APAITORS / APAITANE - apacitance - apacitor types - apacitors in series & parallel - R ircuit harging phase - R ircuit Discharging phase - R ircuit Steady State model - Source onversions - Superposition Theorem - Thevenin Equivalent ircuit Source: Introductory ircuit Analysis Boylestad 10 th Ed. 1
APAITORS A capacitor is a circuit element that consists of two conducting plates separated by a non-conducting, (i.e. dielectric), material. Terminals are connected to the plates. (t) apacitance [Farads] apacitance is a measure of a apacitor s ability to store harge on its plates. A capacitor is said to be a storage device in that it stores energy in the form of electric field. During its normal operation, a capacitor stores charges across its plates. The difference in potential energy across the two plates results in a voltage across the capacitor. Power is delivered to a capacitor and this energy is stored unlike the resistor whereby power is dissipated in the form of heat (power loss). 2
ELETRI FIELD When a potential difference of volts is applied across the two plates separated by a distance of d, the electric field strength, E, between the plates is determined by: Ɛ E d [ ] m d v The electric field strength, is represented graphically by electric flux lines (i.e. lines of force). These lines of force always extend from a positively charged body to a negatively charged body. They extend Perpendicular to the charged surfaces and never intersect. Electric Flux Lines A capacitor stores energy in the form of electrical field (i.e. voltage) that is established by the opposite charges on the two plates. 3
ELETRI FIELD The force exerted between two charged particles is given by oulomb s law: F kq1q 2 d 2 The electric field strength, ε, at a given point is the force, F, acting on a unit positive charge (Q 1 1) at that point: 4
APAITOR OPERATION When the capacitor is in a neutral state, both plates have an equal number of free electrons. (Insulator) 5
APAITOR OPERATION When a voltage is applied across the plates, electrons are attracted by the positive side of the battery and electrons are repelled by the negative side of the battery. For every electron that leaves one side of the plate, an electron travels to the other side of the plate. The electrons can not pass through the dielectric so that plate A starts storing electrons and plate B effectively starts storing positive charges. This process lasts until the plates are fully charged to a potential difference of the source. The result is a positive charge on plate B and a negative charge on plate A. B e - e - (Electric flux lines) A harging phase process 6
APAITOR OPERATION URRENT APPEARS TO BE FLOWING FROM THE NEGATIE SIDE OF THE BATTERY TOWARD THE POSITIE SIDE THROUGH THE APAITOR. IN FAT IT IS NOT SO, BEAUSE NO ELETRONS AN FLOW THROUGH THE INSULATOR MATERIAL OR GAP BETWEEN THE PLATES (DIELETRI) THERE ARE IN FAT TWO SEPARATE URRENTS FROM THE BATTERY TO THE APAITOR FROM THE APAITOR TO THE BATTERY 7
APAITOR OPERATION If the capacitor is disconnected from the source, it retains the stored charge for a certain period of time (depends on capacitor type and amount of voltage). apacitors are not perfect. Ideal insulators have infinite resistance. Actual insulators have some very high resistance. The electrons on plate A will attract to plate B - Leakage current takes place. Once all the electrons have been neutralized the capacitor has lost all of its charge. B A apacitors cannot be used as a replacement of batteries as they slowly discharge over time. However, they are used to provide a short life supply of voltage (i.e. surge protectors, smoke alarms) 8
