AC flowing through a cap. What actually happens?

[MrAl]

Hello again Brownout,

Ok i think we are clear on this now, you dont believe that charge flows through a capacitor. Duly noted.
Of course this could have been also been clear when i said "Ok, i thought you said you thought that charge flows through a capacitor".
But no problem, your intent is clear.

 

[BrownOut]

Hello again Brownout,

Of course this could have been also been clear when i said "Ok, i thought you said you thought that charge flows through a capacitor".

But as you've only just said that since your statement I quoted, then I don't see how that would have cleared anything up. But whatever works in your world, I guess....

 

[BrownOut]

What "displacement current" is from a physical standpoint, is the time rate of change of the electric flux.
Don't most of the standard methods of measuring "conduction current" rely on the effects of fields produced by the flow of the charge carriers? Otherwise, wouldn't we have to utilize some way of actually counting the carriers go by?

Pretty much all current measurements rely on the B or H field, as I've been saying for quite some time. So, in essence, some are saying they want to throw out the current measurements that validate iD, while keeping those that measure conduciton current. It's not the method they reject, it's the current that's being measured, see?

 

[killivolt]

While we are at it. "Speaking of Vacuum Tubes"

And for you amusement, this is how is was done or so to speak.

Hand Making a Vacuum Tube Part 1

Hand Making a Vacuum Tube Part 2

 

[ljcox]

And for you amusement, this is how is was done or so to speak.

Hand Making a Vacuum Tube Part 1

Hand Making a Vacuum Tube Part 2

Thanks kilo, that was fascinating. I don't think I'll try to do it though.

You would need a lot of skill, knowledge & equipment.

PS. Does anyone know how to download youtube videos?

There are some I would like to store on my HDD so I can view them more than once, ie. rather than have to watch them off the internet every time.

 

[Mr RB]

Just add SAVE or KICK after www in the address of the youtube video;
h ttp://saveyoutube.com/watch?v=dXP2GdqYCOM
h ttp://kickyoutube.com/watch?v=dXP2GdqYCOM
Then you can select the format to save the video in. These facilities are provided outside of youtube's control so use at your own risk.

We sort of have a couple different arguments going here, i think that is part of the problem. We are
arguing about:
1. What this 'displacement current' really is.
2. If it really can be called a 'current' at all.

For #1, it appears that what is being called a displacement 'current' is actually just a change in a
field flux, but multiplying that change by the permittivity of free space puts it in units of Amperes.
Thus, this leads people to call it a 'current'.
The other side says that the change in field flux only creates a B field, and that the displacement current
is not a real current by any scope of the word.

For #2, it's almost the same as #1, because before we know what to call it we have to know what it really is.
It has been traditionally called the 'displacement current', because it works out to units of Amperes. But
we know that amps is a flow of charge carriers. So the question becomes: is it really a flow of charge
carriers or not? If we want to say that it is we'd have to figure out how it can get through the
vacuum of space. ...

Thanks for clarifying! Like you said there's been a few people arguing slightly different arguments.

I'm going to take a childishly simple stab at it, hopefully leaving out fields. I think the term "current" requires a flow of electrons. In the case of a "displacement current" there seems to be a flow of electrons into one capacitor plate and out of the other capacitor plate. I don't mind the term "displaced" as there's going to be more electrons on one plate and less on the other so that seems to qualify as "displaced" electrons.

And although there is most likely no actual flow of electrons between the plates (vacuum capacitor), since the plates are an integral part of the process, the process having 3 parts (plate/dialetric/plate) and have current going in and out of each plate I think that qualifies as a type of "current" in reference to the process as a whole.

So to my simple mind that's what "displacement current" is, and yes I think it qualifies to be classed as "current".

 

[killivolt]

Thanks kilo, that was fascinating. I don't think I'll try to do it though.

You would need a lot of skill, knowledge & equipment.

I was just curious about the Vacuum tube construction. I thought obtaining a complete Vacuum is difficult if not impossible. I thought it would be good to understand Vacuum tube construction in order to know what was really happening. Now, I see it is reasonably easy once you have removed most of the air heating the glass would then boil anything remaining in the bulb I would assume. I thought how was it done, in Maxwells day.

For anyone interested in Vacuum tube construction. Apparently they are still making them in America I thought that was a bygone and would only be available in other country's.

