Your SlideShare is downloading. ×
Capacitors
Capacitors
Capacitors
Capacitors
Capacitors
Capacitors
Capacitors
Capacitors
Capacitors
Capacitors
Capacitors
Capacitors
Capacitors
Capacitors
Capacitors
Capacitors
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×
Saving this for later? Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.
Text the download link to your phone
Standard text messaging rates apply

Capacitors

1,356

Published on

Published in: Education, Business, Technology
0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total Views
1,356
On Slideshare
0
From Embeds
0
Number of Embeds
1
Actions
Shares
0
Downloads
96
Comments
0
Likes
0
Embeds 0
No embeds

Report content
Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
No notes for slide

Transcript

  • 1. CapacitorsWhy Are Capacitors Important?What Is A Capacitor?Voltage Current Relationships In CapacitorsEnergy In CapacitorsWhat Is Impedance?Using ImpedanceSome Impedance Laws/CombinationsWhy Are Capacitors Important? The capacitor is a widely used electrical component. It has several featuresthat make it useful and important: A capacitor can store energy, so capacitors are often found in power supplies. A capacitor has a voltage that is proportional to the charge (the integral of the current) that is stored in the capacitor, so a capacitor can be used to perform interesting computations in op-amp circuits, for example. Circuits with capacitors exhibit frequency-dependent behavior so that circuits that amplify certain frequencies selectively can be built.What Is A Capacitor? Capacitors are two-terminal electricalelements. Capacitors are essentially twoconductors, usually conduction plates - but any twoconductors - separated by an insulator - adielectric - with conection wires connected to thetwo conducting plates. Capacitors occur naturally. On printed circuit boards two wires runningparallel to each other on opposite sides of the board form a capacitor. Thats acapacitor that comes about inadvertently, and we would normally prefer that it notbe there. But, its there. It has electrical effects, and it will affect your circuit.You need to understand what it does.
  • 2. At other times, you specifically want to use capacitors because of theirfrequency dependent behavior. There are lots of situations where we want todesign for some specific frequency dependent behavior. Maybe you want to filterout some high frequency noise from a lower frequency signal. Maybe you want tofilter out power supply frequencies in a signal running near a 60 Hz line. Yourealmost certainly going to use a circuit with a capacitor. Sometimes you can use a capacitor to store energy. In a subway car, aninsulator at a track switch may cut off power from the car for a few feet alongthe line. You might use a large capacitor to store energy to drive the subway carthrough the insulator in the power feed. Capacitors are used for all these purposes, and more. In this chapter youregoing to start learning about this important electrical component. Remembercapacitors do the following and more. Store energy Change their behavior with frequency Come about naturally in circuits and can change a circuits behaviorGoals You need to know what you should get from this lesson on capacitors. Heresthe story. Given a capacitor, o Be able to write and use the voltage-current relationship for the capacitor, o Be able to compute the current through a capacitor when you know the voltage across a capacitor. Given a capacitor that is charged, o Be able to compute the amount of energy that is stored in the capacitor. Capacitors and inductors are both elements that can store energy in purelyelectrical forms. These two elements were both invented early in electrical history.The capacitor appeared first as the legendary Leyden jar, a device that consistedof a glass jar with metal foil on the inside and outside of the jar, kind of like the
  • 3. picture below. This schematic/picture shows a battery attached to leads on theLeyden jar capacitor. Although this device first appeared inLeyden, a city in the Netherlands sometimebefore 1750. It was discovered by E. G. vonKleist and Pieter van Musschenbroek. Although ithas been around for about 250 years, it has allof the elements of a modern capacitor, including: Two conducting plates. Thats the metallic foil in the Leyden jar. An insulator that separates the plates so that they make no electrical contact. Thats the glass jar - the Leyden jar. The way the Leyden jar operated was thatcharge could be put onto both foil elements. Ifpositive charge was put onto the inside foil, andnegative charge on the outside foil, then the twocharges would tend to hold each other in place.Modern capacitors are no different and usuallyconsist of two metallic or conducting plates thatare arranged in a way that permits charge to be bound to the two plates of thecapacitor. A simple physical situation is the one shown at the right. If the top plate contains positive charge, and the bottom plate containsnegative charge, then there is a tendency for the charge to be bound on thecapacitor plates since the positive charge attracts the negative charge (andthereby keeps the negative charge from flowing out of the capacitor) and in turn,the negative charge tends to hold the positive charge in place. Once charge gets onthe plates of a capacitor, it will tend to stay there, never moving unless there is aconductive path that it can take to flow from one plate to the other.There is also a standard circuit symbol for a capacitor. The figure below shows asketch of a physical capacitor, the corresponding circuit symbol, and therelationship between Q and V. Notice how the symbol for a capacitor captures theessence of the two plates and the insulating dielectric between the plates.
