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Elect principles capacitance

The document describes the basic operation and construction of a capacitor. A capacitor consists of two conducting parallel plates separated by an insulating material called the dielectric. When a switch is closed, electrons are repelled from the negative plate and attracted to the positive plate, building up opposite charges on each plate. The stored charge remains when the switch is opened due to the conduction block of the dielectric. Capacitance is directly proportional to plate area and inversely proportional to plate separation distance. Energy is stored in the electric field generated by the charge separation in a capacitor.

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The Capacitor A capacitor can be made up of two conducting parallel plates separated by an insulating material called the dielectric. When switch (S) is closed electrons on the positive plate are repelled by the negative charge on negative and attracted to the positive supply terminal causing a deficiency of electrons in the left plate this becomes positively charged. Since electrons cannot pass through the dielectric insulation an electric field is set up within the dielectric causing a charge to exist across the plates. The charge builds up until the potential difference between the plates equals the source (E). - + + + + pd = E S + E -
The Capacitor If the switch is opened the conduction path between the plates is removed, the surplus electrons remain on the right plate (negative plate) while the left plate remains deficient in electrons (positive plate). The parallel plates store electric charge causing a potential difference to exist between the two plates. The stored charge will remain indefinitely until the capacitor is discharged. The charge will reduce in time due to leakage of electrons through the dielectric. - + + + + pd = E S + E -
Construction of a Capacitor silver (plates) ceramic (dielectric ) Circuit Symbols Non polarised (can be connected either way round) + Polarised Electrolytic (must be connected as shown)
Capacitance Capacitance is the ability of a capacitor to store electric charge and is, a. directly proportional to its plate area. (capacitance increases with plate area) b. inversely proportional to the distance between the plates. (capacitance decreases with dielectric thickness) c. dependent on the dielectric material used to separate the plates. (various materials are used to obtain the required capacity and dielectric strength)
The Dielectric The dielectric is the material used to separate the plates of a capacitor. The dielectric constant (k or εr ) also called the relative permittivity of a material is the ratio between the permittivity of a material (ε) and the permittivity of free space (ε0 ). Permittivity relates to a material's ability to "permit" an electric field. The dielectric strength of a material is the maximum electric field strenth that it can withstand without breaking down. k or εr = ε ε0 K or εr Strength (kV/mm) 1.0 3 7500 3 Polystyrene 2.6 16 Paper 2.5 20 Mylar 3.0 24 Glass 6 120 Mica 5 200 Material Air Barium-strontium titanate
Capacitance of a Capacitor The capacitance (C) of a capacitor is dependent upon the plate area the distance between the plates in metres and its dielectric constant. Knowing these parameters it is possible to determine the capacitance of any parallel plate capacitor. The unit of capacitance is the Farad (F). C = kε0 A Farads d where k is the dielectric constant ε0 is permittivity of free space (8.85 x 10 -12) F/m A is the area of each plate in m2 d is the distance between each plates in m.
Capacitance of a Capacitor Activity 1 Determine the capacitance of a 2 plate capacitor having a plate dimensions of 2m x 10cm using 0.20mm of paper as its dielectric material. If the dielectric material was changed to mica what would you expect to happen to the capacitance. If the plate area was to double what would be the effect on the capacitance. Activity 2 Choose a suitable dielectric material and determine the plate area for a 25uF 2 plate capacitor having a 0.1mm thickness of dielectric material, for the smallest case size of capacitor. Also state the working voltage, assume 100% margin for safety.
Capacitance and Charge A capacitor is a component that stores electric charge. The symbol for capacitance is “C”. The unit of capacitance is the Farad “F”. The unit for charge is the coulomb, “C”. The symbol for charge is “Q”. Charge, Q = C x V coulombs where C is in Farads Charge, Q = I x t where t is in seconds also coulombs
Capacitors in Parallel Capacitors connected in parallel effectively increases the plate area and therefore increases the capacitance. The total capacitance of capacitors connected in parallel is the sum of all the individual capacitors. CT = C1 + C2 The total charge drawn from a supply is the sum of the charge stored in each capacitor. QT = QC1 + QC2 + - E C1 C2 = + - E CT
Capacitors in Parallel Activity 1. Three capacitors 10uF, 25uF and 50uF are connected in parallel to a 18 volt dc supply. Determine a) the total capacitance, b) the charge on each capacitor and c) the charge drawn from the supply. 2. An electronic circuit operating from a 12 volt dc supply requires a charge of 200uC. If we already have a 12uF capacitor, determine the capacitance that we must connect in parallel with this to obtain the required charge. It is a good idea to sketch these circuits before starting to calculate your results.
Capacitors in Series The positive charge on the top plate of C1 attracts electrons from the top plate of C2 to the bottom plate of C1. The negative charge on the bottom plate of C2 repels electrons from the top plate of C2. This effect is the same as increasing the space between the plates thus reducing the capacitance. The total capacitance of capacitors connected in series 1 1 1 = C +C CT 1 2 Since the current is the same in all parts of the series circuit, charge drawn from a supply and the charge on each capacitor is the same. However the pd across individual capacitors will vary. QT = QC1 = QC2 = CT E Q + C1 + - E C2 + - VC1 = VC2 = T C1 QT C2
Capacitors in Series Activity 1. A 20uF capacitor and 40uF capacitor are connected in series to an 36 volt dc supply. Determine a) the total capacitance, b) the charge on each capacitor, c) the charge drawn from the supply and d) the potential difference developed across each capacitor. 2. Three capacitors 10uF, 25uF and 50uF are connected in series to an 18 volt dc supply. Determine a) the total capacitance, b) the charge on each capacitor and c) the charge drawn from the supply. It is a good idea to sketch these circuits before starting to calculate your results.
Capacitors in Combinational Networks Capacitors connected in combinational networks are treated using the principles developed for the series and parallel circuits. C2 C1 E + - VC1 C3 VCP 1 1 1 CT = C1 + CP Q T = CT x E QT VC1 = C 1 QT VCP = C P
Combinational Networks Activity 1. A 20uF capacitor and 40uF capacitor are connected in parallel and a 30uF is connected in series with this parallel section. If this network is connected to a 36 volt dc supply determine; a) the total capacitance, b) the total charge, c) the potential difference developed across each capacitor. It is a good idea to sketch these circuits before starting to calculate your results.
Energy Stored By A Capacitor When a capacitor is charging energy is transferred from the source voltage to the capacitor. When the capacitor is removed from the source the potential difference across its plates remains constant. Since no current is required to maintain this potential difference, the capacitance stores electric energy. This energy is stored in the electric field between the plates. The energy stored by a fully charged capacitor, W= 1 CV2 2 The letter symbol for energy is “W” The unit for energy is the joule, “J”
Energy Stored By A Capacitor Activity 1. A 470pf high voltage filter capacitor is charged to 12kV, determine the energy stored. 2. A power supply uses two 4700uF capacitors connected in parallel to provide smoothing in a full-wave rectified dc supply. Both capacitors are charged to a peak voltage of 24 volts. Determine a) the energy stored by the combination and b) the current required to charge the capacitors during the 10mS charge time. It is a good idea to sketch these circuits before starting to calculate your results.