Hi,

I am reading about the capacitors. I have read that the capacitors block DC and behaves as open circuit in DC while short circuit in AC. The reason behind this is the charging and discharging of the capacitor. What is my perception about it, I will mention about it.

"
During the DC voltage the capacitor charges up. During this charge process the negative charges from the battery comes in the wire and then get store on the plates and on the positive side there is a deficiency of charges on the plate as the electrons are attracted by the positve terminal of the battery. Due to
this deficiency of charges an Electric field is generated that opposes further movement of electrons towards the positive side and stop the excess amount of electrons on the negative plate side. When this potential strength is equal to battery potential then the the impact is neutralized and no further movement
happened and that's where the electrons started flowing on the other path and capacitor behaves as open circuit"

Now for the noise filtering capacitor they are actually connected for noise removal so that the current can safely go out. It makes sense that the noise in the form of sudden spikes comes from power supply. What I am confused is the statement about the capacitor behavior for this sudden spikes. I have read the capacitor
gives the path of the Ground towards sudden spikes which is according to me is utterly wrong. Because no electrons pass through the dielectric material.

According to the above explanation for the DC voltage, the capacitor is in stable state. If the sudden spikes of voltage come in (Obviously High Frequency) then the equilibrium of Electric field between the plates and voltage is disturbed and capacitor started charging again with the access charge (More electrons come in from the negative side and more positive charges or electrons departure from the positive plate). But when the spike is over the capacitor is required to discharge in order to reach to that equilibrium again. Now there are two ways of going back towards equilibrium.
1) If the capacitor negative plate loose electrons and positive plate gain some electrons (This is not possible)
2) If the electrons cross the dielectric and reach towards positive terminal of the battery (This is again not possible)

My logic is not conventional flow, it is electron flow. May be I am totally wrong but please help me to solve this dilemma.

 

Hello,

Electrons do not cross the dielectric. According to theory some accumulate on one plate, and some come off of the other plate. This creates the potential difference we measure with a voltmeter.

When the capacitor discharges, some of the electrons that had accumulated come off, and some that were lost on the other side go back.

That's about the simplest explanation i can think of. Add to that the imaginary displacement current and that's all there is to it.

 

RE: DC Filtering

When the spike from the Power Supply arrives at the capacitor ...
a) The capacitor absorbs the excess current and
b) The voltage on the capacitor rises slightly

When the spike has passed ...
a) The capacitor now has a slightly higher voltage than the power supply
b) Now the capacitor feeds the LOAD, temporarily reducing the current from the Power Supply
c) The voltage on the capacitor DISCHARGES back down to the same voltage as the power

Repeat, over and over ...

 

1) If the capacitor negative plate loose electrons and positive plate gain some electrons (This is not possible)

This is what happens.
If 1000 electrons move onto the negative plate then 1000 electrons will leave the positive plate and the voltage across the capacitor will increase.
If 1000 electrons leave the negative plate then 1000 electrons will move onto the positive plate and the voltage across the capacitor will decrease.

 

This is what happens.
If 1000 electrons move onto the negative plate then 1000 electrons will leave the positive plate and the voltage across the capacitor will increase.
If 1000 electrons leave the negative plate then 1000 electrons will move onto the positive plate and the voltage across the capacitor will decrease.

How they will come back on the positive plate. Did you mean they come from the ground.

 

RE: DC Filtering

When the spike from the Power Supply arrives at the capacitor ...
a) The capacitor absorbs the excess current and
b) The voltage on the capacitor rises slightly

When the spike has passed ...
a) The capacitor now has a slightly higher voltage than the power supply
b) Now the capacitor feeds the LOAD, temporarily reducing the current from the Power Supply
c) The voltage on the capacitor DISCHARGES back down to the same voltage as the power

Repeat, over and over ...

I dont understand the statement the capacitor feeds the load temporarily reducing the current from the power supply.

 

Hello,

Electrons do not cross the dielectric. According to theory some accumulate on one plate, and some come off of the other plate. This creates the potential difference we measure with a voltmeter.

