Understanding Capacitor Behavior for Noise Filtering?

[Muhammad Usman]

--> 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.

 

[jim hardy]

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

why not ?

 

[anorlunda]

Trying to understand electricity with electron flow as an analogy to water flow is harmful. Your own logic and puzzlement about capacitors demonstrates that.

My recommendation is to forget that you ever heard of electrons.

A spike can be represented as the sum of several high frequency AC signals. As you said, the capacitor acts like a short circuit for AC, and therefore as a short circuit for noise spikes.

The next step in learning is complex impedance Z, and Ohm's law applied to AC, namely V=IZ. Would you be interested in studying that more? It requires math, not just logic. Electrons and fields will not be mentioned. If there's enough interest, I could write a tutorial for PF Insights.

 

[Muhammad Usman]

Trying to understand electricity with electron flow as an analogy to water flow is harmful. Your own logic and puzzlement about capacitors demonstrates that.

My recommendation is to forget that you ever heard of electrons.

A spike can be represented as the sum of several high frequency AC signals. As you said, the capacitor acts like a short circuit for AC, and therefore as a short circuit for noise spikes.

The next step in learning is complex impedance Z, and Ohm's law applied to AC, namely V=IZ. Would you be interested in studying that more? It requires math, not just logic. Electrons and fields will not be mentioned. If there's enough interest, I could write a tutorial for PF Insights.

Thank you very much, Yes its easy to learn maths. Once upon a time there was a box which was locked. Every body try to open it, they call engineers , physicist, chemist, black smiths and lock makers to open it or break it but failed and then they call mathmetician and when mathmetician come he said "Lets suppose if the Box is opened and assign x = lock". That was on the lighter side. But I am more interested in learning, what is actually physically happening. Please help me in that.

 

[Muhammad Usman]

why not ?

I didnt get you, what you are trying to say the ground will send electrons back to the capacitor plate ?

 

[anorlunda]

But I am more interested in learning, what is actually physically happening.

There is a current through the capacitor, but is not electron flow. Read about it here:
https://en.wikipedia.org/wiki/Displacement_current#Current_in_capacitors

Edit:

Displacement current density has the same units as electric current density, and it is a source of the magnetic field just as actual current is. However it is not an electric current of moving charges, but a time-varying electric field.

 

[jim hardy]

I didnt get you, what you are trying to say the ground will send electrons back to the capacitor plate ?

Here's a link to a page where the math is less, well, less 'burly' .
It may help you visualize current flow around a circuit.

https://www.allaboutcircuits.com/textbook/direct-current/chpt-13/capacitors-and-calculus/

 

[Nik_2213]

Slightly off-topic, but you also have 'Effective Series Resistance', ESR, which is frequency dependent. I'll ignore inductive effects of a large capacitor's leads...

IIRC, this was why you'd often find a big PSU's electrolytic 'smoothing' capacitance ~1000 uF bypassed by a non-electrolytic ~ 0.1uF, with more ~0.1 uF scattered around the circuit.

I remember seeing one high-end 'linear' design with ~5000 uF and ~100 uF electrolytics, plus a 0.1uF, to be sure, to be sure...

There may have been a 'choke' coil included to suppress possible 'ringing', but I was too busy gawping at that row of 'beer can' electrolytics to take heed...

 

[Tom.G]

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)

Since you are using Electron Flow convention, let's assume a negative spike comes along. This results in an excess of electrons on one plate, with the other plate being Grounded. The excess Electrons repel more Electrons from the grounded plate. That's why a capacitor is said to 'conduct AC'. The Electrons don't actually move to the other plate, they just repel more Electrons from it, giving the impression of passing thru the di-electric.

When the Negative spike goes away, the capacitor now has a larger charge on it resulting in a higher voltage. This higher charge is then available to whatever circuit the capacitor is in. If it is in a power supply that is supplying a Negative voltage, then the charge is available to the load; and since the capacitor voltage is (slightly) above the actual power source, that's where the electrons (current) are drawn from.

With a few word changes, the above also works for Positive spikes and Positive supplies. For instance if it is a Positive power supply, the spike will decrease the capacitor charge, thereby drawing extra current from the actual power source until the voltage returns to 'normal'.

Hope this helps the mental image a little, the maths are still useful to learn though.

Cheers,
Tom

 

[Muhammad Usman]

When the Negative spike goes away, the capacitor now has a larger charge on it resulting in a higher voltage. This higher charge is then available to whatever circuit the capacitor is in. If it is in a power supply that is supplying a Negative voltage, then the charge is available to the load; and since the capacitor voltage is (slightly) above the actual power source, that's where the electrons (current) are drawn from.
Cheers,
Tom

Thank you very much Tom your answer really help a lot. "This higher charge is then available to whatever circuit the capacitor is in". If we connected the capacitor between the load and power supply so is it possible that the extra charge will be sent to load, if yes then how it can protect the circuit. Ahh I am very much confused.

