In my opinion it is not correct to state that "in general" there is "90 deg phase shift between current and voltage in the capacitor": it is true only if one is studying a specific dynamical behavior i.e. the one that it is observed when the transient solution becomes negligible and one can observe the steady state solution only in the specific input configuration of a periodic single tone input.

You are right, Federico. I have added this detail in the question above.

Regards, Cyril.

The behavior of the electrical circuits is described either in the time domain or in frequency domain. In the time domain, one determines the dynamic performance of circuits and their elements by applying unit step voltage on them and observing the waveform of the current in every circuit element and the voltage across it. It is the shape of the waveform of response which determines the behavior of elements and circuits. An alternative way is to determine the behavior of the circuits and their elements by applying on them a sinusoidal voltage waveform and observing their performance after reaching the steady state condition. If the circuit is linear such your R-C circuit , the current and the voltage across every element will be also sinusoidal having the same frequency but with different magnitude and phase. Accordingly, the phasor was introduced. To determine the circuit response on any sinusoidal waveform one needs only to determine amplitude and phase of the sine wave since its form is already known. To calculate such phasor , one introduced the concept of the impedance. The impedance Z of an R-C circuit is R+ iX, with X =1/wC, where R is the resistance, Xis the reactance of the capacitor which is inversely proportional to the frequency of the input sine wave signal. As a complex quantity the impedance Z will have a magnitude and phase. By definition the phase is arctan X/R. At very low frequency if w tends to zero the phase of Z will tend to 90 degrees. This because 1/wC will be >> R and the circuit is dominated by the capacitor. On the other extreme, when the frequency w tends to infinity R>> 1/wC and the circuit behaves as a pure resistance. Consequently, the phase shift will be zero.

Therefore the phase shift will vary with frequency from 90 degrees to zero degrees when the frequency changes from nearly zero to infinity. This is because the R-C circuit behaves capacitive at low frequencies and resistive at high frequencies.

This can be verified experimentally by measuring the voltage on the capacitor and the voltage on the the resistor and comparing them with input alternating voltage at different frequencies. The dependence of the phase on frequency is due to the reactive behavior of the capacitors and inductors.

I hope i shed some light on your question.

It's because it takes some time to fill the cap with a restricted amount of charge per time (finite current). When you have passed the max of your input wave, the cap still charges, because the voltage of the cap is still below actual input voltage.

To Supplement the answer given by Abdelhalim let me go into the circuit concept and state as follows:

If we go through the Phasor Diagram of R-C Series Circuit with A.C. Sinusoidal Voltage supplied, we can see that the current Phasor in the circuit will be leading the Voltage Phasor by angle Theta or Phi= tan inverse of ( Xc / R ).

The Voltage drop across R will be in Phase with the current and the voltage drop across the Capacitive Reactance will be lagging behind the Resistance drop of Voltage by 90 electrical Degrees (pye / 4 Radians),if it is a Pure Loss less Capacitor.

If it is a Lossy capacitor with loss angle delta than the angle will be less than 90 degrees by angle delta. This Phenomenon causes the phase shift.

The above holds good for steady state A.C. Circuits.

Refer IMPEDANCE TRIANGLE and also VOLTAGE DROP TRIANGLE or KVA TRIANGLE

In impedance triangle,for a capacitor, the Capacitive Reactance will be lagging Resistance by 90 degrees.

In a capacitive circuit the Current leads the Voltage by Power Factor Angle.

In the voltage drop triangle the capacitance voltage drop phasor lags the resistance voltage drop phasor(which is in phase with the current phasor ) by 90 degrees.

If we consider the transients for response to a step voltage the charging of the capacitor and hence the Voltage across the capacitance will be increasing exponentially with a time constant R*C from time to time till the voltage across the capacitor reaches the steady state value( = supply voltage) in time equal to 5 times the circuit time constant (called Settling time) for the steady state value. After that the charging of the capacitor stops and then no current flows through the capacitor.Hence for D.C. the capacitor acts as a Blocking Capacitor and prevents D.C Signal passing through the capacitor and allows only A.C. signal through it.

Pisupati

when A.C voltage is applied to capacitor voltage will not appear across capacitor immigiatly, charging current will flow from capacitor and voltage starts builting up,or voltage lags current in capacitive circuit for A.C.For D.C voltage capacitor will charged voltage reached applied value, no charging current flows or we can say capacitor blocks D.C signal.

James, I have no access to this book but I have no doubts that the phase lag is perfectly explained in terms of electricity like the Abdelhalim's precise explanations and the Pisupati's thoughts above. But, without offence, I like more the Holger's simple explanation: "When you have passed the max of your input wave, the cap still charges, because the voltage of the cap is still below the actual input voltage." I like even the Bhupendra's attempt at explaining although I cannot agree that "when A.C voltage is applied to capacitor voltage will not appear across capacitor immediately"...

