Silly question: Debuging is still possible if reset pin is disabled?

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Silly question: Debuging is still possible if reset pin is disabled?

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fjrg76
Contributor IV

Hi,

I don't need the Reset pin in my application (LPC804, but I guess it's the same situation for any LPC mcu), so I would like to use it in something more useful. The question is that the debugger needs to reset the mcu for all debugging sessions, so I'm thinking aloud: "No, if you still want to debug, then you do need the reset pin. Don't disable it!". Is this sentence correct?

Thank you!

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BlackNight
NXP Employee
NXP Employee

Hi,

It is not a silly question at all.

You don't need the reset pin for debugging purposes per se. It is more about how the debugger can get control over the target, and using the reset pin is one way to get the hardware in a known state. But the debugger still can use the data/clock JTAG pins to talk to the hardware. But that might not be possible if the device is in a low power mode with say the debug block disabled to safe power. In that case the reset pin is a way to get the device back to life and to a known state.

What I do in my designs where I'm using the reset pin for other functions (say using it as a GPIO pin) is to keep the reset functionality present for some time (say 500-1000 ms) after power-on. That way I have a window right after power-on in which the debugger still can use the reset pin to get control over a target. Still having that reset functionality available can be critical for some applications.

I hope this helps,

Erich

PS: you might find more information on this topic here: Using the Reset Button on the Freedom Board as User Button | MCU on Eclipse 

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fjrg76
Contributor IV

What I understand, and I've already thought (kind of), is that everytime it's needed to debug I must power off the chip and then power it on so that the window you mentioned take place, right? It might be done through an (always set) jumper or normally closed push-button attached to the VCC pin. Of course, the production line will get rid of both of them.

I guess I'll need to sacrifice one chip in order to verify whether this solution works reliably.

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