setting values and percentages

[Scampy]

If B_Max is the value you send to the PWM controller (0-4095) and maxbright is the percentage value (0-100) as entered by the user then yes.
Actually, at 100% it'll give you 4094 so make that B_max = maxbright */10484 instead.

/Henrik.

Yes - B_max and W_max are the values that are sent to the PCA chip (0 - 4095). And maxbright has a value between 0 and 100 as you assumed. But using that line of code the values for B_max and W_max do not change from the pre-programmed default of 4095 after the menu option is used to change the max brightness level

[HenrikOlsson]

Hi,
Strange..... I tested the following code

HSEROUT["Program start",13]
For i = 0 to 5
    Lookup2 i,[1, 10, 25, 50, 75, 100], maxbright
    B_max = maxbright */ 10484
    HSEROUT ["Maxbright: ", DEC3 maxbright, "     B_Max: ", DEC4 B_max, 13]
NEXT

And got the following results:

Program start
Maxbright: 001     B_Max: 0040
Maxbright: 010     B_Max: 0409
Maxbright: 025     B_Max: 1023
Maxbright: 050     B_Max: 2047
Maxbright: 075     B_Max: 3071
Maxbright: 100     B_Max: 4095

Seems to work fine here....

/Henrik.

[EarlyBird2]

the values for B_max and W_max do not change from the pre-programmed default of 4095 after the menu option is used to change the max brightness level

Is the default of 4095 inside the main loop and resetting B_max and W_max after the menu option returns to the main loop.

[Scampy]

Don't know what happened to my post last night, but I found out why it was doing that - It was one of those oversights caused through hours of programming without a break. The menu option saved all the changes when exiting from the menu to the main program, so unless this action was performed all the changes were lost if you remained in the menu section as the diagnostics option reads the values stored in memory and thus read the defaults and over wrote them. This was changed so each section of the menu now writes the new values to memory when exiting the option and reverting back to the menu screen, so switching between menu options and then the diagnostics means that the changes made in any of the other menu sections are read back in.

One thing I would like to know is what the */ function means

[HenrikOlsson]

It's been covered multiple times on the forum and it's covered in the manual.
In short, think of it as multiplying by units of 1/256. 10484/256=40.95.

/Henrik.

[EarlyBird2]

One thing I would like to know is what the */ function means

This is what I understand or misunderstand possibly.

Internally the maths can have a 32 bit result because the PIC has a 32 bit register. But if your maths calculation will fit in a word then using word=byte*byte will work and you do not need ** or */.

If you have wordH=word**word and wordL=word*word then you have all 32 bits stored in two words.

*/ is less obvious but gives the middle word of the 32 bits. Which is the same as word=word*word/256 losing the top and bottom 8 bits.

If as in your case and as you know 4095/100 gives a result of 40 therefore B_max = (4095/100)*maxbright will give 2000 when maxbright is 50.

Whereas 4095*256/100 gives result of 10483 and 50*/10483 gives 2047 because by multiplying by 256 and using the */ to remove the 8 lower bits (or divide by 256 which is the same) increased accuracy is achieved. So where is the precision lost? The division /100 is the source of lost precision so as a rule multiply first and divide last which is exactly what */ does in that your program multiplies by 256 and */ divides by 256 last. As an aside working this way reduces the calculations within the program, there is only a division

A question then. What is the result of 50*4095/100 2000 or 2047? and (4095/100)*50?

[richard]

result 50*4095/100 is 2047
result (4095/100)*50 is 2000

the calc follows typical bodmas rules (brackets, of , divide .....) see precedence in manual

the only 'suprising' bit is that the indermediate result (50*4095) is 204750 way too big for a word var. the pbp compiler code does indeed use 32 bit intermediate results.
the */ and ** operators allow you access these 32 bit results (providing you follow the rules of course)

[richard]

however
if you have a word var say numx
and x =4095
then 50*numx/100 the result is 81
the word var causes a intermediate stage overflow (bits 0-15 of 204750 =8142)
I still say integer math is tricky and results need to be double checked

[EarlyBird2]

however
if you have a word var say numx
and x =4095
then 50*numx/100 the result is 81
the word var causes a intermediate stage overflow (bits 0-15 of 204750 =8142)

I do not understand this.
What is the relationship between numx and x?
Why does an intermediate stage overflow occur and what is an intermediate stage overflow?

[richard]

What is the relationship between numx and x?

poor typing
if you have a word var say numx
and numx =4095
then 50*numx/100 the result is 81
the word var causes a intermediate stage overflow (bits 0-15 of 204750 =8142)
I still say integer math is tricky and results need to be double checked

intermediate overflow
50*4095 =204750 max value for word 65535 therefore an overflow occurs and the result becomes 8142