Automatically discharging capacitor

Question

Say I have a capacitor of 100uF connected to 2Volts battery supply (In real circuit, this will be a solar panel exposed in the sun)... After some time, the capacitor will be charged fully with a voltage of 2Volts across its leads.. Now, How do I automatically discharge the capacitor to the load? I do this because output of capacitor will have increased power output than the power output of the battery... Thus, the load(maybe a motor) will run at greater speed than it would run if I had connected it directly to the battery..

This question is a continuation of the previously asked question

Edit:

Have a look at this circuit:

In this circuit, Initially I close switch1.. the capacitor charges fully.. when it is fully charged, i open switch1, close switch2.. the capacitor disharges to the load.. Then I open switch2, close switch1 again.... the process repeats..

I want to achieve this action automatically.. using BJT's or something.. Hope the question is clearer

Answer 1

If you made this to work, it wouldn't do what you hope. The capacitor won't "concentrate" the energy from the battery / solar cells / any other energy source. Not much anyway, unless your power source is extremely weak and your load is super light.

The energy stored in a capacitor is:

\$ U = \frac{1}{2}CV^2 \$

Let's say you can charge that capacitor up to 5V. Then:

\$ U = \frac{1}{2} 100\mu F (5V)^2 = 1.25 mJ\$

A watt (\$W\$) is one joule (\$J\$) per second. So if your load is \$1W\$, this capacitor can store enough energy to run it for \$1.25ms\$.

Put more intuitively, 1 joule is about the energy it takes to lift an apple one meter up. So, \$1.25 mJ\$ is the energy it takes to lift an apple \$1.25mm\$ or 0.05 inches. This is a truly tiny amount of energy. I'm having a hard time imagining what useful work you could do with so little energy. You will need a much bigger capacitor, or a much higher voltage.

Your inspiration for this idea seems to be a camera flash. Firstly, a camera flash looks bright, but it really does not release that much energy in each strobe. Secondly, it works by charging a capacitor to a very high voltage. You can read about boost converters to get some idea of how this is accomplished, but even though a boost converter can create high voltages, it can't create energy. You almost surely need a different approach to whatever your underlying problem is.

Answer 2

For the long time average the capacitor will do nothing to increase the "power output" of the source.

For short term transients (i.e. current surge demands coming from the load) the capacitor can help to supply current to the load and minimize the voltage sag in the power bus if:

1. The impedance of the connection from the capacitor is lower than the impedance of the connection from the power source to the load.
2. The duration of the current surge is shorter than the time it takes the power source to recharge the capacitor back up to the fully charged state (which really means charged up to the voltage of the source).
3. The internal impedance of the capacitor is low enough such that the voltage drop across that internal impedance due to the discharge current does not add more than the accepted amount of voltage sag.

Edit after Original Question is Edited

In the switching circuit shown you cannot gain any power or energy as a result of switching the two switches alternately. At best with ideal components the transfer of energy from the battery to the load will be 1::1. In real life the transfer will be less - The switches have resistance, the wires have resistance, the battery has internal impedance, the capacitor has a leakage and so we go.

You could add some MOSFET switches to be used in place of the mechanical switches but you would also have to add an additional circuit that was an oscillator with two square wave outputs that would be used to switch the two MOSFETs that substitute for the switches.

Answer 3

Use a bleeding resistor to discharge the cap.