I had a CV of conducting material (file attached), and through current data and charge data, the results obtained are different. Could you please suggest a way to determine the specific capacitance of the material?
mass= 0.01 g
potential window= -0.5 to 0.5 V
Estep= 0.002
scan rate = 0.001 V/s
You can calculate it by using the formula
CS = Area of cv curve/2*m*k*v
Where m= mass of electrode
K= scan rate
And v= potential window
Hi Harish,
The expression for specific capacitance (discharging curve) is:
Cp = A/mk. delta V
A = area under I-V curve; k = scan rate; m = active mass of polymer dipped in electrolyte; and delta V = potential window
*If you consider the total area under the curve, you need to divide the above expression by 2
When capacitor is charging, then the area = A1 can be written as:
A1 = (V2-V1) Cp * m * k ------------- (8)
Similarly, when capacitor is charging, then the area = A2 can be written as:
A2 = (V1-V2) Cp * m * k --------------- (9)
For the calculation of area (A) inside the CV curve, we have to substract equation (9) from equation (8),
A = A1 – A2 = [(V2-V1) Cp * m * k] – [(V1-V2) Cp * m * k]
A = [(V2-V1) Cp * m * k] + [(V2-V1) Cp * m * k]
A = 2 [(V2-V1) Cp * m * k]
A/2 = [(V2-V1) Cp * m * k]
Cp = A/(2 (V2-V1) * m * k)
Where, Cp is the specific capacitance in F/g, A is the area inside the CV curve having units AV, m is the active mass of material, k is the scan rate of CV in volts per second, and (V2-V1) is the potential window of CV.
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