nRF52832 GPIO state at no input power (Vdd)

[Brosnan]

Hello.

I wanted to know what the gpio state when no power is connected to the device.

I have connected a RTC battery to one of the ADC channels via a series 100K resistor, for monitoring the cell voltage in operation. But I noticed when the device is kept off for a long duration the cell voltage drops significantly. I think the gpio is sinking current in power off state. Is this possible?

Thanks

Brosnan

[ketiljo]

There are ESD protection diodes to VDD that will start to conduct and drain the battery trough the resistor. So the GPIO should be considered pulled low when the device is not powered.

[Brosnan]

Thanks for your response Ketiljo,

What should I do to prevent this from happening. I want to monitor the battery voltage when the board is powered. And not drain when the board is unpowered.

I have increased the resistor value to 10M ohms, but was thinking if this will degrade the accuracy of the measurement.

[ketiljo]

The idea is the same, to use a large resistor and charge the capacitor

[Brosnan]

Hi Ketiljo,

Thanks for your help on the original problem. I have a new problem now related to this one. The coin cell I connect to the to the RTC and the ADC (via 10M ohms) is still seeming to be draining fast when the device is not powered.

I put in a fresh cell at 3.28V in a device and within 4 days it had dropped to 3.17V while the device was never used. Have noticed this on 6 devices up till now who all were put with fresh batteries at the same time and they depleted around the same voltage after 4 days.

The cell type is a CR1220 with nominal capacity of 36mAh.

The cell seems to be draining via the ADC path as I have other devices which don't have the ADC connection at all and there the cell voltage is still 3.25V after several months of the board staying off.

What can be happening here?

[ketiljo]

That was strange. With a 10M ohm resistor, even if the pin is shorted to ground, you will have a current of ~900 nA. After 4 days, it has used ~29 µAh of the battery capacity. Are you sure the resistor is 10M ohm and not 10k ohm. 10k ohm will be about right after 4 days. 

[Brosnan]

Yes double checked the value, it is 10M. By the way how did you get 900uA through the 10M if it was shorted. The battery voltage is ~3.2V, hence I was considering 3.2/10000000 = 320nA.

[ketiljo]

Sorry, a typo, it should of course be ~300 nA. 

[Brosnan]

Do you think of anything else happening here?

[ketiljo]

There must be a considerable current draw to drain the battery this fast, and if the resistor is indeed 10M ohm, the current must go somewhere else. 

[Brosnan]

Is there a possibility that the cell discharge characteristics are into play here? Like the battery will discharge from 3.28V to 3.0V faster and then stay at 3.0V for most of it capacity?

[ketiljo]

Does the battery discharge at all after a while?

[Brosnan]

So what is noticed now is the battery discharges quickly to 3.0V from 3.2V and then stays at that voltage. Observing this on all units (after 1+month) with the 10M ohms resistor in series.