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  1. #1
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    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.

    Thank you in advance.

    •   Alt12th July 2012, 16:41

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  2. #2
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    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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    •   Alt12th July 2012, 19:51

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    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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    •   Alt13th July 2012, 11:17

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

    Quote Originally Posted by emontllo View Post
    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. #5
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    Re: Charging/Discharging a supercapacitor.

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


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