Not one electrode capacitance value. I want to calculate, 2 electrodes so a symmetric supercapacitor. Is ıt different or the same formula? How can I calculate my symmetric supercapacitor?
2-electrode cell is equivalent to the following circuit -C-R-C-, where C is your electrodes acting as capacitors, and R is a serial resistance between them. The total capacitance of this cell Ct is:
1/Ct = 1/C + 1/C,
from where : Ct = C/2
So, the total capacitance of your cell is a half of the capacitances of each electrode.
Sergey N Pronkin Thank you for your answer. But I use CV technic my symmetric capacitor capacitance value with a two-electrode system. So I calculate the capacitance value for one electrode C=Idv/2*m*V*v from this formula. Of course for symmetric capacitor capacitance value m*2 because of 2 electrodes. I ask same formula or a different one or different constant for to found the symetrıc capacitor capacitance value.
So, just to be sure about the variables:
Capacitance of a capacitor (by definition):
C = dQ/dV,
Mass specific capacitance:
Cm = dQ/(m.dV)
dQ = int (I.dt) = int(I dE) / v, here v = dE/dt, sweep rate
So, Cm = int (I.dE) / (m.v.dV).
So, you measure CV, integrate it int(I.dE), divide by (m.v.dV), where m is the mass of the whole device, and you get your Cm - mass specific capacitance of the capacitor, doesn't matter either symmetric or asymmetric. No coef.2 intervene, to my understanding.
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