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I found this illustration from the SparkFun website on how a capacitor charges and discharges, but I have some difficulties with it (and capacitors in general): enter image description here

  1. What initially causes the capacitor to accumulate a charge difference? I would expect the current to see this as an open circuit, given that no current flows through the dielectric to ground, and therefore there would be no initial flow.
  2. When we discuss a build-up of positive charge on one of the plates, we are saying the valence electrons have been repelled by the negative charge of the the other plate. Maybe this is a chicken/egg situation, but shouldn't the electrons on the negative side of the plate be repelled by the same force (even more since it's closer) as the positive side, inhibiting the charge buildup from occuring in the first place?
  3. In the picture below, won't the discharge reverse the direction of current? That is, it's going to flow from the negatively charged plate, across the LED, resistor, gate, and then to the positively charged plate?
  4. When the capacitor discharges, the current seems to want to flow to redistribute/equalize charge across the circuit. This is different than just flowing to ground (or the positive terminal on the battery), as I am used to seeing. If when the gate closed to allow discharge there was also a path to the positive terminal on the battery, would this change anything?
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closed as too broad by Eugene Sh., Voltage Spike, uint128_t, Dmitry Grigoryev, jonk Dec 23 '16 at 10:15

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    \$\begingroup\$ I think you have mixed too many different abstraction levels here. First understand the physics of the capacitor (the fields, plates and electrons stuff). Then go back and treat it as a discreet circuit component with a behavior abstracted by a math formula derived from the physics. \$\endgroup\$ – Eugene Sh. Dec 21 '16 at 20:32
  • \$\begingroup\$ Wiki Electric Field, they are everywhere \$\endgroup\$ – Voltage Spike Dec 21 '16 at 21:00
  • \$\begingroup\$ Here a SIM I made to show with more details including series resistance of each part. Never mind that Relay uses more current than LEDs goo.gl/8p1BZf \$\endgroup\$ – Sunnyskyguy EE75 Dec 21 '16 at 22:59
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    \$\begingroup\$ Their animation is wrong. The yellow color is supposed to indicate current, but during "charging," there's no current in the switch or the positive terminals!! Very silly. Try Falstad simulator instead. (I suspect that this artist suffers from the "technician misconception" that "electricity is the electrons." This error causes them to believe that the current within the positive wire must be zero. (It's positive? So, no electrons? No current?) Nope! Try to avoid the crazy stuff found in many technician books/websites, and stick with solid physics and EE, both texts and forums. \$\endgroup\$ – wbeaty Dec 22 '16 at 3:13
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  1. Current does flow through the capacitor, through the dielectric. But only while it's charging.

  2. This discussion of capacitors may help. The charge buildup does repel further charge buildup. That's why the rate of current flow decreases as the capacitor charges. Work must be done on the charge to move it - the integral of that work is the energy stored in the capacitor.

  3. Yes. The animation is very bad at showing this.

  4. Yes, current will tend to flow so as to equalise the potential on both sides. Neither is necessarily related to ground. That's how a Cockroft-Walton multiplier works.

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    \$\begingroup\$ Current doesn't flow through the dielectric. The charging of the capacitor causes a current flow but into the capacitor, not through it. The dielectric is an insulator. \$\endgroup\$ – TonyM Dec 21 '16 at 22:05
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    \$\begingroup\$ @TonyM You've picked up a misconception. Dielectrics support Maxwell's displacement current, which is the basis of all modern EM technology, and covered in physics class. Capacitors are closed circuits. But you're right: in a vacuum-cap, charged particles don't cross the gap, only fields leap the gap. However, with most caps (especially ceramic,) the dielectric is full of mobile electrons, giving a charge-flow which is tens of thousands of times larger than the pure displacement current of a vacuum-capacitor. Heh, the charge in the ceramic is attached to little tiny springs! They bend! \$\endgroup\$ – wbeaty Dec 22 '16 at 3:22
  • \$\begingroup\$ @wbeaty, pjc50, I was sure of my comment but you're convincingly sure of yours...either way, I think I have some capacitor-current reading to do :-) Thanks for the pointers and a very Merry Christmas to you both. \$\endgroup\$ – TonyM Dec 23 '16 at 22:11
  • \$\begingroup\$ @TonyM When I sat down and tried to craft my own explanation of capacitors, I came up with this: the engineer's capacitor amasci.com/emotor/enCap.html Even with no charges crossing the gap, a capacitor acts like a weird sort of conductor, since we cannot force any charges into one terminal alone. Charges must "flow through," much like a resistor or a coil ...even if they genuinely don't flow through! The thin gap is bridged by incredibly intense field-repulsion, and that makes us treat capacitors like we treat inductors: they're 2-terminal devices with an open pathway inside. \$\endgroup\$ – wbeaty Dec 24 '16 at 1:02
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shouldn't the electrons on the negative side of the plate be repelled by the same force (even more since it's closer) as the positive side...

Electrons flow from the battery to build up on the negative plate. This forces electrons away from the positive plate where they flow out the lead and to the battery (+), which is collecting them.

...repelled by the same force as the positive side, inhibiting the charge buildup from occurring

This is what happens when the capacitor is charged. Since the dielectric is an insulator, no electrons flow through it; they just build up until the force from the battery is balanced by the force of the crowd of electrons pushing back.

This is different than just flowing to ground

"Flowing to ground" is a common lazy way to overlook the fact that the current ultimately must flow back to its power source--in this case, the battery, or the capacitor plate. Connect a ground symbol to the bottom of the circuit, and now we can talk that way. But when we get this specific about where the currents actually go, it's more accurate to identify where they go, not just "to ground".

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https://goo.gl/8p1BZf < JAVA Falstad Simulator

Consider charging a Cap like jump starting a dead battery. ( a tiny one) A battery is like a Cap but thousands of times more charge storage in Farads.

Do you understand that rate of change of charge flow in milli-Coulombs/second = milliamps. Since Battery has very low series resistance e.g. 10 milliohm for LiPo and some E-caps are similar but much lower capacitance per unit volume compared to a battery.

I assume you understand polarity of current is not the flow of electrons for electronics by convention rather from positive to negative for logical reasons not physics.

This is to add to the other answers not replace them. enter image description here

Scope traces show max:min only for each node. So Cap shows a steady -ve drain and large peak pulse. such that if ideal Ichg * t integral area under the charge spike = steady drain to LEDs Id * t. for conservation of charge. But I added realistic Rs series values for each part. Since Red has a lower forward voltage it illuminates longer.

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