A negative capacitor is an inductor (coil). One has to be careful with the impedance concept. Some teachers explain that impedance is similar to resistance but for AC signals, and this leads to confusion, misunderstanding and to questions like yours. Impedance is an artificial concept linked exclusively to sinusoidal AC signals (voltage or current) to justify that In the time domain, current and voltage responses in the terminals of a component show a DELAY between them (as observed easily using an oscilloscope). Delay is also a relative concept, and it is referred to the moment at which a signal cross the zero level. If the delay is "before" or "after" respect to this crossing, one speaks about positive or negative impedance. Consequences over energy comsumption are well explained above.

Your question is very long. But if I look at first part, I would say negative resistance requires an active component. Yes it behaves as negative in certain defined range of volatges or currents, that does not mean it produces energy. It consumes energy to keep it in that condition and then while delivering energy to a part of the circuit it consumes energy from another source. It is an effective resistance that for that selective part, for simplifying your understanding and calculations, you can treat this as having a negative resistance.

I can give you three points of reference on this topic which you can look up.

1. Transfer electron effect - also known as the Gunn effect. This is seen in (for example) GaAs devices and is related to carrier mobility in two different conduction 'valleys'

2. Some ferroelectric materials can exhibit 'negative impedance'

3. heating effects in MOSFETs can exhibit a negative differential conductance due to the temperature dependence of mobility.

Hope it helps

Colleagues,

I would like only to remember that there are separate questions about the true (absolute) negative impedance (resistance)

https://www.researchgate.net/post/What_is_negative_differential_resistance_How_is_it_implemented_How_does_it_operate_What_is_its_relationship_with_the_true_negative_resistance,

its implementation

https://www.researchgate.net/post/What_is_the_basic_idea_behind_the_negative_impedance_converter_How_is_it_implemented_How_does_it_operate_What_does_the_op-amp_do_in_this_circuit

and also about the negative differential resistance

https://www.researchgate.net/post/What_is_the_basic_idea_behind_the_negative_impedance_converter_How_is_it_implemented_How_does_it_operate_What_does_the_op-amp_do_in_this_circuit

I have asked the present question with the purpose to clarify what "negative impedance" (capacitor reactance) mean in the area of electrotechnics and what it means in electronics (ngative impedance converters). It is especially interesting to consider the humble capacitor and inductor from this viewpoint.

Regards, Cyril

A negative capacitor is an inductor (coil). One has to be careful with the impedance concept. Some teachers explain that impedance is similar to resistance but for AC signals, and this leads to confusion, misunderstanding and to questions like yours. Impedance is an artificial concept linked exclusively to sinusoidal AC signals (voltage or current) to justify that In the time domain, current and voltage responses in the terminals of a component show a DELAY between them (as observed easily using an oscilloscope). Delay is also a relative concept, and it is referred to the moment at which a signal cross the zero level. If the delay is "before" or "after" respect to this crossing, one speaks about positive or negative impedance. Consequences over energy comsumption are well explained above.

Quote: "A negative capacitor is an inductor (coil)."

Well, that is statement only - without any proof.

What about the frequency dependence of the impedance?

(I am aware that the term "impedance" does apply to sinusoidal signals only).

I would specify this statement:

With regard to the time (frequency) behavior, negative capacitor is an inductor (coil).

It is interesting that from the same "time viewpoint", the resistor in an RC circuit can be considered as a negative capacitor (inductor) as well... and this is used in gyrators (simulated inductors).

About the term "impedance"... In negative impedance converters, this concept is extremely large - everything (including even non-linear elements - diodes) connected between the op-amp output and its non-inverting input in an INIC has impedance.

I will follow up Angel`s answer. The RESISTANCE is term from time domain, and IMPEDANCE is term from frequency domain (Circuit Theory point of view).. IMPEDANCE is defined as input or transfer function for monoharmonic steady state as ratio of two phasors; voltage and current. Therefore, whenever you have OP-amp connected on sinusoidal voltage you can write about IMPEDANCE, with presumption that all elements are linear and time-unchangeable . If is OP-am connected on DC voltage (similarly is with DC converter) then it should be written: RESISTANCE ( i.e. input resistance). Please, I will like to skip explanation when the input source is polyharmonic and one or more elements are nonlinear or time-changeable. :-D

It is obviously for me that IMPEDANCE is complex number. Then it can not be written " negative impedance".On the other side; negative resistance has meaning because it is term from time domain.

