Example - Timer Interrupt duration (SOLVED)

[rajeshkhan808]

I am trying the following example in which the timer interrupt is being called every time and the overflow duration is not being considered at all.
The code is

[Ttelmah]

Start by telling us your clock rate....

However, Timer1 in general, is a 16bit counter. You are feeding it off the CPU instruction clock (OSC/4). It overflows when it reaches 65536 (when it wraps from 65535 back to 0), so to take 8 seconds from start (ignoring your '8000', which makes no sense at all), would require the instruction clock to be running at 8192 instructions per second, and Fosc, to be 32768Hz.
You'd not be able to do serial at any sensible rate from such a slow clock, so I suspect something is deeply 'wrong' in your calculations.
Now, assuming the Fosc is something more reasonable, like 4Mhz, then with the biggest prescaler (/8), the timer is still going to wrap every (4*8*65536)/4000000 second. Just over 0.5 second. To get an 8 second timeout, you'd need to be running off a much slower clock (or just use the timer as a 'tick', and only do your task every sixteen calls for example).

You then set the timer 'to' a value, and talk about it counting to 2000. The value you set it 'to' is just a count. The timer wraps when it reaches 65536, so if you wanted 2000 counts, you'd have to load the timer with 65536-2000. However you'd have to do this every time the interrupt is called, and given that this is happening much faster than you want, you probably don't want to do this.....

Best Wishes

[rajeshkhan808]

Thanks that helped , So while going through the example

[Ttelmah]

Timer0, is an 8 bit counter on most chips. Hence wraps at 255 to 0, and does 256 counts. Even on chips where it offers 16bit operation, it can be turned 'down' to run in 8bit mode. It is running in 8bit mode.
In your code, you are using Timer1, which is 16bit by default.

[rajeshkhan808]

@Ttelmah thanks for clearing that up.