ECE2262 Electric Circuits. Chapter 6: Capacitance and Inductance

ECE2262 Electric Circuits Chapter 6: Capacitance and Inductance Capacitors Inductors Capacitor and Inductor Combinations Op-Amp Integrator and Op-Amp Differentiator 1

CAPACITANCE AND INDUCTANCE Introduces two passive, energy storing devices: Capacitors and Inductors LEARNING GOALS CAPACITORS Store energy in their electric field (electrostatic energy) Model as circuit element INDUCTORS Store energy in their magnetic field Model as circuit element CAPACITOR AND INDUCTOR COMBINATIONS Series/parallel combinations of elements RC OP-AMP CIRCUITS Integration and differentiation circuits 2

6.1 Capacitors Electronic Symbol: 3

A capacitor consists of two conductors separated by a non-conductive region. The non-conductive region (orange) is called the dielectric Charge separation in a parallel-plate capacitor causes an internal electric field. A dielectric reduces the electric field and increases the capacitance. Because the conductors (or plates) are close together, the opposite charges on the conductors attract one another due to their electric fields, allowing the capacitor to store more charge for a given voltage than if the conductors were separated, giving the capacitor a large capacitance. 4

In the hydraulic analogy, a capacitor is analogous to a rubber membrane sealed inside a pipe. This animation illustrates a membrane being repeatedly stretched and un-stretched by the flow of water, which is analogous to a capacitor being repeatedly charged and discharged by the flow of charge https://en.wikipedia.org/wiki/file:capacitorhydraulicanalogyanimation.gif In the hydraulic analogy, charge carriers flowing through a wire are analogous to water flowing through a pipe. A capacitor is like a rubber membrane sealed inside a pipe. Water molecules cannot pass through the membrane, but some water can move by stretching the membrane. The analogy clarifies a few aspects of capacitors. The more a capacitor is charged, the larger its voltage drop; i.e., the more it "pushes back" against the charging current. This is analogous to the fact that the more a membrane is stretched, the more it pushes back on the water. 5

The capacitance describes how much charge can be stored on one plate of a capacitor for a given "push" (voltage drop). A very stretchy, flexible membrane corresponds to a higher capacitance than a stiff membrane. A charged-up capacitor is storing potential energy, analogously to a stretched membrane. C : Capacitance of 1 Farad = 1 Coulomb of charge on each conductor causes a voltage of 1 Volt across the device. 1C = 1F!1V! Q = CV 6

Current! Voltage Relations C! A d The charge on the capacitor is proportional to the voltage across q = C v C : Capacitance in Farads (F) - common values µf, pf( pico = 10!12 ) 7

Example 8

Voltage! Current Relation Since q = C v and i = dq dt then we have i = d( C v) dt = C d v dt. ( ) = C d v ( t ) i t dt 9

( ) = C dv( t) i t dt In circuits with with time-independent voltage ( DC circuits) the capacitor behaves as open circuit (capacitor blocks dc). In analyzing a circuit containing dc voltage sources and capacitors, we can replace the capacitors with an open circuit and calculate voltages and currents in the circuit using our many analysis tools. 10

Example C = 5µF i = C d v dt 11

Current! Voltage Relation i( t) v( t) Since i t ( ) = C d v( t) dt then ( ) t i (! )d! t " = C d v! t d! t 0 " = C dv! t 0 ( ) d! " = = C v t t 0 { ( )! v( t 0 )} v( t ) = v t 0 ( )+ 1 t C " i (! )d! t 0 t! t 0 12

v( t ) = v t 0 ( )+ 1 t C " i (! )d! t 0 for any time t! t 0 v t 0 ( ) - initial value (initial state) representing the voltage due to the charge that accumulates on the capacitor from the past to t = t 0. 13

Example Suppose that C = 1F and v 0 Determine v( t). ( ) = 1V. The current source i( t) = e!t/2, t > 0. i( t) v( t) v( t) = v( 0) + 1 t C i! " ( )d! t = 1+ # e!" /2 d" = 1! 2 e 0 0 {!" /2} t 0 = 1+ 2{ 1! e!t/2 } for t! 0! v( t) = # % $ &% 1+ 2{ 1! e!t/2 } t > 0 1 t " 0 14

v( t) = # % $ &% 1+ 2{ 1! e!t/2 } t > 0 1 t " 0 Note: i( t) has a jump, but v( t) is continuous! 15

Power Delivered to Capacitor Since i t ( ) = C d v( t) dt or v( t) = v( t 0 ) + 1 t C i! t 0 ( )d! " then we have p( t) = v( t)i( t) = C v t ( ) d v( t) dt = i t # $ % ( ) v( t 0 ) + 1 t C i! " t 0 ( )d! & ' ( Note: If v( t) would have a jump then 16 d v( t) =! and p( t) =! but this is dt impossible. Hence, the voltage response of a capacitor must be always continuous!

