5.1 Capacitor Construction & Connections

Why This Matters for the Exam

Capacitors are fundamental in aircraft electrical systems for filtering, timing, tuning, voltage regulation, power factor correction, and decoupling. On the EASA Part-66 exam, questions frequently test factors affecting capacitance, dielectric constants, calculations for series and parallel configurations, and working voltage ratings. Underestimating voltage ratings or incorrectly calculating circuit capacitance represents a serious safety hazard in aircraft maintenance.

Capacitor Operation and Function

A capacitor is a passive electrical component designed to store energy in an electrostatic field. Unlike an inductor, which stores energy in a magnetic field created by current flow, a capacitor stores energy by accumulating electric charge on its plates.

When a direct current (DC) voltage source is connected across a capacitor, a transient current flows as charge redistributes. To understand this process, it is useful to compare conventional flow and electron flow. In conventional flow theory (which assumes positive charge carriers move from positive to negative), positive charge is described as leaving the positive terminal of the source and accumulating on the plate connected to it, while positive charge leaves the opposite plate. In physical reality, governed by electron flow theory, actual current is the movement of free electrons (which carry a negative charge). Electrons flow from the negative terminal of the source and accumulate on the capacitor plate connected to it, giving it a net negative charge. Simultaneously, an equal number of electrons are repelled from the opposite plate and flow back into the positive terminal of the voltage source, leaving that plate with a net positive charge (a deficiency of electrons). This means no electrons physically cross the dielectric spacer; instead, the current consists of charge displacement in the external circuit, establishing an electrostatic field across the dielectric. This displacement current ceases when the potential difference across the plates equals the source voltage.

The fundamental property of a capacitor is its capacitance ($C$), which is defined as the amount of electric charge ($Q$, measured in Coulombs) stored per unit of potential difference ($V$, measured in Volts): $C = \frac{Q}{V}$

The SI unit of capacitance is the Farad (F), named after Michael Faraday. One Farad is an extremely large amount of capacitance; a one-Farad capacitor charged to one Volt stores one Coulomb of charge. In practical aviation electronics, capacitance is measured in smaller sub-units:

• Microfarads ($\mu$F) = $10^{-6}$ F
• Nanofarads (nF) = $10^{-9}$ F
• Picofarads (pF) = $10^{-12}$ F

Factors Determining Capacitance

The physical construction of a capacitor dictates its capacitance. A basic capacitor consists of two conducting plates separated by an insulating material called a dielectric. Three primary factors determine the capacitance of a parallel-plate capacitor:

1. Area of the Plates (A): The capacitance is directly proportional to the physical surface area of the plates. Larger plates allow more electrons to accumulate, increasing the charge-storing capability.
2. Distance Between the Plates (d): The capacitance is inversely proportional to the distance separating the plates. As the plates are brought closer together, the electrostatic attraction between the accumulated positive and negative charges increases, allowing more charge to be stored at a given voltage.
3. Permittivity of the Dielectric ($\varepsilon$): The dielectric material acts to reduce the effective electrostatic field strength between the plates for a given amount of charge, thereby allowing more charge to be deposited. The total permittivity ($\varepsilon$) is the product of the permittivity of free space ($\varepsilon_0$, which is a constant equal to $8.854 \times 10^{-12}$ F/m) and the relative permittivity ($\varepsilon_r$, or dielectric constant) of the material: $\varepsilon = \varepsilon_0 \cdot \varepsilon_r$

These relationships are combined in the parallel-plate capacitance formula: $C = \varepsilon_0 \cdot \varepsilon_r \cdot \frac{A}{d}$

Where:

• $C$ is the capacitance in Farads (F).
• $\varepsilon_0$ is the permittivity of free space ($8.854 \times 10^{-12}$ F/m).
• $\varepsilon_r$ is the relative permittivity of the dielectric (dimensionless).
• $A$ is the area of overlap of the plates in square meters ($m^2$).
• $d$ is the distance between the plates in meters (m).

Relative Permittivity and Dielectric Strength

Different dielectric materials have different insulating and polar properties. The relative permittivity ($\varepsilon_r$) compares the material's permittivity to that of a vacuum (where $\varepsilon_r = 1.0$).

Additionally, each dielectric has a dielectric strength, which is the maximum electric field the material can withstand without undergoing electrical breakdown (arcing or shorting). If the voltage across the capacitor is too high, the electrostatic forces rip electrons from the dielectric atoms, causing a sudden surge of current that permanently destroys the capacitor.

Working Voltage Ratings (WVDC)

Every capacitor is stamped with a Working Voltage DC (WVDC) rating, representing the maximum continuous DC voltage that can be safely applied across the capacitor without risking dielectric breakdown.

In aviation maintenance, safety margins dictate that a capacitor should not operate at more than 50% to 60% of its rated working voltage under normal conditions to account for voltage spikes, temperature fluctuations, and aging.

Furthermore, the working voltage rating decreases as the operating temperature increases. A capacitor rated at 400 WVDC at $85^{\circ}$C might need to be derated to 250 WVDC if operated in an engine nacelle environment at $125^{\circ}$C.

AC vs. DC Voltage Ratings: In AC circuits, the dielectric must withstand the peak voltage, which is $1.414 \times \text{RMS}$ voltage. Additionally, the continuous reversal of polarization in an AC circuit causes dielectric heating, which weakens the physical structure of the dielectric. Therefore, a capacitor rated at 400 WVDC is typically only safe for AC voltages up to approximately 150 VAC RMS.

