AJ Designer

Capacitor Design Calculator

Capacitance equals charge divided by voltage

Solution

—
Share:

Capacitance Equation

The basic capacitance relationship ties capacitance to charge and voltage. A capacitor stores electrical energy by holding opposite charges on two conductive plates separated by an insulator.

C = Q / V

Stored Energy (C, V)

Stored energy depends on capacitance and voltage squared. Doubling the voltage quadruples the stored energy.

U = ½CV²

Parallel Plate Capacitor

Two flat conducting plates of area A separated by distance d with a dielectric of permittivity ε form the simplest capacitor. Larger plates and thinner gaps increase capacitance.

C = εA / d

Cylindrical Capacitor

A cylindrical capacitor consists of two coaxial conductors used for coaxial cables and cylindrical sensor geometries.

C = 2πεL / ln(b/a)

How It Works

A capacitor stores electrical energy by holding opposite charges on two conductive plates separated by an insulator. The basic relationship C = Q/V ties capacitance to charge and voltage. For specific geometries, dedicated formulas (parallel plate, cylindrical) account for physical dimensions and dielectric properties. Three stored-energy forms (U = ½CV², U = Q²/2C, U = QV/2) let you solve with whichever pair of variables you know.

Example Problem

A 470 µF capacitor is charged to 25 V. How much energy does it store?

  1. Identify the knowns. Capacitance C = 470 µF and voltage V = 25 V across the capacitor's terminals.
  2. Identify what we're solving for. We want the stored electrical energy U in joules.
  3. Write the energy formula: U = ½ × C × V². Energy scales linearly with capacitance and with the square of voltage.
  4. Convert capacitance to SI units: C = 470 × 10⁻⁶ F = 4.7 × 10⁻⁴ F. Skipping this step inflates the answer by a factor of one million.
  5. Substitute the known values: U = ½ × (4.7 × 10⁻⁴ F) × (25 V)² = ½ × 4.7 × 10⁻⁴ × 625.
  6. Simplify the arithmetic: ½ × 4.7 × 10⁻⁴ × 625 = 0.146875 J ≈ **0.147 J (about 147 mJ)** — enough energy to deliver a sharp jolt if discharged through a low-resistance path.

If the same capacitor held 0.01 coulombs of charge, the voltage would be Q/C = 0.01/0.00047 ≈ 21.3 V.

When to Use Each Variable

  • Solve for Capacitance (C = Q/V) — when you know the charge stored and the voltage across the capacitor, e.g., measuring capacitance from a charge-discharge test.
  • Solve for Stored Energy (U = 1/2 CV^2) — when you know the capacitance and voltage and need the energy stored, e.g., sizing a capacitor bank for a camera flash.
  • Solve for Parallel Plate Capacitance — when you know the plate area, gap distance, and dielectric permittivity, e.g., designing a MEMS sensor.
  • Solve for Cylindrical Capacitance — when you know the coaxial conductor dimensions and dielectric, e.g., calculating capacitance per meter of coaxial cable.

Key Concepts

Capacitance is the ability to store electric charge per unit voltage. Three equivalent energy formulas (U = 1/2 CV^2, U = Q^2/2C, U = QV/2) let you work with whichever pair of variables you know. Physical capacitance depends on geometry: for parallel plates, it scales with plate area and inversely with gap distance; for cylindrical capacitors, it scales with length and inversely with the logarithm of the radius ratio. The dielectric material between conductors multiplies capacitance by its relative permittivity.

Applications

  • Power electronics: sizing filter capacitors for DC power supplies to smooth voltage ripple
  • Energy storage: calculating the energy capacity of supercapacitor banks for regenerative braking systems
  • Sensor design: using parallel plate capacitance changes to detect displacement in MEMS accelerometers and pressure sensors
  • RF engineering: determining the capacitance of coaxial cables and transmission lines for impedance matching
  • Timing circuits: selecting capacitor values for RC time constants in oscillators and delay circuits

Common Mistakes

  • Forgetting to convert microfarads to farads — 1 uF = 1e-6 F; entering 470 instead of 0.00047 produces energy values that are a million times too large
  • Confusing series and parallel capacitor combinations — parallel capacitances add directly, but series reciprocals add; the total series capacitance is always less than the smallest individual capacitor
  • Using relative permittivity where absolute permittivity is needed — the parallel plate formula uses absolute permittivity (epsilon = epsilon_r times epsilon_0); omitting epsilon_0 gives values that are 8.85e-12 times too large
  • Assuming voltage rating equals maximum charge — exceeding the rated voltage can cause dielectric breakdown and capacitor failure even if the calculated energy seems reasonable

Frequently Asked Questions

What is the difference between a farad and a microfarad?

One farad is an enormous amount of capacitance. Most practical capacitors are measured in microfarads (1 µF = 10⁻⁶ F), nanofarads (10⁻⁹ F), or picofarads (10⁻¹² F). Supercapacitors can reach several thousand farads.

How do capacitors combine in series vs parallel?

In parallel, capacitances add directly: Ctotal = C1 + C2. In series, reciprocals add: 1/Ctotal = 1/C1 + 1/C2, so the total is always less than the smallest individual capacitor.

Why does stored energy depend on voltage squared?

As voltage rises, each additional increment of charge requires more work against the growing electric field. The result is a quadratic relationship: doubling the voltage quadruples the stored energy.