If we encounter circuits containing switches then the idea of continuity of voltage for a capacitor tells us that the voltage across the capacitor just after a switch moves is the same as the voltage across the capacitor just before that switch moves. 17

Energy Stored in the Capacitor Electric Field p( t) = C v( t) d v( t) = C 1 dt 2 = d! 1 dt 2 C $ " ( v2 t) % # & d v 2 ( t) dt! d dt ( t) { v 2 } = 2v t ( ) d dt { v( t) } w( t) = 1 2 C ( v2 t ) (J) Since q( t) = Cv( t) then also w( t) = 1 2 q 2 t ( ) C. 18

Example An uncharged capacitor 2 F is driven by a rectangular current pulse. i( t) 1 t 1 (a)the capacitor voltage v( t) = v( 0) + 1 2 t " i (! )d! = 1 0 2 t # 1( 0! "! 1)d" = 0! " # t / 2 0! t! 1 1/ 2 t > 1. (b) The capacitor power: p( t) = i( t)v( t) =! " # t / 2 0! t! 1 0 else. (c) The capacitor energy: w( t) = 1 2 C ( v2 t) = v 2 ( t) =!# " $# t 2 / 4 0! t! 1 1/ 4 t > 1. 19

Note that power is always positive for the duration of the current pulse, which means that energy is continuously being stored in the capacitor. When the current returns to zero, the stored energy is trapped because the ideal capacitor offers no means for dissipating energy. Thus, a voltage remains on the capacitor after i t ( ) returns to zero. If at some later time an energy-absorbing device (e.g., a flash bulb) is connected across the capacitor, a discharge current will flow from the capacitor and, therefore, the capacitor will supply its stored energy to the device. 21

Example ( ) = C dv( t ) i t dt p( t ) = v( t )i( t ) w( t) = 1 ( 2 Cv2 t ) 22

Energy is being stored in the capacitor whenever the power is positive (the capacitor acts as a load) Energy is being delivered by the capacitor whenever the power is negative (the capacitor acts as a source) 1 "! p( t)dt = 4µJ - energy stored; # p( t)dt =!4µJ - energy delivered 0 1 23

Sinusoidal Input to Capacitor: i t ( ) = C dv( t) dt v( t) = V m cos( 2! f 0 t) i( t) =!CV m 2! f 0 sin( 2! f 0 t) The current i( t) is 90! phase shifted compared to the voltage v( t) The current leads the voltage by 90! 24

v( t ) i( t) 25

i( t) for f 0 = 1,5,10 Hz The capacitor works as an open circuit for dc voltage ( f 0! 0 ) The capacitor works as a short circuit for high-frequency voltage ( f 0! " ) 26

Application RC Low Pass Filter Circuit The reactance of a capacitor varies inversely with frequency, while the value of the resistor remains constant as the frequency changes. At low frequencies the capacitive reactance of the capacitor will be very large compared to the resistive value of the resistor R. This means that the voltage potential, V C across the capacitor will be much larger than the voltage drop, V R developed across the resistor. At high frequencies the reverse is true with V C being small and V R being large due to the change in the capacitive reactance value. 27

6.2 Inductors An inductor, also called a coil, choke or reactor, is a passive two terminal electrical component which resists changes in electric current passing through it. It consists of a conductor such as a wire, usually wound into a coil. When a current flows through it, energy is stored temporarily in a magnetic field in the coil. 28

When the current flowing through an inductor changes, the time-varying magnetic field induces a voltage in the conductor, according to Faraday s law of electromagnetic induction. An inductor is characterized by its inductance, the ratio of the voltage to the rate of change of current, which has units of henries (H). Inductors have values that typically range from 1!H (10!6 H) to 1 H. 29

A capacitor is an integrator of its input current: v( t) = v( t 0 ) + 1 t C i! " t 0 ( )d! L Inductance ( in henrys) An inductor produces a current response that is related to the integral of the voltage applied to its terminals. i( t) = i( t 0 ) + 1 L " t t 0 v (! )d! 30

Current! Voltage Relation L ( ) = L di ( t ) v t dt L : Henry = Volt Amp / sec 31

Voltage! Current Relation L i( t) = i( t 0 )+ 1 L " t t 0 v (! )d! Current must be continuous: i( t 0!) = i( t 0 + ) 33

Power Delivered to an Inductor p( t) = v( t)i( t) = Li t ( ) di ( t ) dt = v t # $ ( ) i( t 0 ) + 1 % L " t t 0 & v (! )d! ' ( 34

Energy Stored in the Inductor Magnetic Field Since p( t) = i t ( ) ( )L di t dt! p t ( ) = 1 2 L ( di2 t) dt! p( t) = d dt! 1 " 2 Li2 t # ( ) $ % & w( t) = 1 ( 2 Li2 t ) (J) 35