Your "electrical" explanations are based on some mathematical abstractions (complex numbers, phasor diagrams, etc.) But I would like to grasp the basic idea of the lag phenomenon at the most general level, to reveal the "philosophy" of the lag. The math model is a quantitative generalization of the phenomena while the philosophy is a qualitative generalization. I think I have understood a circuit if I have realized the basic idea behind it. The basic ideas are nonelectrical; they do not depend on the specific circuit implementations; they are eternal and immortal. Figuratively speaking, basic ideas are the soul and the concrete circuits are the body...

James, thank you for the well-minded answer. It sounds wonderful to wield all this theory but I am too far from it... Nevertheless, I will try to show the potentialities of the intuitive approach when revealing the basic ideas behind circuits at the initial stage of understanding.

This morning, when walking, as usual, in the park and thinking about the lag, virtual ground, RC oscillations and other interesting circuit phenomena, I finally realized what the problem with the humble passive RC circuit was and how it was solved in the active electronic circuits. Let me share my insights with you and the curious contributors here.

I will consider the phase shift between the output and input voltage of an integrating RC circuit (a voltage-to-voltage integrator). It can be thought as of a consecutively connected voltage-to-current converter (the resistor R) and a current-to-voltage integrator (the capacitor C).

If the voltage-to-current converter was perfect, its output current direction and magnitude would not depend on the output voltage across the capacitor and it would change its direction exactly when the input sinusoidal voltage crosses the zero line. So, during the entire positive half wave (even when the current decreases), the current will charges the capacitor and at the moment when the input voltage becomes zero, the voltage across the capacitor will be maximal. So, the phase shift of the perfect integrator will be exactly 90 degrees and will not depend on the frequency (just like in the capacitor).

The problem of the humble RC integrating circuit is that the resistor R is an imperfect voltage-to-current converter whose current direction and magnitude depend on the voltage across the capacitor. The resistor is "stretched" between two voltages (the input voltage and the capacitor voltage) and the current direction and magnitude depend on the difference between them. As a result, the current changes its direction when the input voltage becomes equal to the voltage across the capacitor; the higher the frequency is, the later this moment happens and the phase shift varies from 0 to 90 degrees.

There are two ways to solve the problem (to obtain a phase shift exactly of 90 degrees): the first is to ensure perfect load conditions (short circuit, virtual ground) to the imperfect resistor; the second is to make the resistor perfect (infinite). The first solution gives the op-amp inverting integrator; the second one - the famous Deboo integrator:

http://en.wikibooks.org/wiki/Talk:Circuit_Idea/Deboo_Integrator

Regards, Cyril.

James, thank you that you implicitly concede to some extent the necessity of intuitive explanations. At the least, they give a chance to the mere mortals:) to comprehend (not only to know) great circuit ideas. Only see these touching words of the boy in my old correspondence:

http://en.wikipedia.org/wiki/Talk:RC_circuit#Why_there_is_a_phase_shift_between_the_current_and_voltage_in_a_capacitor

"... Im having a really hard time trying to understand what the phase shifts mean in real life in a circuit (I know it means......but how). I understand them mathematically but not in physicality, mainly because I cant find any explanation..."

"...So what im trying to work out is WHAT causes this OVERALL lag in the circuit of 78 degrees..."

"... I understand fully the lag inside the capacitor or inductor but when it comes to the overall lag in the circuit, what analogy can be used to discribe THIS lag. What is causing it to be -78 the physical happening of it..."

There is some"gap" between the fundamental subjects that are too sterile and the electronic circuitry that is too practical. In circuitry, we have to distinguish the particular elements and their functions in the concrete circuit (usually, one element has many functions - the resistor is a typical example). For the purpose of understanding, there is a need to show the connection between the imperfect passive electrical circuits and their perfect electronic versions. Usually, electronic circuits can be considered as improved passive circuits (as their superstructure) and in every perfect electronic circuit we should see an imperfect passive circuit. So, it is useful to build and reinvent even the humble RC circuit. I have been doing it with my students since the early 90's:

http://en.wikibooks.org/wiki/Talk:Circuit_Idea/How_to_Make_a_Perfect_RC-integrator

http://en.wikibooks.org/wiki/Circuit_Idea/Group_67a (the best student group)

Regards, Cyril.

To show that the problems of the simple passive RC circuit are not so simple, here are more considerations:

We talk about an integrating and differentiating RC circuit but practically they are the same circuit consisting of three elements (a voltage source, resistor and capacitor) connected in series. This is because the resistor can be thought as of a voltage-to-current converter and as of a current-to-voltage converter.

