I purchased 10PCS Socket Adapter plate Board For NRF24L01+ Wireless Transceive​ module and would like to know how much current/power it will draw as I want to run a project on batteries.

A close-up of the unit is below.

enter image description here

Using a multimeter, I measured the following current from 7 of these units (powered from 5.2V source):

4.67, 3.85, 4.70, 4.58, 3.93, 3.78, 3.80mA.

I scratched the power LED off another unit (LED is between GND pin and IRQ pin), and measured 3.28mA.

My question is, is the current I can see (minus the LED current) the quiescent current of the 3.3V voltage regulator? Also, there is no nRF module plugged in to the socket; does the regulator need a load before I can measure its power usage?

I think this is the regulator - Advanced Monolithic Systems AMS1117 1A LOW DROPOUT VOLTAGE REGULATOR, and there's a row in the datasheet with these values:

enter image description here


  • \$\begingroup\$ Sounds reasonable for the (unloaded) regulator quiescent current. But that value will likely increase slightly as the load is added. Add the load, measure the load usage, then the total usage, and subtract to find the regulator-only value. \$\endgroup\$
    – rdtsc
    Apr 16, 2016 at 14:56

1 Answer 1


To measure the quiescent current of your regulator, you would need to measure the input current and the output current (as the input current will be the sum of load current and quiescent current for a linear regulator).

If you take a close look at the regulator datasheet

Min load current

You will note that the minimum load current is specified for the adjustable version only; the true quiescent current is the adjust pin current for this device:

Adjust pin current

This is listed for the adjustable version only (so the designer knows how much current will need to be drawn from the regulator in the case the load may not take the listed minimum).

The conclusion here is that the fixed version (which is what is on your board, AMS1117-3.3) has been provided with a minimum load internally. That load is very likely the current in the voltage set resistors and probably some other physical load provided internally to the devie package.

The adjustable versions list the quiescent current as the minimum load current, but for guaranteed regulation, there may need to be another 5mA of load.

The datasheet shows the typical quiescent current at 5mA, but note 5 states:

Note 5: Minimum load current is defined as the minimum output current required to maintain regulation. When 1.5V \$\le\$ (VIN - VOUT) \$\le\$ 12V the device is guaranteed to regulate if the output current is greater than 10mA even though this note only applies to the adjustable version (the devices will all be identical except for the adjustment resistors and loading).

  • \$\begingroup\$ Thanks, how do you know the AMS1117-3.3 is the adjustable version? Doesn't the "3.3" mean it's fixed? \$\endgroup\$
    – Paul Grime
    Apr 16, 2016 at 15:18
  • \$\begingroup\$ Thanks - You are correct, and I meant to say fixed, and I have corrected that in the answer. \$\endgroup\$ Apr 16, 2016 at 15:19
  • \$\begingroup\$ Thanks. And when you say "has been provided with a minimum load internally", do you mean that load is inside the regulator IC itself, or that load is one (or more) of the other components on the board? \$\endgroup\$
    – Paul Grime
    Apr 16, 2016 at 15:21
  • \$\begingroup\$ The implication is that there is a load provided internally to the IC package. \$\endgroup\$ Apr 16, 2016 at 15:21
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    \$\begingroup\$ I would not be surprised if the manufacturer happened to have several thousand of these very cheap (and abundantly available) devices on hand. Cost is a major driver for something like this. If I were doing a battery powered unit for extended operational life, I would not personally use this regulator. \$\endgroup\$ Apr 16, 2016 at 15:28

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