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For the purpose of a project I wish to power an arduino using a supercapacitor charged to 5V. The supercapacitor will be fed straight into the power Vin and GND terminals on the Arduino. No power plug or USB connected to a computer will be connected, so all power is to be drawn from the supercapacitor. Previously I have powered the board using a 0-12V variable voltage supply, and the arduino draws between 20 and 35 milliamps.

My question is this, if I connect the supercapacitor straight to the arduino board, will the arduino govern how much current to draw from the supercapacitor? Or will I need to place a resistor in series with the supercapacitor discharging in order to regulate current and stop the supercapacitor rapidly discharging and frying the board?

Lastly is this feasible at all? I am aware that as the 5V from the supercapacitor will drop off the period during which the arduino will run is limited.

Any advice is appreciated as I do not wish to fry my Arduino board attempting this.

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    \$\begingroup\$ The arduino will draw the current it requires. It is certainly feasible, you will need to decide what run time you desire, and from that are able to work out how big the capacitor(s) need to be. \$\endgroup\$ – Colin Mar 27 '18 at 11:20
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    \$\begingroup\$ Yes, current isn't 'pushed' into devices really. If it is designed to operate at 5 V and you attach it to a 5 V source it will draw the current from the source it needs. \$\endgroup\$ – Colin Mar 27 '18 at 11:32
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    \$\begingroup\$ You will get more run time if you use a 5V regulator and e.g. a 12V capacitor. Also beware of the in-rush current with a 15F capacitor! \$\endgroup\$ – Oldfart Mar 27 '18 at 11:48
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    \$\begingroup\$ Is there any reason other than "because it's there" why you want to do this? Supercapacitators excel at delivering high currents, but suck in every other respect compared to a cheap rechargeagble battery. An arduino doesn't precisely need lots of current, so... not sure this is a sensible approach at all? \$\endgroup\$ – Damon Mar 27 '18 at 19:37
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    \$\begingroup\$ The reason is basically 'because its there'. Its for a college project taking an existing setup and using just supercapacitors as a power source. An arduino is a part of the setup \$\endgroup\$ – TRJ94 Mar 27 '18 at 20:13
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The voltage on a capacitor is proportional to the charge stored in it.

That means that as the device draws current, the voltage will drop. You have to decide how low is acceptable. Do the math.

For example, let's say the device can still run from 4.5 V. That means the capacitor voltage can drop 500 mV before the system doesn't work anymore. Let's use a 1 F cap as example.

    (500 mV)(1 F) / (35 mA) = 14.3 s

That's how long a 1 F cap charged to 5 V can run your device at the worst case current draw.

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    \$\begingroup\$ I would suggest a boost regulator such as LT3426, see page 9 for circuit examples for 5V <cds.linear.com/docs/en/datasheet/3426fb.pdf> and feed the output into the 5V pin, and not the Vin pin (Vin then goes thru a 5V regulator, you won't get 5V out). \$\endgroup\$ – CrossRoads Mar 27 '18 at 13:16
  • \$\begingroup\$ A one farad cap would be the Godzilla of capacitors. \$\endgroup\$ – zeta-band Mar 27 '18 at 22:02
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    \$\begingroup\$ @Zeta: That used to be true a couple of decades ago, but not anymore. Look up things called "super caps" and "ultra caps". \$\endgroup\$ – Olin Lathrop Mar 27 '18 at 22:05
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First to answer your question (Although that has been done in the comments already): No you do not need a resistor to limit the current. The Arduino will take just what it needs.

Next: Your discharge over a small voltage range 5v down to 4.5V will be ~CV/I. Olin gave a figure for that.

BUT!
You can get a lot more time using a smaller and cheaper capacitor if you would use a 5 Volt regulator and e.g. a 12V capacitor. Let's say you want 10 seconds whilst the voltage drops from 5V to 4.5V. You need a capacitor of 0.035*10/0.5 = 0.7 Farad.

Some assumptions:
We use 12V, have 0.5V drop over the regulator and it uses an extra 1ma. The voltage can now drop 6.5 Volts (and we still have 5V on the Arduino). You now need a capacitor of 0.036*10/6.5 = 0.055 Farad. Yes, you need a higher voltage supercap but it can be much, much smaller.

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    \$\begingroup\$ Most Arduinos have a LDO regulator on board at the VIN pin, so the OP wouldn't even need additional components. \$\endgroup\$ – Sanchises Mar 27 '18 at 12:56
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    \$\begingroup\$ I have just read that the Vin pin regulates to 5v and so can take in a voltage from 5-12V. The capacitor I have has a max voltage of 7V, so I may charge to this max voltage for added runtime (7v down to probably 4.5v) \$\endgroup\$ – TRJ94 Mar 27 '18 at 14:54
  • \$\begingroup\$ the LDO still has some drop-out voltage. so it'll probably drop out around 6V \$\endgroup\$ – user371366 Mar 27 '18 at 15:17
  • \$\begingroup\$ @dn3s In my experience, the voltage drop from the LDO is negligible at the current draw from an Arduino, and it will work fine to less than 5V on VIN (the MCU can also tolerate quite a margin below 5V) \$\endgroup\$ – Sanchises Mar 28 '18 at 7:28
  • \$\begingroup\$ huh i didn't know that was a thing, I was under the impression that LDOs just shut down under the dropout voltage, are you saying they just derate current / drop below the regulated voltage? that's neat. i could see how that might go undocumented in datasheets which would explain me not knowing this \$\endgroup\$ – user371366 Mar 29 '18 at 23:08
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Another option could be to get a DC-DC step up converter which gives a 5V output. This can run on an input of 0.9V - 5V.

Based on your 15F capacitor and a voltage drop of 4V (for a 5.5V supercapacitor charged to 5V) you could get 4000mV * 15F / 41mA (85% efficient) = 1,463s or about 24 minutes run time.

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  • \$\begingroup\$ The 15F capacitor has a max voltage of 7V, so I had planned to charge to 5V \$\endgroup\$ – TRJ94 Mar 27 '18 at 14:52
  • \$\begingroup\$ @TRJ94 - I've changed the calculation to use a supercapacitor charged to 5V instead of a 2.7V one. Ironically, I'd calculated both, and accidentally put the answer for the supercapacitor charged to 5V in. \$\endgroup\$ – Holmez Mar 27 '18 at 22:00

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