From what i see that 3.3Vpin provide 3.3V and 400mA. The fingerprint module i use is AS608. Connecting the fingerprint module power pin to 3.3V of the nodemcu. From the datasheet it only can handle 3.3VDC, operating current<60mA and peak current at <70mA. Do i need to add a resistor to take the current away. Since last time i use another fingerprint module which is R307 connecting to Vin of the nodemcu and now it not working anymore?

  • 1
    400mA is the maximum the regulator can provide. It's not what the regulator does provide. You may have a car with a 1.4l engine. That doesn't mean the fuel tank is only 1.4l as well. The engine only takes what it needs from the fuel tank.
    – Majenko
    Jul 9, 2020 at 10:32
  • uhmm so it mean if i directly plug the fingeprint power pin to 3.3v it will just draw the current it need and wont burn? Jul 9, 2020 at 10:38
  • 2
    Correct. The 400mA basically means that "the sum of all the peak current requirements of all devices connected to 3.3V must not exceed 400mA".
    – Majenko
    Jul 9, 2020 at 10:55

2 Answers 2


Be careful using Vin on the NodeMCU for 3.3V devices! If you are powering it by 5V USB power, this Vin pin will probably have something like 4.76V on that pin. The circuit diagram shows that Vusb is connected to Vin with just a schottky diode, which drops the current about .34 volts and provides some noise isolation from the Vusb supply. If you connected a 3.3V device to this pin at 4.76V, you may have damaged your component.

You'd have to check the datasheet or documentation for the fingerprint module, but it sounds like the fingerprint module will only draw 60-70mA when it's given 3.3 volts. Datasheets often provide application examples and may specify that you do need additional resistors, capacitors, etc. The refrain in cases like this is always read the datasheet.

If your NodeMCU datasheet says it can provide 400mA then it should certainly be able to provide the 60-70mA that the fingerprint sensor needs. I've powered an ESP8266 by connecting its Vin pin to the Vin of another NodeMCU and they seem fairly happy together, although there tends to be quite a bit of noise on the Vin pin.

  • The question clearly mentioned the its 3.3v being connected, this answer is is not wrong, but is unsuitable
    – Coder9390
    Aug 4, 2021 at 0:59
  • Dont want to be bad, but Vin is not a 3V3 output pin. as such the schotky is there just to protect the USB not to drop the voltage.The 3.3 output pis is often marked as 3V
    – Tomas
    Jul 29, 2022 at 23:51
  • You should not use Vin for taking power out
    – Tomas
    Jul 30, 2022 at 0:14

Dont mix your pins. and use 3V pin (NOT Vin), that should be able to output it. AMS1117 used in most nodeMCU's I have seen can reasonably provide 800mA medium lenght periods with voltage drop of 8-9 Volts. The MCU can reasonably take 100mA-350mA. This varies based on the amount of the work it does, running core frequency (I am not advanced enough to mess with that yet), Whether its communicating over wifi and how strong does it have to make the signal (influenced by distence, other wifi devices around). This leaves you with about 450mA to play about. As designers were responsible a good designers 50mA was cut as a margin.

Because of how voltage regulators work, the more voltage you are droping the lover current you can sustain. (worth researching heat decipation) Next is the matter of wether the device is enclosed etc. So 60mA is OK for most nodeMCU as extra load. If you reach few hundred mA and are really concerned you can reasonably control consumption simply by controling whether the nodeMCU is sending any majour communication over or if it does encryption (demanding tasks) etc.

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