Relationship of frequency and capacitance.

[chopnhack]

Hello all, I am reading Electronics for Dummies.... yes, I am that desperate LOL

I am 71 pages in and finally have found something to say, hey, I have a question:

In the discussion of capacitors, it states how capacitors can be used to adjust the frequency of a signal. My thought was that the larger the capacitor in Farads, the larger the "space" to absorb signals thus the signal would have to be large to allow it to pass through and vice versa for smaller signals. A large signal would be one of low frequency - longer wavelength to traverse the "space"

Does this logic make sense? Or am I misunderstanding.

Thanks

 

[LvW]

It seems that you are mixing "large frequencies" with "large signals". This must be clarified.
Furthermore, if the text speaks about "adjusting the frequency" I think that the subject of discussion is a frequency-dependent network (filter, oscillator,....)

 

[duke37]

I think that the book refers to the operation of an oscillator using a tuned circuit.

The frequency is 1/(2*pi*sqrt(L*C)), thus the bigger the capacitor, the lower the frequency.

 

[Ratch]

Hello all, I am reading Electronics for Dummies.... yes, I am that desperate LOL

I am 71 pages in and finally have found something to say, hey, I have a question:

In the discussion of capacitors, it states how capacitors can be used to adjust the frequency of a signal. My thought was that the larger the capacitor in Farads, the larger the "space" to absorb signals thus the signal would have to be large to allow it to pass through and vice versa for smaller signals. A large signal would be one of low frequency - longer wavelength to traverse the "space"

Does this logic make sense? Or am I misunderstanding.

Thanks

No, it does not make sense and yes, you are misunderstanding. Never forget that a capacitor is a energy storage device. It does this by separating charge quantities. When a capacitor is energized by an applied voltage, electrons accumulate on one plate and deplete on the opposite plate. This causes a momentary or transient current to occur. The electrolyte between the plates does not allow charge to pass through the capacitor. So one plate has a surplus of electrons and the opposite plate has a scarcity of electrons. It takes energy to separate the charge quantities, which comes from the voltage supply or signal. The difference in charge between the plates causes a electric field to form, and that field stores the energy. A larger capacitance value means that the cap can store more energy at the same voltage or signal level. A capacitor in series or shunt with a signal will pass all frequencies, but it passes higher frequencies more readily. If the signal frequency is so low that the transient current is becomes negligible, then the capacitor has effectively blocked the signal. I will explain resonance when you digest what I wrote already.

Ratch

 

[chopnhack]

Let me post the section that fired off the few synapses in my brain for clarity

A capacitor in series or shunt with a signal will pass all frequencies, but it passes higher frequencies more readily.

Why? What characteristic of the capacitor makes it such that it passes mostly high frequencies?

 

[AnalogKid]

About the capacitor plates, you have it backwards. All other parameters being equal, the closer the plates are the higher the capacitance (and the lower the voltage rating). This is partly because of the inverse square law. As distance increases, a charge exerts less force. So the thinner the dielectric, the greater the capacitance.

ak

 

[Ratch]

Let me post the section that fired off the few synapses in my brain for clarity

View attachment 21607


Why? What characteristic of the capacitor makes it such that it passes mostly high frequencies?

Its ability to store energy. It works like this. Suppose you apply a relatively low frequency signal to a capacitor. The slow frequency assures that there is time to pump a lot of charge onto one plate and delete it from the opposite plate as the voltage rises. But if there is a point where the cap cannot accept all that charge difference within that long climb to max voltage, then that cap will act as if there were an impediment to the current. The same happens when the voltage falls. The cap cannot sustain the charge needed for the long time it takes the voltage to reach zero. This impediment is called reactive impedance, and it increases as the frequency becomes lower, and decreases as the capacitance becomes larger, and is able to sustain longer periods of energizing and de-energizing. Higher frequencies mean shorter energy transfer times, which allow a cap to better keep up with the signal, and results in a lower reactive impedance. This means the cap will pass higher frequencies better.

Ratch

 

[chopnhack]

Bravo Ratch - that is a brilliant explanation. Someone should sticky that response! @Ian

The reactive impedance is effectively a resistance then that is intrinsic to the material - a heat less resistance.

But if there is a point where the cap cannot accept all that charge difference within that long climb to max voltage, then that cap will act as if there were an impediment to the current.

As the voltage continues to rise on a long wavelength (low frequency) how does the cap "stop" further packets of electrons - is it through repulsion, as if the capacitor has reached a maximum density on the plates and thus that is the "resistance" or impedance?

Thanks again!!

