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Resistance of capacitor

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What is the resistance of a capacitor? From the data of experiement, I was convinced that the resistance tends to zero. Is this the truth?

There's a part of empty space between two parallel plates. Isn't the resistance very large as it seems it's hard for electrons to pass through these spaces. How do you reconcile the two facts?

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I would say that a capacitor would be disigned to have almost zero resistance, because you wouldn't want your capacitors in a ciruit to add noticable resistance to what the resistors may already be providing.

As for the empty space, I'm not to sure - but I would guess that there must be something between or around them that helps electrons pass it with little resistance

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Caps do have a Breakdown voltage, and dielectrics do have a certain (although Very high) resistance.

Air has a resistance if was take a simple air gap capacitor like a Jackson rotary, there will be a voltage where the air breaks down and you`ll get conduction as the resistance is overcome.

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What data leads you to think R=0?

 

In reality resistance is really undefined, and you have to look at the impedance, which has a frequency dependence. In DC circuits you look at the voltage across the capacitor to discern the behavior. But if there were no plates to charge up, the resistance of an open circuit is infinite.

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"Resistance being at ohm" - from a bumper sticker

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"Ohm sweet Ohm" a sign we had above the door in the Electronics Lab :)

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In DC circuits you look at the voltage across the capacitor to discern the behavior. But if there were no plates to charge up, the resistance of an open circuit is infinite.[/Quote]

Do you mean that the capacitor keeps on changing during the discharge of itself in a DC circuit?

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in a DC circuit ie/ you put a cap across a battery, they have a charge Curve, IIRC it`s something like 2/3 charged over a given unit time, and then 2/3 charged of the remaining 1/3 in the same unit time and so on, so it never Really gets to 100% charged.

also capacitors leak charge, some types more so than others, so you charge a cap now, leave it on you desk and check it next week, it will have lost some charge.

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If you want to analyze a circuit with a capacitor in it, you look at the voltages. As YT said, you have a charging (or discharging) curve, which is an exponential with a characteristic time related to RC (since there will always be some resistance somewhere in the circuit, but usually you have an actual resistor). The voltage across the capacitor is Q/C and across the resistor is IR, and the add to be equal to the battery voltage.

 

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capchg.html

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What is the resistance of a capacitor? From the data of experiement, I was convinced that the resistance tends to zero. Is this the truth?

There's a part of empty space between two parallel plates. Isn't the resistance very large as it seems it's hard for electrons to pass through these spaces. How do you reconcile the two facts?

 

All replies so far correct: Summation: Any conductive paths (like 2 parallel wires) constitute a capacitor. The material between them (dielectric), unless vacuum, has resistance, usually very high, which allows some undesirable "leakage current". The data of your experiment is flawed. Initially, as charging begins, capacitor resistance is low; as charge builds, electrons gather in excess on one plate, and are deficient on the other. When capacitor voltage reaches source voltage, current flow is nearly zero, dependent on dielectric resistance (leakage current). Apparent capacitor resistance is then very high. So, the apparent "resistance" of a capacitor is not fixed, but varies between nearly zero and nearly infinity.

 

The capacitor therefore "resists" a change in voltage. In an alternating current circuit, or under any voltage varying periodically, the capacitor circuit current varies with time as a function of the resistance and capacitance in the circuit.

 

I hope this helps. Please ask if it does not. imp

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One way that you can answer that is by transfering everything into the imaginary plane. If that is the case, you can think of capacitor of capacitance C being a resistor with resistance equal to j C. From one point of view it is 0. However from another it gives you exactly the exponential decay solution that you need. Moreover if you add inductors and think of them as being L/j, then you can combine all 3 in whatever series or parallel that they are formed by the same token that you combine resistors, and find out the complex resistance of the whole system, and therefore would completely characterize its behavior by writing out that differential equation.

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In technological capacitors, the leakage current uses to be a far minor cause of losses. It can be significant in very special constructions, for instance in propagation lines for huge power that use pure water as a dielectric.

 

Even in DC or extremely low frequency uses, the volume resistivity of a decent insulator is generally not the primary cause of leaks. Leaks appear first in the external circuit, then at the surface of the capacitor component, then along the surface of the insulator film, and only then through the insulator's volume. So much that measuring the resistance of a good insulator (polystyrene, polypropylene...) is extremely difficult.

 

For practical uses, and in the datasheets, all losses in the capacitor are represented by a resistor (generally in series) or by a quality factor. This resistor is not physical and it varies quickly with the frequency.

 

At low frequencies, the first cause are losses in the dielectric, not due to leakage, but due to the slow and lossy response to the external field. If ions move as a result of the field (and the heat) in electrolytic capacitors and type II and III ceramic, they do it slowly, so the capacitor appears in part as a resistor. This tells that such components work only at low frequency.

 

Polyester film capacitors too use some small atoms movements to increase the permittivity and are limited to audio frequencies more or less. Faster dielectric are non-polar, like polypropylene, polystyrene, polytetrafluoroethylene... though a few polar ones like epoxy, type I ceramic are UHF capable. Non-polar dielectrics absorb also less humidity.

