BTEC NC - Electronic and Electrical Principles - Principles of Capacitors and Capacitance

Brendan Burr

BTEC National Certificate in ElectronicsPrinciples of Capacitors and Capacitance

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Task 11.1Type of Dielectric Vacuum Plastic Film Mica Paper Ceramic Aluminium Electrolytic Tantalum Electrolytic Capacitance Range 10pF - 1000pF 100pF - 2.2uF 5pF - 10nF 10nF - 10uF 5pF - 47uF 1uF - 100,000uF 2.2uF - 3500uF Voltage Rating 5000 - 30000 63 - 1000 60 - 600V 500V 60 - 10000 6.3 - 500 6 - 100 Application Radio transmitters and high voltage power devices. Timing circuits and high frequencies. Tuned Circuits. Used for relatively high voltages. High frequency circuits. Relatively high-current and low-frequency electrical circuits. Popular in miniature applications such as mobile telephones.

1.2 Five functions of a capacitor: Energy Storage: Capacitors are able to store energy when the current is removed from a circuit. They are commonly used in electrical devices where the battery would need to be changed, without losing any information. An example of this would be a Sky Remote Control, it states that the batteries must be changed within 15 seconds otherwise the information of the connection between the sky box and the remote will be lost. Power Conditioning: The component can be used to smooth current flow, in full or half wave rectifiers. They can lessen the power loss allowing a much smoother dc current, giving a better input voltage to components in an ac to dc circuit. Filtering: A capacitor can also be used to filter out radio interference. This can be caused in small scale circuits, where a spark is generated. The spark is not powerful enough to damage the circuit, however it can still cause the radio frequency interference. Signal Processing: Due to binary information being a simple on/off category, the energy stored in a capacitor can be used to represent a 1 or 0. This allows a slight delay on the information given. Tuned Circuits: Variable capacitors can be used in conjunction with inductors to allow information to be accessed from a particular frequency band. An example of this would be a radio, where different stations are broadcasted on different frequencies, and to listen to the station you want, the circuit must be tuned to the correct band.

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Task 2

2.1

2.2

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Task 22.3 a)

Task 2.310 9 8 7 6 5 4 3 2 1 0 0 5 10 Time (secs) 15 20

b) i) The time taken for Vout to reach 6.32V = 2.2 seconds ii) The time taken for Vout to rise from 1V to 9V = 4.8 seconds iii) The time taken for Vout to reach 9.99V = 15.2 seconds c)

Vo (volts)

Table 1.2 Excel (i) Time Constant ( ) (ii) Rise Time (tr) (iii) Settling Time (ts) 8

Simulation 2.2 secs 4.9 secs 13.2 secs

2.2 secs 4.8 secs 15.2 secs

Brendan Burr


2.4

Task 2.410 9 8 7 6 5 4 3 2 1 0 0 5 10 Time (secs) 15 20

b) i) The time taken for Vout to reach 3.68V = 2.2 seconds ii) The time taken for Vout to reach 0.01V = 15.2 seconds iii) The time taken for Vout to fall from 9V to 1V = 4.8 seconds

Vo (volts)

Table 1.2 Excel (i) Time Constant ( ) (ii) Fall Time (tf) (iii) Settling Time (ts) 9

Simulation 2.2 secs 4.9 secs 13.2 secs

2.2 secs 4.8 secs 15.2 secs

Brendan Burr


Task 3 3.1 C1 Vin C2 V2 V1 Vin = 50V C1 = 10F C2 = 40F V1 = UNKNOWN V2 = UNKNOWN

a) Total capacitance CT = . 1 . . 1 . . 1 . 10x10 - 6 + 40 x 10 - 66

CT = 8 x 10 CT = 8F b) Total charge

Q = CT x V Q = (8 x 10 6) x 50 Q = 4 x 10 4 Q = 0.0004 (c) = 0.4m(c) c) Voltages V1 = Q C1 V1 = . 0.0004 . 10 x 10 6 V1 = 40V V2 = . 0.0004 . 40 x 10 6 V2 = 10V V2 = Q C2

Q = Charge in Coulombs (c) CT = Capacitance Total in Farads (F) V = Voltage in Volts (V)

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3.2

Q1 Vin C1 C2

Q2

Vin = 100V C1 = 20F C2 = 30F Q1 = UNKNOWN Q2 = UNKNOWN

a) Total capacitance CT = C1 + C2 CT = 20 x 10 6 + 30 x 10 6 CT = 50 x 10 6 = 0.00005 F b) Total charge stored in Q1 and Q2 Q1 = C1 x V Q2 = C2 x V Q1 = 20 x 10 6 x 100 Q1 = 0.002 (c) Q2 = 30 x 10 6 x 100 Q2 = 0.003 (c) c) Total Energy Total Energy in Joules (J) E = Q.V E = C.V2 E = x 0.005 x 100 E = 0.25 J Check: QT = 50 x 10 6 x 100 QT = 5 x 10 3 (c) = 0.005 (c) Q1 + Q2 = QT 0.002 + 0.003 = 0.005 (c)

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Task 44.1 Vin = 100V C1 = 10F C2 = 20F C3 = 30F V1 = UNKNOWN V2 = UNKNOWN V3 = UNKNOWN

V1

a) Total capacitance CT = C1 + . . 1 1 . . 1 C2 + C3 . . .

CT = 10 x 10 6 CT = 2.2 x 10 5 = 0.022F = 0.000022 b) Total Charge QT QT QT QT

+ . 1 . 1 . . 1 . -6 -6 10x10 + 40 x 10

= CT x V = 0.000022 x 100 = 2.2 x 10 3 = 0.0022 (c)

c) Voltages QT = Q1 + Q2.3 Q1 = V1 x C1 Q1 = 100 x 10 x 10-6 Q1 = 0.001 (c) 0.0022 = 0.001 + Q2.3 0.0022 0.001 = +0.001 + Q2.3 0.001 0.0012 = Q2.3 V2 = Q2 C2 V2 = . 0.0012 .

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Brendan Burr


20 x 10-6 V2 = 60V V3 = Q3 C3 V3 = . 0.0012 . 30 x 10-6 V3 = 40V d) Total energy Total Energy in Joules (J) E = Q.V E = C.V2 E = x 0.0022 x 100 E = 0.11 J

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BibliographyThrough guidance from my lecturer, the following websites and text books I was able to complete this assignment: Books BTEC National Engineering (Mike Tooley & Lloyd Dingle) ISBN: 978-0-7506-8521-4 Higher Engineering Mathematics (John Bird) ISBN: 978-0-7506-8152-0 Websites www.wikipedia.org www.kpsec.freeuk.com/index.htm www.technologystudent.com/elec1/elecex.htm

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BTEC NC - Electronic and Electrical Principles - Principles of Capacitors and Capacitance

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