Question about capacitor coupling

[Firefox123]

This question comes from a thread I started about astable multivibrators...but I think my question might be more of a physics question then an engineering question...Here is the discussion from that other thread..


Hello all...

I am trying to understand astable multivibrators and I keep on reading this phrase in connection to one transistor going into saturation and the other going into cutoff...

"This change in voltage is coupled through C2 to the base of Q1, forcing Q1 to cutoff."

This statement is referring to the fact that when the second transistor conducts, the end of the capacitor connected to the collector rapidly drops from (approximately) the voltage of the source to (approximately) ground.

Im not seeing how this change in voltage is "coupled" thorugh the capacitor to the base of the other transistor.

Any help would be appreciated.


Thanks.

High frequency signals "pass" through capacitors. The signal modulates the voltage on one end of the capacitor, which causes oscillations in the electric field between the capacitor's plates, which causes oscillations in the potential on the other side of the capacitor.

In fact, the higher the frequency of stimulus you apply to a capacitor, the more like a short circuit it will behave.

- Warren

Thanks for the reply...

First question......could you explain the connection between the oscillations in the electric field and the change in potential on the other side of the capacitor?

If I am asking too much, then could you direct me to a good source where I could study this in more depth?

The way I have always thought about a capacitor acting like a "short" at AC was to picture charge being put on the plates and then taken off very quickly...so quickly that there never really is an accumulation of charge or volatge (kind of like a tug of war with charge)...

But with this astable multivibrator I don't have a nice clear path (at least I cant picture it) connecting the plates of the capacitors because of the transistors, so my simple picture doesn't seem to work.

But you are going much deeper into the physics than I was and I would like to try and understand what you are saying, if you will bear with me...

Here is what I am picturing.....before the transistor 'B' is turned on we have one end of the capacitor at the source voltage (say 6 volts) and the other end is at the base voltage of transistor 'A' (which is approximately 0.7 volts since transistor 'A' is on)....

So there is about 6 volts accross the capacitor and....since the volatge across a capacitor can't change instantly...after transistor 'B' turns on we have one end of the capacitor at zero volts and the other end at -6 volts, thus maintaining the rule that the voltage can't change instantaneously.

Is this correct?


Russ


So can anyone either explain this to me or provide me with some resources so I can study it in more depth?

Thanks.



Russ

[pervect]

Lets try the mathematical route.

Charge = capacitance * voltage, therfore

current = capacitance * dV/dt, i.e. the current is proportional to the capacitance times the rate of change of voltage.

Suppose one end of a capacitor is at 5 volts, and the other end of the capacitor is at .7 volts. The voltage across the capacitor is 4.3 volts.

Now, suppose the end of the capacitor that was initally at 5 volts is pulled down to 0 volts. If no current flows, the voltage across the capacitor will still be 4.3 volts. Thefore the other end of the capactitor will be at -4.3 volts.

Even if a large current flows, it will take time to charge or discharge the capacitor. The amount of time it will take is delta-t = C * delta-v / I,

C being the capacitance of the capacitor, I being the current, and delta-t being the time and delta-v the change in voltage.

If you hook a capacitor up to a resistor, you get a differential equation. I could go through the details, but I'm guessing from your question that you haven't had calculus yet.

I will mention one important result without proof - if you attempt to charge or discharge a capacitor through a resistor, the voltage will be a damped exponential,

V = A * exp(-t/RC) + B

R*C has units of time, and is called the "time constant" of theis almost any electric circuit. You should find detailed proofs of this (using calculus) in almost any EE textbook, and quite a few introductory physics textbooks as well.

[Firefox123]

Hi pervect thanks for the reply...

Lets try the mathematical route.

Okay..

Charge = capacitance * voltage, therfore

current = capacitance * dV/dt, i.e. the current is proportional to the capacitance times the rate of change of voltage.

Suppose one end of a capacitor is at 5 volts, and the other end of the capacitor is at .7 volts. The voltage across the capacitor is 4.3 volts.

