A few questions about capacitors

[Guidestone]

Hey guys, I know I have left some topics unanswered. I'm working on reading more about electricity to improve my understanding. Thank you for having helped me in earlier topics, I have certainly begun to comprehend a lot more.

Now, right to the question. So I got a capacitor being charged up by a battery. I know that capacitors are basically composed by two plates separated by an insulator. Once the circuit is closed, one of the plates gets negatively charged and the other one gets positively charged until there is a voltage equal to that of the battery.
Is voltage in the capacitor caused by the electric field between the plates?
If I take the charged capacitor out of the circuit and connect its terminals to a bulb for example, does it get discharged because negative charges flow to the positive plate through the bulb, but not through the space between the plates?
Does current stop flowing in a circuit once the capacitor is completely charged?
Why do electrolytic capacitors are electrolytic? I mean, I know an electrolyte allows current to flow for example in water, by placing salt in it. The electrolyte is between the plates, but there should be an isolator instead, not a conductor.

Thank you in advance. Please correct me if I'm wrong in any of my suppositions.

 

[Hesch]

Is voltage in the capacitor caused by the electric field between the plates?
If I take the charged capacitor out of the circuit and connect its terminals to a bulb for example, does it get discharged because negative charges flow to the positive plate through the bulb, but not through the space between the plates?
Does current stop flowing in a circuit once the capacitor is completely charged?

Yes to all.

The electrolyte is between the plates, but there should be an isolator instead, not a conductor.

The "plates" are made of aluminum. By inducing dc-current through the assembled capicitor, the surface of one of the plates is anodized, which creates an insulating layer ( aluminum oxide ).

 

[Baluncore]

Is voltage in the capacitor caused by the electric field between the plates?

The battery voltage is assumed to be fixed by the internal chemistry and a sufficient quantity of reactants. When you connect the battery to the capacitor, the electric field from the battery is passed along the wires to the capacitor. Current flows as charge is redistributed. When sufficient charge has flowed to charge the capacitor and the capacitance of all the wire surfaces, the current will fall to zero. Once current ceases to flow, the battery terminal, connecting wire, and the connected terminal of the capacitor, together form an equipotential.

 

[Guidestone]

The battery voltage is assumed to be fixed by the internal chemistry and a sufficient quantity of reactants. When you connect the battery to the capacitor, the electric field from the battery is passed along the wires to the capacitor. Current flows as charge is redistributed. When sufficient charge has flowed to charge the capacitor and the capacitance of all the wire surfaces, the current will fall to zero. Once current ceases to flow, the battery terminal, connecting wire, and the connected terminal of the capacitor, together form an equipotential.

So once the capacitor has been completely charged and the current fallen to zero, is it like having an open circuit?
Also, capacitors avoid sudden changes in voltage, right?

 

[Guidestone]

Yes to all.

The "plates" are made of aluminum. By inducing dc-current through the assembled capicitor, the surface of one of the plates is anodized, which creates an insulating layer ( aluminum oxide ).

Oh! Interesting! Thanks! A little bit hard to understand though [emoji29]

 

[meBigGuy]

So once the capacitor has been completely charged and the current fallen to zero, is it like having an open circuit?
Also, capacitors avoid sudden changes in voltage, right?

Kind of, but not precisely stated. If a capacitor is connected to DC and is fully charged, then no current flows.
I wouldn't go further and say it is like having an open circuit because it isn't an open circuit. It is a charged capacitor. Just because they both have 0 current flow, don't say they are "alike".

Capacitors do not "avoid" sudden changes in voltage. The current through a capacitor is I = Cdv/dt, so in order to get a sudden voltage change (high dv/dt) a high current is required. That is not "avoiding". Maybe you can say it is impeding a sudden voltage change by presenting a low impedance to high frequencies.
But even that makes me uncomfortable.

A capacitor requires a current to produce a voltage change. The faster the change the higher the required current.

