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:

Q=CV
ic(t)=C*dv(t)/dt
energy stored in a capac=1/2*C*V^2

 

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:

Q=CV
ic(t)=C*dv(t)/dt
energy stored in a capac=1/2*C*V^2

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.

 

Thank you very much! Im gonna use a battery.