Infinite Capacitance: Uses & Implications in JFET Circuits

[Yura]

Theoretically only, would a capacitor of infinite capacitance be considered an open circuit?
What would it be used for in application if it did exist?

The reason I am asking is because I have a homework problem where it is used with a JFET transistor in a circuit.

If I can't solve the problem I might ask for help later, but I would like to understand if there would be a reason for using an infinite conductance.

 

[TheAnalogKid83]

Theoretically only, would a capacitor of infinite capacitance be considered an open circuit?
What would it be used for in application if it did exist?

The reason I am asking is because I have a homework problem where it is used with a JFET transistor in a circuit.

If I can't solve the problem I might ask for help later, but I would like to understand if there would be a reason for using an infinite conductance.

if you look at the capacitor impedance as a complex number you can see that its impedance is defined as:

z = 1/(c*j*w)

so let c = infinity

you get that its impedance would be 0 from w = infinite down to w approaching DC, so that it would be a short, not an open.

 

[TheAnalogKid83]

I should sum that up, an infinite capacitor would pass all frequencies except the DC component of a signal. The DC component would see the cap as an open, but all other frequencies would see it as a short.

The problem is probably just telling you that the circuit is decoupled from DC and coupled with AC.

 

[Phrak]

I should sum that up, an infinite capacitor would pass all frequencies except the DC component of a signal. The DC component would see the cap as an open, but all other frequencies would see it as a short.

Umm. I believe you have the DC part of it wrong. The voltage across the capacitor would remain constant with any current into or out of it. That would be zero resistance in series with an ideal potential.

 

[TheAnalogKid83]

Umm. I believe you have the DC part of it wrong. The voltage across the capacitor would remain constant with any current into or out of it. That would be zero resistance in series with an ideal potential.

Caps aren't open circuits at DC? You can have voltage across an open circuit . .

First
current in a cap is
i(t) = C*dV(t)/dt

at DC, the voltage is not changing such that dV(t)/dt = 0

so i = C*0

so i = 0, even if DC voltage is 5kV (depends on practical voltage rating of cap in real life though)
use ohm's law V = IR.
5kV/0 = R,
R = infinity which is an open
also you could just use the expression in the laplace domain
1/(Cs) for Zc

its obvious that at s = 0, which is DC, Zc is infinite, which in practical terms is an open circuit, so what do I have wrong?

 

[Defennder]

Think about this, would the voltage across the ends of an infinite capacitor ever be equal to the voltage of the DC source connected in parallel to it?

You're assuming dV/dt = 0, but is this assumption justified for an infinite capacitor?

 

[TheAnalogKid83]

Think about this, would the voltage across the ends of an infinite capacitor ever be equal to the voltage of the DC source connected in parallel to it?

You're assuming dV/dt = 0, but is this assumption justified for an infinite capacitor?

The capacitance is independent of the voltage applied to it and vice versa

dV/dt = 0 is NOT an assumption. . it is the property of DC, and this is the case at capacitor steady state with an applied DC voltage. But really you need to look at the frequency domain to see passed steady state, where 1/(Cs) tells you the impedance to derive your transfer function, regardless of its capacitance and regardless of the applied voltage.

Now you could question the energy required for charging an infinite capacitor, but this is a different discussion all together. And a charging capacitor can be based off the difference between its initial voltage and the applied DC voltage, which is a change in voltage and consequently you're introducing frequencies (a step function for example). This will create an in-rush of current because ideally dV/dt would be infinite with a step function at the time of charging, which would take infinite energy as well. If you have an ideal voltage source, the voltage is going to remain the same at its two nodes regardless of the capacitance quantity of the capacitor, because the ideal voltage source can supply any amount of current.

