Hello,

The imaginary component of the impedance provides a measure of the reactive part of the impedance (e.g. capacitance). Impedance is the vector sum of the capacitance ( imaginary ) and the resistance (real) of the circuit in question. This can be better resolved from the Nyquist plot. You can think of the two components as two vectors which combine to produce the impedance. The angle between these two vectors is the phase angle and it changes over frequencies, thus producing the Nyquist plot. I hope that my explanation is clear.

Thanks.

In general, an alternating current(A.C) impedance value is an interfacial parameter of an interface between a solid and an electrolyte, rather than a bulk parameter of the bulk of the solid. The A.C. impedance can only obtain in a frequency domain, i.e., alternating current (A.C), in which the applied potebtial/current are oscillating with a phase shift difference between the  applied potebtial/current.   As a result, the A.C impedance is a vector parameter which has two components.   One component is a real impedance, resistance.   The other component is an imaginary impedance which  represents the energy storage part of the circuit element, capacitance.  

As the previous two responses stated: impedance that is capacitive is imaginary, but inductive impedance is also imaginary (-90 degrees vs +90 degrees out of phase).  Note also that some real phenomena like diffusion have both real and imaginary components (45 degree phase shift).

To add to the earlier answers. Real and Imaginary components are mathematically convenient ways of expressing a quantity (here, impedance) which has a magnitude and phase. There is nothing special or 'unreal' about the imaginary component. In case of capacitor or inductance, the ac current and ac potential have a phase difference of 90 degree, which comes out as imaginary number when expressed in this form.

Regarding why imaginary component decreases over time, unfortunately there is no general answer. It really depends on the system, and one has to know more details of the experiment and look at the entire spectra, in Nyquist format as well as in Bode format, to get a clear understanding and draw any meaningful conclusion.

At temperature higher than 0K it is not possible to transfer electrical charge through material without losses, which measure is a material resistance. In case of AC it is called impedance. You can look on it from both sides. The impedance is a nice tool to measure all reasons in one go (and like capacitance, inductance got different physical meaning, but the impact on charge transfer is the same-  slow down). Or go opposite and using measured impedance split it in a parameters, finding out teh impact of specific parameter. It was developed for studying processes on interfaces. It works well, if there are few distinct parameters dominating. For example, if the electrodes are immersed in a liquid electrolyte and charge transfer is limited by diffusion of charge carrier species, Warburg behavior is observed adn diffusion coefficients migt be measured. If no limiting process takes place and different processes are mixed most likely the method will be of no use to investigate those processes, just electrical resistance in general. The use of complicated equivalent circuits can not solve the problem, because the circuit is just a mathematcial model without adding any extra parameters for calculation. The method depends on a number of time domains on a real measurement.

Hi, regarding this interesting  question I may refer you to my paper published last year at Journal of Energy.

In the aforementioned paper, a full description of the phenomena occuring during EIS measurement is provided. Thereafter, LIBs using LTO anodes at different aging states are examined and the results are discussed comprehensively. For further questions please dont hesitate to contact me.

Few simple answers. Parasitic effects. The cell set-up. How the contact on interface electrode/eelctrolyte is arranged. Typically, the pressure is changing etc and interface capacitance is changing. Inductance effects might mean the electrode corrosion (interaction with electrolyte), protonic electrolytes are acidic and corrosive sometimes. In order to establish a better contact, the pressure is increased and it will nicely facilitate the solid state reaction. 

I have a question. can I say that, since imaginary impedance is mathematically computed from the equation Z = sqrt(Z'^2 + Z"^2), meaning that even if the imaginary impedance is negative, it contributes to higher magnitude of overall impedance (complex impedance). Therefore, a more negative Z” means higher impedance, not less, right? Z' = real impedance, Z" = imaginary impedance, and Z = complex impedance. How would imaginary impedance be related to energy storage then? Would higher the Z" result in higher or lower energy storage?

1) Yes, Z scales with Z”, for one of the C, (or L) exclusive[1] cases; note a model-C, as a simple case[2] : Q1:...even if the imaginary impedance is negative, it contributes to higher magnitude of overall impedance (complex impedance). Therefore, a more negative Z” means higher impedance.

2) The high(er) C, or L, values, the high(er) energy[3] (stored); Q2:...How would imaginary impedance be related to energy storage then? Would higher the Z" result in higher or lower energy storage?

1. If both C or L are included, then this rule is valid for their Z”-difference; near Resonance, Z” is very low.

2. An ideal C, as a simple lumped (-C, as a) component model, case ( Z = sqrt[0^2 + Z"^2] ).

3. Energy (stored) in an ideal C is : C . V^2 / 2; and energy (stored) in an ideal L is : L . i^2 / 2.

Thank you so much Dr. Samaras!

I still have some questions for #2. Say we are dealing with a model-C, so Z = sqrt(0^2 + Z" ^2) -> Z = Z". Energy stored in this case is C*V^2/2 . 

How does Z" relate to C*V^2/2 ? Right now I'm stuck thinking Z_c = 1/(jwC). If we say the whole system is purely capacitive, meaning that Z" = 1/(jwC), wouldn't increase in Z" mean decrease in C? How is there more energy stored if the equation is C*V^2/2? It looks like increase in Z" will actually cause less energy stored. 

I feel like I must have overlooked something here... please forgive my ignorance. 

For common (cases of interest in) EIS measurements, accepting a model-C, the energy stored, as an internal (VDC+0AC) electric field, is, mainly, constant[1], ~ C*VDC2/2, using[2] a very small AC-amplitude, VAC-->0.

Also, C,  in a model-C, is, surely, a constant parameter, by model's-definition; so, Z" increases, only, with w-1;   Q: If we say the whole system is purely capacitive, meaning that Z" = 1/(jwC), wouldn't increase in Z" mean decrease in C?

1.  In this topic, we exclude other cases, for high amplitude, alternating current (AC) applications :  electronics for the mains, electronics for Ac/Dc converters, etc.., that are not considering a pure model-C, as an interesting case in their focus. There is a (source's) safety issue for such (HV and/ior HF) power source(s) with a model-C, as a reactive "load" (see specs of the power source).

2.  However, you might ask for a related (special) case of an electrical resonance, that occurs in a particular (resonant) frequency, R,L,C case(s), apart a pure model-C.

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Thank you so much Dr. Samaras!