Cyril,what you will say about my viewing? ;)

Denis, very interesting thoughts... you make me realize that actually I don't know what "impedance" is:)... maybe "reactance" is the righter term for my intuitive needs... But what is even more interesting is that the very op-amp also does not know what "impedance" is:); it blindly "copies" and doubles the voltage drop across the "sample" element Z (a resistor - if we want to obtain "negative resistance", a capacitor - if we want to obtain "negative capacitance", etc...) and "inserts" the doubled "copy" into the circuit as a voltage... So inside a NIC is not exactly a negative impedance converter; it is something as a "voltage (drop) converter"...

Sorry, the atached picture represents the dual INIC... just swap the op-amp inputs to obtain a VNIC:)

Quote Cyril: "With regard to the time (frequency) behavior, negative capacitor is an inductor (coil)."

Cyril - sorry, but I cannot follow you.

Neither the frequency behavior nor the step response (current or voltage) of a negative capacitor (realized based on the NIC principle) is identical to the respective properties of an inductor.

Please, can you explain the above quoted statement?

Lutz, I have just added " "With regard to the time (frequency) behavior" to "a negative capacitor is an inductor (coil)"; i.e., I have just tried to clarify this statement... This is an interesting example of an "electrical analogy"...

This topic is more suitable for the future question about the gyrator. IMO here we should clarify what "negative" means, first in electrotechnics, and second, in electronics...

Lutz, what do you think about the impedance? Can be the impedance considered as an instantaneous quantity?

And also, about the NIC - "What does it actually converts?" My last insight is that it does not convert impedance (positive into negative); it converts the voltage drop across a passive element into a voltage across a voltage source... and this source is connected in series to the passive element...

Following Lutz, I rectify (like a diode!) my first statement: "A negative capacitor is an inductor (coil).", because generally speaking conventional capacitors and coils, as derived form Maxwell equations, follow inverse dependence with frequency. It was just a way of saying that "phase" difference (delay) is opposite, so at a certain fixed frequency you can theoretically find the coil with the opposite impedance to a capacitor. Thanks also to Denis for adding new details and nuances.

OK - I agree and I repeat:

There is one single frequency for which a negative capacitor can have the same impedance (magnitude and phase) as an inductor. This sounds logical.

A negative capacitor in the sense of electrotechnics or electronics?

My experience is in electronic circuits..., Thanks, Cyril, for this nuance.

An impedance can be complex value. As any complex number impedances have real and imaginary part or absolute value and argument. The real part of impedance denotes its resistance, the imaginary part denotes reactance. What does it mean negative complex number? Are we talking about negative resistance and any reactance? By the way electronics is a part of electrotechnics. Why should be different definitions in electronics and electrotechnics. They are certainly the same.

Simple.

In DC: R (or Z) =V/I, so if current and voltage have the same sign, the impedance is positive.

In AC: z=dv/di, so if the slope of the i Vs v curve is negative z is negative.

An impedance is a current controlled voltage source: V = Z(I)

The input current is measured in a short circuit between nodes 1 and 2

The output voltage is implied between nodes 3 and 4.

In case of a two-terminal impedance node 3 = node1 and node 4 = node 2

The sign of V may be positive or negative..

Erik, an element with negative impedance can be implemented as a current-controlled voltage source V = Z.I in series or as a voltage-controlled current source I = V/Z in parallel. These sources are connected in such a way as to "help" the input source...

Saying "an impedance is a current controlled voltage source" you probably mean to simulate the impedance by connecting an "opposing" current controlled voltage source in series...

The impedance element is two-terminal (1-port); if it is four-terminal (2-port), we talk about transimpedance...

Cyril, A voltage controlled current source is an admittance I = Y(V) .

I agree that if the elements are 3- or 4-terminal elements we talk about transfer elements.