Now, suppose the end of the capacitor that was initally at 5 volts is pulled down to 0 volts. If no current flows, the voltage across the capacitor will still be 4.3 volts. Thefore the other end of the capactitor will be at -4.3 volts.

Even if a large current flows, it will take time to charge or discharge the capacitor. The amount of time it will take is delta-t = C * delta-v / I,

C being the capacitance of the capacitor, I being the current, and delta-t being the time and delta-v the change in voltage.

If you hook a capacitor up to a resistor, you get a differential equation. I could go through the details, but I'm guessing from your question that you haven't had calculus yet.

I will mention one important result without proof - if you attempt to charge or discharge a capacitor through a resistor, the voltage will be a damped exponential,

V = A * exp(-t/RC) + B

R*C has units of time, and is called the "time constant" of theis almost any electric circuit. You should find detailed proofs of this (using calculus) in almost any EE textbook, and quite a few introductory physics textbooks as well.


This is all true....I have had calculus and I have a Bachelors in EE from Penn State. I am looking for an explanation that explains why this happens.

Since the voltage cant change instantly, we know that the other side of the cap drops in voltage to maintain the original voltage....but what is physically happening here?

When we ground the one end and force it to zero volts, are we removing the excess positive charge from the plate or just changing the potential energy each charge has?

And how does this effect the charges on the other plate? (Im looking at the cap as having both negative and positive charges on both plates, even though in reality we could say the positive charge is a lack of electrons).

Im trying to picture what is physically going on.....the differential equations and circuit theory havent provided me with the answers I seek so I figured I would just ask.

Can you explain physically what is going on and why? How does the voltage drop on one plate affect the other plate?




Russ

[pervect]

If one sticks to low frequencies, with no significant magnetic fields, the physical significance becomes fairly simple.

One defines one particular "node" of the circuit as a ground node. This is usually connected to the case and/or ground plane of the circuit. This node is defined as having a potential of zero volts.

The voltage at any "node" is that voltage that a voltmeter (or an oscilliscope) would measure between that node and the "ground" node.

Things become less simple when high frequencies and/or magnetic fields are involved. Technically, there is what is known as a "gauge degree of freedom" in electromagnetism, which makes assigning a voltage to a specific node theoretically somewhat arbitrary.

Engineers make use of this gauge degree of freedom to simplify the analysis of their circuit. They make the assumption that the case, or groundplane of the circuit has a potential of zero volts - this node then serves as a reference. This engineering approximation really only works well when there are no significant effects of stray electromagnetic coupling to the environment. But a well-designed circuit has to be designed so that it is not sensitive to stray electromagnetic coupling. Otherwise one runs into circuits that will work only if one stands close to them (or perhaps only when one stands far away). This is not a desirable state of affairs.

Much more can be said about this - IIRC one of my E&M books devotes an entire chapter to "lumped circuit approximations" in terms of field theory. But this gets to be a bit arcane, one really only needs to know that a ground node should generally have the property that a circuit will work if it is connected to an "earth ground", and that such a node is used as a practical reference to define the concept of voltage within a circuit.

One might add that the reason this approach works is because electromagnetic fields are kept confined to individual circuit elements as much as is feasible (and when it is not feasible, external shielding is added as needed to confine the electromagnetic fields within the shielded area). In short, good engineering design is part of the reason that the "voltage" concept works.

[add]
I suppose I should say a bit more about the more general case. In the presence of strong changing magnetic fields, one needs to replace the voltage, V, with the 4-potential (V,A), where A is the magnetic vector potential. One can write the fields as


E = \nabla V \hspace{.5 in} B = \nabla \times A


The voltage is only part of the 4-potential. The resulting 4-potential is not unique, it has a "gauge degre of freedom".

When we try to measure "the" voltage between two points in this general case, the reading our voltmeter will measure will depend on how we hook it up to the circuit, i.e. we will have signals induced by magnetic induction in the "leads" of the voltmeter by the changing magnetic flux. Thus the engineering concept of "voltage" as a simple scalar is an approximation, the most general case needs a 4-potential instead.

