Hi,
I am trying to generate two interrupt from "timer 0 " and "pwm".
So, writing below line will enable both the interrupt or there is another procedure.
IER |=M_INT1;
IER |=M_INT2;
OR
IER |=0X0011;
which line will enable interrupt from both group's?
Thank you
Hi Mihir,
mihir dave said:which line will enable interrupt from both group's?
The first method you have described above will set the correct value of the Interrupt Enable Register (IER)
IER |= M_INT1; // Group 1
IER |= M_INT2; // Group 2
You could also do something like this to enable both:
IER |= M_INT1 | M_INT2; // Group 1 & 2
The second method you described could be altered in the following way:
From: IER |= 0x0011; // INT1 and INT 5
To: IER |= 0x0003; // INT1 and INT2
The IER is a 16 bit register as follows:
Don't forget to modify your PIE Control Register afterwards. Also ensure to enable the Global Interrupt INTM (EINT) and global real time interrupt DBGM (ERTM).
Best Regards,
Marlyn
