Powering the arduino Pro Mini 3.3v from 3.7v 18650 battery to vcc pin

Can i power the arduino Pro Mini 3.3v from 3.7v 18650 battery connected to vcc pin?
In this way the current goes directly to the processor without going through the regulator.
or does the 3.3 volt have to be more exact?
I need to minimize the power consumption, because of that a want to:

    -remove the power regulator and leds, due to the highest consumption
    -Use LowPower.h library to sleep the arduino between reading

I appreciate your help.

Rather depends on what your connecting the Pro Mini to.

The so called 3.7v battery is probably a Lithium Polymer type, so it will be 4.2v when charged.

Are all the devices\components your connecting to the Pro Mini able to withstand a 4.2V supply ?

Thanks for the reply Smet
Yes, all the sensor can support this voltage, but my concern is if the microprocessor is going to support it ..

The controller works up to 5.5V, so 4.2V is within the specs. More delicate may be the capacitors on the board, see the circuit diagram and add some safety percents for clones.

I did exactly that with a Pro Mini, and it worked fine. As srnet says, the only potential problem is with sensors or other modules that require 3.3V, and won't work at 4.2V. If you don't have any of those, it should work fine. I removed the regulator and LED, and using a pin change interrupt to awaken the device, I was able to get sleep current down to under 1µA. It's an IR remote powered by an 18650, and I didn't bother to put in an on/off switch. If your wakeup is timer based, it will use more sleep current because you have to leave an oscillator running.

There is no "3.3 V version" of the microprocessor; it operates up to 5.5 V. It's just that you have to use an 8 MHz crystal for 3.3 V to work rather than a 16 MHz one. Of course the 8 MHz version will work perfectly well at 5 V.

It is absurdly improbable that the capacitors on the board would be rated for less than 6 V. The only concern is whether the regulator would be harmed by back-feeding more than 3.3 V to its output. What may happen is that it will draw more quiescent current and in fact if you propose to operate it from a lithium battery you certainly should remove or disable the regulator.

For battery standby operation you also want to remove the "pilot" LED or its resistor. :grinning:

You will need to check the board that you have, but at least some of the Pro Mini's made by SparkFun have a jumper you can cut to isolate the voltage regulator and power LED from VCC without having to remove any components.

Thank you guys for the replies!
I have another question related to this.
Now i want to connect a 2 or 3 soil moisture sensor (soilwatch 10- pinotech) to this device (promini 3,3v 8mhz without regulator and led, powered by one 18650 battery).
I do it some test and a have weird values from de sensor. The manufacturer recommends me to connect the sensor through a transistor, not directly to the pin, because the sensor consumes 24mA and the voltage drop is too much for its normal operation. what do you think?
Does this mean that I must connect the sensor to the battery through the transistor and drive it with the pin?
He also tells me that the voltage has to be stable at the sensor input for better precision. but I removed the regulator ... will I be able to do this without regulator?
Or maybe I should put 2 batteries in series and keep the regulator?
I did this to minimize consumption.

No problem if the sensor is ratiometric (depends strictly on Vcc) because the Arduino ADC also depends on Vcc (default reference voltage). Else you need a stabilized voltage for both the sensor and ADC reference.

Hi DrDiettrich!
That means that i always need a power regulator? that is bad or power consumption...

Depends on what version of the soil sensor you have.

No problem if you have the 1.1volt version.
Then you can switch to 1.1volt Aref in setup().
Leo..

jbollecich:
I do it some test and a have weird values from de sensor. The manufacturer recommends me to connect the sensor through a transistor, not directly to the pin, because the sensor consumes 24mA and the voltage drop is too much for its normal operation. what do you think?
Does this mean that I must connect the sensor to the battery through the transistor and drive it with the pin?

Are you trying to power the sensor from an output pin of the arduino? If you want to power down the sensors while the arduino is sleeping then you will need to make sure the transistor stays in the OFF state when the arduino cannot actively drive its output pin.

The output pins should not change state during power down - Vcc is still present on the controller and the output stage is MOSFET not bipolar.