For some related information, see

http://en.wikipedia.org/wiki/Gauge_invariance


The definition of electrical ground in an electric circuit is an example of a gauge symmetry; when the electric potentials across all points in a circuit are raised by the same amount, the circuit would still operate identically; as the potential differences (voltages) in the circuit are unchanged. A common illustration of this fact is the sight of a bird perched on a high voltage power line without electrocution, as the bird is insulated from ground.

This is called a global gauge symmetryTrefil,1983. The absolute value of the potential is immaterial; what matters to circuit operation is the potential differences across the components of the circuit. The definition of the ground point is arbitrary, but once that point is set, then that definition must be followed globally.

In contrast, if some symmetry could be defined arbitrarily from one position to the next, that would be a local gauge symmetry.

[Firefox123]

Hey pervect...

Thanks for the continued discussion...

I understand what you are saying about how we define "ground".

So in this specific situation we have a capacitor with 5 volts one one plate and .7 volts on the other (all voltages with respect to the point we have defined as the "ground" node)...

Now we pull the 5 volt plate down to 0 volts, thus lowering the potential energy per unit charge. I think this would also set up oscilliations in the electric field like chroot mentioned earlier...

So in my understanding, we havent moved charge to or from the plates, but we have changed the potential energy per unit charge by dropping the plate from 5 volts to 0 volts.

Since we know that a capacitor will not change its voltage instantaneously we know the other plate drops to -4.3 volts..

But my saying that I know the rule that "a capacitor cant change its volatge instantaneously" doesnt really mean I understand the underlying reason why the voltage drops from .7 volts to -4.3 volts on the other plate.

I read the article on gauge symmetry but I still dont know what is physically happening to the capacitor and the associated electric field from the moment you drop the plate voltage to 0 volts that causes the other plate to drop to -4.3 volts.

And even though the ground might really be "arbitrary" the voltage drop definately is physcially real, because one of the transistors (in the astable multivibrator circuit) goes into cutoff mode.


So what am I missing? What is going on here? What are the physics behind the statement "since the capacitor cant change its voltage instantaneously the other plate drops to -4.3 votls"?






Russ

[pervect]

I'm not quite sure what you are looking for, then.

If you understand how to compute voltages in a circuit (with respect to ground), and the currents that flow through the various components, you should be able to understand, for instance, how a multivibrator works


I am trying to understand astable multivibrators and I keep on reading this phrase in connection to one transistor going into saturation and the other going into cutoff...

"This change in voltage is coupled through C2 to the base of Q1, forcing Q1 to cutoff."


You should be in a position to understand why the base voltage of Q1 must drop when Q2 turns on. Let us assume a 5-volt power supply. The voltage at the collector of Q2 drops from 5v to 0 volts (approximately) when Q2 turns on. The base voltage of Q2 was initally at .7 v. Because the voltage across the capacitor doesn't change, you know that the base voltage of Q2 falls to approximately -4.3 volts, as we have just discussed.

This turns Q2 off, because a transistor can only be on if it's base voltage is high. (Assuming that we are talking about npn transistors with a positive power supply).

You can also see that due to other components in the circuit, the capacitor voltage at the base of Q2 will start to change at a rate dv/dt = I. The base bias resistor for Q1 will slowly raise the voltage at the base of Q1, as the capacitor will be charged via the current flowing through the resistor. Eventually, the base voltage of Q1 will raise to the point where Q1 turns on. When Q1 turns on, it will turn off Q2.

It seems like you've gone from asking a straightforwards question about how a multivibrator works, to a rather vague philosophical question about the meaning of voltage.

Ignoring magnetic fields for the time being, and sticking with quasi-static electrostatics, the voltage can be defined as the intergal of the electric field, and the electric field can be defined as the rate of change of the voltage with respect to position, i.e. E = \partial V / \partial x.

The voltage is defined only up to an additive constant, i.e. you can add a constant to the voltage, and it will not change any electric fields. This is a specific example of the "gauge degree of freedom" that voltage has.

So this defines the voltage in terms of an electric field (at least in the quasi-static case).