We can use the voltage across the resistor as an input and the voltage across the capacitor as an output and this will be perfect voltage-to-voltage integrator. Then, we can use the voltage across the capacitor as an input and the voltage across the resistor as an output and this will be a perfect voltage-to-voltage differentiator. But this arrangement is inconvenient to implement for some reasons (the input voltage is not really the input voltage and the one of the elements is not grounded).

So, we usually use as an input the voltage source and as an output - the voltage across the capacitor (RC integrator) or the voltage across the resistor (RC differentiator). But a problem arises - the voltage across the output element. To solve it, in electricity we increase the resistance in the case of the integrator and decrease the resistance in the case of the differentiator. In electronics, we use more clever techniques: in some cases we add voltage to compensate the undesired voltage drop (virtual ground - op-amp inverting integrator); in other cases, we add a current (negative resistance - Deboo non-inverting integrator).

When driving the RC integrator with a sinusoidal voltage, my students are puzzled by the fact that the output voltage continues to increase even when the input voltage decreases (the two voltages move against each other to meet). Obviously, this phenomenon contradicts the human intuition but exactly it causes the phase shift.

There is also another strange (for students) situation: when the input voltage becomes less than the voltage across the capacitor the current begins flowing back from the capacitor to the input source. It seems, the capacitor is a source and the source has become a "load"? I usually note that the input voltage source has to allow the current to flow in the reverse direction. Of course, this situation is typical also for the step input voltage stimulation.

Hi Cyril,

without trying to jump into this very specific discussion I like - from the engineering point of view - to comment on some earlier statements from your side:

Quote: "There are two ways to solve the problem (to obtain a phase shift exactly of 90 degrees): the first is to ensure perfect load conditions (short circuit, virtual ground) to the imperfect resistor; the second is to make the resistor perfect (infinite). The first solution gives the op-amp inverting integrator; the second one - the famous Deboo integrator."

My comment (just for clarification): The above is true only for the non-realistic case of IDEAL opamps (with infinite gain and "true" virtual ground properties). In both cases (MILLER and DEBOO integrator) the finite opamp gain causes a first-order lowpass behaviour with a very low cut-off frequency and severe phase deviations for large frequencies (in the transit frequency region). In fact, the phase shift will be exact 90 deg for one single frequency only !.

Regards

Lutz

Hi, Lutz!

Thank you for the comment. I would like to clarify that when talking about basic concepts I implicitly neglect the op-amp imperfections or leave only the finite gain.

Regards, Cyril.

Cyril, regarding opamp based integrators:

There are some another properties of active RC integrators worth to be discussed.

As an example: As mentioned before, real opamp integrators exhibit a phase shift of exactly 90 deg. at one single frequency only (let`s call it f1).

Nevertheless - an RC oscillator containing two such integrators (also known as quadrature oscillator) is able to oscillate at frequencies below or above f1 (where the sum of both phase shifts does not yield 180 deg.). Violation of the oscillation condition?

Many years back, in the Eighties.when I attended a course on "Systems Engineering" at I.I.T.,Delhi,conducted by Prof.Satsangi, I got fascinated by the School of thought that all Systems.not only different Disciplines in Engineering but all other systems like Medicine,Economics, Finance,Social Sciences etc. all fall under the same category of Systems Engineering.

Because all systems are modeled Mathematically-by one or a set of Linear or Non Linear equations of one or more degrees in the case of Steady State Systems, and one or a set of Differential Equations of one or more order in the case of Dynamic Systems.. Change t he name of the variable and the Boundary Conditions in the case of Partial Differential Equations, a problem in one Field gets converted into a problem in some other Field.

That is why we can say that all systems can be analog-ed with Electrical Systems,for; solutions of Electrical Systems are well established.

Also I am convinced that all Human beings are alike in their Basic approaches, wherever they are.

Hence the tools applied in one Field can be applied in another Field. There is no point in thinking that each one is a water tight compartment.

I am amazed by the way Prof.Cyril approaches the Problem in a different way.

I can say, he is a Real Systems Engineer.. Hats off to him

This is a little bit of Philosophy. People may kindly bear with my Random thoughts.

Let me also think a little about his Physical approach to the Lag, and Lead topics and come out with an explanation.

Dr.P.S.

Lutz, your question exceeds the potentialities of my imagination:( And what do you think about the Bubba oscillator?

http://www.ti.com/sc/docs/apps/msp/journal/aug2000/aug_07.pdf

Regards, Cyril.

Pisupati, thank you for your exclamations about my humble person:) I only rely on the common sense, intuition and my experience... nothing else... You, Lutz and the rest contributors here are just wonderful people; we mutually stimulate the so difficult mental activities - understanding, realizing, grasping circuit ideas...

Regards, Cyril.

Cyril,

as the BUBBA oscillator does not consist of so called "integrating stages" there seems to be no problem to explain the behaviour - it simply consists of 4 passive RC lowpass stages, which always are capable to produce a net phase lag of 180 degrees..

