simulate this circuit – Schematic created using CircuitLab

I have 6 AA batteries that I would love to power my Arduino Nano and HCSR04 ultrasonic sensor. However, for my low power application I want to use the bare minimum power required for the components. Since the Nano can run as low as 4.5 volts and the sensor requires 5 volts, I think the best solution would be 4 AA batteries in series to power the hcsr04 and 3 AA batteries in series to power the nano. Since I have 6 batteries I was thinking 3 AA batteries wired in parallel with 3 AA series batteries. I drew the circuit diagram above and realize something is horribly wrong. So I was hoping someone could point me in the right direction.

  • You've got yourself in a right muddle there. You can't make 3 and 4 or of 6. You need 7. Also there are other factors you need to consider. I'll write a pepper answer when I get to my computer later.
    – Majenko
    Oct 22 '20 at 7:26
  • According to specs, HC-SR04 uses 15mA. I think it is safe to power it through Nano 5V pin. And then you need just to power only your Nano. Oct 22 '20 at 7:33
  • Newer HC-SR04's will run off 3-5V, which makes things easier.
    – Majenko
    Oct 22 '20 at 8:17
  • 1
    Yeah, don't try to get cute with "lowballing" the input to the Nano. The built-in linear regulator is a big power waster. As Majenko says in his excellent answer, you're much better off using an external buck regulator to power both devices at a full 5V. Those are a LOT more efficient than linear regulators (which convert excess voltage to heat)
    – Duncan C
    Oct 22 '20 at 16:06

Wiring 6 batteries up so it gives you both a stack of 3 and a stack of 4 is not easy, and although you can do it you shouldn't do it. The only arrangement is this:


simulate this circuit – Schematic created using CircuitLab

(The grey link is optional and makes little difference if it's there or not).

The reason you shouldn't really do that arrangement is that you have a mismatch in your capacities, as [BAT1+BAT2] || [BAT3+BAT4] has more capacity than BAT5, so the current drawn from the 4.5V output will cause BAT5 to go flat far sooner than the others. This will then cause BAT5 to take a reverse charge as current from the other cells flows through it damaging it permanently (especially bad for rechargeable batteries as you will not be able to recharge it, but pretty nasty for alkaline batteries too).

Another consideration when running from different voltages like this is the communication between the different units. If the Arduino is running at a lower voltage the signals it sends out will be of a lower voltage, and it expects to receive signals at a lower voltage. Likewise with the HC-SR04; it will expect to receive higher voltage signals and will send out higher voltage signals. So you will require some logic level translation between the two devices to correct that mismatch in the signal voltages.

As you can see it's all starting to get a little messy now.

A much simpler solution is to keep everything at one voltage. The ideal voltage is 5V since that is the common voltage that all the devices expect to work at.

Since you are running from batteries you want efficiency. To that end you need an efficient solution for generating a stable 5V. The Arduino's built-in regulator is far from efficient, so that can be instantly discounted.

A more efficient solution is a step-down or "buck" switching regulator. This uses PWM to rapidly turn the power on and off (before then filtering and smoothing it) to effectively convert voltage into current.

You'd run your system off a voltage higher than 5V then use a switching regulator to reduce the voltage down to 5V. Since you have 6 batteries at your disposal the logical voltage would be 6x1.5V = 9V - all 6 batteries in series:


simulate this circuit

If you're running non-rechargeable batteries then you don't need to worry about low-voltage cutoff (you shouldn't over-discharge rechargeable batteries), so this arrangement will get the most out of your battery life. Depending on the "drop out" voltage of the regulator (you can get some with a "zero drop-out bypass" mode) your batteries could get down as far as 0.8V each before the system starts to become unstable (even lower if your HC-SR04 is one of the "upgraded" ones that can run from 3V).

  • Thanks, why will only the 4.5V outupt drain BATT5? The upgraded hcsr04 is intriguing since I need my battery capacity to last longer than AA batteries in series.
    – I Like
    Oct 22 '20 at 15:26
  • 1
    Not only the 4.5V output. But batt5 will be the first to go flat. It has less capacity than the 1/3 or 2/4 pairs, and it is being used by both outputs. It's the one battery that is weakest in the chain. One of the smallest batteries with the most current being drawn from it. Batt6 has the same capacity as batt5 but only a tiny current being drawn from it for the 6V output.
    – Majenko
    Oct 22 '20 at 15:46

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