Capacitor Elements
A capacitor element forms the basic building block of a capacitor unit. It has two metal electrodes. A dielectric medium separates these electrodes. Engineers connect multiple elements in series or parallel to form a capacitor unit’s internal circuit. They join several units in series to match the system’s rated voltage. The required MVAR rating determines the total number of units in the capacitor bank.
Each capacitor element uses a dielectric material, such as polypropylene or paper film, placed between two aluminum foil electrodes. Initially, manufacturers used thin metal sheets; however, they later switched to aluminum foil because it offered better electrical and mechanical properties. Today, aluminum foil has completely replaced tin sheets and now dominates capacitor construction. Furthermore, manufacturers stack and fold alternating layers of aluminum foil and dielectric film to create a compact, efficient, and space-saving element.
It is important to note this point. A capacitor element usually contains multiple dielectric sheets. Manufacturers place these sheets between the aluminum foil layers. A single dielectric sheet may be sufficient under ideal conditions. However, engineers use multiple sheets to reduce the risk of short circuits. Such faults can occur if conductive impurities or particles become trapped within the dielectric layers. Therefore, to enhance reliability and ensure better protection, it is standard practice to use more than one dielectric film.
Types of Capacitor Element
Three main types of capacitor elements are available in the market: internally fused, externally fused, and fuse-less capacitor elements.
Internally Fused Capacitor Elements
A capacitor element often has a series-connected fuse. In some designs, engineers connect several elements in parallel to form a group and add one fuse in series with that group. When the fuse is inside the capacitor casing, the element becomes an internally fused capacitor element.
Externally Fused Capacitor Elements
Engineers connect several capacitor elements in parallel to form a group that handles the unit’s current.
They protect the group with a series fuse. When the fuse is placed outside the casing, the setup becomes an externally fused capacitor unit, and the elements are externally fused capacitor elements. In this case, the fuse protects the entire group and is externally accessible.
Fuse Less Capacitor Elements
We use fuses in capacitor elements to isolate faulty components. When a capacitor element short-circuits due to dielectric failure, the fuse blows and disconnects the damaged part from the rest of the unit. This keeps the capacitor bank operational and avoids immediate maintenance.
Under normal conditions, each element or group has its own fuse to maintain service continuity and improve reliability. However, this setup increases manufacturing cost and complexity because it needs extra space for fuses inside the unit. To overcome these issues, designers developed fuse-less capacitor elements that eliminate the need for individual fuses.
In fuse-less designs, engineers connect all capacitor elements in series to form a vertical column without fuses.
They set the number of series elements based on the unit’s voltage rating. To handle more current, they connect multiple columns in parallel. Each element has a specific voltage and current rating, and engineers combine them in series-parallel to achieve the required overall ratings.
Internal Fuses
Internal fuses protect capacitor units, especially in shunt capacitor banks. They isolate faulty elements and keep the unit operating within the bank. Usually, each capacitor element has a fuse in series, though sometimes a single fuse protects a small group of elements.
These fused elements or groups connect in parallel, and several parallel groups connect in series to achieve the required voltage and capacitance ratings of the capacitor unit.
If a fault occurs in any capacitor element due to dielectric failure, the internal fuse associated with that element or with the group containing that element will blow to disconnect the faulty part from the unit. This prevents the failure of the entire capacitor unit or bank. In this way, internal fuses reduce the risk of catastrophic failures such as explosions or fires, enhancing the safety of the installation. Since an internal fuse isolates the faulty element from the unit, it ensures continuity of service with only a slight reduction in capacitance. Thus, internal fuses help minimize downtime and reduce maintenance requirements.
Engineers place internal fuses inside the unit casing and connect them in series with each element or group.
They design a specific internal layout to fit these fuses within the unit. This increases the design complexity and requires additional space within the casing or container of the unit. However, this is justified by the inherent benefit of isolating faulty elements without interrupting overall service continuity.
Discharge Device
When a capacitor bank is switched off and isolated from the live system, the energy is still stored in each unit of the capacitor bank. It is essential to discharge this energy in view of safety. This is done by a discharge device connected across the terminals of the capacitor unit, inside the casing of the capacitor unit. This discharge device is essentially a resistor.
Standards require capacitor banks below 52 kV to discharge to under 50 volts within 5 minutes.
For units above 52 kV, the voltage must drop to 75 volts or less within 10 minutes after disconnection.
The value of the internal discharge device, i.e., the resistance of the discharge resistor of the capacitor unit, is calculated using the following formula: \[ R = \frac{t}{C \cdot \ln\left(\frac{\sqrt{2}U_N}{U_R}\right)} \]
R = (discharge resistance),
t = discharge time in seconds,
C = capacitance in farads,
\( U_N \) = rated voltage of the capacitor unit in volts, and
\( U_R \) = permissible residual voltage.
Residual voltage is the voltage that remains in the capacitor unit after it has been discharged for 5 or 10 minutes through the discharge resistor, as per the definition.
For example, if C = 10 microfarads, \( U_N \) = 10 kV, \( U_R \) = 50 volts, and t = 5 minutes or 300 seconds, the resistance of discharge device will be calculated to ensure the voltage drops to 50 volts within 5 minutes as follows.\[ R = \frac{300}{10\times 10^{- 6} \cdot \ln\left(\frac{10,000\sqrt{2}}{50}\right)} \approx 5.32 \, \text{M}\Omega\]