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isr component directly detects rising edge. Is this idea going to be work? | Cypress Semiconductor

isr component directly detects rising edge. Is this idea going to be work?

Summary: 2 Replies, Latest post by PSoC73 on 21 Mar 2013 12:36 AM PDT
Verified Answers: 0
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qingshanshan's picture
24 posts

I would like to simplify my hardware design for raising 2 interrupts events with a single digital input pin to detect pulse edges.  So my idea is to use digital input hardware pin to detect pulse falling edge while a isr component to detect a rising edge by connecting to the digital input pin. This trys to raise 2 seperated interrupt events. Make clear that it is not to raise a single event with detecting rising and falling edge. The topdesign is shown as following diagram. Is this going to work? Has anyone have any this experience?


user_14586677's picture
7646 posts

Answered here -


Regards, Dana.

user_119377051's picture
866 posts

Hi qing-san,
Pin-direct isr(isrLick0_rise) and Pins isr(isrLick0), Of course both are working.
ISR property only can select a rise-edge but to through an invert logic can be select fall-edge.

Besides, Pins isr(isrLick0) can reduce interruption resources,
One ISR for multi pins.
[input] tab of PIns property can select rise and fall.
I recommand use Pins ISR and select [derived] in ISR property. That is neat I think.

To use Pins-ISR, need one notice.
Have to ensure clearing ISR in ISR routine.
[uint8 Pin_ClearInterrupt(void);]

That is my experience and the article is here
Please refer bundle design.

Happy Interruption..

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