Dear Bidhan and Adam,

welcome again,

I will try to derive the relation for nonlinear GCD,

At first we have to make some assumptions:

- We are interested in the electrode capacitance C which has an active material mass= m, then the specific electrode capacitance of the electrode is Cs= C/ m,

This capacitance can be measured directly.

The electrode capacitance is nonlinear such that one defines its value at certain given voltage by  C= dQ/dV, With dQ= Idt, then we have 

C= I dt/ dV,

Since the capacitance is nonlinear it value changes with charging and discharging voltage of the electrode. With constant charging current the voltage can change from an initial voltage Vi to a final voltage to Vf. Then one can define an average capacitance during the charging Cav which can be deified as:

Cav =INTEGRA LI dt/ dV,which can be put in the form by multiplying the numerator and denominator by V :   Cav=  Integral  I V dt / VdV,

Integrating we get from Vi to Vf we obtain:

Cav= 2* I integral Vdt / V^2 from Vi to Vf,

Then Cav= 2* I integral Vdt / (Vf^2- Vi^2),

And the integral is the area under the charging voltage curve from t at V i to t at Vf.

We need  only to divide both sides by the active area mass of the electrode m to get the specific average charging capacitance of the electrode; that is:

Cs av= 2* I integral Vdt / m(Vf^2- Vi^2),

This is the formula which you introduced in your question. Now it is fully derived from the first principles.

Best wishes

As for calculating the capacitance  in case of symmetric or asymmetric two-electrode cell you need to decide if you want to present specific capacitance given per 1 electrode or for the full supercapacitor.

For example when you have a formula:

C=I∆t/m(Vf-Vi)

 if m is a active mass of both electrodes then if you multiply current and discharge time by 4 you will get the specific capacitance of assembled SC per one electrode. When you divide your result by 2 you should recive a capacitance of the device. 

I suggest you to look up for the prof. Ruoff work: Best practice methods for determining an electrode material's performance for ultracapacitors. 

Dear Adam Moyseowicz sir,

I read the paper you suggested. actually there device capacitance is calculated as per the electrode capacitance values. Here, I want to calculate device capacitance from the charge-discharge curve of the device which is not mentioned specifically in that article.

as per your suggestions,

specific capacitance of device,C = 2I∆t/m(Vf-Vi)

where, m is a active mass of both electrodes.

is that okay sir?

Dear Bidhan, 

There was a similar question in the researchgate and i gave the following answer:

It is a matter of definition of terms. To be specific and have a rigorous proof of the formula, let us assume that you have two similar super electrodes making a super capacitor. Then it is required to get the specific capacitance Csp of electrode material, then by definition it is equal to the electrode capacitance Celec divided by the mass m of the active material in the electrode, that is  Csp= Celec/melec. 

The capacitor is built from two electrodes , then the cell capacitance Ccell= Celect /2 since the two electrodes are connected in series.

Accordingly, Csp= 2 Ccell/melect,

But the active mass of the two similar electrodes mt = 2melec,

It follows that Csp=4Ccell/mt.

We measure the cell capacitance by Ccell= I dt/dV,

Substituting for Ccell, we get finally 

Csp=  4 Idt/ mt dV,

Which is the formula you brought in your question but this time the times is very precisely

Best wishes 

Why is the formula for calculating capactitance in two electrode system for supercapacitors is Cg=4 I/(m*dV / dt) instead Cg= 2I/(m*dv/dt) ?. Available from: https://www.researchgate.net/post/Why_is_the_formula_for_calculating_capactitance_in_two_electrode_system_for_supercapacitors_is_Cg4_I_mdV_dt_instead_Cg_2I_mdv_dt [accessed Apr 11, 2017].

Dear Bidhan Pandit, the equation you've written is fine, however prof. Zekry has given you a precise explantion how to calculate the capacitances of two-electrode systems.

Best wishes

Dear Abdelhalim Zekry sir,

 for accurate calculation, following formula is used nowadays (non-linear charge-discharge curve),

C=2I∫Vdt/m(Vf-Vi)2

then, ‘m’ signifies mass (g) of active electrodes.

can use CD electrode and device as well. isn't that?

plz check the link below.

best regards,

http://www.nature.com/articles/ncomms3923

Dear Adam Moyseowicz sir,

thanks for ur kind reply.

for EDLC based charge-discharge behavior. thats fine.

but I'm concerning about pseudocapacitive electrode/device and with accurate calculation with minimum error.

can check the link as well.

regards,

http://www.nature.com/articles/ncomms3923

I am familiar with this work of prof. Chen. Their calculations for non-linear GCD curves are probably the most accurate in the literature. But someone with more experience in this field should respond to your question.

Best regards,

Adam Moyseowicz

Thank you Adam sir.

Dear Bidhan and Adam,

welcome again,

I will try to derive the relation for nonlinear GCD,

At first we have to make some assumptions:

- We are interested in the electrode capacitance C which has an active material mass= m, then the specific electrode capacitance of the electrode is Cs= C/ m,

This capacitance can be measured directly.

The electrode capacitance is nonlinear such that one defines its value at certain given voltage by  C= dQ/dV, With dQ= Idt, then we have 

C= I dt/ dV,

Since the capacitance is nonlinear it value changes with charging and discharging voltage of the electrode. With constant charging current the voltage can change from an initial voltage Vi to a final voltage to Vf. Then one can define an average capacitance during the charging Cav which can be deified as:

Cav =INTEGRA LI dt/ dV,which can be put in the form by multiplying the numerator and denominator by V :   Cav=  Integral  I V dt / VdV,

Integrating we get from Vi to Vf we obtain:

Cav= 2* I integral Vdt / V^2 from Vi to Vf,

Then Cav= 2* I integral Vdt / (Vf^2- Vi^2),

And the integral is the area under the charging voltage curve from t at V i to t at Vf.

We need  only to divide both sides by the active area mass of the electrode m to get the specific average charging capacitance of the electrode; that is:

Cs av= 2* I integral Vdt / m(Vf^2- Vi^2),

This is the formula which you introduced in your question. Now it is fully derived from the first principles.

Best wishes