# Brief capacitor charge and discharge through LEDs circuit

Capacitors store energy that was provided by a power source. Charging through an LED, you will see that most capacitors charge pretty quickly because the LED won’t be bright for very long. The stored charge in the capacitor provides the power to light a different LED for a brief period of time while the capacitor discharges through it.

Video below shows an easy way to build this circuit on an electronics prototype breadboard.

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This circuit works better with voltage supply that is slightly higher than a normal LED circuit since both LEDs drop voltage.

## Video

#### Important points:

• Farad is the unit of capacitance used in formulas. 1 farad (a super capacitor) is much larger than is used in basic electronics. Typical values are in the microfarad (uF)/millionths of a farad, nanofarad (nF)/billionths of a farad, and picofarad (pF)/trillionths of a farad.
• 2 closely spaced conductive areas (plates) can easily get an imbalance of charge when a voltage is applied. There are extra electrons on one side (-), and reduced electrons on the other side (+). The more charge imbalance there is, the more voltage that the capacitor will build up. larger plates (higher value capacitors) need to move more charges to get to a certain voltage than capacitors with smaller plates (lower value capacitors).
• Charge imbalance still wants to return to normal. So, once there’s a conductive path between the plates (a short circuit or through resistance), the charges move back to equality on the plates. The voltage across the plates drops while current flows. Avoid making a short circuit discharge of a large value and/or high voltage capacitor. Instead discharge it through a resistor to prevent too much current from flowing.
• A 1 farad super capacitor will gain 1 volt per second if charged at a steady 1 amp of current. It will drop 1V per second if discharged at 1 amp of current.

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