 

[Averagesupernova]

The question really is condensed down to this:
-
Scenario A:
Infinite capacitor leads are connected to voltage source in series with a resistor of value X ohms. The capacitor will charge indefinitely and the voltage across it will remain at zero.
-
Scenario B:
The voltage source in series with a resistor of value X ohms is connected to the capacitor before the capacitor exists and it will not conduct any current since the capacitor was charged from the first moment of its existence . Obviously scenario B is as ridiculous as the notion of a capacitor with infinite capacitance to start with.
-
I recall reading in an ARRL handbook that a transmission line that only had one end and extended into infinity in one direction when hooked up to an ohmeter (DC) the meter would read the characteristic impedance as long as you held the probes on the transmission line. The same thing will apply to an infinite capacitance capacitor.

 

[TheAnalogKid83]

The question really is condensed down to this:
-
Scenario A:
Infinite capacitor leads are connected to voltage source in series with a resistor of value X ohms. The capacitor will charge indefinitely and the voltage across it will remain at zero.

You're throwing in a resistor, which will create an RC time constant of infinity. You're also not saying what the frequency you're putting across the circuit is. You're also not stating initial conditions. Derive the transfer function of the circuit you describe you get:

Vout = Vin/(RCs +1)

so at DC all of the voltage across the capacitor is Vin/(0+1) which is Vin.

So voltage will not remain 0 at DC, it will remain Vin. At any other frequency, it will remain 0, since it is a short at any frequency besides DC with an infinite capacitance. The entire voltage drop will be across the resistor at any frequency > 0.The fact remains that an infinite capacitance capactor is an open circuit at DC (like any capacitor), and a short at all other frequencies (only for an infinite capacitance cap).I remember these problems on opamps and transistors, and the intent of stating that there is an infinite capacitor in series with your voltage source is to imply that the DC component has been blocked. They do this all the time on audio outputs, which is why you see 100uF or bigger capacitors in series to an amplifier stage, its to knock out the DC component and allow for rush currents without loading the source.

 

[Averagesupernova]

Sorry, I suppose I could have specified the voltage source in my hypothetical scenarios was a DC source. I thought it was well enough implied.
-
On to more important things, you're telling me that if I have a capacitor with infinite capacitance and I hook up a DC voltage source to it with no series resistor no current will flow in the circuit? Yet if I do the same thing with a series resistor since I have an infinite RC time constant then current will flow indefinitely? Adding a series resistance will cause more current to flow?

 

[TheAnalogKid83]

Sorry, I suppose I could have specified the voltage source in my hypothetical scenarios was a DC source. I thought it was well enough implied.
-
On to more important things, you're telling me that if I have a capacitor with infinite capacitance and I hook up a DC voltage source to it with no series resistor no current will flow in the circuit? Yet if I do the same thing with a series resistor since I have an infinite RC time constant then current will flow indefinitely? Adding a series resistance will cause more current to flow?

It seemed that charging to a DC level is being confused with DC. If you were implying that hooking up the voltage source to the caps was some sort of voltage change, then yes, that would not be DC and that would cause the cap to short out. That is why I mentioned that you did not point out your voltage source or initial conditions, because your conclusion was that the voltage across the Capacitor would be 0V at DC, and this is not the case. Charging with a DC source is not the DC input to the system. You're changing from initial voltage to final voltage in a time of 0 seconds in the case of a Heaviside step function when charging. This is not DC, in fact, it represents all frequencies, just like a delta function impulse does. This is why you can gain the frequency response of a system over all frequencies by looking at your output with a step function input. CHARGING with DC source != DC input, because DC will not change voltage at any time.

And where did I say current would flow with a DC voltage source if you put in the resistor? I'm saying exactly the opposite, that a capacitor is an OPEN at the DC frequency (0 Hz). Current will not flow in either scenario. Adding a resistor will change nothing about the DC characteristic of a capacitor, and even in this case, current will not flow, such that you will have complete voltage drop across the capacitor, and 0 voltage drop across the resistors since (Vr = R*Ir = R*0 = 0), which is contrary to saying that you will have 0V across the capacitor when applying a DC source. This is why it would take infinity time to charge the cap. The fact remains that an infinite capacitance cap is OPEN at DC and SHORT at f > 0HZ, and such a cap in series with an input to a transistor base or gate implies the DC component is not there, which is important when trying to bias transistors, which is where this probably came up in the original poster's homework problem.