DC Circuits Consider the case of a dc current flowing through an inductor. Since ( ) = L di( t) v t dt we see that the voltage across the inductor is directly proportional to the time rate of change of the current flowing through the inductor. A dc current does not vary with time, so the voltage across the inductor is zero. We can say that an inductor is a short circuit to dc. 36

In analyzing a circuit containing dc sources and inductors, we can replace any inductors with short circuits and calculate voltages and currents in the circuit using our many analysis tools. 37

Continuity of i( t)! p( t) = Li( t) di( t) Due to p( t) = Li( t) di( t) we note that an instantaneous change in inductor dt current would require infinite power. Since we don t have any infinite power sources, the current flowing through an inductor cannot change instantaneously. This will be a particularly helpful idea when we encounter circuits containing switches. This idea of continuity of current for an inductor tells us that the current flowing through an inductor just after a switch moves is the same as the current flowing through an inductor just before that switch moves. dt 38

Example The independent current source generates zero current for t < 0 and a pulse i t ( ) = 10te!5t A for t > 0 in the following circuit (a) i( t) 39

(b) v t ( ) = L di( t) dt = 0.1!10e "5t ( 1" 5t) = e!5t ( 1! 5t), t > 0 At t = 0.2s the voltage changes polarity At t = 0 the voltage has a jump, i.e., the voltage can change instantaneously across the terminals (but not current!) 40

(d) w( t) = 1 ( 2 Li2 t) = 1 2! 0.1! ( ) 2 10te"5t = 5t 2 e!10t An increasing energy curve indicates that energy is being stored. Thus, energy is being stored in the time interval 0 to 0.2 sec. This corresponds to the interval when p t ( ) > 0. (the inductor acts as a load) A decreasing energy curve indicates that energy is being extracted. This takes place in the time interval 0.2 to!. This corresponds to the interval when p( t) < 0 (the inductor acts as a source) Energy is max when i( t) is max: w max = 27.07mJ at t = 0.2 sec. 42

(e) Integrals 0.2!! p( t)dt, " p( t)dt 0 0.2 0.2 p( t)dt = 27.07mJ! 0 The area The area! " p( t)dt = #27.07mJ 0.2 0.2! p( t)dt represents the energy stored in the inductor during 0,0.2 0! " p( t)dt represents the energy extracted 0.2 [ ] Energy Stored + Energy Extracted = 0 44

Example Find the total energy stored in the circuit (DC circuit) w C = 1 2 Cv2, w L = 1 2 Li2 45

DC circuit! capacitors! open circuit, inductors! short circuit We need V c1, V c2 and I L1, I L2 46

KCL at A: I L1 + 3! I L2 = 0! I L2 = I L1 + 3 KVL (big loop):!9 + 6I L1 + 3I L2 + 6I L2 = 0! 6I L1 + 9I L2 = 9! I L1 =!1.2A, I L2 = 1.8A 47

KVL (first loop):!9 + 6I L1 + V C1 = 0! V C1 = 9! 6I L1 = 16.2 V V C 2 = 6I L2 = 10.8 V 48

I L1 =!1.2A, I L2 = 1.8A, V C1 = 16.2V, V C 2 = 10.8V w C1 = 1 2 C V 2 1 C1 = 2.62 mj, w C 2 = 1 2 C V 2 2 C 2 = 2.92 mj w L1 = 1 2 L I 2 1 L1 = 1.44 mj, w L2 = 1 2 L I 2 2 L2 = 6.48 mj The total stored energy = 13.46 mj 49

Sinusoidal Input to Inductor: v t ( ) = L di( t) dt i( t) = I m cos( 2! f 0 t) v( t) =!LI m 2! f 0 sin( 2! f 0 t) The voltage v( t) is 90! phase shifted compared to the current i( t) The current lags the voltage by 90! <----- i( t) crosses zero later 50

i( t) v( t ) v( t ) 51

v( t) for f 0 = 1,5,10 Hz The inductor works as a short circuit for dc current ( f 0! 0 ) The capacitor works as an open circuit for high-frequency current ( f 0! " ) 52

6.3 Capacitor and Inductor Combinations Series Capacitors v = v 1 +...+ v N The charge gained by a plate of any capacitor must come from a plate of an adjacent capacitor, i.e., Q C S = Q C 1 +...+ Q C N 1 C S = 1 C 1 +...+ 1 C N 54

Example Find C 2 C 1 = 30µF v 1 = 8V v = 12V C 2 The charge on both capacitors must be the same! Q = v 1 C 1 = ( 8V )! ( 30µF) = 240 µc! Q = v 2 C 2! C 2 = Q v 2 = Q v! v 1 = 240µC 4V = 60µF 55