I talk about both simulating and measuring. We simulate by means of models.

We "believe" in our models if we see a great agreement between simulations and measurements.

Dear Professor Cyril Mechkov, Ohm's law is a straight line equation. And slope may be negative or positive. In 2-port network theory its very clear when we deal with open circuit and short circuit impedance/admittance.

@Afaq Ahmad

In AC it is the slope you should consider. In DC it is the actual values. Please see my original reply above:

In DC: Z=V/I, so if current and voltage have the same sign, the impedance is positive.

In AC: z=dv/di, so if the slope of the i Vs v curve is negative z is negative.

Dear Professor Fernando Soares Schlindwein, Slope can be for any signal.I mean to say about V-I curve.Yes, I looked your answer.

Fernando, you have said, "In DC: Z=V/I, so if current and voltage have the same sign, the impedance is positive." Now look at the IV curve of an S-shaped negative differential resistor (the attached picture below). In the middle (red) region, both the voltage and current have the same (positive) signs but they say the resistance is negative:)

Cyril Mechkov, I stand to what I wrote. Indeed in your figure your r(diff) is negative but the diff means AC, that is d(diff)=z=dV/dI. As all points of the red part of the curve are in the first quadrant V>0 and I>0, therefore the DC impedance Z=V/I is positive.

Quote Erik L.: "Cyril, A voltage controlled current source is an admittance I = Y(V) ."

Hi Erik - I think, it is more correct to say that a VCCS is not an admittance but has a TRANSFER characteristik that can be called TRANSFER-ADMITTANCE.

However, if we take as an example the classical opamp-based NIC (which is a voltage controlled current source) , the current goes back to the driving source and, thus, we can say: The voltage causes a current through the source (like a resistor), however in a "negative" direction (if compared with a classical ohmic resistor).

An excellent answer, Fernando! Of course, you understand that I have added this comment to make you clarify what is negative and what - positive in this arrangement... and you have done it in a perfect way...

I would add only that, for the purposes of understanding, it is not sufficient to say that the IV curve in this region has a negative slope and the AC resistance is negative; you have to explain WHY the slope (the AC resistance) is negative (although the present resistance, voltage and current are positive)... and HOW it is done...

Lutz, thank you for the perfect INIC explanation! To second you, here is a picture that visualizes this situation (the input voltage is positive but the current goes back to the source)...

... and a picture representing its operation by IV curves superimposed on the same coordinate system. More about this topic can be seen at the link below

https://en.wikibooks.org/wiki/Circuit_Idea/Linear_Mode_of_Current_Inversion_NIC#Positive_input_voltage.2C_negative_output_current

By the way, the input stage of a TTL gate also "pushes" back a current into the input voltage source... but does it exhibit negative resistance?

Quote Lutz v.W.:

"Hi Erik - I think, it is more correct to say that a VCCS is not an admittance but has a TRANSFER characteristic that can be called TRANSFER-ADMITTANCE"

Hi Lutz - If the component is a 4- or 3-terminal (2-port) it is a transfer immittance. If it is a 2-terminal (1-port) it is just an immittance. Control may be from signal or from time-derivative of signal.

Erik, what is the "time-derivative of signal"?

Erik, I know what you mean - and to make it clear we can use the well known OTA as a n example.

* In case the OTA is used as a transfer block (input-output, amplifier) we have the ratio output current-to-input voltage resulting in a transfer-admittance (OK, or immitance). This is usage as a 2-port, of course.

* If we, however, apply 100% feedback to the inv. terminal the OTA behaves (at this terminal) like a grounded resistor. This is a usage as a 1-port.

* When we do something similar (feedback to the non-inv. Y-input) with a positive current conveyor (CCII+ like AD844) we mimic a controllable NEGATIVE resistance.

Lutz, your thought...

"If we, however, apply 100% feedback to the inv. terminal the OTA behaves (at this terminal) like a grounded resistor. This is a usage as a 1-port."