 

[chroot]

The current into a capacitor is proportional to its capacitance. If you have an infinite-capacitance capacitor, then it will draw infinite current with even the slightest applied voltage, and it will continue consuming that infinite current forever. It's a short.

It's meaningless to discuss its frequency-dependent behavior, because it is pathological even at dc.

- Warren

 

[TheAnalogKid83]

The current into a capacitor is proportional to its capacitance. If you have an infinite-capacitance capacitor, then it will draw infinite current with even the slightest applied voltage, and it will continue consuming that infinite current forever. It's a short.

It's meaningless to discuss it's frequency-dependent behavior, because it is pathological even at dc.

- Warren

When you say slightest applied voltage, are you implying that the voltage is changing from one value to another? If so, this is not DC.

At DC it is an open, and this is important when biasing transistors, or amplifying a signal with an opAmp which is where the question came up. My professors gave me these exact same scenarios when they taught transistors. You can ride a sine wave biased by 5V DC, but only the sine wave will get through to the other side of the cap, not the 5V DC bias. By saying the capacitance is infinite is just to imply that all frequencies > 0 Hz will be passed equally, which means that your input AC signal will not be distorted. It's an ideal situation rather than a practical one. Obviously you can never charge an infinite capacitance capacitor, it would be like a black hole.

 

[Phrak]

I surely don't understand the confusion. dv/di=0; zero dynamic impedence. It's simply an ideal voltage souce/sink.

Now, what sort of circuit element has no change in the voltage across it, no matter how much current you force through it?

A short circuit fits the bill.

So does the infinite capacitor with an initial potential of zero volts. That is, it exhibits dv/dt=0, and v(t=0)=0.

With a little bit of imagination you should be able to convince yourself that the voltage across a charged capacitor doesn't change either (V = QC), no matter how much charge, Q moves into one end.

If one takes the trouble to look up a voltage source, one sees that this is exactly what an ideal voltage source does. It can source current as readily as it sinks current, and the voltage will remain constant.

Incidently, none of this depend on the frequency domain. The same argument holds over the entire spectrum until it runs up against another impossibly infinite circuit element.

 

[TheAnalogKid83]

everyone wants to ignore the frequency characteristics, but I guess I'm going to have to show you exactly what the original poster likely has encountered, and it will back up everything I'm saying. No one wants to acknowledge the mathematics, they just argue about CHARGING, which is not the only characteristic of a capacitor. I am done arguing over a concept, when the impedance mathematics are being ignored, therefore, I'll let you argue with sedra and smith instead.

 

[TheAnalogKid83]

Alright, this is from the book used in my microelectronics course, it is the Sedra and Smith book. I'm done arguing my points which are backed up with the mathematics used to describe circuits. I'll let the book do the talking, and if you want to disagree with the book you can try to do some math to prove it wrong instead of talking hypothetically.

PDF attached.

 

[Averagesupernova]

AnalogKid, I think YOU are the one missing the proverbial boat here. I, for one, can agree that a capacitor with infinite capacitance will in fact block DC. BUT, when is the DC applied? To be a true DC voltage, the voltage needs to have been present forever, or at least prior to the creation of the capacitor. This was the whole point of my first post in this thread which you seemed to skim over by arguing about a series resistor which is really irrelevant. The attachment really doesn't tell much. All it does is tell me that the capacitance could be infinite as far as the AC signal is concerned. The capacitor in the attached circuit needs to be small enough so that it can charge at power up of the circuit without waiting an unreasonable amount of time. An infinite capacitance capacitor in that part of the circuit would require you wait an unreasonable amount of time after power up. Like forever. The argument still comes back to what I stated in my first post. How are you defining DC?