 

[Ratch]

Bravo Ratch - that is a brilliant explanation. Someone should sticky that response! @Ian

The reactive impedance is effectively a resistance then that is intrinsic to the material - a heat less resistance.

Not quite. A reactance opposes a current by inducing a back voltage. It does not produce heat because it stores energy in an electric or magnetic field, and gives it back to the circuit later. It is also frequency dependent.

A resistor opposes a current by reducing the energy density of the charge (voltage) by the heat dissipation. It is relatively frequency independent.

As the voltage continues to rise on a long wavelength (low frequency) how does the cap "stop" further packets of electrons - is it through repulsion, as if the capacitor has reached a maximum density on the plates and thus that is the "resistance" or impedance?

If a cap is fully energized at a particular voltage, the only way to unbalance the charge even more is to increase the voltage, Q = C*E. The lack of a higher voltage is what stops the charge imbalance.

Thanks again!!

You're welcome.

Ratch

 

[hevans1944]

I generally like @Ratch 's explanation, however I have a small quibble:

The electrolyte between the plates does not allow charge to pass through the capacitor

I think you must have meant the dielectric between the plates.

In an electrolytic capacitor, the electrolyte does allow charge to pass through the electrolyte, and in fact the electrolyte serves as the cathode or negative terminal of the capacitor. But said charge motion is stopped by the very thin dielectric film (aluminum oxide for aluminum electrolytic capacitors) that forms on the aluminum foil serving as the anode or positive terminal of the capacitor. The electrolyte is the reason that electrolytic capacitors have a polarity. Reversing the polarity applied to an electrolytic capacitor has the effect of removing the dielectric film by electrolysis, thus shorting the capacitor.

As @Ratch mentioned, the key to understanding capacitance (and inductance) is understanding reactance. Reactance is a property associated with the lossless storage of electrical energy, either in an electrical field for capacitors or in a magnetic field for inductors. Capacitive reactance subtracts algebraically from inductive reactance if a capacitor is connected in series with an inductor. For any pair of capacitor and inductor connected in series there will be a frequency where the series reactance is zero, a condition called series resonance. Not quite so obviously, for any pair of capacitor and inductor connected in parallel there will be a frequency where the parallel reactance approaches infinity. In both cases, this occurs when the magnitude of the capacitive reactance is equal to the magnitude of the inductive reactance without regard to "sign" of the magnitude.

Although reactance is measured in ohms, it is non-dissipative: reactive current does not dissipate energy or consume power. The reactance of a capacitor is XC = 1/ (2 π f C), where f is in hertz, C is in farads, and XC is in ohms. The reactance of an inductor is XL = (2 π f L), where f is in hertz, L is in henries, and XL is in ohms. Note that these equations have no meaning at DC because reactance is only defined for sinusoidal AC waveforms. However, the implication is that XC approaches infinity as the frequency approaches zero, and that XL approaches zero as the frequency approaches zero. For any given pair of inductance and capacitance there is always a frequency greater than zero and less than infinity where |XC| = |XL| and that frequency is called the resonant frequency. It is this property of reactance that allows tuned band-pass circuits, peaking circuits (a narrow band-pass filter), and notch filters (a narrow band-reject filter) to be constructed from L and C components. Such circuits can also be constructed in other ways, but generally this requires active components such as op-amps which limit the range of frequencies that can be accommodated.

What characteristic of the capacitor makes it such that it passes mostly high frequencies?

The reactance of a capacitor decreases as the frequency increases. This is a smooth hyperbolic function whose asymptotes are zero at infinite frequency and infinity at zero frequency. Plug in some numbers to the capacitive reactance equation above and you will see why higher frequencies are impeded with less capacitive reactance than lower frequencies. OTOH, the reactance of an inductor is a linear function of frequency so higher frequencies are impeded with more inductive reactance than lower frequencies. However, it is guaranteed that these two functions will intersect somewhere between zero frequency and infinite frequency for any arbitrary values of capacitance and inductance, said intersection occurring at resonance. Now, as to why capacitors and inductors behave this way... go back and read @Ratch 's explanation for perhaps a qualitative understanding. A quantitative understanding requires considerable effort, but it all boils down to the four Maxwell's Equations, which one of my college professors told me were handed down to James Clerk Maxwell by God. Of course this was at a private Catholic university, so there may have been some prejudice on his part, but I have no other explanation for why the Universe is put together and works the way it does. So, good luck with Electronics for Dummies and I hope you graduate to something a bit more in depth. Brush up on your math and physics; it does get easier (and fascinating too) after some basic concepts are nailed down.

I hope this helps!

Hop

 

[Ratch]

I generally like @Ratch 's explanation, however I have a small quibble:

I think you must have meant the dielectric between the plates.