APAITANE apacitance is a measure of a capacitor s ability to store charge on its plates. As such, the amount of charge a capacitor can Michael Faraday store per unit of voltage across its plate is called capacitance (Q, ). A capacitor is said to have a 1 Farad Q Q capacitance when 1 oulomb of charge is deposited on the plates with A 1 volt potential difference across it. Q apacitance FARADS [ F ] Q harge OULOMBS [ ] oltage OLTS [ ] 1 1 F 1 The charge across a capacitor is proportional to the voltage across it: Q ( t) ( t) 9
APAITANE apacitance can be computed as follows: Michael Faraday ε ε r o A d Permittivity for vacuum ε o 8.85 x 10-12 F/m APAITANE [ F ] A AREA OF THE PLATES [ m 2 ] d DISTANE BETWEEN PLATES [ m ] ε DIELETRI PERMITTIITY 10
APAITANE Eq 1: Sub Eq 3 into Eq 2: Sub Eq 4 into Eq 1: Ԑ Ԑ d Q Eq 2: Eq 3: Qd Ad ε Eq 4: Qd εa Q A ε ε A d Michael Faraday Q Ԑ ε A 11
APAITANE Michael Faraday 12
APAITOR OLTAGE BREAKDOWN Every capacitor has a voltage rating. A user must not exceed this voltage rating (i.e. voltage breakdown). oltage breakdown is the voltage required to break the bonds within the dielectric that will create current flow in the dielectric through the plates When breakdown occurs, the characteristic of the capacitor becomes that of a conductor (i.e. capacitor shorted) Lightning is an example of breakdown: The potential between the clouds and the earth is so high that charge can pass from one to the other through the atmosphere (which acts as the dielectric) +++++++ louds Atmosphere Earth 13
8.85x10 12 εr A d a. 3(5 uf) 15 uf b. ½ (0.1 uf) 0.05 uf c. 2.5(20 uf) 50 uf d. (5)(4)/(1/8) (1000 pf) (160)(1000 pf) 0.16 uf 14
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b. or Q 16
Assume the breakdown voltage for this capacitor is 200 /mil. 17
APAITORS There are many types of capacitors and they are characterized by the type of dielectric material used between the conducting plates. eramic Disc Tantalum Polypropylene Film Polyester Film Mylar 18
APAITORS There are many types of capacitors and they are characterized by the type of dielectric material used between the conducting plates. ariable apacitors SMT apacitor Non-polarized Electrolytic Radial Lead Axial Lead ariable Electrolytic 19
APAITORS There are many types of capacitors and they are characterized by the type of dielectric material used between the conducting plates. eramic Disk apacitors 20
APAITORS There are many types of capacitors and they are characterized by the type of dielectric material used between the conducting plates. Axial Lead Plastic Film Dielectric Tubular apacitors 21
APAITORS There are many types of capacitors and they are characterized by the type of dielectric material used between the conducting plates. Tantalum Electrolytic apacitors 22
APAITOR ODES eramic Disc 100,000 pf 100 nf.1uf 23
APAITOR ODES Other capacitors may just have 0.1 or 0.01 printed on them. If so, this means a value in uf. Thus 0.1 means just 0.1 uf. alues marked as 50 or 330 means just pf. Electrolytic and large types of apacitors usually have the value printed on them (i.e. 470uF 25). Most of the smaller caps have two or three numbers printed on them, The alue is in pf. 105 means 10 x 10 5 1.000.000pF 1000 nf 1 uf. Letters added to the value represent the tolerance and in some cases represent the temperature coefficient. A 474J ceramic capacitor means 47 x 10 4 470.000pF and J 5% tolerance. (470.000pF 470nF 0.47uF). 24
