlw 0x30 addwf INDF,W movwf temp btfsc temp,7 ; test if result > 7 movwf INDF ; save as MSD RETLW 0

The easiest (and slowest) method is to implement a binary down counter and a BCD up counter. Load the down counter with the binary data. Clear the BCD up counter. Decrement the down counter to zero. For each decrement, increment the BCD counter by one. ~ For a more elegant solution, see this article: BIDEC - A Binary-to-Decimal or Decimal-to-Binary Converter, J. F. Couleur, IRE Transactions on Electronic Computers, Vol. EC-7, pp313-316, IRE, 1958. Regards, Jon

Divide your number by 10. The remainder is the ones value in BCD.

Divide the quotient by 10. The remainder is the tens value in BCD.

Divide the new quotient by 10. The remainder is the hundreds value in BCD.

The final quotient is either 1 or 0 if you really started from 10 bits. It is the thousands value.

You can carry this procedure on indefinitely, and it is a general method for converting from one base to another (so should you want to display things in base 5, now you know how).

One of the nicest approaches I've seen can be found here:

Once you figure out how it works, it's quite elegant.

Here's a nice page on how to do it with combinational logic...

...Jim Thompson

Thank you guys...I've got the idea...thanks a lot

well, Convert the Integer to ascii-string and then take each char and make a Case-Construckt to a other char that transforms your BCD direct to

Marte

In article , Tim Wescott wrote: [.. binary to BCD ..]

If you don't have a divide but do know how to add BCD values:

Consider your binary number to be all to the right of the binary point.

X = your number Y = zero Loop 10 times: Y = Y + Y decimal adjust Y ; ie make it a BCD add X = X + X ; Shift the MSB above the "binary point" if Y>=1024 then ; thats 2^10 Y = Y + 1 ; Put the bit in the bottom of Y end if end loop

Or, since you have BASIC, you could do the repeated mod 10/divide by 10 loop, as Tim Wescott suggested. It'd be nice if high-level languages had a facility to do an integer divide and get both the quotent and remainder back in one instruction.

Cheers! Rich

BCD?

this means you have to count up to 2^10

An improvement to this method would be to start with the most significant decimal digit, ie while (number is > 1000) sub 1000 from number and inc 4th digit while (number is > 100) sub 100 from number and inc 3rd digit etc...

then you only have to count up to 36

if you have a lot of digits you could use a digit index and a variabvle to hold the 1000/100/10 value etc.

Colin =^.^=

In article , Rich Grise wrote: [...]

Make that 2X 512. The LSB doesn't have to get translated.

BCD?

BCD.

You are a nice guy Tim, but please do not do high school homework for OP that won't do any homework themselves.