When the capacitor discharges, some of the electrons that had accumulated come off, and some that were lost on the other side go back.

Go back to where, come back to positive plate. How is it possible because the positive plate is connected with the Ground, and one more thing to be noticed the electrons that come off definitely go back to the load. Now there is also temporary increase of the load current. the charging time and discharging time is same so dont you think it will create the same impact.

 

Hello,

Have a look at this chapter of the eBook for the capacitor basics:
https://www.allaboutcircuits.com/textbook/direct-current/chpt-13/electric-fields-capacitance/

Bertus

 

Remove thinking about the effect of the ground connection in your analysis. You do not need to consider the effect of ground. Ground provides a 0V reference point in the circuit.

Your power supply is still an essential part of the circuit since it is connected to the positive and negative plates of the capacitor.
Any electrons that leave one side of the capacitor are accompanied by an equal number of electrons arriving on the opposite side. To return to equilibrium (meaning zero volts across the capacitor), the electrons have to flow in the reverse direction.

If there is a load connected across the capacitor, you now have two paths for the electrons to take, the power supply and the load. The electrons will be shared between the two paths.

 

Hello,

Have a look at this chapter of the eBook for the capacitor basics:
https://www.allaboutcircuits.com/textbook/direct-current/chpt-13/electric-fields-capacitance/

Bertus

" Conversely, to release energy from a capacitor, the voltage across it must be decreased. This means some of the excess electrons on the (-) plate must be returned to the (+) plate, necessitating a current in the other direction ".

It doesnt mention how does the electron returned to positive plate, did they across the dielectric material ? Moreover if the electrons discharge from the capacitor it will go to load. Assuming the charging and discharging capacitor time is same so hte noise returned back to load so what is the benefit ? I am very much confused.

 

Electrons don't cross the dielectric. Electrons on one side are accompanied by lack of electrons on the other side. They arrive and return through the connection to the rest of the circuit.

The rise and fall of the voltage is slowed down by the size of the capacitance.
The voltage V across the capacitor is V = Q / C
where C is value of the capacitance
and Q is the amount of charge on the capacitor (number of electrons).

Yes, the charge and discharge times are the same. With no or little capacitance, a small change in Q causes the voltage V to change.

When you increase C, the voltage V is less affected by a rise or fall in the charge Q.

 

Go back to where, come back to positive plate. How is it possible because the positive plate is connected with the Ground, and one more thing to be noticed the electrons that come off definitely go back to the load. Now there is also temporary increase of the load current. the charging time and discharging time is same so dont you think it will create the same impact.

Hi,

I think you might be reasoning that a lack of or abundance of electrons constitutes a voltage.
For example, you might think that if we add 100 electrons to some place (like a plate) that will be interpreted as a voltage.
But that's not exactly how it works.
To measure a voltage we need two points, and in the broadest physical sense this means we have to have TWO different places that have a local difference in electrons. So you cant say that if you loose 100 electrons that the voltage there changes, because you may have lost 100 electrons at the other place too and therefore the voltage stays at zero.
Now if we hold one of those places to be the reference point, then anything else that changes will change the voltage measurement across the two points, and it doesnt really matter if we add or remove electrons in any given place, if the net difference is the same then the voltage is the same, and if the net difference increased then the voltage will increase, etc.
So adding 100 electrons to the bottom plate does not mean that the bottom plate has to change voltage. What changes voltage is the entire difference between the top plate and the bottom plate. So if we add 100 to the bottom and take away 100 from the top we get a net increase in voltage across the two plates, but we dont see any change in voltage of the bottom plate itself because for one thing it is the reference point.

Does this help? If you dont agree or something just yell

 

" Conversely, to release energy from a capacitor, the voltage across it must be decreased. This means some of the excess electrons on the (-) plate must be returned to the (+) plate, necessitating a current in the other direction ".