Vacuum tube - Wikipedia, the free encyclopedia

 

[ljcox]

Just add SAVE or KICK after www in the address of the youtube video;
h ttp://saveyoutube.com/watch?v=dXP2GdqYCOM
h ttp://kickyoutube.com/watch?v=dXP2GdqYCOM
Then you can select the format to save the video in. These facilities are provided outside of youtube's control so use at your own risk.

Thanks for that, I'll try it.

 

[MrAl]

Hi again,


Well Mr RB let me put something else in your field of view here. This is interesting i assure you

We have two tanks of water, first one on the left, second one on the right. They both have 100 gallons of water in them when we start and nothing is moving.
We also have a pipe coming from the second tank going to a pump (which is not running), and the pump connected to another pipe that goes to the first tank.
Now we turn on the pump and pump water into the first tank so water flows through the two pipes and pump, but because of the piping we can only get that water from the second tank. This means that as the first tank gets more and more water, the second tank looses more and more water. Note that we did not have any other connection between the two tanks as they are separated from each other.
Question 1: We only have one pipe connected to each tank, so where is the displacement current?
Question 2: Assuming a flow rate of 1 gallon per second, how much water is pumped into the first tank after 1 second?

 

[ljcox]

I was just curious about the Vacuum tube construction. I thought obtaining a complete Vacuum is difficult if not impossible. I thought it would be good to understand Vacuum tube construction in order to know what was really happening. Now, I see it is reasonably easy once you have removed most of the air heating the glass would then boil anything remaining in the bulb I would assume. I thought how was it done, in Maxwells day.

For anyone interested in Vacuum tube construction. Apparently they air still making them in America I thought that was a bygone and would only be available in other country's.

Vacuum tube - Wikipedia, the free encyclopedia

I thought he used a vacuum pump. Then once sealed, my recollection (from a long time ago) is that there is an element inside called a "getter" I think that is connected to a high voltage & this attracts the remaining air atoms and somehow bonds them to the surface of the metal thus improving the vacuum.

BTW. we call them "valves" in Australia since the diode is equivalent to a fluid valve. ie the current can only go one way through it.

Triodes & pentodes are used for audio amps as some people believe that the "valve sound" is better than the sound from solid state amps.

I think the reason is that the valves introduce third order harmonics which chord with the notes being played & thus improve the tonal quality.

 

[killivolt]

But, the getter may have not been available at the time of Maxwell. I actually was hoping someone would comment on it since I didn't know. Plus if you are using a getter it would only apply to Diodes & Triodes & Pentodes when I read the wiki article, it said, to stabilize performance over time because of the high heat generated from the heater, maintaining a thermal equilibrium & vacuum.

In the instance of the Capacitor this would not apply. It does not require a heater. Only a complete vacuum without assistance from artificial methods to dry them. You wouldn't want to use RF to stimulate the A - B plate either in order to raise the temp forcing any remaining molecules. This process might also produce carbon which might influence the Chamber as well, De-stablizing the performance & altering the characteristics of the Capacitor or so you would think!

Maybe someone could explain more for me as I just don't understand. A bit confusing.

But, again I just don't understand how it was constructed back then, either.

 

[ljcox]

Hi again,


Well Mr RB let me put something else in your field of view here. This is interesting i assure you

We have two tanks of water, first one on the left, second one on the right. They both have 100 gallons of water in them when we start and nothing is moving.
We also have a pipe coming from the second tank going to a pump (which is not running), and the pump connected to another pipe that goes to the first tank.
Now we turn on the pump and pump water into the first tank so water flows through the two pipes and pump, but because of the piping we can only get that water from the second tank. This means that as the first tank gets more and more water, the second tank looses more and more water. Note that we did not have any other connection between the two tanks as they are separated from each other.
Question 1: We only have one pipe connected to each tank, so where is the displacement current? See below.
Question 2: Assuming a flow rate of 1 gallon per second, how much water is pumped into the first tank after 1 second?

The "displacement current" is the air that flows out of one tank & into the other.

However, this is not a valid argument with or without the "displacement current" as it is a completely different situation to that inside a capacitor

 

[ljcox]

But, the getter may have not been available at the time of Maxwell. I actually was hoping someone would comment on it since I didn't know. Plus if you are using a getter it would only apply to Diodes & Triodes & Pentodes when I read the wiki article, it said, to stabilize performance over time because of the high heat generated from the heater, maintaining a thermal equilibrium & vacuum.

