I have a bright white LED from a flashlight. Approximately how long will it light up with a 150 farad 2.5 volt capacitor? Do I need a resistor? If so, how many Ω? The capacitor is a Maxwell 150 farad 2.7 volt boostcap model # "BCAP0150 P300 S17" (currently <$12 for one).

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    \$\begingroup\$ If you give us the datasheet for your LED I can modify my answer and give you the actual numbers for the resistance needed and time. \$\endgroup\$
    – Sponge Bob
    May 2, 2013 at 21:39
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    \$\begingroup\$ That's a fairly huge capacitor? Also, white LEDs tend to have a forward voltage of ~ 3-4V, so it probably won't light at all. \$\endgroup\$
    – pjc50
    May 2, 2013 at 21:43
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    \$\begingroup\$ What's with the close votes? Give the kid a break. This is a reasonable question for his level of knowledge. He has a lot to learn, but that is exactly what he is trying to do. \$\endgroup\$ May 5, 2013 at 15:18
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    \$\begingroup\$ @OlinLathrop - Yes, sadly here on StackExchange down votes and close votes are "awarded" rather randomly. A very harsh community. \$\endgroup\$
    – IceCold
    Jan 23, 2018 at 20:19

3 Answers 3


Direct answer to the question

The direct answer to your question, assuming you intend to just connect the capacitor to the LED with a series resistor is no time at all. That is because a white LED takes more than 2.7 V to light. Check its datasheet. These things usually need a bit over 3 V.

There are two options. The simplest is to use a LED with a lower forward drop. Let's say you try this with a red LED that has a 1.8 V drop at 20 mA. That means at full charge, there will be 2.7V - 1.8V = 900 mV accross the resistor. If you want the maximum brightness at full charge, which we are saying is 20 mA, then you need a 900mV / 20mA = 45 Ω resistor. Let's pick the common nominal value of 47 Ω.

Now that we have a capacitance and resistance we can compute the time constant, which is 150F x 47Ω = 7050 s = 118 minutes = 2 hours. At full charge, the LED will be nearly at full brightness, which will then decay slowly. There is no fixed limit at which it will suddenly go out, so we have to pick something. Let's say 5 mA is dim enough to be considered not usefully lit anymore in your application. The voltage accross the resistor will be 47Ω x 5mA = 240mV. Using the first approximation of the LED having constant voltage accross it, that means the capacitor voltage is 2 V.

The question is now how long does it take to decay from 2.7 V to 2.0 V at a 2 hour time constant. That is .3 time constants, or 2100 seconds, or 35 minutes. The actual value will be a bit longer due to the LED having some effective series resistance too and therefore increasing the time constant.

A better way

The above tries to answer your question, but is not useful for a flashlight. For a flashlight you want to keep the light at close to the full brightness for as long as possible. That can be done with a switching power supply, which transfer Watts in to Watts out plus some loss but at different combinations of voltage and current. We therefore look at the total energy available and required and not worry about specific volts and amps too much.

The energy in a capacitor is:

$$E = \frac{C \times V^2}{2}$$

When C is in Farads, V in Volts, then E is in Joules.

$$\frac{150F * (2.7V)^2}{2} = 547 J$$

The switching power supply will need some minimum voltage to work with. Let's say it can operate down to 1 V. That represents some energy left in the cap the circuit can't extract:

$$\frac{150F * (1.0V)^2}{2} = 75 J$$

The total available to the switching power suppy is therefore 547 J - 75 J = 470 J. Due to the low voltages, the losses in the switching power supply will be quite high. Let's say that in the end only 1/2 the available energy gets delivered to the LED. That leaves us with 236 J to light the LED.

Now we need to see how much power the LED needs. Let's go back to your original white LED and pick some numbers. Let's say it needs 3.5 V at 20 mA to shine nicely. That's 3.5V * 20 mA = 70 mW. (236 J)/(70 mW) = 3370 seconds, or 56 minutes. At the end of that, the light would go dead rather quickly, but you will have fairly steady brightness up until then.


Yes! You need a resistor. If you don't use a resistor the Cap will discharge instantaneously across the LED. The resistance of the resistor will determine how long the LED will stay lit.

You must look at the LED specs to determine how much current is needed to drive it. Once you know that you can calculate the resistance of the resistor needed using V=IR. solving for R=V/I

Once you know the resistance you can find out the current and using the current you can calculate the time it will last using C=It/V. When you solve for t you get t=CV/I which will come out in seconds.

Hope that helps!

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    \$\begingroup\$ Wouldn’t you have to take the forward voltage of the LED into account in your calculations? Furthermore, does’t the capacitor voltage drop while discharging, so \$I\$ is not constant? \$\endgroup\$ May 2, 2013 at 22:01
  • \$\begingroup\$ That is true. The voltage of the capacitor will be equal to Vi*e^(-t/RC), where Vi is the initial voltage. \$\endgroup\$
    – Sponge Bob
    May 3, 2013 at 5:18
  • \$\begingroup\$ Your calculations are incorrect. The diode will turn off when cap voltage will fall down to the forward voltage of the LED. Therefore if the diode voltage is 2.49 it will turn-off almost instantly and if it is 2.1 it will take more time. \$\endgroup\$ May 3, 2013 at 7:25

Alternative approach: I'm going to assume a standard red/green/yellow LED which takes 2V at 20ma: 40mw. The capacitor stores 1/2 CV^2 joules of energy: 300J. That would suggest 300/0.04 = 7500 seconds or about 2 hours. However, in practice you won't get all of the energy out because the voltage will drop fairly quickly below a level that will output light. I'd estimate about half an hour until "dim", and possibly another 15 minutes of faint glow.

Edit: by the way, your capacitor is almost certainly 150mF, which will give you 1/1000 of that, or a few seconds.

(For easy illumination of a white LED, try 3 NiMH AA batteries, which give you almost exactly the right voltage)

  • \$\begingroup\$ OP specifies "bright white LED" \$\endgroup\$ May 3, 2013 at 3:45
  • \$\begingroup\$ The 360F ....yes farad ultracaps do exsist...but only 2.7V. so for most white leds you need 2 in series ... i will look for a link after later... \$\endgroup\$
    – Spoon
    May 3, 2013 at 17:52
  • \$\begingroup\$ www.easby.com/nesscap/medium_snap.php ...these are about flashlighf/torch size \$\endgroup\$
    – Spoon
    May 3, 2013 at 18:02
  • \$\begingroup\$ I got one 4 free here cdiweb.com/ProductDetail/… \$\endgroup\$
    – skyler
    May 5, 2013 at 14:17
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    \$\begingroup\$ What makes you so sure that "your capacitor is almost certainly 150mF"? The linked Maxwell 150F supercap is only $12, so the price isn't an insurmountable barrier... \$\endgroup\$ Jun 20, 2023 at 17:49

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