Yes, you are correct. It is never a good idea to answer a post late at night.

Not quite so obviously, for any pair of capacitor and inductor connected in parallel there will be a frequency where the parallel reactance approaches infinity. In both cases, this occurs when the magnitude of the capacitive reactance is equal to the magnitude of the inductive reactance without regard to "sign" of the magnitude.

For a series resonant circuit, there is only one frequency. For a parallel resonant circuit, some folks and textbooks like to define 3 resonant frequencies. One is where the L and C reactance are equal, a second one where the parallel impedance is the largest, and the third frequency where the phase is zero and the power factor is 1. In a high Q circuit, these three frequencies are very close together. I personally accept only the first definition and regard the second and third definitions as spurious. It is also possible to make a parallel circuit that is resonant at all frequencies by inserting certain resistance values in the C and L legs of the parallel circuit. I can elaborate further on this statement if requested.

Ratch

 

[hevans1944]

@Ratch : Interesting comment. I didn't want to confuse John yet with the concept of Q until he has a firm grasp of what is going on in a capacitor (and an inductor). In practice, Q is of course very important to circuit behavior and it leads to the kind of bogus "definitions" you mentioned. I will also stay with XC = XL as defining resonance, serial OR parallel connection.

I would be interested in seeing how a judicious insertion of resistances in series with the capacitor and inductor of an LC tank circuit results in "resonance" at all frequencies. Please do elaborate!

 

[Arouse1973]

As always Ratch, very good.
Adam

 

[Ratch]

@Ratch : Interesting comment. I didn't want to confuse John yet with the concept of Q until he has a firm grasp of what is going on in a capacitor (and an inductor). In practice, Q is of course very important to circuit behavior and it leads to the kind of bogus "definitions" you mentioned. I will also stay with XC = XL as defining resonance, serial OR parallel connection.

I would be interested in seeing how a judicious insertion of resistances in series with the capacitor and inductor of an LC tank circuit results in "resonance" at all frequencies. Please do elaborate!

OK, attached are a couple of pages from the Schaum's Outline Series, Electric Circuits. The circuit problem 8.12 is rather useless because there is no selectivity and its impedance is somewhat low, but it delivers on being resonant at all frequencies.

Ratch

 

[HellasTechn]

Its ability to store energy. It works like this. Suppose you apply a relatively low frequency signal to a capacitor. The slow frequency assures that there is time to pump a lot of charge onto one plate and delete it from the opposite plate as the voltage rises. But if there is a point where the cap cannot accept all that charge difference within that long climb to max voltage, then that cap will act as if there were an impediment to the current. The same happens when the voltage falls. The cap cannot sustain the charge needed for the long time it takes the voltage to reach zero. This impediment is called reactive impedance, and it increases as the frequency becomes lower, and decreases as the capacitance becomes larger, and is able to sustain longer periods of energizing and de-energizing. Higher frequencies mean shorter energy transfer times, which allow a cap to better keep up with the signal, and results in a lower reactive impedance. This means the cap will pass higher frequencies better.

Ratch

That is why we need large smoothing capacitors to turn 50-60 hz mains AC voltage to DC right ? plus the current being drawn by the load.

 

[Ratch]

That is why we need large smoothing capacitors for to turn 50-60 hz mains AC voltage to DC right ? plus the current being drawn by the load.

Yes, a large capacitance value is need to store enough energy to supply the load for the period of time when the input sinusoidal goes below the desired DC output. The AC is turned into pulsating nonalternating current by a rectifier.

Ratch

 

[Flurng]

Hello all, I am reading Electronics for Dummies.... yes, I am that desperate LOL

I am 71 pages in and finally have found something to say, hey, I have a question:

In the discussion of capacitors, it states how capacitors can be used to adjust the frequency of a signal. My thought was that the larger the capacitor in Farads, the larger the "space" to absorb signals thus the signal would have to be large to allow it to pass through and vice versa for smaller signals. A large signal would be one of low frequency - longer wavelength to traverse the "space"

Does this logic make sense? Or am I misunderstanding.

Thanks

You are correct that a lower frequency does indeed have a longer "wavelength", which does in fact prevent it from passing "through" a smaller capacitor. However, one should always remember that a capacitor does not NEED to be fully charged in order to perform it's function. In fact, for the protection of the capacitor, and other components, it really should NEVER be fully charged, since TOO MUCH charge can damage the device, and under the "wrong" conditions, a larger capacitor can actually explode!

In addition, a common misconception I would like to clear up is the idea that alternating current flows "through" a capacitor, while direct current does not. In reality, NO current flows "THROUGH" a capacitor ( unless it is defective ). Instead, current simultaneously flows INTO one side, while it flows OUT OF the other. Imagine, for example, a "duplex" style home, which is essentially two distinct houses which share a common SOLID wall. Now, imagine the left side of the duplex is filled with party-goers, who are a bit inebriated, so they pour OUT the front door, INTO the door of the right side of the duplex to sleep it off. After a time, they all wake up and, wanting to keep the party going, they flow OUT the front door of the right side, and INTO the door of the left side ( the Fun side! ). From the street, it would appear that current is passing through the home, first in one direction, then the opposite, while in reality, it merely flows INTO or OUT OF either end.

I find it easiest to think of a capacitor as a "bucket" or a "pool" for containing electrical charge. So, for instance, if you have a "bucketful" of charge, you could easily pour that into ( and out of ) a swimming pool, whereas you could NOT pour a "poolful" into ( or of course, out of ) a bucket. Continuing with that analogy, given a garden hose that delivers water at a fixed rate, which could you fill / dump quicker? If we assume that the process of "filling & dumping" the water (charge) represents one "cycle" of alternating current, it should then be rather clear that the "smaller" capacitor ( the bucket ) results in a higher frequency of current, since less time is required between "fill" and "dump" cycles.

Hope this helps, and please feel free to respond if you should have any further questions.

 

[Ratch]

I find it easiest to think of a capacitor as a "bucket" or a "pool" for containing electrical charge. So, for instance, if you have a "bucketful" of charge, you could easily pour that into ( and out of ) a swimming pool, whereas you could NOT pour a "poolful" into ( or of course, out of ) a bucket. Continuing with that analogy, given a garden hose that delivers water at a fixed rate, which could you fill / dump quicker? If we assume that the process of "filling & dumping" the water (charge) represents alternating current, it should then be rather clear that the "smaller" capacitor ( the bucket ) results in a higher frequency of current.

Do hydraulics folks use electrical analogs to study their craft. No? Then why should folks studying electronics use hydraulic analogies for their studies? Why not accept that a capacitor is charged with energy or energized, and study capacitors using energy as a quantity? The bucket brigade does not take into consideration that a capacitor will only accept a limited amount of energy/<charge imbalance>, and then will block the flow of charge or storage of energy; whereas a bucket of water will simply overflow and not block the water flow.

Ratch

 

[Paul Grodkowski]

Hello all, I am reading Electronics for Dummies.... yes, I am that desperate LOL

I am 71 pages in and finally have found something to say, hey, I have a question:

In the discussion of capacitors, it states how capacitors can be used to adjust the frequency of a signal. My thought was that the larger the capacitor in Farads, the larger the "space" to absorb signals thus the signal would have to be large to allow it to pass through and vice versa for smaller signals. A large signal would be one of low frequency - longer wavelength to traverse the "space"

Does this logic make sense? Or am I misunderstanding.

Thanks

When I think about the relationship of capacitance to frequency, I refer to the mathematical formula:

Xc = 1 / 2 Pi f c

From this relationship we see that capacitive reactance is indirectly proportional to frequency.

It helps to look at a graph to see this relationship.

Here are a few short videos that may help:

This should help in your understanding.


Paul Grodkowski
VE3SQM
Toronto, Canada

 

[Flurng]

Do hydraulics folks use electrical analogs to study their craft. No? Then why should folks studying electronics use hydraulic analogies for their studies? Why not accept that a capacitor is charged with energy or energized, and study capacitors using energy as a quantity? The bucket brigade does not take into consideration that a capacitor will only accept a limited amount of energy/<charge imbalance>, and then will block the flow of charge or storage of energy; whereas a bucket of water will simply overflow and not block the water flow.

Ratch

To answer your question, the reason that I ( Me, personally ) think of the classic water analogy, is that it helps me to visualize the problem, which is the way I learn complex concepts best. On the contrary, what does an electrical charge "look" like?
I must agree with your assertion that the water analogy does not directly explain the "blocking" action of the capacitor, however, I would contend that any water/charge/quantity "overflowed" would be lost, and thus "blocked" from re-entering the system. Furthermore, if we are concerned only with the time interval (and thus, the frequency) required to fill/dump (charge/discharge) the device in question (be it a capacitor, a bucket or a pool), then I believe it is clearly implied that one would naturally stop filling once the device is full, or dumping, once it is emptied.
Finally, since Chopnhack seemed to be struggling with what is a relatively simple concept, it appeared somewhat counterproductive to explain a simple concept by using more complex, abstract concepts such as "energy storage" "electrostatic fields" or "charge imbalance".