What is a parallel plate capacitor?

A parallel plate capacitor is two flat conductive plates separated by an insulating material (dielectric). Its capacitance depends on the plate area, the gap between plates, and the dielectric’s permittivity: C = εA/d. It is the most common theoretical model for understanding capacitance.

What are common uses of capacitors?

Capacitors are used for power supply filtering, signal coupling, timing circuits (RC), motor starting, camera flashes, defibrillators, and power factor correction. Decoupling caps near ICs stabilize voltage and reduce electromagnetic interference.

Which energy formula should I use — ½CV², Q²/2C, or QV/2?

All three give the same answer; pick the one matching the variables you have. Use U = ½CV² when you know capacitance and voltage, U = Q²/2C when you know charge and capacitance, and U = QV/2 when you know charge and voltage. They are algebraic rearrangements of each other via Q = CV.

How does the dielectric material change capacitance?

A dielectric multiplies the vacuum capacitance by its relative permittivity εᵣ. Air is ≈ 1, paper ≈ 3.5, polyester ≈ 3.3, ceramic X7R ≈ 1,000–4,000, and tantalum oxide ≈ 25. Higher εᵣ means more capacitance for the same plate area and gap, which is how miniature ceramic and electrolytic capacitors pack microfarads into millimeter-scale parts.

What is ESR and why does it matter?

Equivalent series resistance (ESR) is the small parasitic resistance in series with an ideal capacitance — caused by leads, foils, and electrolyte conduction losses. Higher ESR means more self-heating under ripple current and less effective filtering at high frequencies. Power-supply electrolytics rated for switching converters typically advertise low-ESR construction to limit ripple losses.

Worked Examples

Power Electronics

How much charge does a 470 μF smoothing capacitor hold at 25 V?

Aluminum-electrolytic capacitors on a DC bus typically run at a derated voltage well below their 35 V or 50 V rating. Find the stored charge for a 470 μF cap sitting at a steady 25 V.

  • Knowns: C = 470 μF = 0.000470 F, V = 25 V
  • Formula: Q = C × V
  • Q = 0.000470 × 25

Q = 0.01175 C (11.75 mC)

This is the static charge at the operating voltage; ripple current and equivalent series resistance still matter for thermal sizing and lifetime, which Q alone does not capture.

Photography

How much energy does a 800 μF camera-flash capacitor store at 300 V?

A disposable-camera xenon flash typically charges an 800 μF capacitor to about 300 V before dumping the energy through the flash tube. Find the stored energy that drives the bulb.

  • Knowns: C = 800 μF = 0.0008 F, V = 300 V
  • Formula: U = ½ × C × V²
  • U = ½ × 0.0008 × 300²
  • U = ½ × 0.0008 × 90,000

U = 36 J

Stored energy scales with the square of voltage, so derating from 300 V down to 240 V (e.g., as the cap ages and ESR drifts) drops the available flash energy from 36 J to 23 J — a noticeable underexposure.

Sensor Design

What capacitance does a 1 cm² air-gap touch sensor with a 0.1 mm gap have?

A capacitive touch sensor uses two parallel copper pads of 1 cm² area separated by a 0.1 mm air gap. The vacuum permittivity ε₀ = 8.854 × 10⁻¹² F/m sets the baseline capacitance before any finger or stylus changes the dielectric.

  • Knowns: ε = 8.854 × 10⁻¹² F/m, A = 1 cm² = 1 × 10⁻⁴ m², d = 0.1 mm = 1 × 10⁻⁴ m
  • Formula: C = ε × A / d
  • C = (8.854 × 10⁻¹²) × (1 × 10⁻⁴) / (1 × 10⁻⁴)

C ≈ 8.854 pF

Touch sensors detect the small permittivity change when a fingertip enters the fringing field; the baseline pF-range capacitance is what the sensor IC measures against to set its threshold.

Capacitor Formulas

The capacitor calculator covers five related equations spanning basic capacitance, three equivalent stored-energy forms, and two common geometries:

C = Q / VBasic capacitance (charge per volt)
U = ½ × C × V² = Q² / (2C) = Q × V / 2Stored energy (three equivalent forms)
C = ε × A / dParallel-plate capacitor
C = 2π × ε × L / ln(b / a)Cylindrical (coaxial) capacitor

Where:

  • C — capacitance, in farads (F)
  • Q — stored charge, in coulombs (C)
  • V — voltage across the capacitor, in volts (V)
  • U — stored electrical energy, in joules (J)
  • ε — absolute permittivity of the dielectric, in F/m (= εr × ε₀, where ε₀ ≈ 8.854×10⁻¹² F/m)
  • A — plate area, in square meters (m²)
  • d — separation between parallel plates, in meters (m)
  • L — axial length of a cylindrical capacitor, in meters (m)
  • a, b — inner and outer conductor diameters of a coaxial cylinder, in meters (m)

The basic relation C = Q/V holds for any two-terminal capacitor. The parallel-plate and cylindrical forms assume the dielectric fills the gap uniformly and edge effects (fringing fields) are small relative to the active region. At very high frequency, all real capacitors deviate from these ideal formulas due to ESR and lead inductance — see the FAQ for how to account for parasitics in RF and switching designs.

Related Calculators

Related Sites