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I'm trying to build a battery operated hand lantern. It has the Arduino, an MPU6050 accelerometer, and 4 APA106 (WS2812B-like) LEDs.

If everything's powered through the Arduino's 5V pins, that's 80mA for the Arduino, 240mA for the 4 LEDs, and let's round up to 5 mA for the MPU6050.

If I'm using 5x AA batteries for 7.5V (at full charge) that should be 2.5V * 325mA of power wasted or just under a watt that'll be heating up the regulator? I'm not sure if that's huge or tiny or in between or what.
What's the most sane way to power this device for an absolute newbie?

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Use en external voltage regulator (go switch-mode if you care about battery life) that is adequately cooled.

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  • "Switch mode"? What would I be regulating the voltage down to, since it still has to be fed through the Arduino's linear regulator? – Different55 Jul 5 '19 at 11:33
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    @Different55 Switchmode voltage regulators rapidly switch on and off the current. Together with a capacitor they can provide a stable voltage with a lot less power lost. Regulate to 5V or slightly above that. The arduino can handle a slight undervolt. – ratchet freak Jul 5 '19 at 11:38
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    @Different55 I dare to assume that the answer suggest feeding 5V from the regulator directly to the 5V pin of Arduino, skipping its weak and inefficient regulator altogether. – Dmitry Grigoryev Jul 5 '19 at 12:32
  • Let me know if this is completely stupid, but as I was thinking about regulators and batteries I remembered I have a portable USB power bank. Doesn't the Arduino's USB input also skip the regulator? – Different55 Jul 5 '19 at 15:58
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1W is near the limit of what a small LDO can handle. It will get considerably hot.

I would use 4 batteries instead of 5, and a 3.3V Arduino clone instead of the UNO, and power the LEDs directly from the batteries. You could also use your original UNO, feeding it with 6V on a 5V pin through a simple diode, which would drop it to something like 5.4..5.3V.

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  • How would powering the LEDs from the battery work? They seem pretty demanding of a steady 5v of power, and at least at a full charge 4 of them would be 6V. – Different55 Jul 5 '19 at 11:27
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    @Different55 instead of the positive coming from VCC it comes straight from the + of the battery. The negative of the battery and LED both get connected to arduino ground. – ratchet freak Jul 5 '19 at 11:40
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    @Different55 APA106 have a working range of 4.5 to 6V. – Dmitry Grigoryev Jul 5 '19 at 12:28
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I suggest using a DC-DC buck/boost voltage regulator. Those can take a wide range of input voltages and put out an adjustable output voltage

This one on Amazon, for example, can take an input of 3-40V and provide an output of 1.5-35V(Adjustable). It is a switching power supply, so it should not produce much waste heat. That one is rated at 2A, or 3A if you add a heat-sink.

You could feed the regulated 5V into the USB connector on your Arduino, and to your other components as well. (Just take a USB cable, cut it, and tie the red and black wires to ground and +5V, respectively.

You might need a filtering capacitor to protect the Arduino from voltage variations as the load on the power supply changes.

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  • Is that a special type of capacitor or just a normal one used for that purpose? Will any capacitor work? Does it just store energy and spit it out to smooth out tiny dips? – Different55 Jul 6 '19 at 20:42
  • You want a decent sized electrolytic, between 5V and ground physically near the input to the Arduino. I'm not an EE - I'd have to do some digging to figure out what rating of capacitor is appropriate. Yes, it acts as a filter, and resists changes in voltage. – Duncan C Jul 6 '19 at 20:48

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