... is a very interesting idea for me (as you know, these 2-terminal arrangements are my pursuit:) As I can see, this is an artificial (virtual) "resistor" whose resistance value is determined by the amp gain and whose kind of resistance (positive or negative) - by the kind of the amplifier (non-inverting or inverting). The negative resistor would be voltage controlled (INIC).

My only remark is only about "apply 100% feedback to the inv. terminal of OTA". You probably mean simply to connect the output to the input by a "piece of wire":) But this does not obligatory introduces a negative feedback. For example, in the case of a perfect input voltage source, there is no any feedback; you will have a negative feedback if there is some (internal) resistance between this voltage source and the amp inverting input.

An example of this arrangement is the input (left) part of the simple current mirror but it will act as a nonlinear "resistor".

Regards, Cyril

Cyril - yes, of course I mean a direct connection between output and inv. input.

And - more than that - you are right that such a feedback would be shorted (theoretically) in case an ideal voltage source is connected to this node.

However, who would do such a garbage?

*At first, the whole circuit would not function because the current output is loaded with a zero resistance, and

* secondly, what should be the purpose of such a connection?

Rather, it is the purpose of such an actively simulated (grounded) resistor to be used in a larger circuit that has a certain function (filter, oscillator). And in such a case, this "active resistor" always is combined with other elements.

Lutz, I do not see a problem in this connection - an ideal voltage source across the input and a direct connection between the amp current output and the inverting input. This simply means that we have connected in parallel three 2-terminal "devices" - the input voltage source, the amp voltage input and the amp current output, and simply there is no feedback. The amp current output will serve as a positive/negative "resistor" that consumes/injects a current from/to the input voltage source that is proportional to the input voltage with a coefficient G (R). A possible application can be a "programmable load" if the amplifier has a programmable gain.

The op-amp INIC ("voltage-controlled negative resistor") in the attached picture is driven by an "ideal" voltage source and is exactly the same case (there is no positive feedback since the input voltage source has fixed the voltage of the non-inverting input). Here are some explanations about the circuit.

The op-amp and the voltage divider R1-R2 form a non-inverting amplifer with a gain G = 2; so, this varying "voltage source" produces two times higher than the input voltage. It is connected through the resistor R to the input of the non-inverting amplifier; thus it acts as a varying voltage-controlled current source (VCCS) or a "true negative resistor" (current-inversion NIC or INIC) that "pushes" the same current into the input voltage source what the "positive" resistor R would consume if it was connected to ground; it has inverted the initial current direction...

https://en.wikibooks.org/wiki/Circuit_Idea/Linear_Mode_of_Current_Inversion_NIC#Positive_input_voltage.2C_negative_output_current

....a programmable load - may be, that could be a possible application (battery test?).

Exactly, Lutz! Have you seen the current output in the circuit above (that OA, R, R1 and R2 form a voltage-controlled current source)? Is it clear the connection with the conceptual circuit below?

https://en.wikibooks.org/wiki/Circuit_Idea/Revealing_the_Mystery_of_Negative_Impedance#The_basic_idea_of_the_parallel_NR_compensation

The doubling voltage source BH (producing a voltage VH) on the right represents the noninverting amplifier with gain of 2 that, in combination with the resistor R in series, constitutes the needed current source (INIC negative resistor). It is "helping" since it adds its current to the input current (created by the voltage divider's output voltage across the load Rl); thus it "helps" the input source in its "desire" to pass a current IL = Vl/RL.

This picture shows how simple and clever is the "inverting current" trick - the lower end of the "positive" resistor R is disconnected from the ground and "lifted" by a voltage 2VL; as a result, the current changes its direction (from outflowing it becomes infloing) but retains its value...

... and with even the more general picture below?

https://en.wikibooks.org/wiki/Circuit_Idea/Revealing_the_Mystery_of_Negative_Impedance#Voltage-driven_negative_impedance_elements

In this most general picture, the "helping" current source (INIC negative resistor) is shown by its symbol.