 

[TheAnalogKid83]

AnalogKid, I think YOU are the one missing the proverbial boat here. I, for one, can agree that a capacitor with infinite capacitance will in fact block DC. BUT, when is the DC applied? To be a true DC voltage, the voltage needs to have been present forever, or at least prior to the creation of the capacitor. This was the whole point of my first post in this thread which you seemed to skim over by arguing about a series resistor which is really irrelevant. The attachment really doesn't tell much. All it does is tell me that the capacitance could be infinite as far as the AC signal is concerned. The capacitor in the attached circuit needs to be small enough so that it can charge at power up of the circuit without waiting an unreasonable amount of time. An infinite capacitance capacitor in that part of the circuit would require you wait an unreasonable amount of time after power up. Like forever. The argument still comes back to what I stated in my first post. How are you defining DC?

But you see, you're arguing practicalities over something that is acknowledged in the very beginning to be impractical, infinite capacitance. Let the cap be charged at 0V ok? now put an AC through it, it will still pass through, and DC won't (0V). The original question wasn't to question the charging or series resistance or power up or time involved. It was a question to a homework problem, and I just showed exactly what kind of context this idea comes up in my last post, where I've been trying to show what it means . . and you don't define DC in any other way than it has always been, DC is 0 Hz or A*sin(0) => Amplitude at 0 Hz. Everything I have stated is accurate and I've backed it up with basic EE theory. People here are saying it is a short at DC and this is not true and defeats the whole purpose of the problem the original poster was given, and he would get a wrong answer by letting the DC bias pass through the infinite capacitors. People are also saying to ignore the frequency domain, and this also would result in a wrong answer. You violate KVL by saying that a DC source across an infinite capacitance would be a short. Its funny how such a simple problem and question, with a very simple CORRECT original answer could result in such disagreement.

 

[Phrak]

Thanks for the .pdf AK. There's little better in life than ripping textbook. I'll get to it asap.

 

[Phrak]

[Think of me as the voice of comic relief, as two devoted combatants go at it, head to head.]

I'll paste the text snippet here for all to see.

"An infintite capacitor is used to indicate that the capacitance is sufficiently large that it acts as a short circuit to all signal frequencies of interest. However the capacitor still blocks dc."

As already noted, an infinite capacitor is indistinguishable from an ideal voltage source.

Where the text has gone wrong is in the second sentence due to slightly sloppy English. It's commion enough, but meaningless to say that a capacitor blocks DC. It is intended to imply that when the DC voltage on one side changes, the other side won't change. But there is no such thing as changing DC voltage.

What could have been said was "the initial voltage potential across the infinite capacitor is maintained without change." It's far more accurate, but far less intuitive.

So, indeed, the text does utilize an infinite capacitor from zero Hertz, upward, including zero.

But let's get back to our combatants, shall we? Who will discover first that they are arguing apples against oranges. Who will win round 8?

 

[TheAnalogKid83]

[Think of me as the voice of comic relief, as two devoted combatants go at it, head to head.]

I'll paste the text snippet here for all to see.

"An infintite capacitor is used to indicate that the capacitance is sufficiently large that it acts as a short circuit to all signal frequencies of interest. However the capacitor still blocks dc."

As already noted, an infinite capacitor is indistinguishable from an ideal voltage source.

Where the text has gone wrong is in the second sentence due to slightly sloppy English. It's commion enough, but meaningless to say that a capacitor blocks DC. It is intended to imply that when the DC voltage on one side changes, the other side won't change. But there is no such thing as changing DC voltage.

What could have been said was "the initial voltage potential across the infinite capacitor is maintained without change." It's far more accurate, but far less intuitive.

So, indeed, the text does utilize an infinite capacitor from zero Hertz, upward, including zero.

But let's get back to our combatants, shall we? Who will discover first that they are arguing apples against oranges. Who will win round 8?

ugh

You say the authors are intending to imply that when the DC voltage on one side changes, the other side won't change? This sentence is contradictory. DC voltage is NO CHANGE, by changing your amplitude, you are introducing frequencies besides 0 Hz, this can easily be shown with Fourier transforms. How is the text implying this, or even intending to imply this? It specifically says it will pass any frequency, but block DC . . so how is the text wrong? This is just like any other capacitor, but to say it is infinite will let even the lowest frequencies as they approach zero pass without distortion. My very first post was about this idea, and I showed it mathematically. I don't know why people want to argue with proven concepts that millions of people are taught. And these arguments came up before I even showed what the text said.