First remark: Because we are discussing here some specific active circuit properties I have used quotation marks in the first sentence above because the active circuit called "integrator" does work as a real integrator with some restrictions only.

Second remark: To me, harmonic (linear) oscillators are most interesting because of the following contradictory statement:

"Linear oscillators must contain a certain degree of non-linearity in order to be able to work as a linear device".

Hello Cyril,

only now I had the time to read your wikibooks contribution "how to make a perfect RC integrator". As far as I could understand your explanation is based on the compensation of the undesired voltage drop.

I like to add another simple explanation:

In order to act as an integrator the time constant of a passive RC lowpass should be very large. For example, a pole frequency of 0.1Hz needs R=1.6kOhm and C=1E-3F.

Exploiting the well known Miller effect, we can use a more convenient capacitor C=1nF, which is placed between output and inverting input of an amplifier having a gain of 1E6. Thus, the input capacitance again is 1E-3F. At the same time we have the chance to use the low-resistive opamp output - thereby amplifying the voltage across the input capacitance (which is reduced by a factor of 1E6 if compared with the original passive reference circuit) by a gain factor of 1E6. This is analogous to the compensation of the voltage drop you have mentioned.

Thus this active circuit behaves exactly as the passive reference circuit - but has the advantage of more realistic capacitor values and a low-resistive signal output.

(By the way - this explanation gave rise to the name "Miller integrator").

Lutz

Prof.Lutz's observation regarding Op-amp Integrator and how it can be made to act exactly as a passive R-C Integrator, concluding," Thus this active circuit behaves exactly as the passive reference circuit - but has the advantage of more realistic capacitor values and a low-resistive signal output.

(By the way - this explanation gave rise to the name "Miller integrator")" is Really Interesting.

As a matter of fact Practical Op-amp Integrator is always constructed with a Resistance also between Output and inverting input, in Parallel wit Capacitor and not Capacitor alone.

Professor has also explained the Rationale behind it and how to design the values.

To supplement, I may invite reference to The AC or Continuous Op-amp Integrator in the Link:

www.electronics-tutorials.ws/opamp/opamp_6.html

Dr.P.S.

Here is another tutorial about the Miller integrator:

Prof. P. Subramanyam,

may I add a correction to your contribution?

Quote: "As a matter of fact Practical Op-amp Integrator is always constructed with a Resistance also between Output and inverting input, in Parallel wit Capacitor and not Capacitor alone."

This parallel resistor as mentioned by you is necessary only if the integrator is intended to be used as a stand-alone unit - and this happens not very often!

In this case, the resistor is necessary to provide a certain negative dc feedback for proper biasing the opamp (dc stabilization), thereby turning the integrator into an RC lowpass.

However - in most cases, the Miller integrator is used as an element within a dc stabilizing overall feedback loop (control systems, filters, oscillators).

In all of these cases, such a resistor is not necessary and, of course, will not be used because of it`s damping effect.

Lutz, but if we split the resistor in two and connect a capacitor between the common point and the ground?

I am thinking over your "Miller effect" explanation above trying to extract the essence... Maybe, we have to ask questions about the very Miller effect and Miller theorem. Have you read my (Circuit dreamer) Wikipedia contributions about them?

http://en.wikipedia.org/wiki/Talk:Miller_effect#How_to_modify_impedance

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

http://en.wikipedia.org/wiki/Talk:Miller_theorem

http://en.wikipedia.org/wiki/Talk:Operational_amplifier_applications#List_of_op-amp_circuits_from_this_page_based_on_Miller_theorem

Regards, Cyril

Greg, your frequency-domain explanation is specious and especially useful when explaining the frequency dependant gain of the integrator. This impedance fade is very attractive as an explanation. But if we want to explain the phase shift, IMO the time domain is a more suitable "place" to stay. Regards, Cyril.

Hi Cyril

@question 1: "...if we split the resistor in two and connect a capacitor between the common point and the ground?"

My answer: Very dangerous. The opamp will act as an inverting** amplifier with an additional pole, which reduces the phase margin (worst case: instability).

**Remark: This applies if the output is taken at the common point between both resistors. Instead, the opamp output provides a kind of bandpass response (rising due to the capacitor, which reduces feedback, falling due to the open-loop gain drop).

@question2: "I am thinking over your "Miller effect" explanation above trying to extract the essence".

My answer: For my opinion, the Miller integrator is a classical example to demonstrate the Miller effect. Like in each inverting transistor amplifier, this effect enlarges the capacity between inverting outout and input. This new capacity is effective at the inverting input ("virtual ground", as discussed elsewhere here in RG) and the tiny voltage at this node is amplified by the open-loop gain. For my opinion - that is the essence.