If that's not sufficient, I'm not sure what it is that you want.

[Firefox123]

Hello again pervect...

I'm not quite sure what you are looking for, then.

Sorry...not trying to be difficult....

At times I will say that I "dont understand something" even though the "understanding" I am after is much deeper than what I let on. I tend to overanalyse things at times and search for a complete understanding behind the fundamental physics of things....

I think that you do make a point below that I may have overlooked...

If you understand how to compute voltages in a circuit (with respect to ground), and the currents that flow through the various components, you should be able to understand, for instance, how a multivibrator works



You should be in a position to understand why the base voltage of Q1 must drop when Q2 turns on. Let us assume a 5-volt power supply. The voltage at the collector of Q2 drops from 5v to 0 volts (approximately) when Q2 turns on. The base voltage of Q2 was initally at .7 v. Because the voltage across the capacitor doesn't change, you know that the base voltage of Q2 falls to approximately -4.3 volts, as we have just discussed.

This turns Q2 off, because a transistor can only be on if it's base voltage is high. (Assuming that we are talking about npn transistors with a positive power supply).

You can also see that due to other components in the circuit, the capacitor voltage at the base of Q2 will start to change at a rate dv/dt = I. The base bias resistor for Q1 will slowly raise the voltage at the base of Q1, as the capacitor will be charged via the current flowing through the resistor. Eventually, the base voltage of Q1 will raise to the point where Q1 turns on. When Q1 turns on, it will turn off Q2.

Yes, I do understand all of that. I guess one of my faults is that I never truly feel that I understand something, even if I can explain it, calculate it, derive it, and correctly use it because I always see a deeper layer that I have not yet obtained.

It seems like you've gone from asking a straightforwards question about how a multivibrator works, to a rather vague philosophical question about the meaning of voltage.

My apologies if that is what I am doing...I am not trying to be vague or philosophical...I realize that the questions of what voltage actually is or what an electric field actually is are still not completely answered.

Ignoring magnetic fields for the time being, and sticking with quasi-static electrostatics, the voltage can be defined as the intergal of the electric field, and the electric field can be defined as the rate of change of the voltage with respect to position, i.e. E = \partial V / \partial x.

The voltage is defined only up to an additive constant, i.e. you can add a constant to the voltage, and it will not change any electric fields. This is a specific example of the "gauge degree of freedom" that voltage has.

So this defines the voltage in terms of an electric field (at least in the quasi-static case).

Ah yes....I was trying to get more of a "mental picture" and was overlooking the relationship between voltage, the electric field, and potential energy (or work).

It would appear that the statement that chroot made earlier...

The signal modulates the voltage on one end of the capacitor, which causes oscillations in the electric field between the capacitor's plates, which causes oscillations in the potential on the other side of the capacitor.

is justified by the very definitions.

Not quite the answer I was hoping for, but perhaps there is no deeper answer than that. The voltage and E-field are, after all, intrinsically related so an oscillation in one would necessarily create an oscillation in the other.

If that's not sufficient, I'm not sure what it is that you want.

I guess I was wondering what is happening to the charges present on the two different plates.

When you lower the plate voltage from 5 volts to 0 volts you are lowering the potential energy per unit of charge, since voltage can be defined as potential energy per unit of charge.

Likewise the other plate also has a drop in voltage or potential energy per unit of charge, yet the voltage from one plate to the other has not changed.

If we were to place a test charge and travel from one plate to the other in the instant both before and after the ground connection that test charge would experience the same voltage difference between the plates.

But with respect to the "ground" voltage in the circuit the plates have a lower potential enery per unit charge....Im having a difficult time picturing this.

The closest analogy I can think of is two weights, connected by a rigid bar, at two different heights (so the bar is sloped). We move the higher of the two weights to a lower "platform" and the bar and the other weight are forced to proportionally lower heights as well.
In my analogy the electric field is the "rigid bar" and the heights of the weights are the two plate potentials.

So my purpose was to get a more fundamental understanding of the physics behind the voltage drop.



Russ