CCP / 24 bit timer / 1hz to 150 hz range

[ELCouz]

I will investigate further on this tomorrow too, i will have access to a scope.

Maybe (the problem) it's just fan signal related. We'll see!


Laurent

[ELCouz]

Hi,

I have some good result on the signal side here's the screenshot of the ccp pin (with the 5v pull-up)

Note: the probe was set to X1 instead of X10 , i didnt make the change in the scope setting ,, so its 5V per division instead of 50V as seen on the screenshot.

As you can see the CCP precision is very good!

[PCM programmer]

I don't have any time to work on this until Sunday. I just don't.

[ELCouz]

Dear PCM Programmer,

Take your time! You know, it's already nice that you have taken the time to help me since my first post.

I'm aware that most people on this board do electronics as a hobby when they have time to (including me).

I don't want to rush anybody ... but i have to admit it ,,, that it would be awesome if a fix could be found... I trying as most of my knowledge about CCP know. I have to do trials and errors!

Have a nice day!

Laurent

[deepakomanna]

Dear sir,
I gone through your code but one line not undestand.

[ELCouz]

>>1)How you calculate "0x1000000" value ?

It was already there never touched this...

>>2) ((12000000L + (current_ccp_delta >> 1)) why to shift right ?

I dont know really , i'm new to CCP.

But ...for the correct equation are you running the pic @ 32.768 khz or the timer1 is running at this speed ?

If pic clock = 32.768 khz then timer1 would be running at 8.192khz (clock div by 4) so.... change it to : ((8192L + (current_ccp_delta >> 1))

or if timer1 external clock = 32.768 then ((32768L + (current_ccp_delta >> 1))

in overall its ((XL / Y+ (current_ccp_delta >> 1)

X= timer1 freq
Y = your prescaler but since your have a prescaler of 1 dont put it

i hope i have helped you !

laurent

[deepakomanna]

i am using external clock == 32.768khz & prescaler == 1 only for timer1 & CPU is running on internal oscillator 8 MHz == i.e. 8 Mhz/4 == 2 Mhz.
I did freq. calculation by,

[RLScott]

[Kenny]

Yes, timer 1 is free-running and is not synchronised to the input signal in any way. The input signal edge being detected can occur at any time. The extended timer 1 will wrap from 0xFFFFFF to 0x000000 and continue.

0xFFFFFF is a large value (16,777,199), so at higher input frequencies (ie. low period) for most times the new captured value will be greater than the old one, but if the extended timer wraps before the new value is captured then the new value will be less than the old one. The chances of this occurring increase as the input frequency is lowered and the period increases.

PCM programmer explained why subtracting the two int32 values causes a problem - the extended timer wraps at the 24 bit boundary, not the 32 bit one.

A way to handle this is to test whether the new value is less than the old one, and if it is we know that wrapping has occurred. In that case the real delta value is from the old value to the 24 bit boundary, plus the new value.

An example, using hex. The delta value in a practical application will be much higher, but the example illustrates the method.
old = 0xFFFFFE
new = 0x000002

The delta value is 4 (0xFFFFFE to 0xFFFFFF to 0x000000 to 0x000001 to 0x000002)

current_ccp + (0x1000000 - old_ccp)
is 0x000002 + (0x1000000 - 0xFFFFFE)
= 4

A limitation of the code is that the delta value must be lower than the range of the extended timer. If this exceeded, then the test whether the new value is greater than the old one will foul up. If the shaft is nearly stopped this can occur. To get around this problem I wait for the flag to occur in a while loop, doing other things, and use another timer to do a timeout. There's probably a more elegant way to do it. My programming skills are limited.


Re external clock to timer 1, capture may not work (see data sheet for pic).

Re
frequency = (int16)((12000000L + (current_ccp_delta >> 1)) /
current_ccp_delta);
This is a rounding method where half the denominator is added to the numerator.

[deepakomanna]

Thanks kenny for replying,
But still now i not understand how you calculate the value "0x1000000 ",in your post.

[RLScott]

The problem is that a 24-bit timer wraps at 0xffffff while a 32-bit integer wraps at 0xffffffff. Here is a simple solution. Do the subtraction on 24-bit capture values as follows:

[ELCouz]

Dear RLScott,

Does my problem (0hz reading) because my timer is not syncronised with the input?

Your replies are for deepakomanna problem but what modification i need to do to get it to work ?

[ELCouz]

Tryed at slower clock speed ,,

INTRC_OI @ 8mhz

[Kenny]

[ELCouz]

Dear CCS members,

I have changed my pic18f2550 to an 2553 just to see,

and it's the same problem ... can't be a schematic problem ... i have looked with an oscilloscope and everything was fine ... see the image on the top of the topic...


Anyone ???

Have a nice day,

Laurent

Mtl,Quebec