Incrementing charge/voltage on capacitor with periodic pulses of fixed power

Discussion in 'General Electronics Chat' started by positive8, Oct 2, 2016.

  1. positive8

    Thread Starter Member

    Sep 27, 2016
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    Simply put, if the resistance of a capacitor increases as it charges, can I send periodic DC pulses into it causing the voltage across it to eventually rise to its maximum rated power, thus finally fully charging it?
    For example, in a very simple circuit of a power source, resistor and capacitor, as the relative resistances of the resistor and capacitor change over time, the capacitor should incrementally increase in voltage even though the power source is sending the same peak voltage out with each pulse? (Assume negligible cap discharge and the ability of the power source to match the cap voltage once the respective resistance change sufficiently)>
     
  2. wayneh

    Expert

    Sep 9, 2010
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    The capacitor will charge while the power is on, whether it's pulsed or continuous.

    It's incorrect to use the term "resistance" for the capacitor. The voltage across it changes, but not the "resistance". An ideal capacitor offers zero resistance and a real capacitor offers very little, effectively zero resistance. It's not important except in high frequency applications.
     
  3. positive8

    Thread Starter Member

    Sep 27, 2016
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    But the impedance (I was lazy using the term resistance) changes drastically across the cap even at relatively low frequencies, true?

    Let me rephrase the question more competently (I hope). If I generate 20ms DC pulses from a power source but the time constant does not allow the cap to fully charge (say 25V source only increases cap voltage to 10V first pulse), but the impedance of the capacitor increases due to the frequency of the pulse, will I be able able to charge the cap to maximum given enough additional pulses.
     
  4. #12

    Expert

    Nov 30, 2010
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    Like wayneh said, simply false.
    Definitely.
    Ummm...bad choice of words. Capacitors store charge which can be turned into power, but it isn't power until the amp-seconds are converted to power.

    This thing you speak of is called a charge pump. The ICL7660 chip does this.
    and finally, the impedance of the capacitance decreases as frequency rises, not because of voltage changes.
     
  5. positive8

    Thread Starter Member

    Sep 27, 2016
    38
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    Yes, excuse my terrible terminology. I mean joules not power.But impedance across the cap most certainly does vary according to frequency, from what I've read.
    I'll research the charge pump, but in this case I'm using an incidental analog periodic pulse whose energy I want to store in a capacitor. From what you've told me, it seems that I can create a primitive "charge pump" effect from these pulsing transients which are "additive" in the sense they sequentially raise voltage across the cap as a result of the impedance developed from the pulses.
     
  6. crutschow

    Expert

    Mar 14, 2008
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    You have several muddy concepts about capacitors.

    Capacitors store energy, not power.

    The capacitor voltage increases as you charge it, thus reducing the charge current for a given charge voltage and source impedance as the capacitor charges.
    That's what causes the exponential charge time for a capacitor from a resistive source.

    The AC impedance of a capacitor is inversely proportional to frequency.
    That's unrelated to the frequency of DC charge pulses.
    Longer pulses and/or higher pulse frequency will charge the capacitor faster.
     
  7. wayneh

    Expert

    Sep 9, 2010
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    It's hard to tell what you're actually trying to accomplish. I suggest leading with that to get the best feedback.

    It sounds a little like you want to capture the energy of pulses, although in that case I don't know why you would have a resistor in series. A diode-capacitor "peak detector" can convert a pulse into energy stored in the capacitor. That raises the voltage on the capacitor, though, so subsequent pulses are not fully captured. Eventually after enough pulses, the capacitor shows whatever the peak voltage is, less the diode voltage drop. No additional energy is being captured.
     
  8. positive8

    Thread Starter Member

    Sep 27, 2016
    38
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    Okay, thanks. I was mistakenly assuming the wavelike phase of a half-sinusoidal purely DC pulse would be interpreted by the capacitor as having AC characteristics. See my mistake now.

    Thanks for the nice summation on salient capacitor facts you enumerated , too! It's a very nice encapsulation.
     
  9. positive8

    Thread Starter Member

    Sep 27, 2016
    38
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    Lol, that's exactly what I'm doing.So I a have a Schottky diode in series with the cap (the Schottky does have some resistance, so I just loosely referred to it as a resistor in this dialog). I'm assuming the incoming AC pulse can be eventually fully rectified.
    The reason for my original question was that I was measuring a recurrent 10V pulse on the DC waveform and consequent charging of the 25V cap. THEN, after repeated DC pulses of the same voltage, I saw smaller incremental increases in the voltage across the cap.
    Because I was using a pulsed DC waveform, I wasn't sure if this was different from the standard capacitive charging profile, specifically because I made the mistake of assuming a half-sinusoidal pure DC waveform would be interpreted by the cap as an AC waveform.
    I see my mistake now. Thanks for your cogent response!
     
  10. hp1729

    Well-Known Member

    Nov 23, 2015
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    Is that what we call a charge pump? The way we convert PWM outputs of a microprocessor to an analog voltage.
     
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