I like to calculate the volumetric and areal capacitance of a two electrode system. Can someone suggest how to calculate using a widely accepted formula?
First, the gravimetric specific capacitance of the electrode need to be calculated followed by the volumetric specific capacvitance :
C(F/g) = 2*(I.t/m.V),
where I is discharge current, t is discharge time, m is the mass of active material in the electrode, and V is the potential window.
to calculate the volumetric specific capacitance, just multiply the gravimetric specific capacitance by the density of the electrode which is here should be taken the density of active material of the electrode,
i.e., C (F/cm3) = density*C(F/g), this means the density of the active material and the volumetric capacitance are directly proportional.
Areal capacitances (F cm-2) can be calculated by dividing the single electrode
capacitance by area of the electrode.
Hope this will help!
Formula for capacitance is C_s= (i ∆t)/( ∆V) only
1 specific capacitance C_s= (i ∆t)/(m ∆V) in F/g :Specific capacitance is mass normalized capacitance.
2. Areal capacitance C_s= (i ∆t)/(A ∆V) in F/Cm2 where A is area of electrode material coated
3. Volummetric capacitance is C_s= (i ∆t)/(V ∆V) F/Cm3. V is volume of the electrode mateiral.
To calculate areal and volummetric capacitance, calculation of specific capacitance is not necessary .
Totally agree with Subbukalai Vijayakumar, as no 2 in the eq. mentioned by Girum Ayalneh Tiruye. The 2 in case of 2-electroed setup
Dear Fares Alkhawaja. The capacitance unit is F, but always it represented as specific capacitance in F/g
Dear Fares Alkhawaja, it is a normalization to mass (means how many Farads per unit of mass (1 g)). As we have resistance in Ohm and resistivity in ohm.cm.
Subbukalai Vijayakumar can you elaborate the explanation of areal and volumetric capacitance?
- Badges
- Science topic
- Similar topics
- Engineering
- Electrical Engineering
- Electrodes