PS: I went rather quickly through the links you have given. I have seen that the question "is there a connection between the Miller effect and the Miller integrator" of course was touched. I think, it`s obvious, no doubt about this.

May I propose something?

1.) When we have finished this discussion about phase LAG in an RC circuit - what about the phenomenon of phase LEAD?

2.) And what about the phenomenon of a positive phase response causing a negative group delay?

I would like first to consider thoroughly the Miller effect and its generalization - the Miller theorem, in separate questions.

The concept of the reactance is physical. One can define the reactance , measure it and calculate it. The reactance is the complementary part of the resistance. It is the response of the reactive element on an alternating excitation.

To get the origin of the phase shift we have to go back to the time domain description of a reactive element, here the capacitor as an example. i=C dv/dt, the capacitive current is proportional to the rate of change of its terminal voltage, which describes the physical behavior of the capacitor. Applying sinusoidal voltage on the capacitor and assuming steady state condition, the current will be proportional to the rate the differentiation of the sine wave which is a cosine function divided by the angular frequency. Consequently, the current in the capacitor leads the voltage by 90 degrees and its magnitude is inversely proportional to the frequency of the input sine wave. The time domain description is the most basic one and the other descriptions are derived from it.

The time domain and frequency domain descriptions are equivalent. By definition, the angular frequency w= d phase/dt. If the time delay is T then the phase shift phi will be= w T. This is a well known result: A time delay T leads to a phase shit wT.

Thank you Professor Cyril for positively evaluating my answer. It is your challenging questions that inspired us to provide these answers. My best wishes to you.

There are many paths to the truth and all they are correct... Thank you, Abdelhalim! All we need encouragement.

You have said, "...the current in the capacitor leads the voltage by 90 degrees". Is it more correct than "the voltage across the capacitor lags the current by 90 degrees"? Are these assertions valid in the case of an RC circuit?

Regards, Cyril.

It the the matter of the excitation which is considered as a phase reference, that is of zero phase. IF you applied an AC voltage on an ideal capacitor, the resulting current passing in the capacitor will phase-lead the input reference voltage by 90 degrees.

Alternatively, if we applied an AC current through the ideal capacitor . an AC voltage which lags behind the current with 90 degrees., will be developed across the capacitor. In the first case it is the differentiation action and in the the later case it it is the integration action. The two statements are equivalent and describe the AC capacitor behavior.

When adding a resistor in series with the capacitor to form an R-C circuit, and it with a an AC voltage source as before the current will still basically lead the voltage but with an angle less than 90 degrees.This is also valid in case of the constant current source where the the phase of the phase of the over all voltage will be basically less than 90 degrees. This is because while the voltage on the resistance is in phase with the current the voltage on the capacitor lags 90 degrees from the same current. Summing these two voltages vectorially leads to previously stated result.

In conclusions, the sense of the phase shift is dictated by the capacitor while the addition of the resistance merely decreases the value of the phase shift

Abdelhalim, here are more questions:

1. As the two statements are equal, why do we not use the only one of them?

2. It is interesting to explain how "the addition of the resistance merely decreases the value of the phase shift". What does the resistor actually do to decrease the phase shift?

3. Can we somehow force the voltage across the resistor to act as an input voltage, and thus to make a perfect integrator?

4. Can we force, in similar way, the voltage across the capacitor to act as an input voltage, and thus to make a perfect differentiator?

Regards, Cyril.

As pointed out by Prof.Cyril, both the statements given by Prof. Abdelhalim Zekry are equivalent statements but mean two different actions of Differentiation I = (c*(dv / dt) )

and the other, Integration V = ( 1 / C )*(Integral( I*dt) ).

Hence for Integration we have to consider the statement that " For a pure capacitor Voltage lags the current by 90 degrees" in which case we take the current I as the independent variable.

For Differentiation we have to consider the statement that " For a pure capacitor Current leads the Voltage by 90 degrees" in which case we take the Voltage V as the independent variable.

If the Resistance R is increased the time constant R*C increases.

For Integration of a Rectangular Pulse we should have Larger Time Constant and small Time. That is what is being done in the Integrator.

For Differentiation of a Rectangular Pulse we should have smaller Time Constant and longer Time. That is what is being done in the Differentiators

.

For Differentiators ,as I have already pointed out instead of using an Inductance we obtain the effect of Inductance with Capacitor,by changing the positions of Resistance and Capacitance , as used in Gyrator Networks.

Dr.P.S..

Hi Cyril,

at first, I must confess not to understand the background of your questions No.3 and No. 4.

On the other hand - the answer to question No. 2 seems so simple that I am also not certain what you are aiming at.

Is there any other explanation as:

For a RC lowpass we have the vector sum

Vin=Vr+Vc.

Vr is in phase with the current i and Vc lags by 90 deg. (against the current i).

