Balancing – Theoretical Background

with V_g = V₁ + V₂ as the total voltage. If both capacitance values are equal, the voltage at the terminals of two serial-connected capacitors is equal:

voltage on two capacitors in series with equal capacitance [3]

Thus, the system is balanced and each capacitor is charged at its rated voltage Vr. In the following we may consider the case where C₁ is larger than C₂. With above equations it can be shown that the voltage drop at each terminal is unequal by:

voltage drop on two capacitors in series with unequal capacitance [4]

Using the definition of capacitance C = ∆q/∆V with q as charges at the capacitor interface and V as voltage at the capacitor), the above equation may be rewritten as:

In order to adjust the voltage of each capacitor to V_r = V₁ = V₂ the charge has to be increased at capacitor 1 and decreased at capacitor 2 by the amount of ∆q. Using the definition of electrical current (I = dq/dt) the voltage may be written as:

The current I₁,₂ is interpreted as the electrical current that has to flow for a time period ∆t to equalize this system. The constant current that is required to equalize a voltage difference ∆V in a given time period ∆t is

constant current that is required to equalize a voltage difference on two unequal capacitors in series [11]

Balancing Current and Balancing Time

We may use above equations for the estimation of the current magnitude. In this example we used the full tolerance range of the capacitance, which is 40% (-10%/+30%). Hence, for C_r = 10F we obtain C₁ = 13F and C₂ = 9F. The total voltage of 5.4V provides then a voltage difference ∆V = 0.49V (i.e. at C₂ the voltage drop is V₂ = 3.19V and at C₁ the voltage drop is V₁ = 2.21V). The ∆V ≈ 0.5V is the largest possible imbalance. To illustrate this situation, we use the circuit in Figure 1. The balancing current necessary to balance C₁ and C₂ within 1 sec respectively according to equation [11] are :

Hence, C₁ needs to be charged with I₁ = 6.5A and C₂ needs to be discharged I₂ = 4.5A. The current that has to be provided by the balancing terminal can be calculated with Kirchhoff’s current law. We may consider currents that flow out of the junction as negative and currents that flow into the junction as positive. Since I₁ and I₂ flow out of the junction and the balancing current I into the junction, the balancing current is:

Although the result may vary depending on ∆V and ∆t this example of calculation may show that balancing at the characteristic RC-time requires currents of several amperes. The balancing current, required to balance a strongly imbalanced system of ∆V = 0.5V ( as calculated above) in within ∆t can be estimated with:

So far we have neglected the insulation resistance, which starts to dominate the electrical behavior as soon as the SC is fully charged and the charging current becomes smaller than the leakage current I_leak. Most manufacturers specify a measurement time of 72h at rated voltage V_r to determine I_leak. Under these conditions the capacitor may be simply modeled by an ohmic resistance R_iso = V_r/I_leak. Hence, if a capacitor is fully charged a serial stack of SC may be considered as a stack of serial connected resistances, which constitute a voltage divider.