How does a Capacitor protect against voltage spikes?

[Muhammad Usman]

Hi Everyone !

I have one confusion regarding the role of the capacitor in the circuit. How does the capacitor protects the circuit from the high voltage spikes. If the capacitor is connected from the circuit with battery. And sudden high spikes come then the capacitor is charged and the energy is stored in the Electric field. Now once the spike is over the capacitor is require to discharge. For the discharge it will use the same path back to the load. So how does it protect the Load from the high spikes. May be I am wrong. Please correct me. I have mentioned a little diagram below. Can some one help to figure out.

 

[Rive]

Your schematics is just not really helpful to understand the question. Try with adding a source imperance/resistor, and think about the 'high spike' as an additional high frequency component to the DC!

 

[Muhammad Usman]

hi brother its a simple question doesn't need any schemetic. its just the role of capacitor that's it.

 

[davenn]

How does the capacitor protects the circuit from the high voltage spikes.

capacitors are not very good for spike protection and are rarely used for that purpose, because they cannot respond fast enough.
Capacitors do much better in holding up the voltage during times of voltage "sag" in a circuit where the voltage lowers for periods of time and the capacitor can hold up the supply voltage to the circuit
There are specific components available for spike protection called MOVs Metal Oxide Varistor(s). Most power supplies have these in them these daysDave

 

[Rive]

its just the role of capacitor that's it.

Nope. It is the role of the source resistance. The capacitor is just providing low load resistance at high frequencies.
But your schematics does not even has source resistance. In this schematics the discussed effect does not exist => of course you can't understand how does it work when it just does not work...

 

[Muhammad Usman]

Hi
I have read in the book that capacitor protects the circuit from high spikes by shunting down the energy towards ground.

 

[Muhammad Usman]

As high spikes of voltage or energy are shunted to the ground.

 

[davenn]

Hi
I have read in the book that capacitor protects the circuit from high spikes by shunting down the energy towards ground.

That's what MOV 's do

 

[Muhammad Usman]

https://en.wikipedia.org/wiki/Decoupling_capacitor

In this link it is mentioned that sudden high spikes are stopped by the capacitor. Let's suppose there is a DC voltage. The DC voltage is having some AC component due to non ideal condition. That AC component is considered as the spike or little noise you can say. In the light of circuit above (I know its not perfect or incomplete but I still believe it serve the purpose of conveying where I am confused) that AC component will pass through the capacitor path. In my understanding what happened this spike or sudden sharp voltage will cause the voltage to absorb the energy and then when the sudden spike is over the capacitor will discharge it. And during the process of discharge it will send back the same current or voltage to the load. So how does it protect.

I may be utterly wrong so that's why I came to this forum. Please help me out.

 

[yungman]

Capacitor can protect circuit from spike because by definition of spike, it's very short. You can look at it as a bunch of high frequencies burst for very short period of time. You should know the reactance of the capacitor is 1/(2pi f C). The higher the frequency, the lower the reactance ( or impedance). I am not going to repeat how capacitance smooth out the voltage after the rectifier, you asked in the other thread already.

So the capacitance smooth out the voltage in normal operating condition, when a spike comes in, it sees a very low impedance ( because the capacitance has very low reactance at high frequencies). It will conduct a lot of current to the ground without raising the voltage, thereby smoothing out the spike. Think of it this way, the cap provide a high frequency GROUND at high frequency, so it will short out any high frequency to ground.

Now, this is all theoretical, in real world, you cannot just put a large cap and hope everything will work out. The larger the cap, the higher the ESL it has, at some high frequency, the reactance of the inductance takes over and the cap loses it's ability to absorb the spike. A good design entails a smaller ceramic cap in parallel with the bigger ( assuming an electrolytic cap) so at high frequency, the ceramic cap take over. When I design RF circuits, I even use 3 capacitors. eg, 10uF electrolytic parallel with a 0.1uF ceramic cap then a 1000pF ceramic cap so I cover from low frequency to very high frequency.

Your graph on the right in post 1 is correct, you see a slight bump after the spike. The spike might be shorted out to the ground by the cap, but the charge from the spike will charge up the cap a little, that's why the voltage will show a bump. Then after the spike is over, the voltage will slowly recover back to the normal voltage as shown in the graph.

 

[Baluncore]

How does the capacitor protects the circuit from the high voltage spikes.

Current, I, is the rate of flow of charge, Q.
Capacitance, C,is defined as charge, Q, per volt, V.
C = Q / V.

