BR-549 Stop muddying the waters LOL.
Analog Kid, AC(Ok I won't call it AC Current) is a time varying voltage as applied to a capacitor. As the voltage varies in some time related fashion the voltage applied to the capacitor will vary in a similar fashion. This varying voltage will result in the CHARGE on the capacitor varying also..
If you arrange a slow frequency source(sine wave) and connect a current meter(center zero) in series with the capacitor you will note a current in one direction as the cap charges then a current in the opposite direction as it discharges in time with the source changing polarity.
It LOOKS like it is passing an Alternating current but is only charging and discharging. I never claimed caps pass current. They charge and discharge. Caps do not pass current AC or DC period.
If a cap passes DC what resistance would you expect to read across its terminals with a ohmmeter or megger? Have you measured a good cap with a ohmmeter or megger(allowing for voltage max)?
Consider a tube amplifier which have about 200+volts on the anode and can be coupled to the next stage tube grid with a capacitor. If this cap passed DC you would get 200+ volts applied to the grid of the next tube. What would that do?

 

Further to the above. Have any of you come across a variable tuning capacitor as used in the days of analog radios. The consist of a set of plates, one set fixed, on set movable to intermesh with each other(look it up). They were used to tune radios in to a desired radio frequency in conjunction with inductors forming a resonant circuit. The caps plates were SEPARATED by a small AIR GAP. NO DC could flow between them. I f anything bridged the plates the circuit would not function!

 

Don't know what definition of DC you are using but my definition is that the voltage and current are unchanging (no Fourier component).
If it's a ramp or charging a capacitor than the voltage/current is not strictly DC, it has an AC (Fourier) component and it's that AC component that generates a charge transfer through the capacitor, i.e. a current.
(Note that this definition does not mean the AC component has to go through zero absolute volts twice per cycle).
So yes, capacitors block DC voltage/current and yes, they carry AC current/voltage.
And it's immaterial whether on not the electron going in one end of the capacitor is never the same one coming out, it's still conducting current.

 

crutschow,
I do believe my understanding of dc is essentially the same as yours. Any DC applied to a capacitor will charge it . Doesn't need to be a ramp etc. A simple switch connected from a source to a cap will apply the volts and it will charge. If there is an Ac ripple on top of the DC then the cap will charge and discharge to the level of the AC ripple.
please examine the case of a parallel resonant circuit. The collapsing magnetic field of the inductor will generate an emf to charge the cap. When fully collapsed the cap charge will discharge into the inductor to generate a new field. This is an AC resonant current flowing in the inductor but NO Current flows in the cap. It simply charges and discharges. Of course to maintain resonance it needs a suitable "kick" from a suitable supply at the right time.

 

BR-549 Stop muddying the waters LOL.
Analog Kid, AC(Ok I won't call it AC Current) is a time varying voltage as applied to a capacitor. As the voltage varies in some time related fashion the voltage applied to the capacitor will vary in a similar fashion. This varying voltage will result in the CHARGE on the capacitor varying also..
If you arrange a slow frequency source(sine wave) and connect a current meter(center zero) in series with the capacitor you will note a current in one direction as the cap charges then a current in the opposite direction as it discharges in time with the source changing polarity.
It LOOKS like it is passing an Alternating current but is only charging and discharging. I never claimed caps pass current. They charge and discharge. Caps do not pass current AC or DC period.
If a cap passes DC what resistance would you expect to read across its terminals with a ohmmeter or megger? Have you measured a good cap with a ohmmeter or megger(allowing for voltage max)?
Consider a tube amplifier which have about 200+volts on the anode and can be coupled to the next stage tube grid with a capacitor. If this cap passed DC you would get 200+ volts applied to the grid of the next tube. What would that do?

There is so much fail in this one post it is hard to start.

At heart the issue is quite simple. The current thru a capacitor is equal to the time rate of change of the voltage across that capacitor, times a constant, which we conveniently call the "capacitance". That is it. I = C dv/dt. Anything about how it is constructed or it is not current but charge is hand waving nonsense.

If I can vary the voltage in a linear fashion then dv/dt is a constant, and a constant DC current will flow thru the capacitor.

Conversely, if I flow a constant DC current thru a capacitor then the voltage will change in a linear fashion.

I claim capacitors pass current and I have sources to back me up. Source: Wikipedia Real current. Not via electrons but via the electric field. That does not make it something other than a current. Nothing new here. It was first published in 1861.

Put a capacitor across an ohmmeter. It reads near zero ohms at first, then builds to an open. Why? IT IS CONDUCTING A DC CURRENT. Probably the classic RC time constant exponential decay of current for old school analog meters, perhaps a ramp for a digital meter, but a cap can and will conduct current when connected to a DC source.

Further to the above. Have any of you come across a variable tuning capacitor as used in the days of analog radios. The consist of a set of plates, one set fixed, on set movable to intermesh with each other(look it up). They were used to tune radios in to a desired radio frequency in conjunction with inductors forming a resonant circuit. The caps plates were SEPARATED by a small AIR GAP. NO DC could flow between them. I f anything bridged the plates the circuit would not function!