Thus, the resulting phase angle between Vin and the output voltage Vc is, of course, somewhere between 0 and 90 deg.

Or do you wish another explanation?

To answer the new questions of Cyril:

1- Why use only one of the two equivalent not equal statements when you can use the the two? It is more precise to use them as i indicated in my preceding answer and explained more by Prof. Subramanyam. Bot it is not wrong to permanently use one of the two statements.

2- I explained in my preceeding answer how the phase shit gets smaller than 90 degrees when we add a resistance in series with the capacitor. Prof. Wangenheim added also a valid explanation. ONE can add some details to the previous explanation. Let us make a thought experiment! let us apply an alternating voltage with amplitude Vs on a series combination of a resistor and a capacitor. A current i passes through R and C. The voltage Vs will be divided between between R and C. The voltage drop on R is in phase with the current while the voltage drop on C will lag behined the current by 90 degrees. let us use the vector representation to represent the three voltages. according to angle relation among them they build a right angle triangle with the right angle between the voltage drop on C and R. The total voltage will be represented by the side facing the right angle.Since the the edge representing the R is in the direction of the current then the angle between the Vs and the current is ranging from zero to 90 degrees from the trigonometry. This can be achieved by e. g. varying the frequency.

This what concerns questions 1 and 2

As Prof.Abdelhalim Zekry has pointed out, " the vector representation to represent the three voltages. according to angle relation among them they build a right angle triangle with the right angle between the voltage drop on C and R. The total voltage will be represented by the side facing the right angle.Since the the edge representing the R is in the direction of the current then the angle between the Vs and the current is ranging from zero to 90 degrees from the trigonometry. This can be achieved by

e. g. varying the frequency"

This is what we call Voltage Drop Triangle. in Basic Electrical Circuits.

Similarly we have Impedance Triangle and also K.V.A. Triangle.

These Triangles had been already mentioned by me earlier,if I remember correctly.

From any of these Triangles we can see that unless it is a circuit having only Pure Capacitor with no loss, if the circuit has Resistance in addition to even a Pure Capacitance we will have the impedance angle between zero and 90 degrees for an RC circuit..

Dr.P.S.

To build an ideal integrator we must apply the input voltage to a resistance and then push the current in the resistance to pass in the capacitor. This means to convert the voltage source into current source and then drive the capacitor with the current source.

This can be implemented using op amp in the inverting configuration. Alternatively, In order to implement a NEARLY ideal integrator one can use an R-C circuit with R>> Xc the reactance of the capacitor at the input frequency.

IN order to build an ideal differentiator one changes the positions of R and C in the op amp circuit of the integrator above. Alternatively, one can build a nearly ideal differentiator without op amp by making Xc>> R and taking the output across the resistance.

Without active devices we can not realize ideal integrators and differentiator. This is because we can not satisfy exactly the conditions in your questions 3 and 4.

With this we come to the end of the answer.

It was a long complex question.

THank you CYRIL.

Quote:" Without active devices we can not realize ideal integrators and differentiator."

Just for clarification and to avoid misunderstandings for newcomers:

Even with active devices it is NOT possible to realize ideal integrators/differentiators.

For this purpose we need IDEAL opamps, which are not available.

Hello Cyril,

as to your question no. 1 ("1. As the two statements are equal, why do we not use the only one of them?") I like to state the following:

Up to now - all explanations were given for sinusoidal waveforms and steady-state conditions. This seems to be appropriate if we speak about phase shift.

However, when we try to find an answer which of the respective two possible formulations are to be preferred I think we have to go back to start-up conditions - for example the step response as a typical test signal.

And this viewpoint gives us an indication which kind of exciting source we should consider:

* It makes no sense to connect a voltage source to a capacitor (indefinite current) - and the same applies to an inductor and a current source (indefinite voltage).

* Therefore: To describe the phase shift between voltage and current for a capacitor (inductor) we should start with a current (voltage) source. Thus, in both cases we arrive at 90 deg phase lag, which allows a physical explanation without any logical problems (causality).

I agree wit Prof.Luts, statement, " Quote:" Without active devices we can not realize ideal integrators and differentiator."

Just for clarification and to avoid misunderstandings for newcomers:

Even with active devices it is NOT possible to realize ideal integrators/differentiators.

For this purpose we need IDEAL opamps, which are not available."

Considering Rectangular Pulse Response I think Differentiation and Integration ,both are possible .provided we are careful about the Value of the Time Constant and the time duration of the action.

But if we go for a continuous step response Differentiation is preferable because the action will be over soon and with a very small time constant we can obtain a single sharp pulse for triggering. To safeguard against failure of triggering action we may have to go for a square or rectangular wave as input and to have a specific Positive or negative pulses only, we have to have a one way element like Diode in the output circuit.

