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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:

enter image description here

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

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closed as not a real question by Anindo Ghosh, Dave Tweed, Olin Lathrop, placeholder, The Photon Jan 16 '13 at 4:42

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

    
I don't know what you mean by "power output" of the cap. A capacity of 100µF is not much to supply even a small motor, and usually "too much power" is not the problem for solar panels. My suggestion is to test the system in real conditions, and then look if you really have the problem of excess power, and then maybe limit the capacitor voltage by a Zener diode or sth. similar. –  0x6d64 Jan 14 '13 at 9:24
    
"too much power" from solar panels is not the problem.. I am trying to store the the power from solar panel in capacitor and then use the concentrated power (sudden capacitor discharge! ) –  BLOB Jan 14 '13 at 9:33
    
I think "solarbot" would be a useful search term for this question. –  Brian Drummond Jan 14 '13 at 10:11
    
thank you Brian.. solarbot search has provided some useful links.. working on it –  BLOB Jan 14 '13 at 10:22
    
2 comments on your circuit : (1) your impedance values are (approx) reversed : load is lower impedance than source, and (2) there is no need to open switch 1 (because source impedance is high) which simplifies your task. ps. seen this page? beam-online.com/Robots/Circuits/1381.html –  Brian Drummond Jan 14 '13 at 10:58
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3 Answers 3

up vote 3 down vote accepted

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.

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The symbol for energy is E, not U, I believe. –  Camil Staps Jan 14 '13 at 12:47
    
@CamilStaps: in an electronics context, \$E\$ is usually voltage, as in \$E=IR\$. I've seen \$W\$ and \$U\$ used for energy, but really it doesn't matter, since the text around the equation should define the symbols. –  Phil Frost Jan 14 '13 at 13:02
    
Whoo, never seen that. In my country U is voltage and E is energy. W would be workforce, so I think it's a country depending issue. Sorry for that! –  Camil Staps Jan 14 '13 at 13:23
    
@CamilStaps: yes, I should have qualified my answer with "by US conventions..." –  Phil Frost Jan 14 '13 at 13:26
    
Well I think it's clear :-) –  Camil Staps Jan 14 '13 at 13:34
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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.

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I understood all the points.. But my problem lies elsewhere.. I have edited the question.. plz hav a look –  BLOB Jan 14 '13 at 9:45
    
@BLOB - I added some more comments to my answer. –  Michael Karas Jan 14 '13 at 10:45
    
I know I cannot gain any additional energy.. What I am hoping is to harness the "concentrated energy" from the capacitor instead of "diluted energy" from the batter/solar panel.. Yes.. I am aware this will make the motor run intermittently.. –  BLOB Jan 14 '13 at 10:56
    
@BLOB - Seems to me that you should then experiment with a prototype circuit that uses two push buttons to work just like the circuit that you showed to see if this gives you the type of performance that you would need. I wouldn't mess around with trying to automate the circuit until you have connected this up to your solar cell, capacitor motor and a couple of switches. This way you can evaluate feasibility. –  Michael Karas Jan 14 '13 at 11:38
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Use a bleeding resistor to discharge the cap.

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It says the Bleed resistor is used for 'safety purposes' where we need to safely discharge the capacitor.. But In my case, it is not the case... I want to automatically discharge the capacitor ONLY when it maximum voltage (2V in this case) is reached.. The bleeder resistor values as I see are quite high (100K) which wont be useful for running the motor (load) –  BLOB Jan 14 '13 at 9:01
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