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I was going over the equation for current flowing through a capacitor based on voltage and capacitance (i=C*dv/dt). It makes sense to me in general that you can find current going through a capacitor by multiplying the capacitance by the derivative of the voltage. From this, a constant voltage would result in zero current flow and a sine wave voltage would result in a cosine pattern. What confused me is that this would mean a capacitor with more capacitance would seemingly have more current flowing through it, which didn't seem to make sense. Wouldn't more capacitance mean less current going through a capacitor and more getting stored? Maybe I'm misinterpreting the formula. I'd appreciate help on this.

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  • \$\begingroup\$ Take it to the extremes. A capacitor with 0F capacitance would have no current flow at all, ever, while current would flow into a capacitor of infinite capacitance forever, continually increasing the charge. Since Q=CV, that would be an even larger infinity of charge. You can't store anything without any current flowing. Maybe it's best not to think of current flowing through a capacitor. Don't forget there's a gap in the middle. That's what makes it a capacitance. \$\endgroup\$ – Ian Bland Aug 28 '17 at 23:42
  • \$\begingroup\$ Oh, so could I think of it like there's more rapid charge build up on the two plates from higher current the more capacitance the capacitor has? \$\endgroup\$ – Tom Aug 28 '17 at 23:45
  • \$\begingroup\$ But that still throws me off because that suggests that you can't charge a capacitor at a constant voltage (because the current would equal zero). But that (i) is supposed to be instantaneous current, right? Not current over time. \$\endgroup\$ – Tom Aug 28 '17 at 23:51
  • \$\begingroup\$ So, you apply a DC voltage to a capacitor, and in our ideal model it charges instantly, thanks to the infinite current that flows for zero seconds... \$\endgroup\$ – Ian Bland Aug 29 '17 at 0:22
  • \$\begingroup\$ Wait a minute; current is the derivative of charge. If charge was changing at a constant rate, current would be some constant, and the instantaneous current would be the derivative of that. So, a constant voltage which would produce a constant current would have zero instantaneous current (if I'm correct about instantaneous current being the derivative of current). \$\endgroup\$ – Tom Aug 29 '17 at 0:34
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Wouldn't more capacitance mean less current going through a capacitor and more getting stored?

Current isn't stored in a capacitor, energy is stored. The greater the capacitance, the greater the energy stored for a given voltage across.

Perhaps it's better to start with the physicist's definition of capacitance:

$$Q = CV$$

where \$Q\$ is the amount of electric charge on one plate (there is \$-Q\$ charge on the other plate).

The greater the capacitance, the greater the capacity for charge \$Q\$ for a given voltage across.

So, fixing the voltage, a small capacitance has a relatively small amount of charge \$Q\$ on a plate while a large capacitance has relatively large amount.

In a circuit context, \$Q\$ can only change if there's a current through and so, taking the time derivative of both sides,

$$\frac{dQ}{dt} = i = C \frac{dv}{dt}$$

For arbitrarily large \$C\$, the time rate of change of voltage must be arbitrarily small for finite current, i.e., an arbitrarily large capacitance holds the voltage across fixed for finite current through.

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less current going through a capacitor and more getting stored?

Ah, that's a misconception taught in grade-school.

Actually, it all goes through!

Capacitors are like any 2-lead component: for each electron forced into one end, an electron pops out the other end, and none are stored inside. A fully-charged capacitor contains just as many electrons as a totally-discharged capacitor. So, if we place ammeters in the two capacitor leads, both ammeters read exactly the same value during charge or discharge. Capacitors don't store current (or more specifically, they don't store charge.)

Capacitors do store energy ...but then so do inductors. In other words, inductors store b-fields within their ferrous cores, and capacitors store e-fields in their dielectric layers. But inductors store just as many coulombs as capacitors do: meaning, none at all.

How does the misconception arise? One source is our typical schematic: we usually ground one side of a lowpass capacitor, then pretend that the grounded plate doesn't exist. We can charge the remaining plate positive or negative, as if it were a metal ball hanging in space. But in fact it's a pair of closely-spaced plates, and putting a negative charge on one plate will automatically produce an opposite charge on the second plate, even if the second plate is grounded.

The mechanical analog for capacitor is a spring, or a rubber band. When we stretch a rubber band, how much rubber have we stored? Zero; we've only stored energy, not rubber. The hydraulic analog of a capacitor is a pair of water tanks with a rubber separator. Hydraulic capacitors don't build up any water. Real capacitors only store charge-imbalance. To "charge" a capacitor, we move electrons away from one plate and place them on the other plate (which then exposes the positive protons of copper atoms in the first plate.) No electrons are ever injected or removed from the capacitor as a whole. (If they were, then the whole capacitor would immediately charge up to many billions of volts!) Capacitors only work because we're moving charges around inside. But then, that's how inductors work too: a "charged" inductor contains "spinning charge," while "charged" capacitors contain "stretched" or imbalanced charge.

Charged capacitors: they aren't charged with charge.

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