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I want to use this battery for my Arduino Uno:

http://www.ebay.com/itm/7-4-V-30C-5200mAH-2S-Lipo-Li-Po-Lipoly-Battery-for-RC-Car-Boat-/261351067166?pt=Radio_Control_Parts_Accessories&hash=item3cd9bcce1e

The on board voltage regulator says L117-S50. The writings is alomst wiped out, but through a quick internet search, I think this is the one:

https://www.sparkfun.com/products/595

Now, the minimum voltage the arduino operates at is 7V. While the batteries are at 7.4V. Therefore; after some time of usage, the votage might drop below 7V which might cause issues.

I want to ask is there anyway to overcome this issue without having to buy a higher voltage battery.

Is there a certain module or electronics device that can boost up the voltage to a certain range.

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  • \$\begingroup\$ The device you're after is a DC-DC-converter. \$\endgroup\$
    – user30985
    Apr 10, 2014 at 17:56
  • \$\begingroup\$ Could you refer me to a link, and a schematic of the connection please. \$\endgroup\$
    – Adel Bibi
    Apr 10, 2014 at 17:59

2 Answers 2

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For simplicity you could power the Arduino from a 5V low-drop-out regulator. You'd bypass the normal regulator (it isn't low drop-out) and get more usage from the 7V battery. I don't know what current is required by the uno so I can't definitely recommend something but maybe a 1A LDO like the L4941 would suit. It drops 450 mV at 1A and this means your battery can droop down to 5.45V before things go out of spec. You might need a heatsink - it all depends on how much the uno takes.

Or you could go the whole hog and have a buck-boost regulator like this one: -

enter image description here

This takes any voltage from 2.7V to 15V and gives you 5V and up to 2.5A. Or maybe just a buck regulator would suit. This one will work down to an input voltage of 5.5V and still give 5V out at 2.5A:-

enter image description here

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  • \$\begingroup\$ Thanks alot! Just to make things clear, I will have to connect the output voltage (+5v) to the +5v pin to my arduino and just ignore the on board regulator. BTW, my current will be in normal conditions at around 700mA, and at certain times there will be spikes that reach up to 2Aamp. \$\endgroup\$
    – Adel Bibi
    Apr 10, 2014 at 22:05
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    \$\begingroup\$ Dealing with the spikes can be much helped by using a large electrolytic on the output of the linear regulator. For the switchers, they'll be just fine as shown. You might also want to protect the on-board regulator with a diode across it so the reg input is biased at output -0.7V. It might not need it but read abs max ratings on the data sheet for the part - it's a standard technique should it be needed. \$\endgroup\$
    – Andy aka
    Apr 10, 2014 at 22:11
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7V is a very conservative figure. The LM1117 has a dropout voltage of like 1.4V worst case. So it should continue to regulate just fine down to at least 6.4V and probably down as low as 6.2V practically speaking.

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  • \$\begingroup\$ So do you think using the 7.4v battery will be just fine? \$\endgroup\$
    – Adel Bibi
    Apr 10, 2014 at 18:29
  • \$\begingroup\$ @AdelBibi I think so, Lithium Polymer batteries have a fairly flat voltage decay over their discharge period \$\endgroup\$
    – vicatcu
    Apr 10, 2014 at 21:47
  • \$\begingroup\$ @AdelBibi - FYI - that Lithium battery you linked to will be essentially empty at 6.0 volts, and if you want it to last longer, it would be wise to discharge it to only 6.5 volts or so. It sounds like it will be perfect for your situation! \$\endgroup\$
    – Filek
    Apr 11, 2014 at 4:41
  • \$\begingroup\$ @Filek I will then go for it and get the step down convdverter just in case. Thanks! \$\endgroup\$
    – Adel Bibi
    Apr 11, 2014 at 7:00

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