APAITOR ODES 25
APAITOR ODES 26
APAITOR URRENT & OLTAGE RELATIONSHIP dq( t) i( t) and for a capacitor dt d( v( t)) dv( t) i( t) i( t) dt dt 1 We get : dv( t) i( t) dt 1 v( t) 5v i( t) dt q(t) v(t) If v(t) k i(t) 0, then the capacitor acts like an open to dc (i.e. blocks dc - Steady State). Gnd apacitive coupling circuit For dc circuit analysis, or Steady State circuit analysis, a capacitor is modeled as an open. 27
R IRUIT STEADY STATE ANALYSIS When a capacitor has reached Steady State it is modeled as an open circuit. Find the current I when the circuit has reached steady state. I I 48 v I 8 6k ma 28
APAITOR URRENT & OLTAGE The power delivered is : dv( t) p( t) i( t) v( t) i( t) dt The energy stored in a capacitor is found to be : dw c(t) We know : p(t) dt τ τ dv( t) w ( t) p( t) dt ( ) v t dt dt w w ( t) ( t) 1 2 1 2 v v 2 2 ( t) ( t) v( τ ) v( ) 1 2 2 q ( t) Assuming that J since v(t) v( ) 0 q(t) If v (t) K: o dv(t)/dt 0, p(t) 0 If v (t) is an instantaneous change: o dv(t)/dt and p(t) Power can only be finite, as such the voltage across a capacitor cannot change instantaneously. 29
APAITORS The voltage across a 5-uF capacitor has the waveform shown below. Determine the current waveform. @ 0 i( t) i( t) t @8ms 6ms 6 24 i( t) 5x10 6 @ 6 t 8ms 5x10 t 5x10 6 6 0 2 i( t) 24 2 0 20 ma ma 60 dv( t) dt ma 30
APAITORS - Example w ( t) The voltage across a 5-uF capacitor has the waveform shown below. Determine the Energy at t 6ms. 1 2 v 2 ( t) w (6 1 ms) (5x10 2 6 )(24) w ( 6 ms) 1440 uj 2 31
APAITORS IN SERIES 32 Using KL: ) ( ) ( ) ( ) ( ) ( 3 2 1 t v t v t v t v t v N + + + + dt t i t v i i ) ( 1 ) ( dt t i t v n I i i ) ( 1 ) ( 1 dt t i t v S ) ( 1 ) ( N n I i S 1 1 1 1 1 1 3 2 1 1 + + + + apacitors in series act like resistors in parallel urrent is the same through each element in a series circuit 1 2 2 1 2 1 T +
APAITORS IN PARALLEL Using KL: ) ( ) ( ) ( ) ( ) ( 3 2 1 t i t i t i t i t i N + + + + dt t dv dt t dv dt t dv dt t dv t i N ) ( ) ( ) ( ) ( ) ( 3 2 1 + + + + apacitors in parallel act like resistors in series 33 N N i i P N i i dt t dv t i + + + + 3 2 1 1 1 ) ( ) ( oltage is the same across each branch in a parallel circuit
TESLA OIL A high voltage power supply charges up a capacitor 1. When the capacitor reaches a high enough voltage, the spark gap (switch) fires. When the spark gap fires, the energy stored up in the capacitor dumps into a 1:100 step-up transformer. The primary (L1) is about 10 turns of heavy wire. The secondary (L2) is about 1000 turns of thin wire ( 1:100 ratio). Feed in 10,000 volts, get out 1,000,000 volts. It all happens at a rate of over 120 times per second, often generating multiple discharges in many directions 34
APAITORS IN SERIES & PARALLEL Tesla coil designs require a capacitor with large voltage breakdown. Placing capacitors is series increases the voltage breakdown (add each voltage breakdown). However, placing capacitors in series reduces the total capacitance. This banks of capacitors in series are placed in parallel. Use Power factor correction (PF) capacitors only. These are used to correct the power factor of the A connected to the Neon Sign Transformer (NST). 35
APAITORS IN SERIES & PARALLEL MM: Multi Mini apacitor http://www.teslacoildesign.com/ 36
