Large 2 Farad Capacitor Dissipation

Discussion in 'The Projects Forum' started by kfrazie1, Jan 5, 2010.

  1. kfrazie1

    Thread Starter New Member

    Jan 4, 2010
    Anyone know how to determine how long a 2 Farad capacitor can power a 60 watt load at 12 volts 5 amps, assuming 16 volts is entering the capacitor? Been trying to figure this out for a while using the following equations:

    energy stored in a capac=1/2*C*V^2
  2. steveb

    Senior Member

    Jul 3, 2008
    You can't answer precisely unless you specify the details of how you will power the device. However, you can calculate an upper bound based on energy content. A 60 Watt bulb operating for 4 seconds has an energy dissipation as follows.

    E=P*t=60 W x 4 s = 240 Joules

    The capacitor charged to 16 volts has a stored energy as follows.

    E=1/2*C*V^2=0.5 x 2 F x 16 V x 16 V = 256 Joules.

    Hence, you can't expect more than about 4 seconds of power.

    However, you must realize that a practical circuit is going to do worse than this. For example, you would need a DC/DC converter to maintain 12 V while the capacitor is discharging. Let's say the DC/DC converter operates with an input voltage of 8 V to 16 V, and let's say the converter efficiency is 80 %, then you need to subtract the ending capacitor energy from the starting capacitor energy to get the usable energy, as follows.

    E=16^2-8^2=256-64=192 Joules

    Now factor in the efficiency.

    E=198 J x 0.8=154 Joules.

    The time of discharge will then be as follows:

    t=E/P=154/60=2.56 s

    The answer under these assumptions is about 2.5 seconds. You can do a little better with a DC/DC converter that operates below 8 V input and has greater than 80 percent efficiency, but this gives you an idea how to do the calculations.
  3. kfrazie1

    Thread Starter New Member

    Jan 4, 2010
    Thank you very much! Im gonna use a battery.