From Regulus Berdin
Given a number in bin, from 1 to 30, this code will drive that number of LED's on.
; 1-30 to 30 leds ;input: bin ;output: led1,led2,led3,led4 convert: clrf led1 clrf led2 clrf led3 clrf led4 incf bin,f loop: decfsz bin,f ;do not include 0 goto $+2 return decf bin,f setc ;turn on led rlf led4,f rlf led3,f rlf led2,f rlf led1,f goto loop ;-------------------- I have another solution, using the fact that 1-30 bar is equal to 2^y - 1. This routine is isochronous and faster on some values than the above. Here goes: convert: clrf a1 clrf a2 clrf a3 clrf a4 movlw a4 ;2^bin btfsc bin,3 movlw a3 btfsc bin,4 addlw -2 ;corrected from 'movlw a1' movwf FSR movlw B'00000001' btfsc bin,1 movlw B'00000100' movwf INDF btfsc bin,0 addwf INDF,f btfsc bin,2 swapf INDF,f movlw 1 ;subtract by 1 subwf a4,f skpc subwf a3,f skpc subwf a2,f skpc subwf a1,f
You don't need a 30bit number just because you want to drive a 30led bar :) All you need is a number from 0 to 30, which boils down to a simple division, if your representation is linear. 0->0%->no led, 65535->100%->all (30) leds. that's 30 intervals for 65535, hence each interval (corresponding to one led more) is 65535/30... you divide your load by this number, and get the number of leds to switch on.
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