There is a Single Board computer that gives a range of input supply as xvolts - yvolts.

Let's say we add a few more additional components like webcam or additional storage drive.

Taking these into consideration, how to determine the needed input voltage?

  • \$\begingroup\$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. \$\endgroup\$
    – Community Bot
    Commented Oct 28, 2021 at 18:41
  • 2
    \$\begingroup\$ SBC's commonly use 12V, 5V and 3.3V Choosing peripherals often is best around the supply chosen. PSU's (power supply units) are low budget now due to volume and are much cheaper than the SBC to handle almost any requirement. 350W $0.10 to $0.20 / W link the SBC in your question and all options needed. \$\endgroup\$ Commented Oct 28, 2021 at 18:46
  • 3
    \$\begingroup\$ In a nutshell, the input voltage stays the same but the current capability of the PS needs to increase to handle the additional peripherals. That is, unless the PS had unused capacity. \$\endgroup\$
    – SteveSh
    Commented Oct 28, 2021 at 18:52
  • 3
    \$\begingroup\$ "... how to determine the needed input voltage?" The operating voltage of each component will be listed on their datasheets. It's simplest if you choose components that all have the same voltage requirement otherwise you need multiple voltage rails from your power supply. \$\endgroup\$
    – Transistor
    Commented Oct 28, 2021 at 19:00

2 Answers 2


Each of those devices has a maximum and minimum voltage requirement. You must find the highest of all the minima, and lowest of of all the maxima, and those two values represent a voltage range that satisfies the requirements of all devices.

For example, if your SBC requires between 4.5V and 5.5V, your disk drive requires between 4.7 and 6.0V and your camera needs from 4V to 5.2V:

$$ \begin{array}{lll} &Device &Min &Max \\ \hline &\text{SBC} &4.5 &5.5 \\ &\text{Disk} &4.7 &6.0 \\ &\text{Camera} &4.0 &5.2 \\ \hline &\text{Everything's happy} &4.7 &5.2 \\ \end{array} $$

By that logic, a supply of exactly 5.0V will do the job. Of course all this assumes that they are being supplied by the same voltage source. If a device is being powered from some other source, like a USB port or external power supply, then you can't apply this logic.

Assuming you have such a 5V supply, what's of more concern to you is the current required from that 5V supply. The total current required is the sum of all the individual currents required by each device, again assuming they are all powered from that same supply. If you have an SBC rated for 2A, a disk drive requiring at most 1A, and a camera requiring up to 500mA, that's a total of 3.5A, which your supply must be able to provide.

That doesn't mean they will all be drawing this current all the time, but there is a chance that all 3.5A be needed simultaneously. So the power supply must be up to the task 100% of the time, even if the full 3.5A is needed only 0.0001% of the time. And it needs to be able to do this while staying within whatever the acceptable voltage extremes are, as defined by the devices, and as I explained above.


It's not a matter of voltage.

It's a matter of the max current that the whole system requires once you add the external components.

You have to make sure that the voltage regulator is able to source all the current needed.

Usually, voltage regulators of single board computers are able to source an extra current of 30/50% with respect to the maximum current required by the board.

Verify also the voltage compatibility between the output voltage of the voltage regulator of the SBC and the input voltage of all the external peripherals.


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