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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.

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    \$\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
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    \$\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

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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.

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  • \$\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
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    \$\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

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