In the instance of the Capacitor this would not apply. It does not require a heater. Only a complete vacuum without assistance from artificial methods to dry them. You wouldn't want to use RF to stimulate the A - B plate either in order to raise the temp forcing any remaining molecules. This process might also produce carbon which might influence the Chamber as well, De-stablizing the performance & altering the characteristics of the Capacitor or so you would think!

Maybe someone could explain more for me as I just don't understand. A bit confusing.

But, again I just don't understand how it was constructed back then, either.

I don't know if vacuum pumps were available in those days, but I think they were.

If not, they would have used air as the dielectric.

The relativity permitivity of air is 1.00053, so it is very close to that of a vacuum.

Besides, I believe Maxwell's work was theoretical, I expect that he would have used the results of experiments done by others to develop his theories.

 

[BrownOut]

Hi again,
Question 1: We only have one pipe connected to each tank, so where is the displacement current?

There is no displacement current, because this has nothing at all to do with electro-magnetics. Displacement current is valid only for electro magnetic phenomona.

Question 2: Assuming a flow rate of 1 gallon per second, how much water is pumped into the first tank after 1 second?

Ask a plumber.

 

[BrownOut]

But, the getter may have not been available at the time of Maxwell. I actually was hoping someone would comment on it since I didn't know. Plus if you are using a getter it would only apply to Diodes & Triodes & Pentodes when I read the wiki article, it said, to stabilize performance over time because of the high heat generated from the heater, maintaining a thermal equilibrium & vacuum.

In the instance of the Capacitor this would not apply. It does not require a heater. Only a complete vacuum without assistance from artificial methods to dry them.

Vacuum tubes were developed in the 20th century. Maxwell died in 1879.

 

[killivolt]

There is no displacement current, because this has nothing at all to do with electro-magnetics. Displacement current is valid only for electro magnetic phenomona.Ask a plumber.

Apart from the phenomena, I'm not sure this is established yet in my mind. I'm not satisfied with either result. Apart from the math and physical results it is still an anomaly. It's a physical reality to someone who has been through the rigger of education and has a sound relative perspective and education have it. But, I have not arrived to that result in my mind.

I am getting a really good schooling on capacitors though

 

[Jon Wilder]

AC doesn't really "pass through" a cap like you think. Any sort of constantly changing voltage on a cap causes the cap to charge, then discharge, then charge again. In the case of a coupling cap on an amplifier stage, the DC on the DC side of the cap alternates in value proportionately with the input signal. This changing DC voltage (I call it "alternating DC") causes the cap to charge, then discharge repeatedly with input signal. It is this charging/discharging action of the cap that becomes the AC on the other side, while the static DC on the DC side of the cap is blocked.

 

[killivolt]

Thank you Jon, We accept all who want to contribute this thread.

Thank you, please follow.

By the way just so you know, I was sucked into this thread just like you were. Peace, and thanks again.

Edit: So, in the case as you were saying the a Ac is blocked and the DC build begins somewhat of a charge phenomena. I'm not sure what your saying ?

Edit: Edit: I re-read your post and I think I understand.

It is this charging/discharging action of the cap that becomes the AC on the other side, while the static DC on the DC side of the cap is blocked.

 

[Jon Wilder]

Let's take a look at Example 1 -

**broken link removed**

Just a basic series circuit with two resistors on each side of a cap and a DC voltage across the entire series string. R1 and R2 limit the charge current. We are interested in the voltage drop across R2.

Keep in mind that the sum of all voltage drops in a series circuit must equal the supply voltage.

Now...assuming C1 is fully discharged, C1 will draw a charge current from the battery through R1 and R2 once the battery is connected. This causes a voltage drop to appear across R1 and R2 that is product of the value of R1 and R2 and the amount of charge current being drawn (E = I x R...simple Ohm's Law...with me so far?). We are only concerned with the voltage drop across R2 as this is our output voltage. Upon power being applied, the voltage drop across R2 will be maximized. As C1 charges up, the charge current will decrease, which in turn will also decrease the voltage drop across R2. Once C1 is fully charged up to the supply voltage, charge current will cease to exist, thus the voltage drop across R2 is zero (well...no cap is perfect and there will always be leakage current but it is such a small amount that we are not concerned with it). Therefore, the full supply voltage will be dropped across C1 while R1 and R2 will have zero voltage drop.

