I am trying to use Timer0 to control a blinking LED, but I am having some difficulties... the LED never turns on.

I referred to this tutorial to set things up, but I still can't understand it, could you please tell me where the problem is?

#include "mcc_generated_files/mcc.h"
#include <stdint.h>
#define SBIT_PS2 2

volatile unsigned char value = 0;

void timer_isr(void)
  if (TMR0IF == 1)
    value =~ value; // complement the value for blinking the LEDs
    TMR0 = 101; // Load the timer value. Note: Timervalue is 101 instead of 100 as Timer0 needs two instruction cycles to start incrementing TMR0
    TMR0IF=0; // Clear timer interrupt flag 

Main application
void main(void)
  // initialize the device

  // When using interrupts, you need to set the Global and Peripheral Interrupt Enable bits
  // Use the following macros to:
  // Enable the Global Interrupts
  // Enable the Peripheral Interrupts
  // Disable the Global Interrupts
  // Disable the Peripheral Interrupts
  uint16_t dutycycle;
  TRISD=0x00; // Configure PORTD as output to blink the LEDs

  OPTION_REG = (1<<SBIT_PS2); // Timer0 with external freq and 32 as prescalar
  TMR0 = 0; // Load the time value for 1ms delay
  TMR0IE = 1; // Enable timer interrupt bit in PIE1 register
  GIE = 1; // Enable Global Interrupt
  PEIE = 1; // Enable the Peripheral Interrupt 
  while (1) {
    WB_PORT = value;
    LR1_PORT = value;

Supplement: my pin settings:

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Because there is too little information about PIC in Chinese...


2 Answers 2


It could be the classic missing volatile bug. Change the declaration to volatile unsigned char value = 0;

You should also make a habit of using stdint.h and uint8_t instead. The char type in C has implementation-defined signedness, so it is unsuitable for binary arithmetic and also non-portable.

  • \$\begingroup\$ I've modified and tried it, but it still doesn't work, thanks for your answer! \$\endgroup\$
    – Judy11432
    Aug 16, 2022 at 9:39
  • \$\begingroup\$ @Judy11432 Is the interrupt firing at all? Did you try to set a breakpoint inside the ISR? \$\endgroup\$
    – Lundin
    Aug 16, 2022 at 9:55
  • \$\begingroup\$ Hi Lundin,I don't understand how to interrupt firing, sorry I just learned this stuff \$\endgroup\$
    – Judy11432
    Aug 19, 2022 at 0:56
  • \$\begingroup\$ @Judy11432 If you don't know how to use a debugger, you are probably better off learning the basics on a PC before moving on to embedded systems. But it isn't hard to learn, every debugger out there works similar no matter target. \$\endgroup\$
    – Lundin
    Aug 19, 2022 at 6:24

timer_isr() isn't set up as interrupt service routine.

You need to add the interrupt keyword:

void interrupt timer_isr(void)

Without the interrupt keyword, timer_isr() is a regular function, which is never referenced in your program.


void interrupt timer_isr(void) is the correct syntax.

Deviations like void __interrupt() timer_isr(void) and void __interrupt() INTERRUPT_InterruptManager(void) are wrong.

The error message main.c:49:: error: (1506) multiple interrupt functions (_timer_isr and _INTERRUPT_InterruptManager) defined at interrupt level 1 states that the timer interrupt service routine has already been defined somewhere else.

I've never used the MPLAB Code Configurator (MCC) before, but I strongly suspect that the corresponding code is being included with #include "mcc_generated_files/mcc.h". Have a look at that file to see if you can find the other definition of the timer interrupt service routine

  • \$\begingroup\$ I tried chang to void interrupt timer_isr(void) and void __interrupt() timer_isr(void) The following error occurs : main.c:49:: error: (1506) multiple interrupt functions (_timer_isr and _INTERRUPT_InterruptManager) defined at interrupt level 1 So I chang to : void __interrupt() INTERRUPT_InterruptManager(void) this error will occur : main.c:49:: error: (1098) conflicting declarations for variable "_INTERRUPT_InterruptManager" (mcc_generated_files/interrupt_manager.c:52) I have checked a lot of information, and my understanding is that variables cannot be added to it, right? \$\endgroup\$
    – Judy11432
    Aug 19, 2022 at 0:46

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