For the voltage across a capacitor to change, a charge must be delivered to the capacitor.
The current available to charge the capacitor will limit the height of the voltage spike that can pass the capacitor.

 

[Muhammad Usman]

Hi,

If we read the theory of the capacitor, it shows that during the flow of AC there are two cycles that impact capacitor and reason behind its behavior of short circuit in case of AC is charge distribution and Electric and voltage creation in result of that. For example in case of positive half cycle the capacitor charge distribution is changed and its continuously changed and in the negative half cycle capacitor supply back that energy which stored in the positive half cycle. That is why it is considered as short circuit (Correct me If I am wrong). In case of the noise the component is very sharp so the capacitor charge but again after change it has to discharge and there is my question. During the discharge process it has to supply back the same energy to the load so how does actually it protects the circuit as during discharge it will send back ?

 

[anorlunda]

A pulse (or spike) can be modeled as a sum of high frequency AC signals. We told you that in the previous thread.

In this picture, you see two spikes. The first pulse is positive and you can see that it is the sum of AC signals that are both positive and negative. The sharper the spike, the more high frequency content.

 

[yungman]

Hi,

If we read the theory of the capacitor, it shows that during the flow of AC there are two cycles that impact capacitor and reason behind its behavior of short circuit in case of AC is charge distribution and Electric and voltage creation in result of that. For example in case of positive half cycle the capacitor charge distribution is changed and its continuously changed and in the negative half cycle capacitor supply back that energy which stored in the positive half cycle. That is why it is considered as short circuit (Correct me If I am wrong). In case of the noise the component is very sharp so the capacitor charge but again after change it has to discharge and there is my question. During the discharge process it has to supply back the same energy to the load so how does actually it protects the circuit as during discharge it will send back ?

You are making things too complicated, I notice that even in the other post. You have to keep studying, don't just stop at the first two pages of the book and try to figure everything out. The impedance of the capacitor Z=1/(2pi X f X C). Use this equation and you will see at high frequency, the Z is very low and consider it's a short circuit. It' is just that simple. Then you look at the charge where it will raise the voltage in equation V= C/Q where Q is the total charge.

It's is not that complicated, just read through the book, work on some exercise problems and you'll get it. After you get familiar, then worry about I=C(dV/dt) and all that, one step at a time.

Are you studying this in your electromagnetic class, that is NOT a good start, you are expected to have some common sense knowledge before you jump into the Maxwell's equations. Get some basic books. You don't learn how to fly before you learn how to walk.

Get a book like Electronic Principles by Malvino, get a used one, even the older version is good enough, it's less than $10 on Amazon. You learn to look at circuit ( like capacitor) in DC and AC environment, you just sum the two together to get the complete analysis. Start thinking a cap is an open circuit in DC and a short circuit in AC( generally speaking). In your question, the cap acts like an AC short (Z=1/(2pi f C)). The DC voltage is using DC analysis.

 

[anorlunda]

For example in case of positive half cycle the capacitor charge distribution is changed and its continuously changed and in the negative half cycle capacitor supply back that energy which stored in the positive half cycle.

That too is wrong. In the first half cycle of a sin, the capacitor starts at zero, it charges until the peak of the sin, then it discharges and reaches zero again at the half cycle. During the negative half cycle it does the same in the opposite direction.

Follow the advice of @yungman . If you keep trying to visualize electric circuits one electron at a time, you will fail. It is better to forget that you ever heard of electrons, charge, or fields until your basic circuit courses are complete.

 

[yungman]

Don't try to visualize electronic circuits ( including capacitor) in the eye of physics, that will drive you crazy and going in circles if you are just starting out. Learn it the easy way using DC and AC analysis, learn the common sense how the circuit works. Then when you get good at it, then, move to more advance way of analyzing.

To be frank, you might think you want to get to the truth, but if you really want to get down to it, thinking about electron movement is even wrong! If you really want to be correct, circuit and signal DON'T even run with electrons, signal travels as EM wave, electrons and current are the consequence of the EM wave as shown in Maxwell's equations. Physical electrons move TOO slow to be of any use!

Most people don't design circuits from EM point of view unless in special cases. You'd pull your hair out if you try to. My degree was not in EE, electronics was and is my passion, I studied all on my own. My first book I studied from cover to cover is the Electronic Principle by Malvino in 1979. Yes, I studied back all the electronics, physics and math to post grad level through out the years. BUT when I design circuits, I still use a lot of the basics learned from the Malvino. Believe it or not, I spent a few years designing analog IC that was basically transistor circuit design. I use Malvino more so than the gold book of IC design by Grey and Mayer. Yes, I studied that book too, but still find Malvino more useful. It's NOT that complicated, most of the time, you get by with simple assumptions. Then when it calls for, bring in the EM theory and all the advanced theory.