My radio is separated from the transmitter by a very extremely large air gap. If nothing conducts between them then how do ever I hear anything?

 

crutschow,
I do believe my understanding of dc is essentially the same as yours. Any DC applied to a capacitor will charge it . Doesn't need to be a ramp etc. A simple switch connected from a source to a cap will apply the volts and it will charge. If there is an Ac ripple on top of the DC then the cap will charge and discharge to the level of the AC ripple.
please examine the case of a parallel resonant circuit. The collapsing magnetic field of the inductor will generate an emf to charge the cap. When fully collapsed the cap charge will discharge into the inductor to generate a new field. This is an AC resonant current flowing in the inductor but NO Current flows in the cap. It simply charges and discharges. Of course to maintain resonance it needs a suitable "kick" from a suitable supply at the right time.

Hi,

In conclusion they are both AC currents but one looks more like DC than the other so we consider it to be DC for a limited time. I'll explain here.

There are a couple ways of defining a DC current when using a circuit and especially for this question about a capacitor and a DC current.

First, we have the coupling capacitor. This is a capacitor that passes AC current and blocks DC current.
Second, we have just a lone capacitor being charged from a constant DC current source where the current does appear to pass through the cap.

So which is right?

The answer is that they are both right, but they are used in a different manner and the way the "DC" is being applied is different.

In the first case, we are considering a DC current based on time approaching infinity.
In the second case, we are considering a DC current that is based on some finite time that is NOT approaching infinity.

The example of the coupling cap is a good one for the time toward infinity because we consider all exponentials to have died out.
The example of the constant current and cap is a good one for a finite time period since the turn on of the circuit because then we ONLY consider the exponential part of the response.

So in the first case the cap charges and then that's it, then after that only AC components come through.
In the second case however we are constantly charging the cap, and in doing that the voltage is continuously rising, so we can only consider a finite time period because if we go to time infinity we blow up the cap because the voltage rises too high. Still, we can do this as long as we dont charge it for too long.

To be perfectly succinct, in the second case the current is not exactly a "DC current". It is a DC current that has been switched on at t=0 and has been completely off before that and thus it has an AC component, but we often refer to it as just a DC current. So we cheat a little in the definition of the second case but once that current is flowing, it is a DC current and does not change until we shut off the circuit again, and we must turn off the circuit at some point, making it really a pulse of current not a DC current strictly speaking. or else the voltage would climb too high and thus blow out the cap. Theoretically we could leave it on indefinitely but then we would have to conclude that the circuit was unstable because of that infinite response.

So in conclusion they are both AC currents but one looks more like DC than the other so we consider it to be DC for a limited time.

It should also be noted that the definition of a DC current does not change, but the way we apply it to a circuit does change so instead of stating that all the time we just keeping saying it is just a DC current. To understand completely though we have to look into when that current is applied (and/or turned off) over all time.

 

I claim capacitors pass current and I have sources to back me up. Source: Wikipedia Real current. Not via electrons but via the electric field. That does not make it something other than a current. Nothing new here. It was first published in 1861.

I would only note that the current is rightly explained as a " fictitious displacement current ID "flows" in the vacuum equal to the "real" current in the wires

 

The collapsing magnetic field of the inductor will generate an emf to charge the cap. When fully collapsed the cap charge will discharge into the inductor to generate a new field. This is an AC resonant current flowing in the inductor but NO Current flows in the cap. It simply charges and discharges.

That is contradictory.
Any current that flows in the inductor has to flow in (through) the capacitor.
Charging and discharging a capacitor requires a current. A ammeter in series with the capacitor will show this current.
What's actually happening inside the capacitor to allow this current is not relevant as to whether it's really a "current".
If you can measure current, than it's definitely current.

 

What 'flows through' the capacitor is the electrical energy of the circuit and electrical energy is not charge. There is absolutely no need for actual current in the operation of a capacitor (the Vacuum of space has a permittivity value) as the moving electrical energy interacts with charges for current (as structures of the physical capacitor) or propagate across space as pure fields (as the storage medium for the physical capacitor). As usual the electrical energy is in the form of a field. Nobody seems to bat an eye if we can current doesn't pass (Galvanic isolation) through a transformer that uses magnetic fields for energy transfers but capacitors can also be used for Galvanic isolation. For the (dis)charging capacitor it's a changing electric field that also gives rise to a changing magnetic field that we normally call the fictitious displacement current needed make the circuit theory concept of voltages and current consistent.

 

It might be fields......but we have to shake it with current.
Electronics....and fields from them are mechanical processes.
It's moving mass. It has inertia. It reacts.

 

What 'flows through' the capacitor is the electrical energy of the circuit and electrical energy is not charge. There is absolutely no need for actual current in the operation of a capacitor

That makes no sense.
So what is measured when you put an ammeter is series with a capacitor as it charges and it shows a current?