Example the capacitors have been charged before they were connected Q 1! Q 2! Q 3 (!voltage polarization is different) 1 Equivalent capacitance: = 1 + 1 + 1! C eq = 1µF C eq C 1 C 2 C 3 { } = 31µJ Total energy stored: w == 1 2 C V 2 1 C1 + C 2 V 2 + C V 2 C 2 3 C 3!v + 2! 4!1 = 0! v =!3V! w Ceq == 1 2 C eq v2 = 4.5µJ 56

Parallel Capacitors The total stored charge: Q T = Q 1 +...+ Q N! vc T = vc 1 +...+ vc N C p = C 1 +...+ C N 57

Example Find each capacitor voltage 400V 20µF + V 1! 40µF 30µF +! + + V V 2 9µF 3 70µF V 4!! 20 40! C 1 = 20 + 40 = 60µF C 2 = ( 30! 70) 9 = 30! 70 30 + 70 + 9 = 21 + 9 = 30µF The equivalent capacitors C 1 and C 2 have the same charge because they are in series with the source. 58

400V C 1 = 60µF 20µF + V 1! 40µF 30µF +! + + V V 2 9µF 3 70µF V!! 4 C 2 = 30µF 60! 30 C eq = C 1! C 2 = 60 + 30 = 20µF! Q = C eq eq V 1 = 8mC 60µF = 133V V = 8mC 2 30µF = 267V! V = ( 20 )!10"6 ( 400)=8mC charge on( 30µF! 70µF ) = Q eq! charge on 9µF = 8! 267V " 9µF = 5.6mC V 3 = 5.6mC 30µF = 187 V, V 4 = 5.6mC 70µF = 80 V 59

Series Inductors di( t) v( t) = v 1 ( t) +...+ v N ( t) = L 1 dt di( t) +...+ L N dt ( ) di( t) = L 1 +...+ L N dt L S = L 1 +...+ L N 60

Parallel Inductors i( t) = i 1 ( t) +...+ i N ( t) = i 1 ( 0) + 1 t v! L 1 i t # $ % { } + 1 t $ v! L 1 ( ) = i 1 ( 0) +...+ i N ( 0) # % " 0 & # ( )d! ' +...+ $ i N 0 ( % ( ) + 1 " ( )d! +...+ 1 t & v (! )d! 0 L 0 N " ' ( t & v (! )d! L 0 N " ' (! " # i( t) = i( 0) + 1 +...+ 1 L 1 L N $ % & ( t 0 v (' )d' 61

1 L p = 1 L 1 +...+ 1 L N 62

Example Initial values: i 1 0 The voltage at the terminal v t ( ) = 3A, i 2 ( 0) =!5A ( ) =!30e!5t mv for t! 0 (a) Equivalent inductance: L eq = L 1! L 2 L 1 + L 2 = 60! 240 60 + 240 = 48mH (b) Initial current: i( 0) = i 1 ( 0) + i 2 ( 0) = 3+ (!5) =!2A (current goes up) 63

(c) Find i( t): i( t) = i( 0) + 1 t ( ) v! L 0 eq " d! =!2 + 1 t!30e!5# "10!3 48 "10 $ { } d# =!2 + 1!3 0 8 e!5t!1 ( ) =!2.125 + 0.125e!5t A, t! 0 64

(d) Find i 1 ( t), i 2 ( t) i 1 ( t) = i 1 ( 0) + 1 t v (! ) L 1 " d! = 3+ 0 1 t "30e "5#!10 "3 60!10 $ { } d# "3 0 = 2.9 + 0.1e!5t A, t! 0 i 2 ( t) = i 2 ( 0) + 1 t v (! ) L 2 " d! =!5 + 0 1 t!30e!5# "10!3 240 "10 $ { } d#!3 0 =!5.025 + 0.025e!5" A, t! 0 65

6.4 Op-Amp Integrator and Op-Amp Differentiator 68

Op-Amp Integrator ( ) = C d v( t) i t dt v s Summing the currents leaving the inverting input: 0! v s ( t) + C d ( 0! v ( t) o ) = 0! dv t o R dt dt v o ( t) =! 1 t RC # v s (" )d" + v o 0 0 69 ( ) ( ) =! 1 RC v s ( ) - the initial voltage across the capacitor v o 0 ( t)

Op-Amp Differentiator ( ) = C d v( t) i t dt v s Summing the currents leaving the inverting input: ( ( t) ) C d 0! v s dt + 0! v ( t) o R = 0 v o ( t) =!RC dv s t dt 70 ( )

More realistic circuit R i v s R i R i C dv 0 dt ( t) + v 0 t ( ) =!RC dv s ( t) 71 dt

SUMMARY 73

In dc steady state, a capacitor looks like an open circuit and an inductor looks like a short circuit. The voltage across a capacitor cannot change instantaneously The current flowing through an inductor cannot change instantaneously.! Example 75