I'm working on a project using a nodeMCU using a ICR14500(3.7V and 700mAh) battery and I'm experienced some issues when I try to estimate the total current consumption. Firstly I experienced some problems with the power supply (here is the previous question NodeMCU powered with battery and use WiFi) but this problem is now solved.

At this moment if I measure the current consumption connecting the amperimeter in serie between the battery and the nodeMCU input voltage pin I see a current consumption of 140mA continiously (ignoring the deep sleep mode) and I experienced some strange behaviours in my NodeMCU, even the code doesn't work. If I remove the amperimeter from my set up all works smooth and my source code works perfectly.

In order to validate my amperimeter I modified a usb cable and I added the amperimeter in the middle of the wire and I measured the current consumption powering me nodeMCU from the computer and I saw a current consumption of 80mA when the code is run and 14mA when the device is in deep sleep. This behaviour seems the expected one.

Are there any colateral effect measuring the battery current consumption which I haven't considered?

  • Make sure the battery is fully charged. That might be enough, so the burden-voltage described by Majenko isn't a problem.
    – Gerben
    Commented Apr 28, 2017 at 14:06

2 Answers 2


An ammeter is basically a small resistor in series over which the voltage drop is measured. That resistance will play a part in the effective output impedance of your power source (battery).

This increased impedance will reduce the overall amount of current the battery can provide before the voltage drops to an unacceptable level.

For instance, I just did a crude measurement of the resistor used for the 200mA range of one of my DMMs and came up with a value of 1.2Ω. At 140mA that would give a voltage drop of (V=IR = 0.14 × 1.2) 0.168V. That's 0.168V that your NodeMCU won't be seeing. Add to that the fact that the WiFi will be creating peaks of current usage up to, say, 500mA (probably less, that's just a wild guestimate), the voltage drop will increase to (0.5 × 1.2) 0.6V. From a battery giving 3.7V nominal that would plummet to 3.1V as seen by the NodeMCU.

Depending on what kind of power regulation you are using that will most likely cause brown-outs of the power and the MCU will reset or crash.

When powered by 5V that 0.6V drop would not be as noticeable, since 4.1V is still well above what is needed to power the chip and the voltage regulation can still work properly.

So how can you measure the current with the battery? Well, there's two options:

  1. Use a much smaller resistance (0.1Ω, for example) and measure the voltage across it - then use Ohm's Law to calculate the current.
  2. Add a large electrolytic capacitor to the input of the NodeMCU after your current measurement to act as an energy source for sustaining the peaks of current required by the WiFi. By big I mean something like 3300µF:


simulate this circuit – Schematic created using CircuitLab

  • A very good explanation, I hadn't experienced issues due to the ammeter internal impedance. The idea of the filter using the capacitor is very good. Thanks!!
    – rdiaz82
    Commented Apr 30, 2017 at 19:19

The NodeMCU has an on-board voltage regulator, that needs excess voltage to supply the 3.3V. So if the voltage for the regulator is too low, the voltage for the ESP8266 may become too low too or it can’t draw enough current as it needs. (During boot or wifi operation it’s drawing up to 200mA peak current). If that happens, it’ll be stuck in the boot process, in which it’s drawing the 140mA.

So, when you’re running from battery, be sure to power the 3.3V supply of the board, not the 5V input that goes through the regulator first. When your battery is fully charged it’ll give 4.2V, which may just be enough for the regulator still. With the ampmeter in between it’ll definitely be too low.

Note that with the NodeMCU board you’ll never get the advertised 20μA deep sleep current of the ESP due to the onboard voltage regulator (unless you modify the board).

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