 

[anorlunda]

Don't try to visualize electronic circuits ( including capacitor) in the eye of physics, that will drive you crazy and going in circles if you are just starting out.

Thank you @yungman . That is what I've been trying to say.

It's even more difficult than you said because the true bottom of the pile is not EM, it is QED.

 

[osilmag]

You could use a voltage regulator instead of a capacitor.

A low pass filter rejects high frequency because the high frequency waves encounter less impedance on the way to ground.

 

[Rive]

A low pass filter

Low pass filter would be one of the valid explanations, but unfortunately there is no resistor present to make it a low pass filter

This thing really needs the source resistance/impadance to make it work. At this state it is just... source for confusion, at most. You just can't buffer/decouple/protect again spikes if your source has zero impedance => can provide infinite current to fill a capacitor

 

[essenmein]

Low pass filter would be one of the valid explanations, but unfortunately there is no resistor present to make it a low pass filter

This thing really needs the source resistance/impadance to make it work. At this state it is just... source for confusion, at most. You just can't buffer/decouple/protect again spikes if your source has zero impedance => can provide infinite current to fill a capacitor

Really that's the root of the issue, the OP has not considered the source impedance of the spike source. If spike sources truly had zero output impedance then even a MOV would just explode.

The reality is all sources of spikes or voltage changes have (or should have) a clearly defined source impedance, everything from ESD (eg 330pF charged to 15kV discharge through 330Ohm ISO10605) to line disturbances to transient over voltages, if the source impedance is not defined, then how can a designer make the circuit suitable, or the test engineer make a test to verify?

So if you know the source impedance, voltage and duration you can then make an educated decision if a cap is enough or you need something else (MOV etc), eg ESD strike a ceramic cap and BLM is often good enough. Things where there is already very large capacitance (eg motor drives) laugh at ESD strikes even sometimes can ride through load dump.

 

[jim hardy]

And during the process of discharge it will send back the same current or voltage to the load. So how does it protect.

During the discharge process it has to supply back the same energy to the load so how does actually it protects the circuit as during discharge it will send back ?

If you want an informal answer --

It gives the energy back at a pace that is comfortable for the load.
It absorbs the big gulp of energy from wherever it came,
then let's the supply take a breather while it metes out that energy to the load.
That way the load doesn't have to withstand the "spike"

 

[yungman]

Well, it's obvious that this is the first time OP looking at capacitor, we should not be too nick picking as it's going to be more and more confusing for OP.

Of cause, we need to look at source impedance, the spike energy or the output impedance of the device that introduce the spike. That will make things much more complicated and won't serve the purpose to explain this. If you really want to be exact, the schematic drew by OP has a voltage source driving the cap and the load. By definition, voltage source has zero output impedance, no spike will change the voltage with or without the cap!. But we all know and assume it has finite output impedance and need a cap to give a low impedance path to ground( or return path to be exact). That's the reason the drawing show a bump in voltage after the spike and slowly bleeding away. If you want to be exact, there will be no bump as the voltage source will absorb the spike.

Bottom line, yes, a cap can smooth out a spike if it is done right. Transorb or other MOV won't do a good job in a lot of ways because they come in set voltage and cannot be exact in clamping the voltage. What if your voltage is 5.5V, you likely cannot fine a 5.5V MOV and even if you can, there is a range of voltage it will start clamping. What if it start clamping at 5.4V? then it will short out the 5.5V supply and start smoking. Transorbs and MOV are more useful for protection to guaranty the spike will not cause the voltage to rise beyond the components can withstand. It CANNOT smooth out the voltage like you want it to.

I think we should just limit to discussion on impedance of the cap and will smooth out the charge introduced by the spike...IF the value of the cap is big enough.

 

[Baluncore]

During the discharge process it has to supply back the same energy to the load so how does actually it protects the circuit as during discharge it will send back ?

You considered a symmetric sinewave, but then you try to apply that analysis to an asymmetric spike or pulse.

Energy delivered in a short time by a high amplitude pulse is returned to the circuit over a longer period of time and so has a lower amplitude.

 

[Rive]

it's obvious that this is the first time OP looking at capacitor ... but we all know and assume...