When EET11 TRANSIENT ANALYSIS - APAITOR HARGING PHASE t 0 v (0 ) 0 @ t 0 + switch is closed v (0 + ) 0 @ t I R > ( t) ( t) ( t) 0 E R Ee E e t R t R E 1 e R t R ( t) + ( t) The R ircuit provides a 1 st order Differential equation. Solving this equation results in: v (0 ) 0 apacitor is modeled as a short apacitor is modeled as an open 37
TRANSIENT ANALYSIS - APAITOR HARGING PHASE apacitor harging E E/R voltage current time time
TRANSIENT ANALYSIS - APAITOR HARGING PHASE t 0 ( t) E 1 e t R E harging Phase v (0 ) 0 E/R I ( t) E R e t R τ R [s] R ( t) Ee t R E 39
When EET11 TRANSIENT ANALYSIS - APAITOR HARGING PHASE t 0 v (0 ) i @ t 0 + switch is closed v (0 + ) i @ t I R > ( t) ( t) ( t) 0 E i R E E ( E ) t R ( E ) e i e R ( t) t R i e + t R ( t) The R ircuit provides a 1 st order Differential equation. Solving this equation results in: v (0 ) i I E i R E i i apacitor is modeled as a battery apacitor is modeled as an open 40
TRANSIENT ANALYSIS - APAITOR HARGING PHASE or ( t) E ( t) E + ( E ) ( E) i i e e t R t R E t 0 I ( t) E R t i R e E i R v (0 ) i E i R τ R R ( t) t R ( E ) e i 41
TRANSIENT ANALYSIS - APAITOR HARGING PHASE R follows this curve Steady State Definition f( ) f(5τ ) R, I 42
TRANSIENT ANALYSIS - APAITOR HARGING PHASE 43
TRANSIENT ANALYSIS - APAITOR HARGING PHASE 44
TRANSIENT ANALYSIS - APAITOR DISHARGE PHASE t 0 v (0 ) 0 ( t) Ee t R E t t o I c ( t) E R e t R @ P1 harging phase @ P2 Discharging phase E R Note the current reversal t t o R ( t) Ee t R E Note the voltage reversal 45
TRANSIENT ANALYSIS - APAITOR DISHARGE PHASE - 46
TRANSIENT ANALYSIS - APAITOR DISHARGE PHASE If the charging phase is disrupted before reaching the supply voltage, the capacitive voltage will be less then E. We call it f. 47
SOURE ONERSION A voltage source can be converted to a current source and vice-versa producing equal behaviors across its load. 48
SOURE ONERSION 49
TRANSIENT ANALYSIS 50
TRANSIENT ANALYSIS 51
TRANSIENT ANALYSIS 52
TRANSIENT ANALYSIS 53
SUPERPOSITION Given a linear circuit, (i.e. described by a set of linear algebraic equations), the superposition analysis technique provides a mean to determine a voltage drop or current by calculating the contribution of each source acting independently and algebraically adding each contribution. Procedure: 1. Remove all sources except one of them by replacing current sources with an open replacing voltage sources with a short retaining all internal resistance 2. alculate the desired voltage drop or branch current from that source paying close attention to polarity or direction 3. Repeat steps 1 and 2 for each additional source acting independently 4. Algebraically add each sources contribution 54
SUPERPOSITION Example 1 For the circuit below, find A using superposition: A S1 ontribution: A + ' A (2.7k)(1.8k) 10 2.7k + 1.8k 1.2k + 1.08k ' A + 4.74v 55
SUPERPOSITION Example 1 For the circuit below, find A using superposition: A S2 ontribution: '' A (2.7k)(1.2k) 36 2.7k + 1.2k 1.8k + 0.83k A - '' A 11.37v 56
SUPERPOSITION Example 1 ' '' A A + A 4.74v 11.37v 6. 63 v 57
SUPERPOSITION Example 2 For the circuit below, find A and B using superposition: A B 58
SUPERPOSITION Example 2 S1 ontribution: R2 // R4 4705 Ω RL + 4705 Ω 6705 Ω R3 // 6705 Ω 5021 Ω A B ' A 40(5.021k ) 4k + 5.021k ' B 22.26(4.705k) 2k + 4.705k ' A 22.26v ' B 15.62v 59
SUPERPOSITION Example 2 S2 ontribution: R1// R3 3.33k Ω RL + 3.33k Ω 5.33k Ω R4 // 5.33k Ω 4997 Ω A B '' B 25(4.997k) 5k + 4.997k '' A 12.49(3.33k) 2k + 3.33k '' B 12.49v '' A 7.80v 60
