There is always IR drop of charge-discharge curve whether analysing the electrochemistry of supercapacitor electrode. It signifies for internal resistance of the electrode.
But IR drop changes with the current density applied for a fixed potential window. Many research articles suggest the IR drop increases with the increase of current density.
But some articles suggest that IR drop is negligible/minimised/omitted with the increase of current density as here fast charge-discharge is involved.
So, what is the actual behaviour of IR drop with current density? and electrochemical interpretation?
Dear Bidhan Pandit,
...the internal (equiv. circuit) resistance(s) obtained from EIS are NOT[1] always the same R (for the cause of IR drop). Only in some linear and simple systems, there are (model) values very close to R. In short, EIS give us a hint about the order of R values (and the iR drop).
1. Note: In common EIS measurements iDC is near 0. Also, in common (low amplitude) EIS iAC is actually near 0 (very small).
If you assume that your system looks like a Randle cell, then there is a solution resistance between working electrode and the reference electrode. Now, the iR drop, which is a voltage, comes from the fact that there is a voltage drop due to this resistance. Remember: u=IR
We want to measure the WE interface potential. But, as soon as there is a current, the WE interface potential+i*R will be measured.
By measuring the solution resistance between WE and RE by EIS, the maximum size of the i*R-drop can be estimated. If the value is less than 1mV at all times, it can be neglected.But, if it is larger, it must be taken into account. One way would be to use iR-compensation that most potentiostats would have.
Dear Gustav Karl Henrik Wiberg sir,
then whenever the current increases, the IR drop increases as it is proportional to i*R?
isn't it?
regards,
Yes, IR drop is basically a voltage drop during the initial stage of charge and discharge in capacitive devices, As the voltage is propotional to the current applied, therefore, it should increase propotionally with current densities. However, when researchers report that they have low IR drop even at high current densities, it simply implies that they have high power electrode material (As IR drop is related to power density of the capacitor). Basically, IR drop (int terms of energy) is the energy desipated during the flow of electrons due to any kind of resistance in the devise. Imagine, the electron transfer between current collector and your electrode material is hindered due to the resistive nature of your electrode material. In this case, you will have higher IR drop as compare to the electrode material where electron transfer is easier at the junction as well as through the material itself. (bearing in Mind you have every other factors like electrolyte and composition to be constant in both cases)
Dear Gustav Karl Henrik Wiberg sir,
thanks for your reply. one more request. plz can you give some reference articles related to this. so that I can get more theoretical aspects related to IR drop.
regards,
Electrochemical interpretation is much more difficult in cases of more than one (nearby) thermodynamic states, due to a high kinetics (high IR drop) contribution.
Dear Umair Gulzar sir,
the internal resistance obtained from EIS is the same for the cause of IR drop?
Dear Bidhan Pandit,
...the internal (equiv. circuit) resistance(s) obtained from EIS are NOT[1] always the same R (for the cause of IR drop). Only in some linear and simple systems, there are (model) values very close to R. In short, EIS give us a hint about the order of R values (and the iR drop).
1. Note: In common EIS measurements iDC is near 0. Also, in common (low amplitude) EIS iAC is actually near 0 (very small).
Dear Ioannis Samaras sir,
The internal resistance includes various resistive parameters. In various articles,it is given. thats why I was thinking about the above. now it is understood as per ur answer. thanks once again.
and now I'm searching for the previous question which I've asked u. remember? If find, I'll back to u.
In this article there is some related information
Article CaO impregnated highly porous honeycomb activated carbon fro...
There are two reasons for IR increase with current density. First one is that for a given R, if you increase I, then the product IR is higher and consequently the observed voltage drop. However, it may happen than R also increases with I. A higher current I involves a faster ion diffusion. If ions cannot diffuse that fast, your system will be limited by mass transport, and therefore, R, which is the sum of electronic and ionic resistances, will increase due to the formation of depleted layers of ions, i.e. ionic resistance will be higher.
Thank you Gomaa A. M. Ali for the attached paper.
I'll definitely go through it.
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