It doesnt mention how does the electron returned to positive plate, did they across the dielectric material ? Moreover if the electrons discharge from the capacitor it will go to load. Assuming the charging and discharging capacitor time is same so hte noise returned back to load so what is the benefit ? I am very much confused.

To really understand how a capacitor works you need think less about electrons moving in circles and more about how electrical energy moves. I'm not trying here to explain, just trying to give a good framework to build a deep understanding of what's happening.

By analogy electrons are like the links of a bicycle chain/sprocket in a circuit, they go round and round but energy flows one way (source to load). The individual links don't carry the energy of the circuit. That energy is in the form of tension (electrical potentials) at points in the chain as a whole.

I know it seems counter-intuitive but the important property is local charge Q in the capacitor moving in response to a force (electrical energy in the spike) that separates charge from equilibrium in a conductor or across a dielectric across in space. The charges in the capacitor move in response to the electrical energy of the spike, this transfers spike electrical energy in and out of the stored capacitor electrical energy path (low impedance) where the spike energy is dissipated by that circuit path resistance or removed from the circuit by EM radiation instead moving to the higher impedance load path much like a damper in a timing chain circuit between the gears reducing vibration.

 

This is where sometimes the reservoir capacitor as a tank of water analogy works.

Imagine a falling drop of water into a small glass (small capacity) filled with water. The water drop will cause ripples on the surface which can be observed at the edges where the surface meets the glass. Furthermore, the level of the water will rise (though by a very small amount).

Now add a drop of water into a large bucket or tank (large capacity) with water. The ripples are less pronounced on the surface at the edges of the bucket or tank. The rise in level of the water will be next to impractical to detect.

 

Exactly, it has to do with how much "energy" is in the spike, assuming a short 50 volt spike will charge the capacitor to 50 volts is incorrect.

Example:

Suppose a capacitor has a constant voltage on it of say 12 volts, then it receives a short spike of 50 volts, the capacitor absorbs the spike, but the voltage on the cap only rises to 12.1 volts, so the load only sees the extra .1 volts… not the 50 volts.

Of course the capacitor must have a much lower series resistance than the load for this to work.

 

Here is a 10 minute tutorial on how capacitors work. It's visual and perhaps explains it in a way you can understand. I say "in a way you can understand" because we all learn things a little differently. It's not derogatory against you, as we all were beginners at one time. And I'm not very advanced beyond the beginner state, so even I found this video useful. Just click on the "Here" at the beginning of this reply.

 

Here is a 10 minute tutorial on how capacitors work. It's visual and perhaps explains it in a way you can understand. I say "in a way you can understand" because we all learn things a little differently. It's not derogatory against you, as we all were beginners at one time. And I'm not very advanced beyond the beginner state, so even I found this video useful. Just click on the "Here" at the beginning of this reply.

+1 to that physics video set.

 

+1 to that physics video set.

They have a lot of videos that explain things pretty clearly. Worth checking out other video's by that producer.

 

They have a lot of videos that explain things pretty clearly. Worth checking out other video's by that producer.

I've seen most of them and have used them here to explain concepts.
https://forum.allaboutcircuits.com/threads/i-wanna-be-destin.128281/page-3#post-1334528

 

Hi,


So adding 100 electrons to the bottom plate does not mean that the bottom plate has to change voltage. What changes voltage is the entire difference between the top plate and the bottom plate. So if we add 100 to the bottom and take away 100 from the top we get a net increase in voltage across the two plates, but we dont see any change in voltage of the bottom plate itself because for one thing it is the reference point.

Does this help? If you dont agree or something just yell

I got it. But the question is about shunting to the Ground. Additional spikes will go towards the Ground. Even if we make positive plate as a reference point and make the changes only on the negative plate, in that case, in order to reach to the ground or a reference point the additional electrons are still required to cross the dielectric (Which is not possible). So if we see this analogy when the capacitor is required to discharge then it has to supply the electrons back to the circuit. Now it will supply the electrons towards the same circuit for which it is required to protect (As we know the rate of charging and discharging same). Ahhhhh I am so confused.