I have a project I am looking to power where the secondary load is more than my Arduino can provide amp-wise. I need to know the simplest way to power the whole system. I will be powering the Arduino and a 5v 1.2amp load. Diagrams I have seen show a common ground with two power supplies but I don't know the math for anything beyond a simple circuit. Links to other resources are welcome with all answers. I am looking to learn more about power and circuits in general so more info is better.


If I use one supply, how should I design the circuit? e.g. Series or Paralell. Also, how would I go about determining the requirements for that power supply?

  • \$\begingroup\$ What is the secondary load? \$\endgroup\$
    – tyblu
    Commented Jan 16, 2011 at 16:33
  • 1
    \$\begingroup\$ I got a good answer on why you need the grounds connected when using two supplies here: electronics.stackexchange.com/questions/7018/… \$\endgroup\$
    – Dean
    Commented Jan 16, 2011 at 16:43

2 Answers 2


Without knowing more specifics about your particular set-up, here are some ideas:

The best and safest option would be to use one strong (1.5A or 2A) regulated power supply that powers both the microcontroller (arduino) and the heavier load.

It would also be possible to use two separate supplies, but then, you would have to make sure that the stuff connected to the microcontroller will not back-supply the I/O pins, possibly causing damage to the microcontroller (see: clamp diodes), i.e. you would have to make sure that both supplies provide almost exactly the same voltage (within +/- 0.3 V).

  • \$\begingroup\$ Thanks zebonaut. Back supplying was something I was worried about. When looking for the supply should I think of the two devices as in series or parallel? Do you know of any links that help explain this. \$\endgroup\$ Commented Jan 16, 2011 at 19:37
  • \$\begingroup\$ @JustSmith One way of thinking about the whole series-parallel issue, it may be useful to keep this in mind: parts/devices in series share the same current, and parts in parallel share the same voltage. For IC's with the same voltage requirement, you want to share the same power supply VOLTAGE, so they will be in parallel. \$\endgroup\$
    – W5VO
    Commented Jan 17, 2011 at 3:51
  • \$\begingroup\$ @W5VO thanks that clears up my understanding of of the answer. \$\endgroup\$ Commented Feb 14, 2011 at 15:03

You usually need a common ground but don't want to connect power rails, even if they are the same potential (don't tie two separate 5V rails together, for example). There are some exceptions. You can communicate across isolated grounds using opto isolators and there are some regulators designed to be safely used in parallel, such as Linear Technology's LT3080. Otherwise, there is not much math. You just need to make sure you can provide enough current for each of you rails' load requirements. The only other thing I would mention is that if you do use separate rails of the same potential, and communicate between them, make sure you can't violate any I/O specs of your devices. If you are using two separate 5V rails, for example, each +/-10% , one rail could be 4.5V and the other 5.5V. Outputting a 5.5V signal to an input pin with a 4.5V supply could potentially damage the IC. This is why, when using the same potential across multiple devices that communicate with each other, it's usually preferred to use a single supply rail for all devices. This isn't always easy to do when connecting kits, however, such as the Arduino to some external device.

Edit, if you use one supply, tie all loads to it in parallel. If you were to tie two loads in series to a 5V supply, for example, each load wouldn't be getting 5V. If the two loads were exactly equal, they'd be seeing 2.5V each. You design the supply for the rated load, taking into account heat considerations. 1.5A is a large load which will generate a lot of heat with a linear regulator. You'd need to make sure that A) your regulator can provide at least your maximum load (with some room to spare) and B) can radiate the heat generated by the voltage drop. This means you may require a heat sink. Almost all regulator's data sheets will tell you how much heat is generated for a given amount of power. The National Semi LM317, for example, says it's TO-3 package increases 39 degrees C for each Watt of dissipation. If you are regulating 5V off a 9V rail at 1 Amp, the package would be dissipating (9V-5V) x 1A which equals 4W, or 146 degrees C over ambient temp! Well beyond the maximum spec'd temp. This is where heat sinks come into play, or the use of a switching regulator which are much more efficient, but probably too advanced for your first project. The other thing to be concerned about for regulators is dropout voltage. The supply rail needs to be some amount of voltage greater than your regulated voltage. This is all assuming you have a AC to DC power supply to give you your main power rail. But if this is the case, why not just buy a regulated AC to DC supply at the voltage or voltages you require with the current capability you require. You can buy cheap wall warts with regulated output voltages in most common voltages and with various current capacities.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.