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I'm trying to make a battery powered arduino, so I bought a pro mini, that can be powered by 3.3V. I'll be using rechargeable batteries and found that AA Ni-MH are the most efficient / cost-effective / easy-to-use. All of my research made me realize that a voltage upscale is not that easy (if you prove me wrong on this, I'll thank you eternally), so I'll be using 3 of them. That makes a theoretical 3.6V with a range of 3.24V when depleted and 3.96V when fully charged.

I want the voltage output to be constant at about 3.3V, so my best results lead me to this voltage regulator. I understand that it will reduce the voltage to 3.3V and the rest of the input will be wasted as heat, so my 3 battery plan had the least energy waste.

The problem became obvious when the batteries dropped to just 3.8V: the voltage regulator dropped its output to 3.0V.

Now, keeping in mind that the power requirements will differ by the time, since there will be two tiny motors that won't be running constantly, is there any way for me to provide a constant 3.3V from the least amount of batteries and without spending a fortune on the regulator (that is up to 3€, else I just add one more battery)?

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    \$\begingroup\$ Doesn't the pro mini already have an LDO onboard? \$\endgroup\$ Dec 15, 2017 at 10:00
  • \$\begingroup\$ NOTE: Vin can be between 7 and 12 V. \$\endgroup\$ Dec 15, 2017 at 10:16
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    \$\begingroup\$ @HarrySvensson makes a good point; why not use the MIC5205 3V3 LDO voltage regulator that's on the Arduino board? That LDO has a maximum dropout voltage of 275 mV at 150 mA at 25C; so (at room-ish temperature) you need to ensure the battery's voltage does not drop below 3.3 + 0.275 = 3.575 VDC. IOW, if you add another battery to your battery pack, you can use the MIC5205 LDO regulator that's already on the Arduino board. \$\endgroup\$ Dec 15, 2017 at 10:16
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    \$\begingroup\$ Note that you do not have to supply a very constant voltage to the Arduino. The ATMega MCU on the Arduino can run on 1.8 V to 5.5 V (and that is even guaranteed if you keep the clock at 4 MHz). So you can run the Arduin directly from 2 NiMh cells in series, without an LDO. Since you have motors I would add some extra supply decoupling and maybe a series resistor in the supply of the Arduino. This will save you from the power consumption of the LDO. \$\endgroup\$ Dec 15, 2017 at 10:44
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    \$\begingroup\$ Do note that not all Arduino boards can run at 1.8 V supply by default, their "brown out" detection is set to reset the chip below 2.7 V but this can be re-programmed or disabled using the fuses. I have an Arduino running on 2 AA cells in series with the brown-out disabled and this works fine. No LDO is used, I even removed it from the PCB to save power. \$\endgroup\$ Dec 15, 2017 at 10:48

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Unfortunately an LD33V (real name LD1117V33) is a lowish dropout regulator, not a low dropout regulator. It's advertised as being better than a 7805, which it is, 1v dropout plays 3v, but the 7805 bar is pretty easy to get under. You can get down to 150mV dropout if you look, certainly beat 1v. Go onto CPC, RS or Mouser, and do a parametric search. MIC5205, ADP1752, BU33TD3, the list isn't endless, but goes on and on.

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  • \$\begingroup\$ Thanks for you answer Neil_UK! datasheets are a bit confusing to me though (they say stuff like operating voltage -0.3V to 6.5V). Can you help me on reading the dropout voltage? \$\endgroup\$
    – Dimitris K
    Dec 15, 2017 at 10:09
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    \$\begingroup\$ a '-0.3 to 6.5' spec is not an operating voltage, it's an absolute maximum beyond which damage could occur. Dropout voltage is usually called 'dropout voltage', but it will vary with supplied current. Have you read the comments to your OP, that you don't actually need 3.3v for the arduino itself, it has a very wide operating range, so 2 or 3 NiMH will power it directly. You only need a tight voltage range if you're powering other ICs that need (say) 3.3v+/- 10%. Post a link to a datasheet and I'll try to help. \$\endgroup\$
    – Neil_UK
    Dec 15, 2017 at 10:47

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