I am developing a hot air blower made out of 40w heater, 25mm fan and 12v power supply. I want to be able to keep the fan working in case that the power supply is cut off, so in case the heater is still hot, the fan will cool it. This should be 5-10 seconds fan operation. Is that possible with a capacitor or should I use a battery? The fan's specs is 12 volts, 0.08A Thank you

  • \$\begingroup\$ When you say: "in case that the power supply is cut off" do you mean only the 12 volt supply, or do you also mean the AC mains? \$\endgroup\$
    – EM Fields
    Jan 26, 2015 at 15:35
  • \$\begingroup\$ The 12 volt supply. I supply it through a 12 V power source. \$\endgroup\$
    – ogold
    Jan 26, 2015 at 18:58
  • \$\begingroup\$ Curiosity: why the forced cooling? Won't the heater stop working too in case of power failure? It should be in operational temperature range and naturally go down. \$\endgroup\$
    – Kahler
    Jan 27, 2015 at 0:29
  • \$\begingroup\$ Sure it will, but the heater is located inside of close polymer case. once the power is down, the heat will still remain and could harm the case. this is the reason I need some time to cool the heater. \$\endgroup\$
    – ogold
    Jan 27, 2015 at 7:15

4 Answers 4


The required energy, in joules, to power the fan for 10 seconds can be calculated:

9.6 joules = 12 volts * 0.08 amperes * 10 seconds

The required capacitance needed to store such amount of energy in a capacitor:

0.133 farads = 2 * 9.6 joules / sqr(12 volts)

The capacitance needed is equivalent to 133 millifarad or 133,000 microfarad and the closest E10 value is 150 mF or 150,000 uF. Using a radial aluminum electrolytic capacitor capable of handling 16V, the capacitor physical dimension is 18mm in diameter by 36mm in length.

To answer your question: Yes, it's possible to use a capacitor to store energy for powering the fan. A capacitor solution is simple as you just need to connect it across the load (fan) and no complicated charging circuit is required. However, your may be constrained by the following:

  • high capacitance capacitor requires a large space
  • high capacitance also means a longer charge time: the fan may not be powered for the full 10 seconds if power is lost while the capacitor is still charging
  • a big capacitor charging will cause a high inrush current at power on: your supply must be designed to handle that
  • a capacitor will not provide a constant voltage as it is discharging: you may need to use a higher capacitance so the full 10 seconds discharge will be within an acceptable voltage range for the fan
  • big capacitors are not easily available: may require a special order
  • a battery module may be cheaper than a big capacitor

I leave to you the evaluation of the cost-benefit tradeoff between a capacitor and a battery.

  • \$\begingroup\$ Thank you; I'm convinced , do you have any recommendation for battery . Better a disposable one. \$\endgroup\$
    – ogold
    Jan 26, 2015 at 18:37

The short answer is it is possible with a capacitor, but I would probably go with a battery.

The longer answer:

If you use a capacitor, it will discharge during this time. This means the voltage across the capacitor will drop. So you will have to understand what the acceptable starting voltage and ending voltage could be.

To a first order approximation, if you assume 80 mA current is constant, then the change in voltage per second is the current in Amps divided by the capacitance in Farads.

So for example, let's say you want to run for 5 seconds and have the voltage only droop from 12V to 10V, i.e. a 2 V drop:
You expect 2V / 5s = 0.4 V/s drop. So 0.4 = 0.08/C => C = 0.2 F.

This is a pretty large capacitance for 12 V. Also, note that this was a 1st order approximation, meaning I'm simplifying a bit. When you decrease the voltage on the fan, it will lower the current somewhat, and this will cause both a slower voltage discharge and less cooling. So the answer would require more information.

As I suggested, I would probably use a battery. You might be able to charge it during operation so it would not need to be replaced.

  • \$\begingroup\$ Thank you , for the clear answer. would battery should be connected directly to the fan , and can I use a disposable battery instead of a charging one ? \$\endgroup\$
    – ogold
    Jan 26, 2015 at 18:36

Anything is possible with the appropriate circuit.

In your case, however, the actual circuit depends on the tolerance of your fan to the supply voltage, since both your power supply and the fan are 12V.

With a simple charge/discharge circuit, the capacitor will charge to the supply voltage (12V). After the PS is disconnected, the capacitor will discharge, but as it does so, its voltage goes down as well.

If the fan can tolerate a lower supply voltage, e.g. 10V, you can use a simple charge/discharge circuit, with a really big capacitor - big enough to hold enough charge to keep the fan running for 10sec, as the capacitor's voltage goes down from 12V to \$V_{min}\$, the fan's minimum usable voltage (e.g., 10V). You can use the usual capacitor discharge formula, \$V(t) = V_0 e^{-\frac{t}{\tau_0}})\$, where \$\tau_0=RC\$. In your case, \$V_0=12V\$, \$R=(12V/0.08A)\$ and you need to find a value of \$C\$ that will provide you \$V(t)\geq V_{min}\$ when \$t=10sec\$.

If the fan's tolerance is too small to allow usage of a real-life capacitor, then you'd need a more complex circuit that will always provide 12V. For instance, use a voltage pump to charge the main reservoir capacitor to a higher voltage (e.g. 24V), followed by a voltage regulator to achieve 12V again.

  • \$\begingroup\$ The Fan does tolerate "abuse" with the voltage down to 3-5 V although slow but works. \$\endgroup\$
    – ogold
    Jan 28, 2015 at 7:43
  • \$\begingroup\$ What will be the size of the capacitor I will neec ?Thanks \$\endgroup\$
    – ogold
    Jan 28, 2015 at 7:59
  • \$\begingroup\$ You can use either the formula I specified in my answer, or the formula in caveman's answer. The formula I provided assumes that the current changes linearly with the voltage (so the current drops as the voltage drops). caveman's formula assumes the current is constant, regardless of the voltage. The actual situation is probably somewhere in the middle, but if you want to be on the safe side, use caveman's formula, which will produce a larger capacitance value than my formula. \$\endgroup\$
    – Sagie
    Jan 29, 2015 at 20:50

Ultracapacitors would be suitable here - 2.7V and 5.5V are typical voltages, so you'd need to run several in series to get your desired voltage.

I use ultracapacitors in small boost power supplies for microcontrollers (as a reservoir for the boosted voltage).

A single representative 5.5V 1.5F ultracap is about 20mm diameter x 6m thick (quite close to the size of two 2032 coin cell batteries stacked). In that small space, it packs 22.6875 Joules. Three of these in series would be capable of taking a maxiumum 16.5V charge and delivering a whopping 68 Joules of energy. Obviously, your circuit doesn't need quite that much juice, so you could get lower rated ultracaps - 0.22F caps are a bit more compact, and three of those would provide just shy of 10 Joules total Energy capacity. 12.7mm diameter by 17.5mm height for a representative 0.22F cap I have here, though I know there are much smaller ones (less than 1/2 the height).

If you go with a battery solution, you'll be looking at replacing the batteries periodically because they'll eventually not hold the charge, or will fail (often damaging your equipment). You'll also need to provide some sort of controlled charge circuit for the battery. Charging a capacitor on the other hand is quite straightforward. An ultracapacitor can be recharged hundreds of thousands of times - it will far outlast any rechargeable battery in terms of total lifespan, which means so long as you use the correct specification and stay within device ratings, you shouldn't need to replace the "backup power" portion of your circuit anytime in your lifetime.

Your muffin fan bearings are a different matter.


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.