Capacitor

Introduction

Capacitors are passive electrical components that can store energy in the form of electric charge.

Theory of operation

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On the left you can see a sketch of a plate capacitor. It consists of two metal plates which are separated by an insulator called dielectric, (e.g. Air or ceramic).

A capacitor is charged by storing opposing electrical charges onto the plates and as such creating an electric field. In this field the energy used to charge the capacitor is stored.

Every capacitor has a maximum breakdown voltage (see also dielectric strength) defines the maximum voltage that the capacitor can be applied. When a capacitor is applied higher voltage the the dielectric layer breaks down and the metal plates are short-circuited.

Capacity \( C \)

The capacity of a capacitor defines how much charge it can store at the voltage \( 1   V \) and has the unit farad. The capacity can be determined as follows:

$$ C = \dfrac{Q}{U} \qquad \qquad \mathrm{Units:} \qquad \left[ 1   F = \dfrac{1   C}{1   V} \right] $$

A charged capacitor at the voltage \( U_1 = 5   V \) with the charge \( Q_1 = 5 \cdot 10^{-4} C \), has the following capacity:

$$ C = \dfrac{Q_1}{U_1} = \dfrac{5 \cdot 10^{-4}   C}{5   V} = 10^{-4}   F = 100   \mu F $$
Calculation by plate area and distance

The capacity of a plate capacitor is dependent of the surface \( A \) of the plates and their distance \( d \). The greater \( A \) and the smaller \( d \), the greater is the capacity \( C \).

Furthermore it is important for the capacity which material is used for the dielectric layer. The relative permittivity \( \epsilon_r \), indicates the factor by which the storage capacity of a capacitor increases by the use of the dielectric. For air it is \( \epsilon_r = 1 \). Special ceramics however increase the storage capacity of a capacitor by the factor \( 100 - 10 \, 000 \).

$$ C = \epsilon_0 \cdot \epsilon_r \cdot \dfrac{A}{d} $$ \( \epsilon_0 \) = permittivity of free space / electric constant, \( \epsilon_r \) = relative permittivity, \( A \) = area, \( d \) = distance

Relative permittivity of some materials:

Material \( \epsilon_r \) Material \( \epsilon_r \)
Amber 2,8 Polystyrene 2,6
Glass 5 ... 16 Porcelain 4,5 ... 6,5
Synthetic resin bonded paper 3,5 ... 5 Transformer oil 2,5
Sepcial ceramics 100 ... 10.000 Vacuum 1
Air 1,0006 Water 81
Paraffin 2,3

Field energy \( E \)

During charging electric charge is added to the capacitor plates. This creates a electric field which energy is increased by each charge change \( \Delta Q \):

$$ \Delta E_i = \Delta Q \cdot U_i $$

The addition of these energies gives the field energy:

$$ E = \dfrac{1}{2} \cdot Q \cdot U $$
and with \( Q = C \cdot U \)

$$ E = \dfrac{1}{2} \cdot C \cdot U^2 $$

Charging / Discharging

The following experiment shows a voltage source, a resistor, a capacitor and a switch that controls the connection to the voltage source. Dependant on the switch the capacitor is charged by the voltage source or discharged. (Start the simulation and click on the switch to charge/discharge the capacitor.)

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When charging the voltage increases rapidly but then slows down. That is because the field that is created by the capacitor works against the charging. As such with increasing voltage of the capacitor more energy is needed for further voltage increase.

After charging the whole energy is saved in the electric field. This energy can be used by discharging the capacitor.

When discharging the voltage descreases rapidly but then slows down. That is because the field of the capacitor is getting weaker during the discharge process.

Calculations

When charging the capacitor the voltage and current are described by the following exponential functions:

$$ U_{\mathrm{C}} (t) = U_0 \cdot \left(1 - e^{- \frac{t}{\tau}} \right) = U_0 \cdot \left( 1 - e^{- \frac{t}{R_{\mathrm{C}} \cdot C}} \right) $$ $$ I_{\mathrm{C}} (t) = \dfrac{U_0}{R_{\mathrm{C}}} \cdot e^{- \frac{t}{\tau}} = I_0 \cdot e^{- \frac{t}{R_{\mathrm{C}} \cdot C}} $$
\( e \approx 2,71 \) (Euler's number)

When discharging the following equations hold:

$$ U_{\mathrm{C}} (t) = U_0 \cdot e^{- \frac{t}{\tau}} = U_0 \cdot e^{- \frac{t}{R_{\mathrm{C}} \cdot C}} $$ $$ I_{\mathrm{C}} (t) = -\dfrac{U_0}{R_{\mathrm{C}}} \cdot e^{- \frac{t}{\tau}} = -I_0 \cdot e^{- \frac{t}{R_{\mathrm{C}} \cdot C}} $$
\( e \approx 2,71 \) (Euler's number)
Time constant \( \tau \)

The time constant \( \tau \) is determined as follows:

$$ \tau = R_{\mathrm{C}} \cdot C $$

For the animation above:

$$ \tau = R_{\mathrm{C}} \cdot C = 1000   k \omega \cdot 2   \mu F = 2 s $$

  \( t \) \( U_{\mathrm{C}} \)
\( 1 \cdot \tau \) \( 0,632 \cdot U_0 \)
\( 2 \cdot \tau \) \( 0,865 \cdot U_0 \)
\( 3 \cdot \tau \) \( 0,950 \cdot U_0 \)
\( 4 \cdot \tau \) \( 0,982 \cdot U_0 \)
\( 5 \cdot \tau \) \( 0,993 \cdot U_0 \)

After charge time of \( 1 \cdot \tau \) the capacitor has the voltage \( 0,632 \cdot U_0 \) and after charge time of about \( 0,69 \cdot \tau \) it has reached 50% of its final voltage. After a charge time of \( t_{\mathrm{C}} \approx 5 \tau \) its charged 99 % that's why its practically assumed that it is charged fully after that time.

The capacitor in the animation above will be charged after \( t_{\mathrm{C}} \approx 5 \cdot \tau = 5 \cdot 2   s = 10   s \) which can be seen on the diagram.

Determining the charge/discharge time for a specific voltage

To determining the charge/discharge time for a specific voltage the charge/discharge formulas are solved for \( t \).

\begin{aligned} U &= U_0 \cdot \left(1 - e^{- \frac{t}{\tau}} \right) \\ & \\ \dfrac{U}{U_0} &= 1 - e^{- \frac{t}{\tau}} \\ & \\ \dfrac{U}{U_0} - 1 &= - e^{- \frac{t}{\tau}} \\ & \\ 1 - \dfrac{U}{U_0} &= e^{- \frac{t}{\tau}} \\ & \\ \ln \left( 1 - \dfrac{U}{U_0} \right) &= - \frac{t}{\tau} \\ & \\ - \tau \cdot \ln \left( 1 - \dfrac{U}{U_0} \right) &= t \\ \end{aligned}
\( e \approx 2,71 \) (Euler's number)

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