# Balancing – Theoretical Background

- Posted by doEEEt Media Group
- On February 9, 2023
- 0

**Imbalance of Serial Connected Supercapacitors**

A parallel connection of an R-C unit and an insulation resistance may model a capacitor. We neglect the insulation resistance and consider a series stack of two capacitors with capacities C1 and C2 – see Figure 1.

The conserved quantity in such a stack is the condensed charge q at the capacitor, i.e. at its internal interfaces. Using the conservation of charge V_{1,2} = q/C_{1,2,} the voltage drop over each capacitor is:

with V_{g} = V_{1 }+ V_{2} as the total voltage. If both capacitance values are equal, the voltage at the terminals of two serial-connected capacitors is equal:

Thus, the system is balanced and each capacitor is charged at its rated voltage Vr. In the following we may consider the case where C_{1} is larger than C_{2}. With above equations it can be shown that the voltage drop at each terminal is unequal by:

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} = V_{1} = V_{2} 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 I_{1},_{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

**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 C_{1} = 13F and C_{2} = 9F. The total voltage of 5.4V provides then a voltage difference ∆V = 0.49V (i.e. at C_{2} the voltage drop is V_{2} = 3.19V and at C_{1} the voltage drop is V_{1} = 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 C_{1} and C_{2} within 1 sec respectively according to equation [11] are :

Hence, C_{1} needs to be charged with I_{1} = 6.5A and C_{2} needs to be discharged I_{2} = 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 I_{1} and I_{2} 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.

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