I am making a standalone battery powered arduino project which uses nRF24L01+ for radio communication. nRF draws 20 mA current max. Can I use 3xAA cells along with 2 diodes (which drop 0.6*2 = 1.2 V) to get 3.3 V source for my nRF module?

I will be keeping my microcontroller and nRF module in sleep mode most of the time. All it has to do is to wake up after a certain interval of time (maybe 2 to 5 seconds), communicate to a central radio server and go back to sleep.

  • 3
    \$\begingroup\$ Why not use a $1 low-drop 3.3V regulator? With a 0.5V drop-out your batteries will actually be allowed to deplete a little, in stead of at 1.4V per cell all kinds of stuff cutting out. \$\endgroup\$
    – Asmyldof
    Jun 17, 2015 at 18:37
  • 1
    \$\begingroup\$ Plus, diode-drops are not linear with respect to current and temperature. So "idle" current will produce a different drop, as will cold and hot temperatures. \$\endgroup\$
    – rdtsc
    Jun 17, 2015 at 18:41
  • \$\begingroup\$ @Asmyldof - Small pcb size was the first reason. However I also wanted to keep the cost low and I was worried about power wastage in 3v3 regulator. Can you also suggest me a boost converter which converts 3-4.5V to 10-12V with a current of about 600 mA max. \$\endgroup\$ Jun 17, 2015 at 18:51
  • \$\begingroup\$ @Asmyldof - Can I use 3v3 regulator with input connected to Arduino pin. That way when I make that pin high, my nRF will switched on and I can do all the operations. Once my job is done, I will pull the pin LOW and save on the quiscent current. What do you say? (arduino pin can output 40 mA current) \$\endgroup\$ Jun 17, 2015 at 19:05
  • \$\begingroup\$ Arduino's use Atmels, they can't source 40mA, even if they could, it wouldn't be at 5V. There's tiny regulators for 100mA max output that take only 0.1mA and/or have an enable input. Taiwan Semiconductor isn't expensive. For your boost converter you need to start a new question detailing what you want exactly \$\endgroup\$
    – Asmyldof
    Jun 17, 2015 at 19:29

1 Answer 1


enter image description here

According to the data sheet (https://www.sparkfun.com/datasheets/Wireless/Nordic/nRF24L01P_Product_Specification_1_0.pdf), it looks like you could simply use 2xAA batteries for 3V and it will be sufficient. No need to use the extra battery and waste it across diodes.

  • \$\begingroup\$ I need 4.5V for other reasons, specifically a hall sensor and in order to run my Atmega at 16 MHz. \$\endgroup\$ Jun 17, 2015 at 18:46
  • \$\begingroup\$ Ahh, in that case, Asmyldof's suggestion would probably prove best. \$\endgroup\$
    – TronicZomB
    Jun 17, 2015 at 18:47
  • \$\begingroup\$ Thanks for your time though. Anyways, can you suggest me a boost converter which can convert 4.5V to 10-12 V and 600mA - 1A. I need to run a DC motor. \$\endgroup\$ Jun 17, 2015 at 18:55
  • 2
    \$\begingroup\$ @Whiskeyjack Three AA alkaline cells cannot supply that much power (6-12W) for more that a few seconds, if at all. \$\endgroup\$ Jun 17, 2015 at 20:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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