Quote Cyril Mechkov:

"-Erik, what is the "time-derivative of signal"?-"

A capacitor is a current source controlled by the time-derivative of the imposed voltage:

IC = sC*VC = C*d(VC)/dt (input signal: voltage, output signal: current)

An inductor is a voltage source controlled by the time-derivative of the imposed current:

VL = sL*IL = L*d(IL)/dt (input signal: current, output signal: voltage)

Very interesting... "functional" (time-dependent) current sources? But they do not keep up the current very well... Then? Maybe the ubiquitous op-amp will help?

The capacitor is an admittance element. The coil is an impedance element.

No, I think both they are immittance elements:) By this logic, what is the resistor - a resistance element or a conductance element:)?

Erik, there is a problem with your assertion that "a capacitor is a current source"; I will explain why.

Saying "current source" we mean "constant current source" or "varying current source" but in both the cases we suppose the current does not depend on the voltage across the load (it should depend only on the input voltage and the immitance of the element in series). And to see if this is so, we begin varying the voltage across the load...

If this was a resistor-type current source, and the voltage across the load was negligible, the current would stay relatively constant when you wiggle the load voltage. But in your case even when the load voltage is negligible, the current would change significantly when you wiggle (quickly enough) the load voltage since the voltage across the capacitor does not immediately change... Am I right?

A capacitor is not a current source.

A capacitor provides a voltage between its plates - and this voltage, of course, can drive a current, which DEPENDS on the connected load (and, of course, the time).

Thus, it does not fulfill the definition for "current source".

In general, I think that a formula that connects voltage, current and load cannot say anything about "cause and effect".

(See for example our discussion about Ic=B*Ib. Some people think that this formula would imply that Ib is the cause of Ic.)

Basic circuit theory ;-) :

An immittance is

either

an admittance Y = I/V = I(V) i.e. a current source

or

an impedance Z = V/I = V(I) i.e. a voltage source

A source may be independent or controlled.

An independent source may be a function of time.

An n-terminal circuit is a super-node for which

Kirchhoffs current law must be fullfilled

An n-terminal circuit may be modeled by means a

number of coupled 2-terminals (elements).

An arbitrary node (terminal) may be used as reference (ground).

The variables of the circuits are the element currents and the element voltages.

We model our circuits by means of differential equations and algebraic equations.

Instead of element voltages and element currents (2B approach)

we may use node voltages and impedance element currents

as variables (MNA approach).

Time is an independent variable. ( We do not know what time is ;-/ )

I would add only that the "voltage across the capacitor will drive a current" when the input (excitating) voltage decreases; when it increases, the input voltage source drives the capacitor by a current. The "current-creating" voltage is the difference between the input voltage and the voltage across the capacitor (V - Vc).

It is still interesting to see why "a capacitor is not a current source" in this arrangement (a capacitor connected in series to a voltage source)...

It is because the building "formula" of such a simple current source is

"voltage source" + "constant-current element" = "current source",

while here our building "formula" is

"voltage source" + "constant-voltage element" = "voltage source"

So, connecting a capacitor in series to a voltage source, we obtain again a voltage source (indeed, time-varying but still voltage source). An example is the blocking capacitor in an AC amplifier

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

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

"Thus, it does not fulfill the definition for current source" (Lutz) but sooner for a voltage source...

We have introduced the Dirac impulse signal (time width=0, size=infinite and area 1)

We have introduced the ideal step signal (time width=0, size=1)

These signals are mathematical fictions which create problems.

Remember TTT = Things Takes Time !!!

The ideal op.amp. has infinite gain. We have no access to

the value infinite in our computers so we use zero instead

(of course without divide by zero) in our model. In a sense

infinite and zero is the same. Unfortunately we do not understand

the concepts infinite and zero.

Quote: "An immittance is

either

an admittance Y = I/V = I(V) i.e. a current source

or

an impedance Z = V/I = V(I) i.e. a voltage source"

It seems I am getting old - however, is there somebody who can explain to me WHY an admittance is a current source?

( May be, my engineering point of view is too limited; that means: I try to find an answer without Dirac, etc...)

I think that Erik means transadmittance Y = I/V = I(V) i.e. a voltage-controlled current source and transimpedance Z = V/I = V(I) i.e. a current-controlled voltage source... or he just considers I as an output and V as an input variable in the first case, and V as an output and I as an input variable in the second case...