Please pardon me that I've edited your post in this quote, but this way it is more clear where the problem lies...

Energy delivered in a short time by a high amplitude pulse is returned to the circuit over a longer period of time and so has a lower amplitude.

I think it is wort noting that the relevant voltage drop (V_in-V_out) here belongs to R1 (source impedance). So while the capacitor is still the key point of the circuit, the work (more than just by one sense) is actually done by R1.

 

[sophiecentaur]

hi brother its a simple question doesn't need any schemetic. its just the role of capacitor that's it.

That's ignoring the way that a Capacitor acts in a circuit. If there is no source impedance, the Capacitor will do nothing to suppress spikes.

Hi
I have read in the book that capacitor protects the circuit from high spikes by shunting down the energy towards ground.

As high spikes of voltage or energy are shunted to the ground.

If you are after a proper answer to the question then you have to be precise. A Capacitor merely stores Energy. The energy will be returned to the circuit when the driving voltage drops. There has to be a resistive component in a circuit for Energy to be dissipated. So you either rely on the source resistance of the supply or you have to insert your own in the form of a series resistance. As in nearly all electrical problems, a Shematic Diagram is essential - particularly if you are discussing it with someone else.

(Correct me If I am wrong)

Are you sure that your mind is actually open to 'corrections'. The frequency selectivity of a circuit with a Capacitor, depends upon the time constant RC of that capacitor and an appropriate resistor. Eliminating or reducing interference spikes requires Damping.

 

[AVBs2Systems]

Hi
One way to understand this, is to see the step response of RC and RL systems. In the case of RC systems it opposes sudden changes in voltages, and in the case of an RL network it opposes sudden changes in current.
Check out the step response of an RC network:
$$ \textbf{Step response of an RC network, with output voltage at the capacitor}$$
The differential equation where $x(t)$ is the input voltage, and $y(t)$ is the voltage at the capacitor, where $\tau = R \cdot C$
$$
\dfrac{x(t)}{\tau} = y'(t) + \dfrac{y(t)}{\tau}
$$
The transfer function is $$ H(s) = \dfrac{ \dfrac{1}{\tau} }{ s + \dfrac{1}{\tau} }$$
The impulse response is hence the inverse laplace transform:
$$
h(t) = \dfrac{1}{\tau} e^{-\frac{t}{\tau} } \,\,\,\, (\text{V})
$$
The step response:
$$
\dfrac{H(s)}{s} = G(s) = \dfrac{ \dfrac{1}{\tau} }{s ( s + \dfrac{1}{\tau} ) }
$$
Taking the inverse laplace transform:
$$
g(t) = \epsilon(t) \cdot \big( 1 - e^{ -\frac{t}{\tau} } \big) \,\,\,\, (\text{V})
$$
Hence we can see that for a step voltage:
$$
k \epsilon(t) = \begin{cases} k \,\,\,\, \text{V} & \text{ $t \gt$ 0} \\ 0 \,\,\,\, \text{V} & \text{ $t \lt $0 } \end{cases}
$$
The response is:
$$
\textbf{T}\Big \{ k\epsilon(t) \Big \} = k \epsilon(t) \cdot \big( 1 - e^{ -\frac{t}{\tau} } \big)
$$
The network was inputted by a sudden rising voltage, instantaneous, and its response was not the same but a smoother exponential rise. This is one way of understanding how an RC circuit protects against voltage spikes. Capacitors and inductors oppose sudden changes in voltage and current respectively.
Hence, if you put your $\textbf{Output voltage}$ at the capacitor, you do protect the network at the output from sudden steps.

 

[sophiecentaur]

One way to understand this, is to see the step response of RC and RL systems

+1
The impulse response can also be useful where appropriate.

 

[Charlie Cheap]

I am reminded of an instructor's comment while in school. "Gentlemen, what I am about to explain is NOT what actually happens, but it is the only way I can explain it so you can understand, with what little knowledge you have now." This was Elkins Institute in 1969 (YES, I am old), and we understood the equation...though not technically correct. We understood mechanical actions because we dealt with them every day, but were not yet able to see electrons...which required an oscilloscope. A device we were not allowed to even turn on for weeks.

 

[sophiecentaur]

an oscilloscope. A device we were not allowed to even turn on for weeks.

HAHA. Sounds like parade ground drilling with wooden rifles. I remember similar need for deferred gratification in some of my late fifties Science Lessons.
It totally put me off RTFM for ever.