The PIC is supplied at 3.3V and clocked at 16kHz. For this part of the design I need the external interrupt to trigger on the falling edge of a 100Hz square wave signal on RB0 and toggle the state of RD7.

In the figure below we can see that the interrupt works as intended when I'm testing it with an SPST switch. The following image is the schematic when using the SPST switch (ignore the LM324 and attached components)

For Trace 1:

  • Channel B (Blue) - SPST Input

For both traces:

  • Channel C (pink) - RD7 Voltage Level

enter image description here enter image description here

However in when I attach the square wave (shown in the figure below), it seems to be triggered on the rising edge of the square wave. For Trace 2:

  • Channel B (Blue) - 100Hz Signal

enter image description here enter image description here

At first I thought that for some reason the NE555 was loading the input of RB0 so I added a buffer but that didn't help. I also tried pulling up RB0 but that failed as well. I dug around in the manual and found this:

For external interrupt events, such as the INT pin or PORTB change interrupt, the latency will be three or four instruction cycles.

So it might be that the interrupt is being triggered too quickly? Even at 16kHz however 4 instruction cycles will be 0.001s and the signal period is 0.01s. Does anyone know what might be causing this and how to fix it?

Software used: Proteus Professional 8

EDIT: Code and image Added enter image description here

                        LIST    p=16f877
                    INCLUDE <P16F877.INC>

W_TEMP              EQU     0x20
STATUS_TEMP         EQU     0x21
time_hi             EQU     0x22
time_lo             EQU     0x23
time_cp             EQU     0x24
timerCount          EQU     0x25

                    ORG     0x00
                    goto    main
                    ORG     0x04
                    goto    ISR
                    ORG     0x20

                    BANKSEL TRISD
                    bcf     TRISD,RD7   
                    call    interruptSetup

                    goto    loop

;--------------------------------Configure Interrupts-------------------------------                        

                    call    rb0FallingEdgeInt

                    call    enableInterrupts


                    bsf     STATUS,RP0
                    bcf     STATUS,RP1

                    bsf     TRISB,RB0
                    bcf     OPTION_REG,INTEDG
                    bsf     INTCON,INTE



                    bsf     INTCON,PEIE
                    bsf     INTCON,GIE


                    movwf   W_TEMP      ; SAVE  W AND STATUS REGISTERS
                    swapf   STATUS,W
                    movwf   STATUS_TEMP
                    bcf     INTCON,GIE

                    btfsc   INTCON,INTF
                    call    fallEdgeIntHandler

                    bsf     INTCON,GIE
                    swapf   STATUS_TEMP,W   ; RESTORE W AND STATUS REGISTERS
                    movwf   STATUS
                    swapf   W_TEMP,F
                    swapf   W_TEMP,W


;---------------------------ISR HANDLERS------------------------------
                    BANKSEL PORTD
                    movlw   B'10000000'
                    xorwf   PORTD,W
                    movwf   PORTD
                    bcf     INTCON,INTF

  • \$\begingroup\$ A quick test would be to slow the 555 way down (e.g., 10 Hz) to see which input edge the output follows. Also, keep in mind that "interrupt latency" refers to how long it takes to fetch the first instruction of the ISR -- the number of instructions in the ISR determines how soon the pin toggles after that. Show us your code. \$\endgroup\$ – Dave Tweed Apr 6 '18 at 11:49
  • \$\begingroup\$ @DaveTweed what do you mean by "...which input edge the output follows". Also made an edit to include the code. \$\endgroup\$ – Simeon R Apr 6 '18 at 15:18
  • \$\begingroup\$ @S.Ramjit Your top trace has an inconsistency - the third rising edge of the red C trace isn't associated with an edge in B. I'd investigate that. \$\endgroup\$ – awjlogan Apr 6 '18 at 15:28
  • \$\begingroup\$ @awjlogan if you're referring to the last image (10Hz) then it seems the third rising edge of the red C trace is associated with a falling edge in blue B trace. Which is the way it's supposed to work, just it doesn't do that at 100Hz \$\endgroup\$ – Simeon R Apr 6 '18 at 15:44
  • \$\begingroup\$ @S.Ramjit No, your first image, which I think is with the switch. Apologies if I've misinterpreted something! \$\endgroup\$ – awjlogan Apr 6 '18 at 16:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.