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I need to power a miniature LED light show that will draw a few hundred mA, or possibly more, and requires less than 5 Vin, and a microcontroller board that wants 7-21 Vin.

I was thinking the easiest solution would be to use 3 AA alkaline cells in series, to provide 4.5 V, which is fine for the light show, and then 3 cells in series with that to provide 9 Volts, which is good for the microcontroller board, which will only draw 20-30 mA. Then I wouldn't need to worry about a voltage regulator, and wouldn't have to do worry about heat dissipation inside the electronics enclosure, I've heard that those Voltage regulators dissipate heat under heavier loads.

So, the two systems will have a common ground connector, and two different V+ connectors.

But many places I've heard that it's bad to combine more depleted batteries with less depleted batteries, and since the first three cells will have more current draw than the next three, that will eventually happen. But I don't understand it completely, because surely not all batteries will be exactly equal in terms of capacity, so sooner or later one battery might get discharged while the others still have charge, even for conventional setups?

So will the above setup work, and is it OK practice? Or am I being paranoid regarding the voltage regulator solution? The thing is that I'm not completely sure about the power draw of the light show, because it will also depend on the contents of the show itself. I only know it's within acceptable limits for alkaline AA cells.

I see a similar question has been asked, but it focused on rechargeable batteries, and the answers focused on problems related to charging, while my question is specific to alkaline cells that are not going to be recharged.

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    \$\begingroup\$ Any microcontroller board that accepts 7Vin already has a regulator on it. You could look and see if you can bypass that regulator and operate the processor on 4.5V. \$\endgroup\$ – JRE Apr 15 at 14:10
  • \$\begingroup\$ Thank you. The board is an Arduino Nano Every. I looked at Arduino message forums, and see that this is possible but not recommended. Maybe someone with more experience than me could make it work, but I'll pass. \$\endgroup\$ – Balthazar Apr 15 at 14:29
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Can you do this? Yes.
Is it done often? Yes.
Should you? Probably not unless you understand how it behaves.

As the power draw drains the lower half of the battery pack, there will be voltage issues for the microcontroller, possibly reverse voltage into the drained batteries, and battery acid leaking. Not horrible failure with AA batteries but messy and annoying.

Depending on the current required, you could use the Arduino nano every's regulator. It has a compact single ic switching regulator MPM3610 with a max output of 1.2A at 90% efficiency. Not saying that you can or should pull 1.2A, but you said a few hundred mA. Especially with a low input voltage. The efficiency penalty is very low. Arduino says 7V minimum but its probably down to 6V minimum for proper 5V regulation so a 5x AA would power it just fine.

For longer run time just throw 6x AA at the arduino and power the leds from the regulator. The power cost vs runtime for the higher voltage is in your benefit.

Alternatively you can run the arduino at lower voltages. At 4.5V you skip the regulator and power directly. As voltage drops the max internal crystal and cpu speed are affected but you can switch it to 8 mhz via a setting in the arduino software, problem solved.

Or you can just throw an extra 1 or 2 batteries at it. 5x AA for the microcontroller and 3x for the LEDs.

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  • \$\begingroup\$ Thank you. Actually, considering the low price of 9V batteries, I could just use a 9V for the Arduino, since it won't draw much current. My concerns with the power regulator was whether I had to be concerned about heat. So I looked up the MPM3610, and found this data sheet. monolithicpower.com/en/documentview/productdocument/index/… ... to be continued ... \$\endgroup\$ – Balthazar Apr 15 at 18:26
  • \$\begingroup\$ ... on page 6 there is a diagram called Case temperature Rise vs Output Current, second row to the right. From this I gather that with a voltage below 9V and current below 600 mA, heat generation inside the electronics enclosure won't be a concern. Am I right? Also, just to be certain, it's the Arduino Nano Every, which is half the price of the regular Nano we're talking about. In that case, using the onboard regulator is clearly the solution. \$\endgroup\$ – Balthazar Apr 15 at 18:29
  • \$\begingroup\$ At 500 mA (vin 8 vout 3.3) you get 7 degrees Celsius above ambient at the case. Thats minimal. A 9V battery can work but keep in mind it has a low capacity. And yes I looked at the nano every for the stats. \$\endgroup\$ – Passerby Apr 15 at 18:53
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    \$\begingroup\$ Thank you. I have accepted your answer. Paranoid as I am, I will still test it when it's assembled, to see if there is some temperature rise that might be cause for concern. Regarding the 9V battery, that was for the option of powering the Arduino and LEDs separately, one 9V battery for the Arduino only would be enough capacity. \$\endgroup\$ – Balthazar Apr 15 at 19:10

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