1
\$\begingroup\$

The title says it all.

I'm currently working on an Arduino project. The project involves connecting 3 - 6 different shields non which is a "Motor Shield".

They are VR, brain wave reader, GSM, GPS, TFT screen, and a camera.

What is the right way to calculate the power needed to be supplied to the total project by batteries?

What I know is that I have to calculate the amound of current withdrawn from each module in their highest peaks. Accordingly, I will choose the capacity of the battery that can operate the the amount of time I want.

Is this the way it is done? Or am I missing something? Is there any other concerns I have to keep an eye on?

Regarding the voltage, It shouldn't matter right? The shields are made specifically for Arduino, so if I powered it in the recommended range of voltages (7-12) I should be fine right?

\$\endgroup\$
9
  • \$\begingroup\$ You don't want to supply the 7-12V, since it goes through a linear regulator. Supply 5V directly instead. \$\endgroup\$
    – Ignacio Vazquez-Abrams
    Commented Jan 10, 2014 at 9:29
  • \$\begingroup\$ You probably want average current not peak for estimating battery life. Unless you know how to calculate that for each shield just measuring how long it takes to flatten a battery of a known capacity and adding a bit of margin is probably a decent way to go about it. \$\endgroup\$
    – PeterJ
    Commented Jan 10, 2014 at 9:32
  • \$\begingroup\$ I was referring to the power jack itself. I will be powering (7-12)v to the power jack directly. The shields are mounted on the the arduino, they will take the power from the (5v) pin or directly before the regulator as specified from their data sheet. At the end, I will have to supply it through the power jack. \$\endgroup\$
    – Adel Bibi
    Commented Jan 10, 2014 at 9:33
  • \$\begingroup\$ @PeterJ So is this the only thing I have to worry about? Any other relevant when connecting a battery? Regarding adding the margin, what is normally a good percentage I have to multiply with after calculating the average current? Is 1.3 good enough? \$\endgroup\$
    – Adel Bibi
    Commented Jan 10, 2014 at 9:35
  • \$\begingroup\$ Also keep in mind the C-rating of the battery. Pulling more current than it can supply will hurt the voltage. \$\endgroup\$
    – Ignacio Vazquez-Abrams
    Commented Jan 10, 2014 at 9:36

1 Answer 1

Reset to default
3
\$\begingroup\$

Current

Yes the right way to calculate current required is to work it out for each module and add them up. However, if you are combining Arduino modules, a practical way is to assemble them, power from a relatively high power source and measure the current. The only danger here is that you'll overload the Arduino's 5V or 3.3V regulators. I'd check that the specific Arduino has overcurrent or short-circuit protection.

Rather than measuring for all 5 daughterboards at once, you may be able to devise test programs for using them one at a time. Make measurements, then check the power supply path is adequate before combining all of them.

Voltage

The voltage regulators on a typical Arduino are not expected to supply much current. See Arduino and/or regulator data sheets. As Ignacio commented, it may be advisable to provide for an off-board regulated 5V supply.

\$\endgroup\$
7
  • \$\begingroup\$ It's definitely advisable. Shoving 7.4V into a 5V linear regulator means you waste about 1/3 of your battery. \$\endgroup\$
    – Ignacio Vazquez-Abrams
    Commented Jan 10, 2014 at 9:40
  • \$\begingroup\$ As for current, I'm using an arduino mega. I couldn't find any thing relating the (overcurrent) except for the usb connection. Where they say the following: "The Arduino Mega has a resettable polyfuse that protects your computer's USB ports from shorts and overcurrent. " is this the same max overcurrent the power jack or (vin) can handle? Let me get this staright, when you say to use an off-board regulated 5v supply, do you mean I use a (for exapmle) 7.4v power supply connect it to a (High current) regulator and the out put connected to (+5). Sorry I'm new to this! \$\endgroup\$
    – Adel Bibi
    Commented Jan 10, 2014 at 9:46
  • \$\begingroup\$ At the end this won't solve the problem of high pwoer dissepation, but will only make sure the mega is safe. Did I get that correctly? One last thing, what if the are to operate higher than (+5v)? \$\endgroup\$
    – Adel Bibi
    Commented Jan 10, 2014 at 9:48
  • \$\begingroup\$ @AdelBibi what if the are to operate higher than (+5v)? Please rephrase the question. If you mean what are the alternatives to generate 5v you can use a switching regulator like this (uses a LM2596) that has a high efficiency. \$\endgroup\$
    – alexan_e
    Commented Jan 10, 2014 at 10:24
  • \$\begingroup\$ Sorry I missed typing that. I meant, what if the modules operate to a higher volrage that +5v. When I connect the off-board regulator to the (+5v) pin, that will only provide me with a 5v source. What if one of the modules operate at 7v? @alexan_e \$\endgroup\$
    – Adel Bibi
    Commented Jan 10, 2014 at 10:39

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.