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I have an IOIO board, a 13V 10A Battery (when I tested with my multimeter). For the IOIO board, 5-15V, ~10mA is enough. Can a 13V 10A Battery do wrong to the board or is it okay?


marked as duplicate by Passerby, Olin Lathrop, W5VO Apr 12 '13 at 15:43

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    \$\begingroup\$ .. eh? I think the answer to your question is "a battery with a higher current rating will not damage the thing it's driving, the important thing is that the voltage rating is compatible" but I'm not sure if that's what you're asking. \$\endgroup\$ – pjc50 Apr 12 '13 at 10:39
  • \$\begingroup\$ @pjc50 I think it is, you should make it an answer. \$\endgroup\$ – user17592 Apr 12 '13 at 10:50
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    \$\begingroup\$ I think you generally got your answer regarding how the battery will not supply more current that the load presented by your IOIO board. However be aware that YES the battery very well could damage the circuit if it is connected incorrectly to the IOIO board. For example if you connected + of battery to GND of module and - of battery to VCC of module you may very well fry the module. Similarly if a digital output pin of the module were connected to + of battery and GND of module were connected to - of battery it is likely that the output pin circuit could be severely damaged. \$\endgroup\$ – Michael Karas Apr 12 '13 at 11:24
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    \$\begingroup\$ Most devices are designed to be powered by approximations to voltage sources such that the supply current capability is not relevant as long as it is sufficient. However, you occasional find things (LED flashlights?) where the internal impedance of the battery is what regulates the current to a safe level (behaving somewhat like a current source) - in that case, replacing the battery with a stiffer supply could lead to overcurrent. \$\endgroup\$ – Chris Stratton Apr 12 '13 at 14:42
  • \$\begingroup\$ In general, equipment will not rely upon its power supply behaving as anything other than an ideal voltage source, but some devices which have high inrush currents may be damaged if powered by a very stiff voltage source. For example, a device which turns on and off a motor whose stall current is much higher than what the expected batteries can produce might potentially be damaged if it were connected to a power supply which could output that full stall current without flinching. \$\endgroup\$ – supercat Apr 12 '13 at 16:06

The current rating of a power source (your battery) is the maximum current the supply can provide. The current rating of a power load (your IOIO board) is the amount of current it can demand, for proper operation.

Thus, if the IOIO board just requires 10 mA, and operates at a specified voltage range (5-15 Volts as mentioned), then the power supply can not force more current through the board than demanded - it will merely supply as much current as is demanded, not the maximum that it can supply.

So if the voltage values in the question are correct, then the IOIO board should be perfectly safe to operate with the battery.

  • \$\begingroup\$ awesome!what an answer man! thanks for your valuable Answer , so +1 me toooooo.... ! – \$\endgroup\$ – 7-isnotbad Apr 12 '13 at 15:02
  • \$\begingroup\$ i have a doubt now how can i reduce the 13V10A to 13V1A ? \$\endgroup\$ – 7-isnotbad Apr 12 '13 at 17:25
  • \$\begingroup\$ @IOIOMAD You don't need to: That's the point of my answer above - the battery current rating is irrelevant as long as it is higher than 100 mA or so (for the IOIO). \$\endgroup\$ – Anindo Ghosh Apr 12 '13 at 18:00
  • \$\begingroup\$ Thnk u but i got that already , and this is for some sort of knwlg@ \$\endgroup\$ – 7-isnotbad Apr 12 '13 at 19:05

What tells you 13V-10A is that the maximum current that the battery can deliver is 10A. Either way I don't think you have problems by overcurrents. Anyway on the Internet you can find many current limiting schemes.

  • \$\begingroup\$ While this initially looks like a poor answer, there's actually a very important point in it: "10A" is not a battery rating. Hopefully it is a mis-stated capacity in amp-hours rather than amps. Possible, it's an unwisely measured short-circuit current at a particular state of charge. \$\endgroup\$ – Chris Stratton Apr 12 '13 at 14:40

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