Now let's look at Example 2 -

**broken link removed**

Here we have the exact same circuit, however we now have a basic transistor voltage amplifier connected to it to drive it. In this example, current through the transistor is in a quiescent state (i.e. idle state) as no input signal exists yet. The resistors in the transistor circuit are sized such that the collector voltage is pulled down to about 1/2 the total supply voltage. This means that C1 now charges up to 1/2 the supply voltage while R1 has the other 1/2 of the supply voltage dropped across it. Again...the voltage drop across R2 is still zero since the cap is fully charged.

Now we have Example 3 -

**broken link removed**

Same as Example 2, however we now have an AC input voltage applied to the base of the transistor. As the AC input voltage alternates, it causes the bias voltage on the base to fluctuate.

On the negative swing of the input signal, this makes the base voltage less positive, which reduces current through the base-emitter junction of the transistor. This decrease in emitter-base current causes a large decrease in collector-emitter current, which also flows through R1. This decrease in current through R1 causes R1 to drop less voltage, which pulls the collector voltage up toward the positive supply voltage. Since C1 is connected to the collector, this also pulls one side of C1 up to the positive supply voltage and allows the cap to draw charge current through R2. This increase in charge current through R2 causes a corresponding increasing voltage drop across R2.

Things get interesting on the positive swing. On the positive swing of the input signal, this makes the base voltage more positive, which causes an increase in base-emitter current. This increase in base-emitter current causes a large increase in collector-emitter current. In order for collector-emitter current to increase, the collector-emitter resistance must decrease, which in turn decreases the voltage drop across the emitter resistor (R3) and the transistor. This effectively pulls the collector down to ground. Remembering that the collector side of C1 has a positive charge while the R2 side has a negative charge that the cap acquired when it was pulled up to the positive supply voltage, pulling the collector side of C1 to ground effectively connects C1 across R2, but in reverse polarity, R2 now draws current from C1, but since the polarity is now reversed, this current that R2 draws is a negative going current, which causes a corresponding negative voltage drop across R2.

These alternating positive/negative voltage drops across R2 are your AC output signal.

In short, you're essentially switching the collector side of C1 between the positive supply rail and ground but in a linear fashion. Pulling the collector side of C1 up to the positive supply rail allows C1 to charge through R2 and R1, which causes an increasing voltage drop to appear across R2 the closer C1 is pulled to the positive supply rail. Once the input signal swings positive, the collector pulls the collector side of C1, which has a positive charge on it, down towards ground, where it then discharges across R2, but in reverse polarity relative to ground since the collector side of C1 has the positive charge on it while the R2 side of C1 has a negative charge on it.

 

[MrAl]

The "displacement current" is the air that flows out of one tank & into the other.

However, this is not a valid argument with or without the "displacement current" as it is a completely different situation to that inside a capacitor

Hello there,

Well, that's an interesting idea with the 'air' but then how about if we use a large plastic plate that extends to infinity in both y and z directions, that puts an end to the 'air' flow.
Alternately we might do something similar, but the main idea is that one tank empties and one tank fills, so it sort of emulates the capacitor plates where charge on one side is depleted while charge on the other side is accumulated and we would like to attach a physical significance to the so called displacement current.
I know it's not the exact system, but many principles in fluid flow are often compared to current flow in electrical circuits. It's generalized more or less as a "flow field".

About the vacuum tubes....
Im not sure if this is what you guys are talking about, but the vacuum tube operates on very different principles than the vacuum capacitor. There's thermionic emission in the tube but none in the capacitor. Same with the CRT tube, where we have a beam of electrons. Can we call these current flow? We dont usually do that because the principles are different. This is one point i was trying to make. The other is just what displacement current is physically.
The tube and CRT bring out the point of calling their electron movement an electrical current or not. I just felt that whatever displacement current is, if it is not charge flow, it should be called something other than 'current'. If it is charge flow, then perhaps 'current' is appropriate.

There is another interesting point but i have to wait to bring that up because i wanted to 'reprove' it first.