If I have, for example, an IC that draws 10mA @ 3.3V, how large of a capacitor would I need to power it for 2 minutes? 5 minutes?

I understand that using capacitors to power an IC is not the most practical, since they discharge very quickly, but am curious if the required capacitance is within the range of those commercially available.

I would like to understand the math behind calculating a problem like this as well.

  • \$\begingroup\$ You can control how quickly a capacitor discharges by adding a well dimensioned resitor to the discharge circuit. You just think they discharge fast because the ones we usually see have low capacitance. So, if you get a really big capacitor, like the one Spehro just calculated for you, and pair it with the right resistor, it will discharge at the right pace. And capacitors can be pretty large. \$\endgroup\$ – Ricardo Feb 12 '14 at 1:38

The voltage on a capacitor will drop at a rate of \$ \dfrac{dv}{dt} = \dfrac{-I}{C}\$, assuming a constant current \$I\$ and capacitance \$C\$.

So if the IC is guaranteed to work down to 2.90V (you have determine this value from the datasheet), the capacitance required to get it to work for 'x' minutes with 10mA draw will be:

$$C = \frac{I \cdot T}{\Delta V} = \frac{0.01A \cdot (60 sec/min) \cdot (x\ min)}{0.4V} = 1.5 Farads/min \cdot x min $$

So for 5 minutes, you'd need 7.5F of capacitance.

While that's a prohibitive amount of capacitance for a conventional electrolytic capacitor, it's not impossible for a "double layer" capacitor such as these.

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The cost of such a part is prohibitive for many consumer electronic devices.

  • \$\begingroup\$ Supercapacitors... That is all. \$\endgroup\$ – Soviero Feb 12 '14 at 2:02
  • \$\begingroup\$ Double-layer capacitor aka Supercapacitor aka Ultracapacitor. \$\endgroup\$ – Spehro Pefhany Feb 12 '14 at 3:37
  • \$\begingroup\$ Did you mean to divide by delta(V) instead of multiply? Also, I assume I can increase delta(V) if I charge the cap to the max voltage that the IC can handle (e.g. 3.9V)? \$\endgroup\$ – elin05 Feb 13 '14 at 19:56
  • \$\begingroup\$ Yes, thanks, I'll fix that. And yes, the time is proportional to the delta-V allowable. \$\endgroup\$ – Spehro Pefhany Feb 13 '14 at 20:13

Look for "memory backup capacitors." These typically come in 1.0F to 1.5F capacities, with max voltage ratings of 5.5 Volts. They cost about $3 in singles.

Here's an example: https://www.jameco.com/webapp/wcs/stores/servlet/Product_10001_10001_2162599_-1

Note that these supercaps are not great for high-current loads (or filtering) -- they have significant internal resistance (ESR.) Trying to drive too much current may blow the supercap.

Also, I disagree with the use of a current limiting resistor for backup supercaps. That will just waste energy in the resistor. The load you have presumably already draws the "right" amount of current, assuming you feed the right voltage. The math for that was shown in the reply above.


Super-caps are used as alternatives for coin-cell backups in RTCs all the time. Very low current draw for a long time. Google "supercap rtc" for many whitesheets on them. And then there are incredibly large capacitors commercially available. We are talking about 1+ Farad capacitors used for car speakers. 1F. Not nF or pF, F. 20 to 30 bucks in any retail car audio store.

Even better, Maxim has a calculator for Supercap sizing. It's geared for RTC applications but it's good for your question, providing the math behind it as well.


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