SUPERPOSITION Example 2 A B ' ' A 22. 26v B 15. 62v '' '' A 7. 80v B 12. 49v ' '' ' '' A A + A 30. 06v B B + B 28. 11v 61
SUPERPOSITION Example 3 For the circuit below, find A using superposition: A 62
SUPERPOSITION Example 3 A IS1 ontribution: R2 // (R3+R4) 172 Ω 20.18 ma I A I ' A 100(172) 680 + 172 I ' A 20.18mA 63
SUPERPOSITION Example 3 S2 ontribution: R1 // (R3+R4) 367 Ω A I I '' A '' A ' A 680 20(367) 367 + 220 680 18.38 ma I A I '' A 18.38mA 64
SUPERPOSITION Example 3 ' I A 20. 18mA '' I A 18. 38mA A 1.224 v A I I I ' A + I '' A 20.18mA 18.38mA A 1.80 ma(680ω) I 1.80mA A 1.224v 65
This image cannot currently be displayed. EET11 THEENIN S THEOREM A two-terminal linear network connected to a load can be replaced with an equivalent circuit consisting of an independent voltage source called TH in series with a resistor R TH such that the current-voltage relationship at the load is unchanged. Procedure: 1. Identify and remove the load 2. Label the load terminals 3. Look in the load terminals and calculate TH 4. Remove all sources by replacing: voltage sources with a short current sources with an open If the source has an internal resistance, keep the resistance in the circuit 5. Look in the load terminals and calculate R TH 6. reate a series circuit consisting of TH, R TH, and the load 7. alculate the load current or voltage as desired 66
This image cannot currently be displayed. EET11 THEENIN S THEOREM Example 1 Identify Load, remove load, label terminals, & find TH a TH 10(8.2k) 3.9k + 8.2k b TH 6.77v 67
This image cannot currently be displayed. EET11 THEENIN S THEOREM Example 1 Replace voltage sources with a short, current sources with an open & find R TH a R TH 3.9k(8.2k) 3.9k + 8.2k + 4.7k R TH 7.34kΩ b 68
This image cannot currently be displayed. EET11 THEENIN S THEOREM Example 1 onnect the load to the equivalent circuit across terminals a & b 7.34 kω a 6.77v b 69
This image cannot currently be displayed. EET11 THEENIN S THEOREM Example 1 7.34 kω a 6.77v RL b 6.77(3.3k) 7.34k + 3.3k RL 2.10v 70
This image cannot currently be displayed. EET11 THEENIN S THEOREM Example 1 7.34 kω a 6.77v I RL b 6.77 7.34k + 3.3k I RL 636.27uA 71
This image cannot currently be displayed. EET11 THEENIN S THEOREM Example 2 I I R3 R3 27mA(2k) 7k + 2k 6 ma TH1 6mA(3k ) 18v 72
This image cannot currently be displayed. EET11 THEENIN S THEOREM Example 2 TH 2 24(3k) 2k + 3k + 4k TH 2 TH 8.0v TH1 + TH 2 18 + 8 26v 73
This image cannot currently be displayed. EET11 THEENIN S THEOREM R TH 6k(3k) 6k + 3k + 1k R TH 3kΩ 74
This image cannot currently be displayed. EET11 THEENIN S THEOREM 3kΩ 26v 75
This image cannot currently be displayed. EET11 THEENIN S THEOREM 3kΩ 26v RL 26(3.3k) 3k + 3.3k RL 13.61v 76
This image cannot currently be displayed. EET11 THEENIN S THEOREM 3kΩ 26v I RL 3k 26 + 3.3k I RL 4.12mA 77
R IRUIT THEENIN EQUIALENT 78
R IRUIT THEENIN EQUIALENT 79
R IRUIT THEENIN EQUIALENT 80
R IRUIT THEENIN EQUIALENT 81
FORMULAS E Ɛ d Q ε r ε o A d v i dt p ( t) i( t) v( t) 1 w ( t) v 2 ( t) 2 1 n 1 I S 1 N P i i 1 i E τ R ( t) ( t) R + harging Phase: @ v (0 ) 0 ( t) E 1 e E I ( t) e R R ( t) Ee t R t R t R harging Phase: @ v (0 ) i R ( E ) e the voltage across a capacitor cannot change instantaneously ( t) E ( t) E + I R ( t) ( t) Steady State: t 5τ ( E ) ( E) i E R @ Steady State is an open circuit i i e e t i R e t R t R t Discharging Phase: I ( t) c ( t) R ( t) Ee E R t R e Ee t R t R 82