 

[Tom.G]

"This higher charge is then available to whatever circuit the capacitor is in". If we connected the capacitor between the load and power supply so is it possible that the extra charge will be sent to load, if yes then how it can protect the circuit. Ahh I am very much confused.

Yup, I made a few unstated assumptions to make the main idea clear. One assumption I made was that the capacitor had a high enough value that its voltage increase would be 'small' for the energy supplied by the noise spike. Let's continue with a negative supply and a negative spike; again, things work equivalently with supplies and spikes of either polarity.

The actual voltage rise can be calculated by knowing the energy in the spike and the capacitor value. The spike energy is proportional to its current and to its duration. For instance using a 100 000uF capacitor and a 1 second pulse of 1Amp will add 10V to the capacitor voltage. If the pulse is only 1millisecond long, the capacitor voltage will rise by 10mV.

If the power supply has a regulated output voltage, it will see that 10mV rise and, generally in a fraction of a milliSecond, reduce its output. From there, the load will continue to draw its current, reducing the charge (voltage) of the capacitor. When the voltage gets low enough, the power supply regulator will again start to supply the load current.

This will happen even if the power supply is un-regulated, but it will probably take longer.

Cheers,
Tom

 

[Muhammad Usman]

The actual voltage rise can be calculated by knowing the energy in the spike and the capacitor value. The spike energy is proportional to its current and to its duration. For instance using a 100 000uF capacitor and a 1 second pulse of 1Amp will add 10V to the capacitor voltage. If the pulse is only 1millisecond long, the capacitor voltage will rise by 10mV.This will happen even if the power supply is un-regulated, but it will probably take longer.

Cheers,
Tom

Thanks, I now get some understanding. So If I Summarize it, it will be like below:-

"
Capacitor's both plates are required to be connected with the two opposing terminals of the battery. One side of the battery will attract the electrons and the other side of the battery will supply electrons. Initially when the DC voltage is supplied the electrons from the positive terminal of the battery will suck all the electrons and come to the same reference level as Ground and on the other side it will supply the electrons to the plate. Once the electrons are abundant on the negative plate, due to difference in the charge between the plates the electric field exist or potential difference exist between the plates. This force neutralize the force coming from the battery.
Now comes the spikes case. In case of spikes the voltage spike comes in and cause more electrons to move into the charge as the force is not balanced between the charge plate potential and battery potential. This cause more electrons to come into the negative plate, but the electrons on the positive plate remain same making as a Ground potential or zero potential. So when the spike period is over then the capacitor is discharged and it supplies the charge back to the load ". The battery adjust accordingly so if we make the nodal analysis by KCL then the current going out of the capacitor is in different direction then the current coming in of the battery. So the currents are in opposite direction, causing the current to supply to load a little less "

Please check if there is any thing I miss. One thing I don't understand is how the power supply adjust itself accordingly . For regulated it could be understood but for non regulated, I don't understand if its just three volts battery. Also I don't understand if the battery don't adjust where does the extra current go.

 

[davenn]

. Initially when the DC voltage is supplied the electrons from the positive terminal of the battery will suck all the electrons and come to the same reference level as Ground and on the other side it will supply the electrons to the plate.

No, definitely not

electrons don't "suck" other electrons from anywhere

Once the electrons are abundant on the negative plate, due to difference in the charge between the plates the electric field exist or potential difference exist between the plates

not quite

An energised capacitor will have an equal number of negative charges (electrons) on one plate as there are positive charges on the other plate.
It's the increasing negative charge on one plate that repels negative charges off the other plate (no sucking from the battery)
This is what sets up the electric field between the plates. The energy is stored in the electric field

This cause more electrons to come into the negative plate, but the electrons on the positive plate remain same making as a Ground potential or zero potential

No, for every negative charge (electron) that enters one plate, one will leave the other plate. The overall charge on a capacitor will always remain zero

Also I don't understand if the battery don't adjust where does the extra current go.

what extra current ?

 

[Muhammad Usman]

Hi Davem,
If you can help me out with below mentioned scenerio.

" Once the capacitor connected as spike protector (Spikes in current), then how does the electron movement will happen "

 

[Muhammad Usman]

Hi Davem,
If you can help me out with below mentioned scenerio.

" Once the capacitor connected as spike protector (Spikes in current), then how does the electron movement will happen "

In case of.spike. Because I.read capacitor protect the circuit from spikes by shunting the current towards ground.

 

[anorlunda]

In case of.spike. Because I.read capacitor protect the circuit from spikes by shunting the current towards ground.

That question was answered carefully, yet you circle back and ask it again.

I fear that you have profound misunderstandings about the nature of electricity. They will not be resolved in this thread. I recommend that you use textbooks or conventional university courses to learn the subject.

Thread closed.