 

Electron trampoline.

 

That makes no sense.
So what is measured when you put an ammeter is series with a capacitor as it charges and it shows a current?

You are seeing the current of the conductor(s) as it affects the plates of the capacitor while they store/transfer electrical energy across to the other side as an electric field that in turn affects the charges on that other side of the capacitor for a current in the conductor. Obviously you don't need a series physical attachment to measure the magnetic field of the conductor and by measuring the field with a Hall probe or other device you can calculate the current in the wire from that magnetic measurement. Sure you can also use a series connection to measure the voltage drop across a small resistance and then calculate the current from that.

Yes, it's true, a electronic capacitor because of the usual very close proximity of the plates acts much like a strange conductor so I really don't have a problem with 'current flows through a capacitor' but the actual physics are important too when you get beyond circuit theory in circuits.

http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.205.6223&rep=rep1&type=pdf

Now Roche’s reply [3] takes issue with me, while stating that our differences are slight. It is a fact that both of us and the vast majority of physicists and engineers living today do not think that the unfortunately named ‘displacement current’ is a true current. I believe that it is also widely understood how Maxwell, by choosing to work in the radiation or Coulomb gauge (∇ · A = 0), was led to speak of the total current as the sum of the conduction current J and the displacement current ∂D/∂t. Roche and I agree that the Ampere–Maxwell equation is better written ` as ∇ × H − ∂D/∂t = J, not as ∇ × H = Jtotal = J + ∂D/∂t. The external charge and current densities are the true sources for the fields

J D Jackson University of California, Berkeley, and Lawrence Berkeley National Laboratory, Berkeley, CA 94720, USA

 

the actual physics are important too when you get beyond circuit theory in circuits.

That is true if you're interested in the theory of the devices.
I'm, rather a black box sort of guy and am only really concerned in the external characteristics of a device that affects circuit operation.
What the electrons (and holes) are doing inside of a device is only of minor interest to me.

It's like some of the endless discussions about whether a bipolar transistor is really voltage or current controlled according to the device's internal quantum physics.
Circuit wise it really makes no difference what it does inside, as it is can be considered as both externally.
Thus it usually is easiest to consider it as a current controlled device for large-signal or bias calculations, and voltage-controlled for small-signal calculations.

 

That is true if you're interested in the theory of the devices.
I'm, rather a black box sort of guy and am only really concerned in the external characteristics of a device that affects circuit operation.

I agree for 'regular' slow speed circuits or circuit boards but there are wide areas of electrical science (that indirectly impact many other areas) beyond creating the electronic devices themselves that require a more detailed understanding to troubleshoot and repair systems. One of the reasons that RF, EMI, EMC or high speed digital circuits seems likes 'black magic' to some people is that lack of understanding that the 'device' that modifies signals can exist external or internal to discrete components on a board. When the signal phase angle change (the rise and fall times of signals on a wire are long compared to the time it takes for an EM wave to travel) between points on a board of discrete component is small you can usually ignore the effects but when it's not (causing transmission line (wave) effects ), hold on to your butts.

 

It might get to the point where we might excite insulators......for power and info transfer.

 

Nsaspook, glad to see that you understand what I was trying to say. The current measured by a meter in series with a cap is the current charging the field of the cap. This "energy" is stored in the cap as an "electric field" between the "positive" and "negative" plates(relative terms). Wether AC or DC is to some extant immaterial. In DC it is stored till cap is discharged and in AC it charges and discharges at applied frequency.

Mr Al, I am not sure if I confused you with my examples. Neither of my instances are AC at all.
My example of a blocking cap was meant to show that such a cap cannot pass DC since it would affect(destroy) the tube of the following stage(refer to capacitor coupled tube amplifiers).
My second example was meant to show that a cap takes a CHARGING current(if discharged) at instant of switch on to a DC source. This current charges the "field" between the cap plates till cap is fully charged then dies away to effectively zero. It does not continue till voltage rises to destroy the cap(depends on volts applied and cap volts rating of course). Any cap that continues to draw a DC after charging is faulty and should be replaced!

 

Mr Al, I am not sure if I confused you with my examples. Neither of my instances are AC at all.

I'm pretty sure Mr Al is not confused with your examples.

 

Copied from Post 26:
So we cheat a little in the definition of the second case but once that current is flowing, it is a DC current and does not change until we shut off the circuit again, and we must turn off the circuit at some point, making it really a pulse of current not a DC current strictly speaking. or else the voltage would climb too high and thus blow out the cap. Theoretically we could leave it on indefinitely but then we would have to conclude that the circuit was unstable because of that infinite response.

Oh really???

Ok, I won't bother to pursue this any further since i see so many strange understandings of how a capacitor actually behaves in DC and AC circuits.

 

Hi,
I think this issue should closed now. A lot of above explanation will help you a lot.

END OF ISSUE.