To change the voltage across a capacitor you must put energy into it. 1/2 CV^2 if I remember correctly. The same amount of energy is required whether the voltage change is sudden or gradual. If it is sudden, then the energy is required "suddenly" also, meaning high current.

Hope that helps.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html
https://www.khanacademy.org/science...c/circuits-with-capacitors/v/energy-capacitor

 

[Baluncore]

Also, capacitors avoid sudden changes in voltage, right?

Yes. Because the voltage is the integral of charge, even with an instant step rise in current to the capacitor it can only have a ramp in voltage.
Q = I * T and C = Q / V.
Similarly, with inductors a step change in voltage results in a ramp in the current.

 

[Guidestone]

So, what's the point in having a device that will eventually just stop the current flow within a circuit? I have had lessons about them when it comes to circuit analysis but I still don't understand their utility and they just seem to be everywhere. The only useful thing about it was the rectification of AC current.

 

[Baluncore]

Capacitors couple AC signals between circuits that have different DC bias voltages.
Capacitors provide a low impedance path to ground for AC signals independent of DC voltage.

A capacitor is a negative reactor, while an inductor is a positive reactor. Together they can make a resonator at the frequency where their reactance sum is zero. Neither wastes energy like a resistor.

Because capacitors are charge integrators they are very useful for processing AC signals.
When reactors are used as potential dividers with, or without, resistors they make frequency selective circuits.

 

[meBigGuy]

So, what's the point in having a device that will eventually just stop the current flow within a circuit?

There are very few circuits where they "just stop the current flow". In fact, probably just the one you described. There is not much point in putting them across a battery.

But, they have current flow in any circuit where the applied voltage is varying.

 

[Hesch]

So, what's the point in having a device that will eventually just stop the current flow within a circuit?

The purpose could be to keep a DC-voltage stable. In the real world a DC-voltage contains AC-voltage as well. Then a decouple capacitor is connected between the noisy node and ground, thereby short circuiting the AC-voltage, but not the DC-voltage that must supply IC's, opamps, etc.

In many electronic diagrams, these decouple capacitors are not shown, but the should be there to avoid hum and undesired oscillation. But you may, in the corner of a diagram, see 20 capacitors in parallel to be spread out in the diagram, just placed in the corner to make the diagram more clear.

But an example:

Here all the capacitors are used for decouple purpose, C2+C4 ( 100μF ? ) to decouple hum ( 60Hz ) and C1+C3 ( 100nF ? ) to decouple higher frequencies.

In power supplies, huge capacitors are inserted to smooth the DC output voltage.

 

[meBigGuy]

https://en.wikipedia.org/wiki/Low-pass_filter#Electronic_low-pass_filters

 

[leright]

Essentially, most of the applications of capacitors in electronics are filters of some sort, since capacitors do not allow current to flow due to DC and lower frequency voltage waveforms but allow current to flow for higher frequency voltage waveforms. This is because of the fact that for a capacitor i(t) = Cdv(t)/dt, where i(t) is the current through the capacitor at time t and dv(t)/dt is the "slope" of the voltage waveform at time t. You can see that rapidly changing signals (or higher frequency signals) have larger slopes than slowly changing signals (or lower frequency signals). For instance, if you have a voltage waveform Asin(wt) applied across a capacitor, where w is the angular frequency of the signal, its current will be C*A*w*cos(wt), which is the same as C*A*w*sin(wt + theta) where theta is the phase shift from the voltage waveform of the resulting current. So the capacitor has two effects on a signal. For small w a capacitor "blocks" or does not allow much current but for large w the resulting current is quite large. Remember, f = w/(2*pi). The capacitor also causes a "shift" in the resulting current's phase.

Also, as mentioned above, due to their filtering characteristics they can block DC components of signals but pass AC components. Also as mentioned, things like resonators can be built when using a capacitor in conjunction with an inductor, which is just a device that passes a narrow range of frequencies.

 

[Guidestone]

Thank you for your answers guys, I will check out the material you posted on this thread. I got another question, can there be a current division caused by a capacitor? As you said above, there can be current in a capacitor when it comes to AC voltages as long as there are high frequency voltages. If I got a capacitor in parallel with a resistor, can we say the current is going to be divided as two resistors in parallel do?

