I am trying to make a emergency button with Arduino Nano. it should work with and without input voltage. So that I made this circuit.

enter image description here

if there is a input voltage 5 volt NANO should directly suppling from input voltage. Meantime battery can charge itself. but if electricity has gone Arduino Nano should suppling from li-ion battery. Not long time 30 minute is enough for me. So that I have added a diode.

well if I look sch. of Nano it looks 5v is directly connected to Atmega328p chip.

Part of Arduino Nano schematics

and at docs of Atmege328p it says operating voltage 2.7-5.5 Volt. I can achieve 5 volt while input voltage and 3.6 volt while if only li-ion battery

So at this point everything looks fine but at Arduino Nano products page there is NOTE!

If supplied with less than 7V, however, the 5V pin may supply less than five volts and the board may become unstable.

and this

Supplying voltage via the 5V or 3.3V pins bypasses the regulator, and can damage your board. We don't advise it.

In this case, I wasn't sure myself. will it be a problem if i feed Arduino Nano in this way? I have made a prototype and working almost a hour with blink example and it look working fine. if I plug in 5 volt or plug out input voltage. system not stopping.

  • 1
    \$\begingroup\$ Aside from the main topic of question, as much as I've searched, that particular battery in the picture (forte 14500 lithium battery) is no a lithium-ion battery thus, not rechargeable. so watch out for them. \$\endgroup\$ Dec 31, 2021 at 22:09
  • \$\begingroup\$ Nothing in Google says that the Li-SOCL2 Battery is rechargeable. \$\endgroup\$
    – Audioguru
    Jan 1 at 3:20
  • \$\begingroup\$ @TirdadSadriNejad yes you are right image is wrong. i will change \$\endgroup\$
    – mehmet
    Jan 1 at 7:51

1 Answer 1


It won't be a problem as long you don't feed more than 5V through that pin. They don't advise using it to power just because Arduino are supposed to be powered though it's on-board voltage regulator. But as you can see, USB +5V power are connected directly there too. So it should be safe.

Please note that MCU maximum possible MHz are dependent on supply voltage. You can see this in "Electrical characteristics" part of the datasheet:

Vcc vs MHz slope

As Arduino Nano runs from 16 Mhz crystal you should be fine while you battery supplies 3.6V. Li-Ion batteries are supposed to work in the 3-4.2V range. But as long as the battery discharges and it's voltages drops you can become out of that "Safe Operating Area". Honestly, just from personal experience, ATmega328P will still work at 16Mhz down to 3V. But it's not guaranteed by the vendor. You have several options here - use a step-up DC-DC or re-programm ATmega fuses to enable Brown Out Detection or switch to internal 8 MHz RC, or change the on-board crystal to lower value.

PS: Also as others noted in comments above, make sure you pick a rechargeable Li-Ion battery compatible with that TP4056 charger board. The one you have on your diagram aren't a rechargeable one. Common rechargeable Li-Ion chemistry are rated for 3.7V nominal.

  • \$\begingroup\$ thanks for info, you said "But as long as the battery discharges and it's voltages drops you can become out of that "Safe Operating Area". but is tp4056 overdischare is not enough. this documents says "Over-Discharge Detection Voltagw 2.4 V ±100 mV". addicore.com/TP4056-Charger-and-Protection-Module-p/ad310.htm \$\endgroup\$
    – mehmet
    Jan 1 at 8:06
  • \$\begingroup\$ @mehmet I wasn't talking about charging battery. See the graph in my answer. \$\endgroup\$
    – NStorm
    Jan 1 at 8:13
  • \$\begingroup\$ yea yea I understand, you say that if battary goes lower than 1.8 V (from graph), it makes unsafe area. But DW01A has a overdischarge protection at 2.4V. So never battery enter the unsafe area. i mean \$\endgroup\$
    – mehmet
    Jan 1 at 8:24
  • \$\begingroup\$ @mehmet note the Y axis too. Your Arduino runs @ 16Mhz meaning safe operating voltage are roughly ends down between 4.5 & 2.7V (20 Mhz & 10 Mhz safe voltage). It's linear and it would be ~3.8V for 16 MHz. \$\endgroup\$
    – NStorm
    Jan 1 at 8:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.