 

At VHF and beyond, type I ceramic coexist with capacitors made by the printed circuit, of glass-epoxy or (SHF+) PTFE. Due to skin effect (1µm @5GHz for Cu), losses arise much at the conductors then. That's why SMD capacitors for VHF+ have terminations coated with expensive palladium instead of the ferromagnetic nickel used at lower frequencies.

 

Losses, together with the self-resonant frequency, determine which capacitor technology fits a given frequency. The available capacitance and quality factor constrains heavily the design of a VHF+ circuit. Even at audio and metrology frequencies, detailed knowledge of capacitors capabilities (that's more than losses) is necessary.

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In technological capacitors, the leakage current uses to be a far minor cause of losses. It can be significant in very special constructions, for instance in propagation lines for huge power that use pure water as a dielectric.

 

Even in DC or extremely low frequency uses, the volume resistivity of a decent insulator is generally not the primary cause of leaks. Leaks appear first in the external circuit, then at the surface of the capacitor component, then along the surface of the insulator film, and only then through the insulator's volume. So much that measuring the resistance of a good insulator (polystyrene, polypropylene...) is extremely difficult.

 

For practical uses, and in the datasheets, all losses in the capacitor are represented by a resistor (generally in series) or by a quality factor. This resistor is not physical and it varies quickly with the frequency.

 

At low frequencies, the first cause are losses in the dielectric, not due to leakage, but due to the slow and lossy response to the external field. If ions move as a result of the field (and the heat) in electrolytic capacitors and type II and III ceramic, they do it slowly, so the capacitor appears in part as a resistor. This tells that such components work only at low frequency.

 

Polyester film capacitors too use some small atoms movements to increase the permittivity and are limited to audio frequencies more or less. Faster dielectric are non-polar, like polypropylene, polystyrene, polytetrafluoroethylene... though a few polar ones like epoxy, type I ceramic are UHF capable. Non-polar dielectrics absorb also less humidity.

 

At VHF and beyond, type I ceramic coexist with capacitors made by the printed circuit, of glass-epoxy or (SHF+) PTFE. Due to skin effect (1µm @5GHz for Cu), losses arise much at the conductors then. That's why SMD capacitors for VHF+ have terminations coated with expensive palladium instead of the ferromagnetic nickel used at lower frequencies.

 

Losses, together with the self-resonant frequency, determine which capacitor technology fits a given frequency. The available capacitance and quality factor constrains heavily the design of a VHF+ circuit. Even at audio and metrology frequencies, detailed knowledge of capacitors capabilities (that's more than losses) is necessary.

 

Unfortunately there is also no idea capacitors, resistors etc. Physics, unlike math, never claims to be exact but an approximation of reality that is good enough at the scale that you define your world. Just a side note!

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"Ohm sweet Ohm" a sign we had above the door in the Electronics Lab :)

 

That's from Kraftwerk.

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capacitors normally pass only an AC signal, and block any DC component. In a typical circuit, the larger the cap value, the lower the frequencies that can be passed efficiently to the next stage. That is for signal transfer. In a power supply, a capacitor is generally much larger, and is used to smooth out a pulsating DC voltage into the more pure DC required to power the ICs and other components in a device...all caps "leak" a small DC current due to theoretical limitations of the insulating material, usually a type of plastic or ceramic in small caps, and an oxide layer in large caps called an electrolytic type capacitor...and I still am unclear as to if the charge is actively stored in the dielectic (insulator) or on the plates themselves.

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+1, hoola, your answers are improving.

 

In answer to your question:

 

The charge resides on the plates. There should be none in the dielectric, any there would constitute leakage.

 

But charge is not energy and the energy of separation of this charge (if any) is held in the electric field in the dielectric.

This appears as a voltage, since voltage is energy per unit charge.

 

Does this help

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where does the charge exist within the plate metal? I was under the impression that the charge was maintained in the elongation of the electron shells of the atoms comprising the dielectric material...and a breakdown was caused by excessive elongation allowing dielectric electrons to break free from their atoms and head for the positive plate. With the circuit current available on the negative metal plate supplied by the attendant circuitry, a much larger current follows that initial breakdown path in a sort of secondary avalanche failure mode...you seem to equate charge with electron flow, and if charge was present within the dielectric, that would constitute leakage...that is puzzling. It seems charge must be present within the dielectric as both plates must "see" each other across it...unless you might think of a charge maintenance mechanism "tunneling" across the divide. That would seem to be the charge field, and responsible for the signal transfer itself...

Edited by hoola

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where does the charge exist within the plate metal? I was under the impression that the charge was maintained in the elongation of the electron shells of the atoms comprising the dielectric material...and a breakdown was caused by excessive elongation allowing dielectric electrons to break free from their atoms and head for the positive plate. With the circuit current available on the negative metal plate supplied by the attendant circuitry, a much larger current follows that initial breakdown path in a sort of secondary avalanche failure mode...you seem to equate charge with electron flow, and if charge was present within the dielectric, that would constitute leakage...that is puzzling. It seems charge must be present within the dielectric as both plates must "see" each other across it...unless you might think of a charge maintenance mechanism "tunneling" across the divide. That would seem to be the charge field, and responsible for the signal transfer itself...

