The energy density and power density should be calculate using the mass of active electrode material only or considering all the mass of current collector,electrolyte and packaging elements. Is there any way of calculation energy density and power density by three electrode system testing?
it is not correct to calculate the energy density and power density in three electrode system because three electrode system is only for measurement of maximum capacitance of the electrode. Two electrode configuration is real and practical system and it will have lower capacitance than three electrode system. energy density and power density should be calculated and compared in real cell system only.
As you know that three electrode value was 4 times higher than 2 electrode ful cell. if You want to calculate the energy density with three electrode system ,divide the three electrode capacitance value by 4.then calculate the energy with normal calculation methods.
The above answer is correct in general, however, for symmetrical capacitors, it is possible to use the results from the 3-electrode cell to derive the specific energy and power (Wh/kg or kW/kg) (Note: energy or power density has the unit of Wh/L or kW/L, respectively). In such a case, you need to measure the specific capacitance (F/g, for the material) and the potential range (U), and then use the data to derive the specific energy,
W = CU2/8
(C is specific capacitance, and hence no mass appears in this equation).
Specific power can in principle be derived from specific energy divided by the time for completing the potential change. Or if you may workout the iR drop (i.e. ESR) in the 3 electrode cell, the specific power is
P = U2/(8*ESR*m)
(m is the mass on the working electrode).
Dear prof. George Zheng Chen, can u give any reference paper for this calculations.......
I guess this calculation can be found in the literature, although I did not copy it from anywhere. I am writing up a paper in which the use of the specific capacitance of an electrode material to calculate the specific energy of a symmetrical cell is given.
Derivation is simply as follows, although I thought you should be able to do it yourself.
Csp,cell = C sp,material /4 (This equation is derived below)
Wsp,cell = Csp,cell*Ucell2/2 = (Csp,material/4)*Ucell2/2 = Csp,material*Ucell2/8
*******************
Csp,cell = Ccell /(m+ + m-) = Ccell/(2*m)
where m+ and m- are the masses of the active materials on the positive and negative electrodes (positrode and negatrode), and m+ = m- = m for a symmetrical supercapacitor.
1/Ccell = 1/C+ + 1/C- = 1/(m+*Csp,material) + 1/(m-*Csp,material) = 2/(m*Csp,material)
So, Ccell = m*Csp,material/2
And thus, Csp,cell = Ccell /(2*m) = (m*Csp,material/2)/(2*m) = Csp,material/4
I would like to help you with this paper
Article Ragone Plots for Electrochemical Double‐Layer Capacitors
If you require any further information, feel free to contact me
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