We are interested in how a capacitor behaves under a variety of circumstances. In particular, we seek out the current–voltage behavior of a capacitor under different conditions. This section explores these aspects. We begin with the expression for current from a capacitor (see Chapter 6):

(6.6)

This equation describes an ideal capacitor, where the capacitance, C, is a constant and there is no internal resistance. Suppose for a moment that we would like to charge or discharge the capacitor galvanostatically. According to Equation 6.6, a constant-current charge corresponds to a constant value of dV/dt. Therefore, the voltage would increase linearly with time as the capacitor was charged. Similarly, during constant-current discharge, the potential across the capacitor decreases linearly. This behavior highlights an important characteristic of capacitive energy storage, namely, a voltage that changes significantly with capacity. In Figure 11.8, we compare the behavior of an ideal battery to that of a capacitor for discharge at constant current. In most cases, a sloping potential with discharge capacity is undesirable. For instance, the changing potential at constant current leads to a steady decrease in the power available during discharge. On the positive side, one advantage of capacitors is that the state of charge is easy to determine since it is directly proportional to the voltage across the capacitor.

Next we will examine cyclic voltammetry, which is similar to the constant-current behavior just discussed. Recall from Chapter 6 that for cyclic voltammetry the potential is swept in a triangle wave, where ν is the sweep rate in [V·s⁻¹]. This process is illustrated in Figure 11.9, for which the sweep rate is a parameter. The sweep rate for each experiment is

(11.15)

We see immediately that under these conditions the current in the capacitor is constant and directly proportional to the sweep rate. Consequently, if the potential is swept at a constant rate to a specified value and then reversed to return at the same rate to the original potential, the current is as shown in Figure 11.9. On a plot of current versus potential, the response is a rectangle for the ideal capacitor. Note that the direction (sign) of the current changes immediately with the change in sweep direction. This response is different from the response of a faradaic reaction to a reverse in the sweep direction and is explored in greater detail in Problem 11.11. Once again, constant current charge or discharge of a capacitor is associated with a linear change in the potential across the capacitor with respect to time.

Another common operation is a step change in potential from an initial potential, V_i, to a final value, V_f. As seen in the definition of capacitance, the charge, Q, is proportional to the potential across the capacitor. In practice, we cannot move the charges instantaneously; that is, there is some resistance to current flow. Therefore, we will introduce a resistor in series with our capacitor. As before, let the charge on the capacitor be Q, and let Q_f be the final value of the charge equal to CV_f, the charge at steady state. At time equal to zero, the potential applied to the circuit is changed to V_f. After some time, the capacitor will be fully charged or discharged, depending on whether the voltage is increased or decreased, and the current will be zero. The potential drop across the capacitor is simply Q/C. We can use Kirchhoff’s voltage law to get

(11.16)

Equation 11.16 is integrated to give the following relationship for the charge on the capacitor as a function of time:

(11.17)

where Q_i is the initial charge on the capacitor at t = 0, and Q_f is the final value (CV_f). The quantity R_ΩC has units of seconds and is a time constant related to the time required to charge or discharge the capacitor. Since the capacitance is the proportionality constant between charge and voltage, Equation 11.17 can also be written as

(11.18)

The current is obtained by differentiation of Equation 11.17,

(11.19)

The sign of the current is positive for charge and negative for discharge. As evident from Equation 11.19, in response to a step change in potential, the current decreases exponentially to zero with a time constant τ = R_ΩC. Zero current is the endpoint associated with a constant voltage across the capacitor. A large resistance implies that the current is small and, therefore, the time required to charge or discharge the capacitor is large. Similarly, as the capacitance increases, it will also take longer to charge or discharge the capacitor.

ILLUSTRATION 11.3

A 25 cm² electrode has a capacitance of 900 F m⁻² (superficial area). Initially, the capacitor is completely discharged. A step increase in potential results in a 25 A current step. If the resistance is 4 mΩ, how long will it take the potential across the capacitor to reach 95% of its steady-state value?

First, calculate the time constant for an R_ΩC system:

Initially, all of the potential drop is across the resistor as the discharged capacitor offers no resistance to the flow of current. Therefore, ΔV = ΔIR_Ω = 100 mV, which is the voltage drop across the resistor. At steady state, no current flows through the capacitor. From Equation 11.18, the variation of potential with time is

Solving for t gives 27 ms. As can be seen from the solution, the time required does not depend on the magnitude of the voltage step.