 

[Hesch]

If I got a capacitor in parallel with a resistor, can we say the current is going to be divided as two resistors in parallel do?

Yes you can. But now you must calculate with complex values:
The resistance of a resistor = R.
As for a capacitor the corresponding impdance, Z_C = 1/jωC.

So if you supply a voltage V to a resistor and a capacitor in parallel, you will get the currents:

I_R = V / R
I_C = V / ( 1 / jωC ) = V*jωC

The sum of the currents will be I_RC = V*( 1/R + jωC ) which is a complex current: The voltage V has not the same phase as the current I_RC.

The resistor will conduct the real part of the current, and the capacitor will conduct the imaginary part. You can measure both currents with an amp-meter. ( The imaginary current is not that imaginary ).

Voltage Current

 

[meBigGuy]

Capacitors can be used to halve or double voltages by switching them (voltage doublers). They can also be used as, essentially, variable resistors (switched capacitor filters).

Here is a whole bunch of switched capacitor application notes.
https://www.maximintegrated.com/en/app-notes/index.mvp/id/725

When driven with AC waveforms they can do all sorts of "dividy" things.

 

[sophiecentaur]

Why do electrolytic capacitors are electrolytic? I mean, I know an electrolyte allows current to flow for example in water, by placing salt in it. The electrolyte is between the plates, but there should be an isolator instead, not a conductor.

I don't think this has been dealt with at an introductory level and it could help the OP. The Capacitance of a Capacitor is inversely proportional to the spacing of the plates. There is a limit to how thin you can make a mechanical dielectric film of uniform thickness so it's necessary to make very large Capacitances with a big plate area, which makes them inconventiently big. In a electrolytic Capacitor, the dielectric is made electrolytically and can be much thinner, giving a many times higher Capacitance for the same component volume (but their breakdown voltage can be relatively low). When the Capcacitor is formed, it's a bit like a reverse biased diode, which will not conduct but the dielectric that's formed from the electrolytic action is fairly stable and insulates as long as there is never a negative bias across the plates. So electrolytics need to be placed in a circuit with the + terminal more positive than the - terminal or they can depolarise and start conducting. If left unused for a long time, the early electroytics needed to be rejuvinated with a DC voltage for a while. Modern ones probably don't need this treatment.
Hope that's not too much of a noddy description.

 

[nsaspook]

So electrolytics need to be placed in a circuit with the + terminal more positive than the - terminal or they can depolarise and start conducting. If left unused for a long time, the early electroytics needed to be rejuvinated with a DC voltage for a while. Modern ones probably don't need this treatment.

If you need a non-polarized electrolytic in a purely AC circuit you use two in series (externally or internally) with the capacitor polarities reversed so one is always reversed while the other is conducting.

http://www.cde.com/resources/catalogs/AEappGUIDE.pdf

If two, same-value, aluminum electrolytic capacitors are connected in series with the positive terminals or the negative terminals connected together, the resulting single capacitor is a non-polar capacitor equal in capacitance to half of the rated capacitance of either of the original pair. The two capacitors rectify the applied voltage and act as if they had been partially bypassed by diodes. However, at all levels of AC voltage the apparent capacitance is half of the rated capacitance as you would expect for capacitors connected in series. The nonlinear performance produces distortion where non-polar aluminum electrolytic capacitors are used in audio AC applications. In non-polar aluminum electrolytic capacitors and motor-start aluminum electrolytic capacitors a second anode foil substitutes for the cathode foil to achieve a non-polar capacitor in a single case. These are available for momentary-duty AC applications like motor starting and voltage-reversing applications, but the high DF of aluminum electrolytic capacitors – from 2% to 150% – causes excess heating and short life in most AC applications. Aluminum electrolytic, motor-start capacitors are non-polar and designed for intermittent operation in starting single-phase induction motors or for other brief AC applications such as motor-run capacitors in electric door openers.