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Discharging one side of a capacitor only

  1. Jan 26, 2015 #1
    Hey PF,

    I have a thought experiment that's been driving me nuts, here it is:

    I take a parallel plate capacitor charge it up to a steady state and then isolate / disconnect it from the electrical system leaving me with two charged parallel plates. I then hook up a power supply to two corners of the positively charged plate, still leaving the negative plate isolated and try to run a current through the positive plate and into the ground.

    Will the positive plate act as a resistor because of the electrostatic field of the nearby negatively charged plate? Would the positive plate simply become neutralized by electrons from the power supply?

    The overall objective here is to produce a system with a negative net charge, but I'm a bit flummoxed about how to go about it.

    Cheers
     
  2. jcsd
  3. Jan 26, 2015 #2

    analogdesign

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    If you magically isolate the capacitor you will have an open circuit and the charge will stay "forever" on the capacitor. If you contact the two corners of one of the plate, the conductor making up that plate will act as a resistor in that it will let you conduct current through it but the net charge on it will remain unchanged. It's a zero-sum game at this point because you only can pull charges off one corner that you inject in the other corner. Nothing net changes.

    What you're describing is actually done in high-speed, high-accuracy analog-to-digital converters, like those used for example in digital cameras. The technique is called "bottom-plate sampling". Basically you have a switch from the input to the top-plate of the capacitor, and then another switch from the bottom plate to ground. The sampling instant is when you open the bottom plate switch, not when you open the top plate switch. This is important in practice because it means that the signal-dependent charge injection from the front-plate switch doesn't end up on the capacitor (it ends up being absorbed by the signal source since, like I said, an open-circuit capacitor looks like a high impedance).

    So to come back to your question, what you've got is a kind of bottom-plate sampling system. Therefore you can't get a negative net charge, sorry.
     
  4. Jan 26, 2015 #3
    Darn. OK well thanks for the answer, from the absence of information on the topic I figured that it was a bit unlikely.
     
  5. Jan 26, 2015 #4

    Svein

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    upload_2015-1-26_20-46-21.png
    I am not quite sure what you are talking about, but I can at least show how to generate a negative voltage. All it takes is a square wave generator (easily constructed with digital or analog circuitry) and feed it into the circuit shown here. The unloaded output voltage is close to the amplitude of the square wave.

    There are several possible variations of this circuit. One interesting variation is as a voltage multiplier - it is possible to get higher voltages if desired (but at the cost of increased output impedance).
     
  6. Jan 26, 2015 #5
    I was referring to producing a system with a charge imbalance. Considering a parallel plate capacitor you load it up with free electrons on the one side, which displaces the electrons on the other plate, I was wondering if there was a way to force some free electrons back into the positive plate without losing electrons on the other side, thereby producing a system with an overall negative charge.

    If I'm reading you correctly the system you're talking about here would produce a negative DC voltage?
     
  7. Jan 26, 2015 #6

    analogdesign

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    Yes, Svein's circuit produces a negative voltage. It is typically called a "charge pump" in the literature. It's quasi-DC as it will have pretty brutal ripple on it (depending on the value of C and the square wave frequency). If a stable voltage is needed it can be regulated or filtered.
     
  8. Jan 30, 2015 #7
    In the schematic one of the diodes is reversed and the circuit will only produce about 0.7V output. Otherwise the circuit can possibly double the negative input amplitude if the output common is connected to the generator common. Greater than double the voltage can be achieved by installing more components.
     
  9. Jan 30, 2015 #8

    jim hardy

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    What if....

    you made a parallel plate capacitor out of, say, a pair of musician's brass cymbals
    with facing sides insulated
    and insulated handles

    placed them against one another with insulated sides together, and charged them to 9 volts with a transistor radio battery
    then removed the battery
    then separated the cymbals by a few inches?
    Would one cymbal have a few extra electrons and the other a few too few?



    what would happen to the capacitance of your contraption as you separate the cymbals?
    What would happen to the voltage between the cymbals as they are separated?
    What would happen to the energy stored in the capacitance ??
     
  10. Jan 30, 2015 #9

    Svein

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    Oops (red face). Corrected circuit follows (with voltage doubler inserted).
    upload_2015-1-30_21-45-57.png
     
  11. Jan 31, 2015 #10
    Now you have a voltage quadrupler.....
     
  12. Jan 31, 2015 #11

    Svein

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    No, it's a voltage doubler. See the simulation below (I used a sine wave instead of a square wave because it was handy).
    upload_2015-1-31_8-50-38.png
     
  13. Feb 3, 2015 #12
     
  14. Feb 3, 2015 #13
    I was referring to the circuit and not to how you used it. After correcting the placement of the diode the circuit would be a standard voltage doubler for an alternating current. The second drawing with 4 diodes is clamping the input voltage to ground. In the simulation the entire input is positive. I can only go by your drawing errors. It doesn't matter anyway because it doesn't seem to be related to the question at the beginning.
     
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