I'm not arguing apples and oranges, I'm arguing the usefulness and correctness of the information people are giving the original poster. Some of it is just completely wrong, and some of it is correct, but is being used to shoot down what I said when it doesn't even apply.

Zc = 1/(jwC) I think that is where the idea of infinite capacitance in these problems comes from, and I see no reason to argue with this. Now I'm really done, and will go back to asking noob questions about everything I'm more uncertain of.

 

[Phrak]

The idealism works and it behaves as a short to DC because of presumptions made on initial conditions. Your text goes at it like every other text. There is a hidden assumption in that the capacitor is charged to your DC bias conditions. These are the apples. Without this presumption the DC bias is loaded. These are the oranges.

So you're both right. And I'm out of here too.

 

[fedaykin]

I agree with you Phrak, for a very short time depending on capacitance, capacitors will act as a short circuit for a given power supply.

Seconds = Ohms * Farads, if Farads goes to infinity and ohms is constant, must not seconds also go to infinity?

 

[deakie]

and the second question that the student had...

What would it be used for in application if it did exist?

perfect energy storage unit...could problably store loads of charge...

 

[deakie]

I agree with you Phrak, for a very short time depending on capacitance, capacitors will act as a short circuit for a given power supply.

Seconds = Ohms * Farads, if Farads goes to infinity and ohms is constant, must not seconds also go to infinity?

and considering that the impedance is zero, then the voltage would be instant and the cap becomes a perfect source on the otherside...

so apart from being a good storage unit...it has no use as a capacitor otherwise...
could do without it in the circuit to tell the truth...

 

[Phrak]

1. An infinite capacitor is escentially no more, and no less than an idea emf source. That emf source can be zero volts, as much as it can have any other value. Note that the zero volt EMF source is equivalent to a short.

2. When the textbook uses ideal capacitors, they are implicity charged to the DC bias conditions of the circuit in question. Try it otherwise and see what you get. It would serve equally well to consider them ideal sources of EMF, whos voltages are also determined once the DC bias conditions are found.

3. The reason a text introduces these sources of EMF as ideal capacitors is help you conceptualize them as capacitors.

4. There was misunderstanding over whether or not an infinite capacitor was a short to DC. It was basically caused by differences in how one decides to the initialize the voltage across the cap.

In some retrospect there could be cases in which the ideal source of EMF may break-down. What would happen, for instance, should your bias supply voltages ramp up to double?

 

[TheAnalogKid83]

Why did we have to bring this back? So we can start reintroducing more false ideas and incorrect concepts?

 

[TheAnalogKid83]

1. An infinite capacitor is escentially no more, and no less than an idea emf source. That emf source can be zero volts, as much as it can have any other value. Note that the zero volt EMF source is equivalent to a short.

2. When the textbook uses ideal capacitors, they are implicity charged to the DC bias conditions of the circuit in question. Try it otherwise and see what you get. It would serve equally well to consider them ideal sources of EMF, whos voltages are also determined once the DC bias conditions are found.

3. The reason a text introduces these sources of EMF as ideal capacitors is help you conceptualize them as capacitors.

4. There was misunderstanding over whether or not an infinite capacitor was a short to DC. It was basically caused by differences in how one decides to the initialize the voltage across the cap.

In some retrospect there could be cases in which the ideal source of EMF may break-down. What would happen, for instance, should your bias supply voltages ramp up to double?

Derive the transfer function of such a circuit with an infinite capacitor and then try it with a DC voltage source in place of the cap -.- . You will see there is NO DC bias for the cap, hence why they put it there in the first place. Putting an ideal EMF source is NOT equivalent to an infinite capacitor. Caps BLOCK DC in the discussed circuit setup of this subject (in series between a source and a load) they do not source it.