I understand that to get low power I need to cut the link between the two 3.3V pads to isolate the DC/DC 5V->3.3 V circuit. If I do this, do I have to reconnect the pads each time I want to reprogram a sketch? Or can I load a new sketch through the USB provided I also provide 3.3V power?
Is there an article somewhere that summarises how to get the lowest possible power in idle? I want only a few uA if possible.
cosine32:
I understand that to get low power I need to cut the link between the two 3.3V pads to isolate the DC/DC 5V->3.3 V circuit.
That is correct.
cosine32:
If I do this, do I have to reconnect the pads each time I want to reprogram a sketch? Or can I load a new sketch through the USB provided I also provide 3.3V power?
If you provide power to the 3.3V pin you should be able to reprogram the Arduino.
The nRF has a separate power pin to power the USB peripheral. This pin is connected to the USB 5V directly. I could not find any information in the datasheets regarding any requirements to supply power in a specific order.
I would connect the 3.3V first and then connect to USB. This seems to be more common way a user would use a USB system and therefore the expected power up sequence.
However because we do not know what the Arduino software (bootloader or your sketch) does when it starts up and does not find the USB connected you might need to do it the other way around or press reset when you connected everything.
cosine32:
Is there an article somewhere that summarises how to get the lowest possible power in idle? I want only a few uA if possible.
There are a few other posts in this forum about this topic.
https://forum.arduino.cc/index.php?topic=691488.0
https://forum.arduino.cc/index.php?topic=689256.0
https://forum.arduino.cc/index.php?topic=665272
Thanks for the links. All very helpful but I am disappointed that I cannot get down below 0.9mA without entering system off state. I have tried using delay() inside loop() as recommended with this board, but it makes no difference. A blog entry here claims a blink example is using 4.5uA.
How have they done that? If I take the basic Blink code, add the power saving tips, including cutting the 3.3V jumper I am down to only 1.7mA when the LED is off.
Are they using a sleep mode and a watchdog timer? Or system off with an external RTC? I can only get down to around 5 uA in system off mode.
I understand that with the link between the DC/DC converter and 3.3 Volt still intact the current drawn is always of the order of 10mA minimum.
I do not understand this.
The converter is a MPM3610 which has a quiescent current of 200uA.
And the schematic uses the diagram directly out the MPM3610 datasheet.
So what is wrong here?
Thanks in advance, Sietse
To answer my own question.
The nano 33 ble does not draw that much power when in rest, but it critically depends on the actual voltage at Vin!
For example, using the example that puts the cpu into sleep using NRF_POWER->SYSTEMOFF=1, the system will draw
20 mA when Vin is 6 Volt, but
0.25 mA when Vin is 10 Volt or higher!
The last number is in line with the spec of the DC/DC converter which states 0.2 mA.
As that spec shows, the efficiency of the conversion is lower at lower input voltages, only it is a bit more extreme than I would have thought.
In my use case the system, which uses a supply voltage of 10 to 14 Volt with 2100mAh, it could run for almost 3.5 months, which is sufficient for me.
So cutting the 3.3 Volt line is not needed then.
Sietse.