Output of the Integrator for response to a step input of amplitude A is a continuous ramp and the output becomes unbounded ( A*t ). But, for a Square or Rectangular Wave input,the output of an integrator gives a Triangular Wave.

Dr.P.S.

Pisupati, Lutz, Abdelhalim, thank you for your profound answers to my last series of related questions. I consider your answers as different viewpoints at the same phenomenon helping to assemble the whole picture of the truth. About the first question, I would like to say that although all the answers are comprehensive and correct, I have found the Lutz's comments to my intuitive mind, and I will comment them first.

Since my student years, I have ever problems with understanding the "lead" phenomenon in differentiating devices - I cannot imagine how the differentiator's output quantity can be ahead of the input one. I know, from my experience and common sense, that the input causes the output... and this needs time (sometimes negligible but still time)... so the output quantity should be always behind the input one. Thus I cannot imagine how the output current flowing through a capacitor can lead the input voltage across it and how the output voltage across an inductor can lead the input current passing through it especially in the initial instant... I am able to imagine only the integrator's output quantity lags the input one...

So, I share the first Lutz's advice to pay attention to start-up conditions ("...when we try to find an answer which of the respective two possible formulations are to be preferred I think we have to go back to start-up conditions - for example the step response as a typical test signal..."); the steady-state conditions somehow hide the causality. Then I share his next advice to use the integrator arrangements as a base for understanding and explaining the phase shift between the voltage and current in the capacitor and inductor ("...to describe the phase shift between voltage and current for a capacitor (inductor) we should start with a current (voltage) source") since this does not contradict the human intuition ("...it allows a physical explanation without any logical problems - causality").

I would propose another remedy to solve this "logical problem". In practice, the resistance R always exists as some output resistance of the input voltage source and as some input resistance of the next current input device (e.g., an ammeter); so we always have RC or RL networks. The whole networks are driven by the input voltage and both the partial voltage drops across the two elements can serve as complementary output voltages (integral or differential). So, the logical succession is: the input voltage is the first (the cause) and the partial voltage drops are the second (the result); the input voltage came first and the drops came second; the input voltage was in the past while the voltage drops are in the present... so they can interrelate through time as we want (to lead or lag each other)... Am I right? Or it sounds too "philosophical" in the field of electricity?

Regards, Cyril.

Hi Cyril,

I think your "practical" approach is a very good one - because it is based more or less on an engineering view of the problem. You circumvent mentioning the term "current" because you are going to explain the effect with voltages only. Thus - there is no phase lead anymore.

(By the way: In case of a phase lead by 90 deg on could imagine that it is a phase lag of 270 deg. I think, this is allowed because all these terms are valid under steady-state conditions only. This view is appropriate for the inverting integrator function.)

Lutz, you are absolutely right about my "circumvent mentioning the term 'current' (explaining the effect with voltages only)". I am sure you and all the contributors here will agree with me that the current still explicitly exists:); only, it is represented by the voltage drop across the resistor. So, once we introduced (seen, accepted...) the resistance (desired or undesired), we can work with the more convenient voltage instead the "awkward" current. A remedy: if there is no resistance concentrated in an isolated resistor, we can use the conductor's resistance measuring the voltage drop across it (the Ohm's idea).

Your remark about the equality between the 90 deg lead and the 270 deg lag is exactly what we needed to use as an argument in favor of the assertion that the two definitions are equivalent. BTW I use this basic property of the periodic functions to show what a two's complement code is (I will do this with my students this Monday in the microcomputer laboratory). In your example, you have expressed the negative number -270 (the lag) with the positive number 90 (a lead) or v.v.; so 90 is the 360's complement of the number 270. And if you add some number (e.g., 300) to both the negative and complement codes, you will get (after ignoring the carry) the same result: 300 + (-270) = 300 + 90 - 360. It is interesting to add such a question about the basic idea behind the two's complement code...

About your emphatical "...there is no phase lead anymore..." Should they remove the "lead" explanation from all the electricity textbooks (I think it has repeated millions of times:)?

Regards, Cyril.

Pisupati, I would like to comment your genius speculation below:

"...For Differentiators ,as I have already pointed out instead of using an Inductance we obtain the effect of Inductance with Capacitor,by changing the positions of Resistance and Capacitance , as used in Gyrator Networks..."

It seems, the RC network is both integrating (if we get the voltage across the capacitor as an output) and differentiating (if we get the voltage across the resistor as an output). So, as you said, we can use the resistor as an inductor thus obtaining a simulated inductor (gyrator). See the attached picture and the link below. Is this your idea to create a virtual inductor?