Capacitors in Series

Connecting capacitors in series places them in a single line, which effectively increases the total dielectric thickness. This setup reduces the total capacitance but increases the overall voltage rating of the network.

The formula for the total capacitance ($C_{total}$) of capacitors in series is: $\frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + \dots + \frac{1}{C_n}$

For exactly two capacitors in series, this simplifies to the product-over-sum formula: $C_{total} = \frac{C_1 \cdot C_2}{C_1 + C_2}$

• Charge in Series: The electric charge ($Q$) stored on each capacitor in a series network is identical, regardless of their individual capacitance values. $Q_{total} = Q_1 = Q_2 = Q_3$
• Voltage in Series: The total voltage ($V_{total}$) across a series network is divided among the capacitors. The voltage drop across any individual capacitor is inversely proportional to its capacitance. The smallest capacitor will drop the largest portion of the voltage: $V_x = V_{total} \cdot \frac{C_{total}}{C_x}$

Capacitors in Parallel

Connecting capacitors in parallel places their plates side-by-side, which effectively increases the total plate surface area. This setup increases the total capacitance, while the voltage across each capacitor remains equal.

The formula for the total capacitance ($C_{total}$) of capacitors in parallel is: $C_{total} = C_1 + C_2 + \dots + C_n$

• Voltage in Parallel: The voltage across each parallel capacitor is equal to the source voltage: $V_{total} = V_1 = V_2 = V_3$
• Charge in Parallel: The total charge stored in the network is the sum of the charges stored on each capacitor: $Q_{total} = Q_1 + Q_2 + Q_3$

Worked Calculation Scenarios

Scenario 1: Complex Series-Parallel Network An aircraft radio filter uses three capacitors: $C_1 = 12\,\mu$F (rated at 100 V), $C_2 = 6\,\mu$F (rated at 150 V), and $C_3 = 8\,\mu$F (rated at 200 V). $C_1$ and $C_2$ are connected in series, and this series combination is then connected in parallel with $C_3$. If the entire circuit is connected across a 120 VDC supply, calculate the total capacitance and determine if any capacitor exceeds its working voltage rating.

1. Calculate the series combination of $C_1$ and $C_2$ ($C_{12}$): $C_{12} = \frac{C_1 \cdot C_2}{C_1 + C_2} = \frac{12 \cdot 6}{12 + 6} = \frac{72}{18} = 4\,\mu\text{F}$
2. Calculate the total capacitance ($C_{total}$): Since $C_{12}$ is in parallel with $C_3$, we add them: $C_{total} = C_{12} + C_3 = 4\,\mu\text{F} + 8\,\mu\text{F} = 12\,\mu\text{F}$
3. Analyze voltage drops:

• The parallel branch containing $C_3$ is directly connected across the 120 VDC supply. Therefore, the voltage across $C_3$ is $120$ V. Since $C_3$ is rated at 200 V, it is operating safely.
• The series branch containing $C_1$ and $C_2$ has a total voltage of $120$ V across it. The voltage drops across $C_1$ and $C_2$ are calculated as: $V_1 = V_{total} \cdot \frac{C_{12}}{C_1} = 120 \cdot \frac{4}{12} = 40\,\text{V}$ $V_2 = V_{total} \cdot \frac{C_{12}}{C_2} = 120 \cdot \frac{4}{6} = 80\,\text{V}$
• Now, check the ratings: $C_1$ drops 40 V (rated at 100 V - Safe). $C_2$ drops 80 V (rated at 150 V - Safe). None of the capacitors exceed their working voltage ratings.

Scenario 2: Capacitance Calculations from Dimensions Calculate the capacitance of a parallel-plate capacitor with a plate area of $0.02\text{ m}^2$, separated by $0.5\text{ mm}$ ($5 \times 10^{-4}\text{ m}$) of ceramic dielectric with a relative permittivity of 120. $C = \varepsilon_0 \cdot \varepsilon_r \cdot \frac{A}{d}$ $C = (8.854 \times 10^{-12}\text{ F/m}) \cdot 120 \cdot \frac{0.02\text{ m}^2}{5 \times 10^{-4}\text{ m}}$ $C = 1.06248 \times 10^{-9} \cdot \frac{0.02}{0.0005} = 1.06248 \times 10^{-9} \cdot 40 = 42.499 \times 10^{-9}\text{ F} \approx 42.5\,\text{nF}$

Common Exam Traps

• Confusing Resistor and Capacitor Formulas: This is the most common error. Remember that capacitors in parallel combine like resistors in series, and capacitors in series combine like resistors in parallel.
• Series Voltage Distribution: Technicians often assume that identical-looking capacitors share voltage equally in series. In reality, the voltage distributes according to the capacitance values. If one capacitor has suffered degradation and its capacitance drops, its voltage drop will rise, potentially exceeding its WVDC rating and initiating a domino-effect failure.
• AC Voltage Peak Neglect: Always remember that AC voltages are quoted in RMS (Root Mean Square). A 115 VAC aircraft bus has a peak voltage of $115 \times 1.414 = 162.6$ V. A capacitor rated at 150 WVDC will quickly fail if connected to this bus, even though 115 is less than 150.