Why? Temp conversion C to F code from Rentron (Bruce)

[Heckler]

The first IF condition in the original displays 1 more digit prior to the decimal point.

Yes, I could see that in the original code, the first debug was there to display one more digit for temp>99.

That is why I thought that I could remove the second "TempF = TempF + 3200".

But it appears that I need to keep both of them in the code to get the correct temperature conversion??

Or am I missing something?

My goal is to have a section of code that will return degrees F for at least a range of say -20 F to +120 F (-30 to +50 C) I dont care about fractional digits. I am also trying to keep it small, as I am running out of room in my PIC. I don't need extreme accuracy +- 1 deg or so is ok.

Here is my(Bruce's) complete subroutine which seems to work correctly:

 
'===========subroutine==========================
GetTemp:
OWOUT comm_pin, 1, [$CC, $4E, 0, 0, DS18B20_9bit] 'set resolution 9 bit Start_Convert:
OWOUT Comm_Pin, 1, [$CC, $44] 'Skip ROM search & do temp conversion
 
Wait_Up:
OWIN Comm_Pin, 4, [Busy] ' Read busy-bit
IF Busy = 0 THEN Wait_Up ' Still busy..?, Wait_Up..!
OWOUT Comm_Pin, 1, [$CC, $BE]
OWIN Comm_Pin, 2, [R_Temp.Lowbyte, R_Temp.Highbyte] ' Read two bytes
GOSUB Convert_Temp
RETURN
'================================================================================ 
Convert_Temp: ' +32.0 to +257 F 
IF Cold_Bit = Real_Cold THEN Yikes ' If Cold_Bit = 1, it's below "0" deg C
Sign = "+"
Dummy = 1125 * R_Temp
TempF = DIV32 100
IF TempF >6795 THEN ' Over 99.5 deg F..?
TempF = TempF + 3200
ELSE
TempF = TempF + 3200
ENDIF
tempf=tempf/100 'keep only whole digits
RETURN
'-----------------------------------------------------------------------------
Yikes: ' Display full range -C to -F conversion
Sign = "-" ' Display - symbol for negative temp
TempF = ~R_Temp / 16 ' Begin conversion from -C to deg +/-F
IF TempF >=18 THEN ' Check for -degrees F "-18 C = -0.4 F"
TempF = ((((TempF + 50) * 9) /5) -122) ' -C to -F below -17 deg C
ELSE ' Else result = +deg F
TempF = ((((-TempF + 50) * 9) /5) -58)
ENDIF
RETURN
'==================================================

But It seems that I still need both instances of... TempF = TempF + 3200
and that is what is puzzling me.

Isn't it the case that only one or the other will be executed? Yet if I remove one of them then I get the wrong temperature??

Possibly(likely) I need to go back and do some more testing.


Thanks