Capacitors as a Storage for Electricity

Question

everyone. I am researching into the "holding" of electricity. I found out about capacitors, but I am not sure that I understand. If you charge a capacitor with a battery in a circuit like this:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

, disconnect the capacitor, then connect the capacitor into a circuit like this:

schematic

^{simulate this circuit} will the capacitor power the LED? If not, then how so? Any answers are appreciated.

"(please disregard the charges of the battery, capacitor, and LED)" ... What? — Ignacio Vazquez-Abrams, Aug 29 '13 at 01:56
A suggestion: Add a current limiting resistor both for charging the capacitor and for driving the LED, and try this for yourself. It is a trivial experiment. You might be surprised at how long the LED stays lit with even a small capacitance. — Anindo Ghosh, Aug 29 '13 at 08:06

helloworld922 · Accepted Answer · 2013-08-29T02:25:52.683

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Yes, it will light the LED up but for a very short time.

The amount of energy stored by a capacitor is:

\$E=\frac{1}{2} C V^2\$

E is the energy stored, C is the capacitance of the capacitor, and V is the voltage across the capacitor.

For a 1uF capacitor charged to 9V, that is 40.5 uJ.

In addition to this, typical LED's have a forward current drop of ~2-3V, and capacitors typically have lower internal resistances compared to batteries. This means that you're going to get a very high current for a very short time, which could damage your LED.

Suppose you did limit the output current to 10mA (typical for small LED's), and the LED has a forward voltage drop of ~2V.

That means the LED would stay "lit" for (assuming we could extract every last bit of energy at the same rate from the capacitor):

\$ \frac{40.5uJ}{2V \cdot 10 mA} = 2.025 ms \$

Which is a very short time indeed.

edited Aug 29 '13 at 02:25

answered Aug 29 '13 at 02:17

helloworld922

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what does the e c and v stand for in your equation for the energy of the capacitor – Justin Chang Aug 29 '13 at 02:20
E = energy, C = capacitance, V = voltage – helloworld922 Aug 29 '13 at 02:26
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Key point: A typical indicator type LED with a nominal 20 mA current will stay quite visibly lit all the way down to 0.5 mA or less. See [my answer here](http://electronics.stackexchange.com/a/80368/14004), where I have a shot video showing how much longer an LED stays lit than calculations would indicate. – Anindo Ghosh Aug 29 '13 at 08:05
True, but even still the time the LED could theoretically stay on is on the order of hundredths of seconds. Still a very short time. – helloworld922 Aug 29 '13 at 16:25

score 0 · Answer 2 · answered Aug 29 '13 at 02:13

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The capacitor will power the LED until it no longer has enough voltage to do so, unless the current being drawn from the capacitor is sufficient to destroy the LED.

answered Aug 29 '13 at 02:13

Ignacio Vazquez-Abrams

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But will the capacitor carry charge from the battery even after a while from being disconnected? How would you slow down the discharge from the capacitor? – Justin Chang Aug 29 '13 at 02:15
Yes, until the capacitor's leakage drains it. Same way as with a battery: use a resistor. – Ignacio Vazquez-Abrams Aug 29 '13 at 02:16

Capacitors as a Storage for Electricity

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