I have a 5V Arduino Pro Mini that I want to connect with this 3.3V nRF24L01+ module. (On the other end I have the same nRF24L01+ module connected to an Arduino Uno which has a 3.3V pin that I know is safe to use.)

The nRF24L01+ modules I have are these.

The Arduino board this question relates to is this one.

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    \$\begingroup\$ The short solution is to power the RF module with two AA batteries. Long term, I'm going for 3.3V Arduinos instead of 5V. \$\endgroup\$ Commented Jan 8, 2013 at 21:06

2 Answers 2


You could use a 3.3V voltage regulator. See DigiKey parts here. It looks like all of them accept a 5V input and I set a 20mA output minimum (the max supply consumption of your module is 13.5mA)

  • \$\begingroup\$ Thanks, this I see as a proper solution. In the end I opted for 3.3V Arduinos instead. \$\endgroup\$ Commented Jan 8, 2013 at 21:07
  • \$\begingroup\$ Is there a reason you couldn't just add a resistor to bring down the voltage? \$\endgroup\$
    – DominicM
    Commented Jan 18, 2014 at 22:45
  • \$\begingroup\$ @DominicM You can't just add a resistor to bring down the voltage because the current consumption is not constant - it goes from almost 0 (in which case the voltage drop over the resistor would be 0 as V=IR). A red LED plus a ballast resistor to ensure 1mA minimum current might work, as the LEDs are typically specified as 1.8V drop and 20mA current. \$\endgroup\$ Commented Apr 14, 2016 at 17:55

I've supplied my 3.3V GPS module from a 5V ATMega328 by using two 1N4001 diodes in series. It was not an optimal solution at 3.0V, but it was cheap and I had the diodes laying around.

The same solution may possibly work as a temporary hack if you need something immediately and don't want to wait on an order from a supplier.

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    \$\begingroup\$ At 10 to 20ma max, the drop should be ~0.72v on each, so it should give 3.5~3.6v. Which is perfect for OP. \$\endgroup\$
    – Passerby
    Commented Jan 8, 2013 at 1:49
  • \$\begingroup\$ Nice quick solution. I opted for 2 AA batteries though. \$\endgroup\$ Commented Jan 8, 2013 at 21:08

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