Theoretically, there is no limit. Eventually, the electric field across the capacitor will cause arcing and breakdown of the dielectric.

Almost sounds like a contradiction. So in the real world, there is a practical limit, correct? Will that limit be indicated in the voltage rating of the capacitor?

You charge a capacitor by injecting current, not by applying voltage. Semantics, yes.

I think the difference between current and voltage isn't semantics. My understanding is that current flows only when there's a voltage differential. So you cannot product a current to charge a cap unless you present the cap with a voltage differential. No?

If you apply a voltage, the capacitor cannot charge instantly. You need to know the series resistance in order to determine the charging current.

Understood, doesn't change what i just said above. And doesn't help me understand how a capacitor responds differently to unipolar vs bipolar.

My current understanding is that, as long as the limits of the cap are respected, any amount DC bias or offset will be removed from the input with no difference in behavior, output voltage, efficiency, frequency response, or distortion. (assume non-polarized).

BTW. which institution is this homework for ?

BTW i am not a student.

 

Wow, I had no idea! I thought it's capacitance refers to how much charge it can hold. Like a battery.

My understanding from the video above is that the way to force more electrons into the capacitor is by increasing the input voltage. Capacitors have a voltage rating. Therefore, I would think that the charge limit of a capacitor is based on its voltage rating, no?

An ideal capacitor has no voltage rating. A real capacitor does. But it is not related to how much charge it can hold, it is related to the physical ability of the device to withstand the stresses induced by the voltage differential.

A crude analogy might be a very tall water tank (with vertical sides and, for simplicity, we will assume are circular). Double the height of the water, and you double the amount of water in the tank. The ratio of the amount of water to the height of the water, which might have units of kilograms/meter, is analogous to the capacitance of a capacitor, which has units of coulombs/volt. Saying that a particular tank is characterized by 1000 kg/m tells you nothing about how much water the tank can ultimately hold, only that if you tell me how deep the water is, I can tell you how many kilograms of water is in there (or vise versa).

But a given tank also has a limit on the stresses it can withstand, and the deeper the water, the greater the pressure at the bottom of the tank. At some point, those stresses get too large and the tank fails. Thus the tank is rated for a maximum water depth. Yes, this max water depth translates into a maximum charge, but it has nothing to due, fundamentally, with the capacity of the device, with it somehow being "full" at that point. It is merely the point below which the maker of the capacitor is willing to guarantee that it won't fail due to excessive voltage causing excessive stresses. In practice, it will likely continue to operate just find for quite some amount above that, but you would be operating in increasingly dangerous territory to do so.

 

Almost sounds like a contradiction. So in the real world, there is a practical limit, correct? Will that limit be indicated in the voltage rating of the capacitor?

No contradiction. It's the difference between a theoretical situation in which limits are ignored because we are assuming that everything obeys the basic constitutive equations, and the real world in which practical limits exist on everything. Think of Ohm's Law. In theory, you can put any voltage across it and determine the current through it using Ohm's Law. In the real world, there is a maximum voltage and current that it can withstand. That limit is set by a number of effects, such as the device catching fire from the heat dissipated to arcing across it's terminals due to excessive voltage. The ratings for the device are dictated by which effect causes it to fail first.

I think the difference between current and voltage isn't semantics. My understanding is that current flows only when there's a voltage differential. So you cannot product a current to charge a cap unless you present the cap with a voltage differential. No?

Ever hear of a superconductor?

A superconducting magnet can have current flowing in it for literally decades without any voltage being applied. The residual losses can be down in the 0.1%/year range. MIT had a magnet that maintained its field for over twenty years.

Understood, doesn't change what i just said above. And doesn't help me understand how a capacitor responds differently to unipolar vs bipolar.

There is no simple answer. It depends on the dielectric material involved. If the dielectric has significant hysteresis, the changing polarity of the voltage will likely cause greater losses then the same peak-to-peak voltage applied with sufficient DC bias to prevent the total voltage from every reversing polarity. But if the dielectric is nonlinear (and all dielectrics are nonlinear to some degree), then the DC bias can shift the operating point to a region of either greater or lesser loss. This is sometimes done in precision applications to tune the dielectric properties.

My current understanding is that, as long as the limits of the cap are respected, any amount DC bias or offset will be removed from the input with no difference in behavior, output voltage, efficiency, frequency response, or distortion. (assume non-polarized).

Whether or not the DC bias is removed depends on exactly which part of the signal you are referring to.

On paper, and in many applications in the real world, the difference in behavior is small enough that it would be difficult to measure and/or can be safely ignored while maintaining acceptable performance. But at some level, there probably is going to be some difference, and that difference can, for some applications, be significant enough that it can't be ignored.

EDIT: Fixed up quote tags.

 

merely the point below which the maker of the capacitor is willing to guarantee that it won't fail due to excessive voltage causing excessive stresses.
In the real world, there is a maximum voltage and current that it can withstand.

My original question is all about the real world.

Ever hear of a superconductor?

This question is about capacitors in everyday circuits, not magnets, superconductors, or esoteric research.

In everyday capacitor circuits is it correct or not to say that charging a capacitor requires a voltage differential?

If the dielectric has significant hysteresis, the changing polarity of the voltage will likely cause greater losses then the same peak-to-peak voltage applied with sufficient DC bias to prevent the total voltage from every reversing polarity. But if the dielectric is nonlinear (and all dielectrics are nonlinear to some degree), then the DC bias can shift the operating point to a region of either greater or lesser loss. This is sometimes done in precision applications to tune the dielectric properties.

