[sdiy] Will clamping diode work when power is off?

[Brian Willoughby]

Someone already answered your exact question, but I wanted to add...

Yours is not the only circuit to mysteriously power up with unexpected inputs. Many MCU chips have similar clamping diodes on all GPIO pins, but that allows uncontrolled inputs to power the chip. Sometimes, the power is enough to turn the chip on, but not enough for everything to operate correctly; or perhaps enough that the chip never does a proper power-on reset. I've seen USB devices stay on when their power switch and main USB cable were disconnected, because an accessory cable feeding an A/D input (GPIO) had enough voltage to keep the chip going. Some of these MCU chips - like those from Microchip - can run on anything from 3 V to 5 V, but the clock speeds are reduced at 3 V along with certain other capabilities. It can be very frustrating to diagnose if you don't realize that the chip is half-way powering up.

Typically, the solution for this is to add something like 100 Ω in series with any input. That's not much resistance, but it's enough to drop 5 V to below 3 V with any more than 20 mA draw. Not all inputs will work with series resistance, so you have to adapt to the situation.

Brian


On May 28, 2021, at 09:57, ShedSynth wrote:
> I have an Arduino-based module, with a GATE input protected by 1K input resistor then clamping diodes to 0V & +5V.
> https://shedsynth.files.wordpress.com/2021/05/adsr-a-schematic-gate-to-d2.png
> I find that with power off to that module any GATE or CV to the input can pull up the module's +5V through the diode and even power the whole case's +5V rail.
> The clue was seeing the case’s +5V LED flashing in time with the GATE input.
>  
> And I think that if the input is more than +5V, the un-powered voltage regulator will not be able to regulate so digital components (and perhaps the voltage regulator itself) might be damaged.
> Have I understood this correctly?
>  
> I'm wondering whether to put a 5.1V zener across the module's 0V and +5V to catch any over-voltage.
> Have I understood THAT correctly?
>  
> Thanks in advance,
> Al