Is a 470uF Capacitor too weak to light an LED for more than a few seconds?

Question

I'm doing this for a science class and I can't tell if my capacitor is too weak or if something is wrong with it. Thanks

On the side of the capacitor it says 25v 470uF. I tried running the current through several resistors, and yes, it did increase the amount of time the light stayed lit, but it still was only lit for 2 or 3 seconds.
I used a 9v battery to "charge" the capacitor. Hope that helps.

Answer 1

"Weak" is not the appropriate term. But, indeed, the capacitor is too small (in terms of electrical capacity) for what you intend to do. Even a capacitor ten times larger (4 700 μF, I do not mean physical size) won't provide more than 25 s of operation. Capacitor voltage, by the way, decreases exponentially, not linearly.

The LED is a nonlinear device, so a detailed analysis is rather complex; besides, it is hardly justified. Nevertheless, since the capacitor is discharged through a resistor, its voltage will decay exponentially, at least as a first approximation.

When charged to 9 V, the charge stored in a 470 μF capacitor is Q = CV = 470E-6 × 9 = 4.23E-3 C = 4.23 mC (milliCoulomb). Now, Q = ∫ i dt, where " i " stands for current. Assuming LED current is around 10 mA, and that it remains constant (which it isn't), this latter equation may be simplified to Q = I t. Thus, you may approximate LED operation time as t = Q/I, or t = CV/I.

As noted above, LED current is not constant, but decays exponentially in time. This means time will be underestimated in the formula above; but error will not amount to much, anyway; a good "rule-of-thumb" would be t = 2 CV/I.

Incidentally, have you seen one of those novel "Faraday" flashlights? These don't get their power from batteries, but by rectifiying a.c. emf generated by shaking a powerful magnet inside a coil. At least some of them use this power to charge an electrolytic capacitor, which stores enough energy to keep an ultra-bright LED lit for no more than 10~15 min.

Guess which capacitance this capacitor has? ONE FARAD. No error, no kidding. Draw your own conclusions...

Answer 2

It all depends on how much light you want. You can easily make it last several minutes if you don't need it really bright. For example, you could put the LED inside a dark box and just have a hole to look through for viewing it.

At 470 uF and 9 V, you should get .00423 Coulombs (C times V) of charge in the capacitor. With a 1K series resistor you should get an initial current of 5.5 to 7 milliamps depending on the color of LED you're using. This should be easily visible but will get much dimmer after a few seconds.

If you want a couple of minutes and don't mind a dim bulb, then try a resistor around 100K. This is where you'll probably need the box to see that it is still lit.

It will eventually go completely dark when the capacitor voltage drops below the required LED voltage of around 2V for red or 3.5V for blue or white.

If you or your science class are easily entertained, you could try a 10 Meg resistor on a super bright amber or red LED. It should stay lit for hours, although it will be very dim, even at the beginning.

Answer 3

What voltage are you charging the capacitor to and what resistance do you have in line with the LED? If you are charging only a couple of volts above or at the normal supply voltage for the LED, then the LED is going to go out as soon as the voltage drops below a cut off point. Voltage falls off linearly, not like a battery where it stays up for a long time then drops abruptly. If you are charging the capacitor to its rated voltage and that is higher, say 10 volts, then you will have to put a resistor in line to limit the current flowing to the LED, but you will have a much longer interval of light.

Answer 4

The energy stored in a capacitor is 1/2 CV^2. So if you give it a 10 volt charge it will hold 0.0235 joules of energy. If you put a 1K resistor in series with the LED, you should keep it glowing for a minute or two. As someone pointed out, the current will drop linearly, so it will steadily get fainter. I'm assuming it's an electrolytic capacitor which won't take a higher voltage.

Answer 5

You could connect the relay across the capacitor so that it will remain de-energized until the cap has charged up to the relay's coil voltage. During the first RC time constant, the cap will charge to 63% of the supply voltage. During the second RC time constant, it will charge 63% of the remaining 37% for a total of 86% of the supply voltage. The third, fourth, and fifth time constants will continue the charging at 63% of the remaining voltage until, after 5 time constants, the cap can be considered fully charged. Since the exact voltage that will energize the relay varies a bit from one relay to the next, I don't know how much precision you can get from such a simple circuit. You may need to experiment a bit using different values in the circuit. To calculate one RC time constant, multiply the capacitance, in Farads, times the resistance in Ohms. The time will be in seconds. .

Answer 6

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Answer 7

idk