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Capacitor voltage in charge pump

  1. Nov 3, 2009 #1
    Hi all,

    I have the feeling I'm missing something big. Consider first a capacitor having +9V on both terminals, in this case nothing happens. Now I set the negative terminal to 0V, and make sure a current is able to flow; the capacitor will now charge up to 9V, right?

    Ok, now I make sure the capacitor is unable to discharge in any way; so it has its positive terminal always 9V higher than the lower, right?

    Here's the thing: If I now connect the negative terminal to a +9V potential again, then the positive terminal has +18V (of course, all relative to 0V ground). I understand this has to be the case in terms of the above boldface line, but I don't see how this works, physically.

    I have the feeling I'm almost there; in terms of energy stored per coulomb etc. The battery is still at a +9V relative to ground, but the same capacitor connected to it is now at +18V relative to ground..?
  2. jcsd
  3. Nov 3, 2009 #2


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    That's the thing: think of the capacitors as two batteries. You have two 1.5V cells, and then when you put them in series, you have a 3V battery!

    Your confusion stems from the part about having both capacitors relative to ground. You charge them up that way, connecting the negative terminal of both to ground, and positive of both to whatever supply is available (let's say 1.5V). Now, you disconnect the negative terminal of one (such that it's no longer connected to ground) and connect it to the positive terminal of the other. Since you charged up both [STRIKE]batteries[/STRIKE] capacitors, you measure 3V at the positive terminal of the capacitor you moved around.

    A charge pump just does all the moving around for you (electrically, and with a small switching network).

    When you charge up a capacitor, it acts exactly as a battery. We used to charge up (largeish) capacitors to a few volts and throw them around at each other in some of the labs. Yeah, kinda childish, but nobody ever got hurt from the small voltages and still-relatively small capacitors we had available (somebody stuck an electrolytic one in the mains, but that just exploded, as you might expect).

    With a large enough (capacitance-wise) capacitor and high enough potential, you can actually arc-weld them to metal surfaces (or at least leave some scorch marks).
  4. Nov 3, 2009 #3
    Thanks for the quick reply!

    Ok, that part is very much clear to me now. I have attached a quick drawing of the circuit which made me think about this, I know it is not the nicest drawing but ok.. The text
    that is with this image is somewhat as follows:

    The inverters work as follows: input 9V -> output 0V and input 0V -> output 9V

    1) At the beginning, there is no voltage at the input of the inverter 1, so its output is 9V. The positive (in this case upper) terminal of C2 is also at 9V.

    2) A current flows through the resistor, charging up C1. Once C1 is charged to 9V, the inverter
    output is now 0V. This means again a current flows, this time charging up C2 (this is the capacitor I meant in my previous post).

    3) Capacitor C2 is now unable to discharge, because of the diode and the 9V at its upper terminal. So, C2 now maintains a 9V potential difference.

    4) Since C1 is discharging again, the inverter switches to 9V and so the negative terminal of C2 is now at 9V. The upper terminal is now at 18V (relative to ground).

    I can explain this to myself by saying that the capacitors upper lead always has a potential higher by 9V than the lower. But the battery is connected to the same capacitor, and supplies only 9V.. which confuses me because then I would say the positive terminal of C2 is at 9V..

    Further, saying that the capacitor has a 9V potential difference, means that the charges
    at the positive side have 9 Joules/Coulomb more energy than those at the lower, right? Then how does connecting the negative lead make the charges at the positive terminal have 18 Joules/Coulomb (since the lower now has 9 Joules/Coulomb)?

    I realize that my question is not very clear maybe, which probably reflects the confusion I have.. I'm sorry if I completely missed the point in your reply!


    Attached Files:

  5. Nov 3, 2009 #4
    This circuit charges up C3 at first as Inv2 output is 0V. Once C3 is charged there is no discharge path due to the diode. The diode seems to be where you are going wrong. The diode allows the capacitor to charge up but in theory seperates it from the battery once charged.
    The rest of the circuit does as it says and finally changes the output of Inv2 to 9V.
    Therefore C3 still has 9V charge with respect to Inv2 output terminal.
    Therefore you have 9V between Inv2 output terminal and 0V (battery -ve) and then in addition you have the 9V charged in C3. 9V + 9V = 18V.
    Hope this helps.

    http://www.calibrepower.co.uk" [Broken]
    Last edited by a moderator: May 4, 2017
  6. Nov 5, 2009 #5
    Thanks to both of you for the replies. I think i've found my problem and solved it; It comes down to the electrical field inside the capacitor, storing the energy. Since the electric field is the gradient of the potential, I can always just add a constant to the potential (in this case just the 9V) without changing the electric field and hence the energy the capacitor has to store.

    Further, it is the diode that enables the differences in potential between the battery (9V) and the capacitor (C2 at 18V), since this just means current cannot flow in reverse and therefor the potential remains.

    Thanks again!
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