Regards, Cyril.

http://en.wikipedia.org/wiki/Talk:Gyrator#What_is_actually_a_gyrator.3F

The picture of the simulated inductor presents:

* The varying voltage source Vs (a voltage follower) "copies" the voltage drop across the resistor R (current-to-voltage converter) thus setting the voltage across the virtual inductor

* The voltage across the resistor Rl (voltage-to-current converter) is set equal to the voltage across the capacitor C thus setting the current through the virtual inductor

* As a final result, the voltage and current across/through the initial element (a capacitor) are swapped in the new virtual element (an "inductor")

* The inductance is simulated by the opposing voltage source Vs that changes its voltage through time in the same manner as the voltage across the resistor R of the differentiating network CR changes and as a real inductor would change (if it was connected in the place of the whole "bridge" circuit)

Once we began talking about the causality in circuits, let's start with the very beginning - the Ohm's law:

https://www.researchgate.net/post/Are_there_causality_relationships_in_Ohms_law_If_so_which_is_the_cause_and_which_is_the_effect

Prof. Cyril,

you have yourself given the reference and the action of the Gyrator

"What does a gyrator actually do?

A gyrator does only a simple donkey work: it continuously "observes" the instantaneous values of the voltage across and the current through the original element (model, sample, pattern, load here...) and makes its own instantaneous output voltage proportional to the initial current and its current proportional to the initial voltage. Thus, the combination of the gyrator and the connected actual load acts as a dual 2-terminal virtual element. Shortly, a gyrator is a dynamic voltage source with swapped initial voltage and current."

"a gyrator can "invert" not only classical impedance elements (capacitors and inductors); it can invert elements of all kinds including non-linear ones. For example, a gyrator can convert a diode or a varistor (voltage-stable elements) into a transistor or a baretter (current-stable elements) and v.v. I suppose a gyrator can invert a negative impedance (negative capacitor into a negative inductance) and this extremely odd combination will have some fancy name (e.g., "injectoplactor":) It turns out a gyrator can convert almost everything."

The following link of Texas Instruments (Chapter 6 on Gyrators)may be helpful.

http://www.ti.com/lit/an/sboa093a/sboa093a.pdf

Normally, for Differentiators we can have an Inductor as the Feed Back element and it is difficult to prepare necessary Inductor and the course I underwent at I.I.T.,Bombay on Networks came to my rescue as I learned about Gyrators in Network Synthesis when I was doing Sequential Summer School there during the summers of 1970,'71 and '72

I really feel Honored on receiving Compliments and appreciation from eminent Professors like you and Prof. Lutz..

This is all due to Lord Almighty and my GURUS., so many that I am not able to list now and to Department of Technical Education of Govt. of Andhra Pradesh,India and later, to Jawaharlal Nehru Technological University,Hyderabad who deputed me, With Salary to Institutes of Higher Learning in India like Different I.I.Ts, I.I.Sc,Bangalore etc.for many such courses.

Dr.P.S.

@ Dr. P. Subramanyam: Thank you for your nice words.

Regarding the last subject of discussion (gyrators) I feel not able to contribute new insights. In this area, I feel, think and work more or less as an engineer only who is using these nice units (NIC, GIC) for filters and oscillators.

@ Cyril: As a reminder, we should not forget to come back soon to another question you have raised about RC oscillators and the validity of Barkhausen`s oscillation condition. For my opinion and according to my experience, there are really some open questions which need to be answered (necessary vs. sufficient condition).

Lutz, I have not forgotten these pending questions. I intend to concentrate on the visualizing the operation of the Wien bridge oscillator and particularly, on your powerful idea about the appearance of a negative voltage across the bottom capacitor. Do you mean a split or single supply, or it is no matter?

Cyril, I think it is easier to assume a split supply in order to be independent on additional circuitry for proper bias points. However, in principle, it does not matter, of course. Let me formulate one simple idea:

There is no principle difference between LC and RC oscillators.

Both versions use a filter network (mostly bandpass) as a frequency-determining path.The only difference is that LC filters allow Q values above 0.5 and passive RC circuits only below 0.5 (thus, requiring some more gain from the amplifier). That´s all. (I know this is not the right place to further discuss this subject) .

Expressed in the time domain:

Passive LC circuits can be designed as underdamped systems (showing decreasing oscillations) whereas passive RC circuits always are overdamped.

I have asked a separate question about the structure and the operation of the Wien bridge oscillator:

https://www.researchgate.net/post/What_is_the_basic_idea_of_Wien_bridge_oscillator_How_does_it_operate#share

More than one year later, 51 answers and 12,000 views I am not sure if the answer of the question, "How do we make the phase shift in an RC circuits exactly 90 degrees?" is clear... If not, the answer can be found here:

https://www.researchgate.net/post/How_do_we_convert_the_imperfect_passive_RC_integrator_into_an_almost_ideal_op-amp_inverting_integrator_What_does_the_op-amp_do_in_this_circuit