The impedance changes with the applied potential (DC). Simply, you can think in this way that you are measuring resistance by applying DC potential using Ohms law. The impedance arises when you perturb the system by AC potential with a very small amplitude.

For supercapacitor, AC potential should be 5 to 10 mV while DC potential should vary from 0 V to any +/- potnetial.

Sujoy,

DC should vary from 0V to any +/- potential--- What is the logic?

5-10mV of AC perturbation-- won't the phase difference between the applied V and measure I vary  with different AC-voltage!!

The word 'vary' implies that you can choose any potential including 0 gives you same result unless you deviate from the rectangular shape of the CV curve.

It seems that you did not perform any proper AC impedance experiment before commenting. ''Electrochemical impedance is normally measured using a small excitation signal. This is done so that the cell's response is pseudo-linear. In a linear (or pseudo-linear) system, the current response to a sinusoidal potential will be a sinusoid at the same frequency but shifted in phase.''

Hi Sujoy,

I am only worried about the DC voltage {my question is not about range of voltage- it is about using exact voltage}! Say I vary my DC voltage from 0 to the maximum rated potential. In case of VERY IDEAL Supercapacitor there won't be any  change in impedance. In any other case there will be a deviation in the impedance with increase in the DC voltage. 

What is the reason for this? Is is just because there would be some mass transfer limitation? If yes, from what test can I find out the limitation point?!

Thanks

dear Deepak Sridhar,

DC voltage stands for the working point of the function :VDC=V(q), where q is the charge. The low amplitude value VAC is used for the impedance measurement : Z(q)=(VAC/iAC).ej(phase)

PS : The low amplitude value VAC keep average q the same and Z(q) pseudo-linear vs AC current.