Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Why use capacitors to store energy?

  1. Feb 3, 2010 #1
    Why are capacitors used to give pules of very high potential difference (for e.g. x)?...what's the advantage?

    The source charging the capacitors should have potential difference less than x if capacitors have an advantage here; for e.g. with a source x/2, we can getting a potential x...is this happening here?
    Last edited: Feb 3, 2010
  2. jcsd
  3. Feb 3, 2010 #2
    I'm not sure I really understand your question, but maybe the capacitor's ability to release energy quickly is a part of the answer?

    EDIT: In response to your edit, are you talking about a http://en.wikipedia.org/wiki/Charge_pump" [Broken]?
    Last edited by a moderator: May 4, 2017
  4. Feb 3, 2010 #3
    May be it's ability to release a very high amount of current is the answer. The source might just be inept of giving a high amount of current at that potential, thus capacitors are charged first with a resistance in series to lower the current rate. Then the high current output is taken from the capacitors.

    Any other ways to step up the voltage using capacitors other than charge pumps?
    Last edited by a moderator: May 4, 2017
  5. Feb 3, 2010 #4


    User Avatar
    Science Advisor
    Gold Member

    Any actual source is going to diverge from the behavior of the ideal sources that we use in circuits. The first rough estimation of a more realistic source is to assume that there is an internal resistance. This would be modeled as having the source have a resistor in series (for a voltage source) or in parallel (for a current source I think) between the source and the output terminals. In this way, we see that power gets siphoned off from the source and expended into the internal resistance. It is thus easy to see that this will restrict high current draws on a voltage source for example. That is, if we place a low resistive load across the voltage source terminals, the effective resistance will be dominated by the internal resistance. So even as we place a smaller and smaller resistive load, the drawn current will change very little due to the comparatively larger internal resistance.

    Another effect of the internal resistance is that it behaves as a dampener in a LC circuit. If I place a function generator that has an output impedance of 50 ohms onto a LC circuit, its behavior will be like an RLC circuit, where the R element comes from the output impedance/internal resistance of the source. This affects the supplied voltage/current relationships of the LC circuit. For example, it would shift the resonant time constant of the circuit from what it would have been with an ideal source. So the internal resistance can also affect the transient behavior of a circuit.

    This is one reason why fast transient and high current draws can be better taken off of capacitors since they can be precharged with enough charge to drive the necessary fast current spikes. Capacitors also are simple filter elements. Not only can they provide current compensation when the source lacks the current for a sudden transient, they also can act as a simple low pass or high pass filter at the same time. In a simple power supply circuit that uses a transformer to rectifier to filter to voltage regulator, the filter stage is just a resistor and capacitor. Both elements work double duty as the capacitor and resistor first act as low pass filters to help smooth out the rectified AC wave from the rectification stage. In addition, the capacitor also helps by being a charge resevoir and the resistor controls the bias current that biases the zener diode that acts as the voltage regulator.
  6. Feb 4, 2010 #5
    Ok...thanks for the explanation. Sort of like maximum power theory.
  7. Feb 4, 2010 #6
    In automobile ignition systems up to about 1970, the energy was stored as current in the ignition coil, and when the coil current was suddenly interrupted, there was a 300-volt spike on the coil primary, leading to a 30,000 pulse on the coil secondary.

    Bob S
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook