# Charging/Discharging a supercapacitor.

1. ## Charging/Discharging a supercapacitor.

Let's say I feed a series of supercapacitors from a PV panel with a 5V and 5mA average.
With the use of a boost regulator after the supercapacitors I want to to have an output
of 5V and 500mA.

How can I find how much will the discharging time of the supercapacitors will be
when I want them to produce 5V along with 500mA output?

Please ignore the balancing circuit for the supercapacitors or the input regulator after the
the PV panel.

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2. ## Re: Charging/Discharging a supercapacitor.

A 1 farad capacitor will loose 1V if the current drain is 1 A/Second, so your one will loose .5V /second. But the current will change as the voltage on the cap decays so as your load is 5/.5 = 10 ohms, once you have decided on the capacitor value, you can work out the time constant and apply the equation.
Frank

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3. ## Re: Charging/Discharging a supercapacitor.

According to your explanation, you will be taking a constant power from the supercapacitors (considering an ideal boost). The calculation can be easy:

Energy available in the capacitors = 1/2 * C * ( (5 V)^2 - (Boost_Min_Input_Voltage)^2).

Time = Available Energy / (Power/boost_efficiency)
(you will be draining 2.5W/boost efficiency power from the supercapacitor)

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4. ## Re: Charging/Discharging a supercapacitor.

Originally Posted by emontllo
Time = Available Energy / (Power/boost_efficiency)
(you will be draining 2.5W/boost efficiency power from the supercapacitor)
I didn't understand the last formula. Is it (Energy*boost efficiency)/Power?
Supposing that I have an energy of 60 Joules and the boost efficiency is 75%
then time=60*0.75/2.5 =18 secs?

Thank you :)

5. ## Re: Charging/Discharging a supercapacitor.

Exactly. Therefore, if you improve boost efficiency you will increase the available time.

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