This thread seems to have been started 7 years ago . The original proposer has not been active for 7 years . With 35,000 views . It is well overdue an answer ?

 

Mike

Edited by Mike Smith Cosmos

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a few months ago, I raised this question about charge storage mechanism and the general consensus didn't explain why happens clearly, at least to me...I do remember my idea of charge storage within the dielectric was deemed incorrect and that the charge is stored by a field between the metal plates. I don't think any response listed precisely how or if the phenomena of charge manifests any atomic changes within (or on the proximate surfaces) of the plate structures, as I presume there must be some, and the extent of that change signifies field strength. If charge causes no change in the dielectric, then shouldn't there be some change in the plates that secondarily manifests this field..?

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a few months ago, I raised this question about charge storage mechanism ....... If charge causes no change in the dielectric, then shouldn't there be some change in the plates that secondarily manifests this field..?

I have raised this issue of charge on a gigantuant scale just recently on this forum .

 

I think there might be something quite fundamental to be had here .

 

The entire standard model of atomic particles is absolutely ridden with charge . I look forward to further discussion with you on ' charge '

 

 

Tesla was a good man who investigated charge on a humungous scale .

 

post-33514-0-72777400-1414745794_thumb.jpg

 

Link :

http://en.m.wikipedia.org/wiki/File:Tesla_colorado_adjusted.jpg

 

Mike

Edited by Mike Smith Cosmos

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a few months ago, I raised this question about charge storage mechanism and the general consensus didn't explain why happens clearly, at least to me.

 

Electrons are added to one plate (and removed from the other).

This thread seems to have been started 7 years ago . The original proposer has not been active for 7 years . With 35,000 views . It is well overdue an answer ?

 

It looks like it has been answered.

 

Ideal capacitors don't have resistance, they have impedance.

 

Real capacitors have some leakage current, which can be modelled as a parallel resistor (which should be very large), and some internal resistance, which can be modelled as a serial resistor (which should be very small). Real components obviously have some inductance, as well.

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yes, electrons are added to one plate and pulled from another, but where do the extra electrons sit on the negative plate? What physical position within the atomic structure of the metal atoms? And what is happening to the metal atoms on the positive plate as this process occurs? What determines the maximum gain/loss of these mobile electrons on the respective plates?

Edited by hoola

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yes, electrons are added to one plate and pulled from another, but where do the extra electrons sit on the negative plate?

 

I assume they are as close as possible to the edge facing the dielectric (attracted there by the positive charges on the other side).

 

What physical position within the atomic structure of the metal atoms?

 

The outer electrons of the atoms in metals form a sort of "sea" of free electrons - this is why they are electrically conductive. For quantum mechanical effects that I don't fully understand. the available energy levels become so close togehter that they are effectively continuous, giving the electrons freedom to move around.

 

And what is happening to the metal atoms on the positive plate as this process occurs?

 

 

There will be "holes" as positive charge carriers. Just a reduction in the number of electrons.

 

What determines the maximum gain/loss of these mobile electrons on the respective plates?

 

The breakdown voltage of the dielectric. Which depends on the material and the thickness.

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a few months ago, I raised this question about charge storage mechanism and the general consensus didn't explain why happens clearly, at least to me...I do remember my idea of charge storage within the dielectric was deemed incorrect and that the charge is stored by a field between the metal plates. I don't think any response listed precisely how or if the phenomena of charge manifests any atomic changes within (or on the proximate surfaces) of the plate structures, as I presume there must be some, and the extent of that change signifies field strength. If charge causes no change in the dielectric, then shouldn't there be some change in the plates that secondarily manifests this field..?

 

 

Strange has already given you a pretty good answer to your question, but since is seems to question something I said, I would like to add the following.

 

I don't recall the thread you refer to, perhaps you could find the link (look in your content in the drop down box under your name), but I suspect that you must have misunderstood the underlined part.

 

Charge is not stored in a field. Energy due to the interaction between a charge and another charge or charges is stored in the electric field that exists between those charges.

 

If that field passes through dielectrics ( and remember that empty space is a dielectric) then the charge densities and directions may be modified.

 

Charge is a physical entity that you can add to other physical entities (the plates in this case). Both physical entities are complete in themselves and have their own existance.

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so, the charge field is an effect of the electron imbalance between the 2 plates, not the primary energy storage, but a component mechanism of the entire capacitor system. Isn't some energy "borrowed" from the voltage potential in the creation of this separate field (hence some portion of the initial charge current), that is returned upon the discharge of the cap from the field back to the plate system? This is presuming the field physically exists within the domain of the dielectric between the 2 plates...

Edited by hoola

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