*** This might be the kind of info I'm looking for.

at some level, there probably is going to be some difference, and that difference can, for some applications, be significant enough that it can't be ignored.

*** This might be the kind of info I'm looking for. What kind of applications?

 

In everyday capacitor circuits is it correct or not to say that charging a capacitor requires a voltage differential?

IMHO this kind of thinking is the direct result of lazy-sloppy academics neglecting to teach the difference between "voltage" and "emf".

To "charge" a capacitor you must "force" electrons onto one plate and off of the other plate, the result is a development of "voltage".

 

That seems irrelevant to the question. If the capacitor does not meet the requirements of the job, then it will not work with any kind of waveform. And if it does me the requirements of the job, then the question remains: which is better, unipolar or bipolar?

If the signal across the capacitor is bipolar (changes polarity, ie, AC signal or bi-polar DC) then the cap should be non-polar to prevent damage to the cap. A non-polar cap can be made from two polarized caps by connecting them in series (the capacitance will "half").
If the signal across the capacitor is unipolar (does not change polarity, ie, DC or half wave AC) then the cap can be a polarized cap.

 

Hi ElectricSpidey
johny posted:
In everyday capacitor circuits is it correct or not to say that charging requires a voltage differential?

As you explain it:
To "charge" a capacitor you must "force" electrons onto one plate and off of the other plate, the result is a development of "voltage".

So as johny states correctly: to charge a capacitor requires a voltage differential across the capacitor to create a "force" that moves electrons onto one plate and off of the other plate.


E

 

So as johny states correctly: to charge a capacitor requires a voltage differential across the capacitor to create a "force" that moves electrons onto one plate and off of the other plate.

Yet, a "charged" capacitor has a voltage differential, but does not continue to charge in the absence of a current.

Hmmmm.

(Hint: it's not the differential that charges the cap -- it's the charge that causes the differential.)

 

Electromotive Force EMF, Potential Difference & Voltage » Electronics Notes

 

Hi joey,
I am using the term ‘differential voltage’ based on this definition, not a ‘voltage difference’.

Differential voltage is fundamentally linked to the concept of dV/dt, which represents the change in voltage over a change in time.

E

 

hi ES,
Please copy and paste the relevant text, from that web page, that covers our discussion.
E

 

The entire text of the site relates to my post, but here...

"EMF and potential difference have many similarities, but they also have some significant differences. Essentially the EMF is the driving force ina circuit, whereas the potential difference is the result of the EMF within a circuit to which the source is connected.

Understanding where to correctly use each of the terms is of great benefit as it can give weight to any spoken or written material. If used incorrectly it can indicate that the terms are not correctly understood and may lessen the impact of anything else that is said or written."

 

You charge a capacitor by injecting current, not by applying voltage. Semantics, yes.

I think the difference between current and voltage isn't semantics. My understanding is that current flows only when there's a voltage differential. So you cannot product a current to charge a cap unless you present the cap with a voltage differential. No?

By semantics. I mean it is a chicken and egg problem. Which comes first, voltage or current?
You need voltage to push current. But it is current (charge flow) that charges the capacitor.

C x V = Q

A capacitor can hold that much charge at a given voltage. If you keep the voltage constant, the capacitor will be "full" with charge eventually. Current will stop flowing because there is no voltage difference.

If you increase the voltage the capacitor will take more charge because you have created positive charge flow.

The current does not stop flowing because the capacitor is "full".
The capacitor stops charging because the current is zero.
Yes, semantics matter. Correct language is important to convey precise understanding and communication.

 

Hi ES,
The term EMF usually relates to voltage on an open circuit voltage source, ie: not loaded.

A Potential difference, is usually related to a loaded Voltage source measurement.

How do these relate to your post to johny?

E
BTW: I am not trying to be confrontational with you, I am aiming to help the TS with the correct information.

 

It relates to my post as in EMF in the cause of current and voltage is the result.

As stated in the link I provided.

 

IMHO this kind of thinking is the direct result of lazy-sloppy academics neglecting to teach the difference between "voltage" and "emf".

hi ES.
The statement 'johny' made not sloppy thinking, he is correct, in saying:-
In everyday capacitor circuits is it correct or not to say that charging a capacitor requires a voltage differential?

Definition
Differential voltage is fundamentally linked to the concept of dV/dt, which represents the change in voltage over a change in time.

E

 

Hi joey,
I am using the term ‘differential voltage’ based on this definition, not a ‘voltage difference’.

Differential voltage is fundamentally linked to the concept of dV/dt, which represents the change in voltage over a change in time.

E

I have never heard the term "differential voltage" used to describe the rate-of-change of voltage with respect to time.

I guess I should get with the times and start accepting the alternate and sundry definitions of common terms.

 

LOOK, LET'S BE CLEAR HERE!

I never said the TS made a sloppy or lazy statement, I accused academia of being lazy and sloppy.

We are done here.

 

I have never heard the term "differential voltage" used to describe the rate-of-change of voltage with respect to time.

I guess I should get with the times and start accepting the alternate and sundry definitions of common terms.

Voltage is always a voltage difference, i.e. a potential difference, or voltage with respect to a reference, e.g. altitude.

Maybe he meant derivative, dv/dt.

 

*** This might be the kind of info I'm looking for. What kind of applications?

Precision high-Q resonators and other RF circuits are the one that come to mind.