How to measure an Ultracapacitor’s Specific Capacitance using a PGSTAT128N

A ideal capacitor makes a rectangular "volt-ammogram" because that's how capacitors work. Look at the equation of current through a capacitor as a function of voltage and you should be able to see this.

First, let's clarify what graph you are talking about especially since you are using a term not usual to electrical engineering. I've heard it before from electrochemistry folks, but it took me a while to realize what they were actually saying. You are slowly sweeping the voltage up from some start point to a end point, then slowly back down to the start point. The X axis is voltage and the Y axis is current. Since you are plotting volts versus amps, you mumble these two together into "voltammogram".

If for example a resistor was being measured, then the rising-voltage part of the plot would be a straight line with current proportional to the voltage according to the resistance. As the voltage was swept back down to the original value, the plot would retrace the same line it went up. Not terribly exciting.

Interesting stuff happens when electrochemical reactions are involved. For example, imagine a battery being tested as opposed to a resistor. The battery is being charged as the voltage increases, then discharged as the voltage decreases. It is not going to follow the same path forward as backward. In fact, the area inside the curve is a rough indication of electrochemical activity. Basically anything with "memory" is going to have a non-zero area inside the voltage low-high-low loop.

Now let's consider a capacitor being measured. The current through a capacitor is proportional to the derivative of its voltage:

  A = F V/s

Where A is current in Amperes, F capacitance in Farads, V electromotive force in Volts, and s time in seconds. So now you should be able to see that if the voltage is increased at a steady rate (V/s fixed), then there will be a constant current. On a voltammogram this means a horizontal line. Now when the voltage is decreased the same thing happens but the sign of the current is flipped. This is again a horizontal line but at some negative current (below 0 on the plot) whereas the first line was above zero. The current switches instantaneously from positive to negative as the voltage is changed from increasing to decreasing. The current suddenly changes but with little or no change in voltage, resulting in vertical lines. Put it all together and you have a box for an ideal capacitor.

Under a given voltage the ideal capacitor obeys the following relationship:

\$Q = CV\$ where Q is the charge, V is the voltage applied and C is the capacitance. If we sweep the voltage with time then we can rewrite this as,

$$\frac{dQ}{dt} = C \frac{dV}{dt} $$

where \$dQ/dt\$ is the current and \$dV/dt\$ is the scan rate. It turns out that an ideal capacitor would be a rectangle during a CV scan. Unfortunately resistances present in the system lead to non-ideal behavior and cause rounding of the corners for our perfect square. The equation above is really just what the paper uses but with different symbols; upon rearrangement we see that we can get $$C = \frac{ \frac{dQ}{dt} }{ \frac{dV}{dt} } = \frac{I}{s} $$

So how do we use this technique to get our capacitance? Easy, we'll take the current starting from 0 current to either a maximum cathodic or anodic current going in one direction (in either a positive direction or a negative direction) and divide that by the sweep rate to get the capacitance of the device. In an ideal case it shouldn't matter which direction you pick, either the anodic or cathodic, but non-idealities may make one current give a higher capacitance. You can tell by looking on figure 3 that it is not perfectly symmetrical about the 0 current horizontal line, therefore the integrated area will be larger in one direction rather than the other.

For the second part of your question, recall that an ideal capacitor should give completely flat horizontal lines once charged, but in this experiment there is a slope, but for the equation used we depend on the capacitance being constant no matter the applied voltage once charged: so the authors have simply forced their data to fit to an equation for the ideal capacitor. They are probably just taking the average of the current for the voltage span of interest and then plugging it into the equation.