hello
in this post
stated that maximun voltage for Vdd pin can be 3.7v according ASR6501 datasheet:
or 3.5v according Heltec info.
I would like to clarify this - is it possible to power CubeCell AB01/02 via VDD from fully charged LiFePo4 battery which is 3.65v?
Any issues - with LoRa or something else?
Yup, you didn’t read the whole post:
And as you are using the 3.3V pin and the maximum is 3.5V …
But the ASR6501 has a maximum rated voltage of 3.9V but recommends 3.7V.
If you try to charge the battery, you may well exceed the VDD spec. Perhaps use the board as it was designed for a LiPo on the Vbat input. Or create a separate battery / charger circuit with an intervening LDO or buck convertor to provide the 3.3V
Reality is that it will likely work, but with a reduce life as it’s being stressed above the suggested 3.3V. Unless it lets out the magic smoke, which this month is a nice pinkish hue, then the board will be dead.
YMMV
ok got it…
Thank you for explanation.
in this case board consume too much power during sleep mode - 31uA… and Im looking for way to decrease it…
btw - its interesting situation - we have ability to power board via Vdd 3.3v but this voltage should be stable and it means that you should use some LDO which will consume some power as well, which means that you still will lost some power ))
31uA isn’t amazing but it’s not that bad. But I’m wondering how you think changing the battery will reduce the consumption of the board.
It’s called physics, best to learn to live with it before gravity bites you in the ar5e!
Various not cheap but not expensive buck convertors will get close to 95% efficiency but overall, unless you invent a perpetual motion machine or get a fusion reactor working, this is where all the rest of us are at.
With a LiPo you can plug a solar panel on the device and if it’s the right size, it will compensate for all your losses.
LiFePo4 is 3.65 - so its might be fit for Vdd, LiPo is 4.25 - and this voltage doesnt fit at all, so I thought to change battery and check.
but yes - powering from Vdd not the variant - there is another drawback - to check battery level you would need separate voltage divider and GPIO to control it which is also will consume some power.
I just wondering how it is possible to get ~3uA here:
I know its about AB02 but still…
That’s not how electricity works. The board consumes power which is defined as Watts = I (current) x V (voltage). If you reduce the voltage, the current will go up and vice-versa. You have to change the sources of power consumption.
Perhaps in the linked thread ask@UniquePete for the code. Usually it’s about setting the IO in to the best mode possible for its design - sometimes as an analog input - but with consideration to what other items are attached to each pin on the board.