  • 4. Now, consider a capacitor that starts out with no charge on either plate. Ifthe capacitor is connected to a circuit, then the same charge will flow into oneplate as flows out from the other. The net result will be that the same amount ofcharge, but of opposite sign, will be on each plate of the capacitor. That is theusual situation, and we usually assume that if an amount of charge, Q, is on thepositive plate then -Q is the amount of charge on the negative plate. The essence of a capacitor is that it stores charge. Because they storecharge they have the properties mentioned earlier - they store energy and theyhave frequency dependent behavior. When we examine charge storage in acapacitor we can understand other aspects of the behavior of capacitors. In a capacitor charge can accumulate on the two plates. Normally charge ofopposite polarity accumulates on the two plates, positive on one plate and negativeon the other. It is possible for that charge to stay there. The positive charge onone plate attracts and holds the negative charge on the other plate. In thatsituation the charge can stay there for a long time. Thats it for this section. You now know pretty much what a capacitor is.What you need to learn yet is how the capacitor is used in a circuit - what it doeswhen you use it. Thats the topic of the next section. If you can learn that thenyou can begin to learn some of the things that you can do with a capacitor.Capacitors are a very interesting kind of component. Capacitors are one largereason why electrical engineers have to learn calculus, especially about derivatives.In the next section youll learn how capacitors influence voltage and current.Voltage-Current Relationships In Capacitors There is a relationship between the charge on a capacitor and the voltageacross the capacitor. The relationship is simple. For most dielectric/insulatingmaterials, charge and voltage are linearly related.
  • 5. Q=CVwhere: V is the voltage across the plates.You will need to define a polarity for that voltage. Weve defined thevoltage above. You could reverse the "+" and "-". Q is the charge on the plate with the "+" on the voltage polarity definition. C is a constant - the capacitance of the capacitor. The relationship between the charge on a capacitor and the voltage acrossthe capacitor is linear with a constant, C, called the capacitance. Q=CV When V is measured in volts, and Q is measured in couloumbs, then C has theunits of farads. Farads are really coulombs/volt. The relationship, Q = C V, is the most important thing you can know aboutcapacitance. There are other details you may need to know at times, like how thecapacitance is constructed, but the way a capacitor behaves electrically isdetermined from this one basicrelationship. Shown to the right is a circuitthat has a voltage source, Vs, aresistor, R, and a capacitor, C. If youwant to know how this circuit works,youll need to apply KCL and KVL tothe circuit, and youll need to knowhow voltage and current are relatedin the capacitor. We have arelationship between voltage andcharge, and we need to work with it to get a voltage current relationship. Welllook at that in some detail in the next section. The basic relationship in a capacitor is that the voltage is proportional to thecharge on the "+" plate. However, we need to know how current and voltage are
  • 6. related. To derive that relationship you need to realize that the current flowinginto the capacitor is the rate of charge flow into the capacitor. Heres thesituation. Well start with a capacitor with a time-varying voltage, v(t), definedacross the capacitor, and a time-varying current, i(t), flowing into the capacitor.The current, i(t), flows into the "+" terminal taking the "+" terminal using thevoltage polarity definition. Using this definition we have: ic(t) = C dvc(t)/dt This relationship is the fundamental relationship between current and voltagein a capacitor. It is not a simple proportional relationship like we found for aresistor. The derivative of voltage that appears in the expression for currentmeans that we have to deal with calculus and differential equations here - whetherwe want to or not.QuestionQ1 If the voltage across a capacitor is descreasing (and voltage and current aredefined as above) is the current positive of negative? This derivative kind of relationship also has some implications for whathappens in a capacitor, and we are going to spend some time exploring thatrelationship. Clearly, we need to understand what this relationship implies, and thenwe need to learn how it affects things when we write circuit equations using KVLand KCL. Well start by considering a time varyingvoltage across a capacitor. To have somethingspecific, lets say that we have a 4.7 f capacitor,and that the voltage across the capacitor is thevoltage time function shown below. That voltagerises from zero to ten volts in one millisecond, thenstays constant at ten volts. Before you go on try todetermine what the current through the capacitorlooks like, then answer these questions.
  • 7. QuestionsQ2. Is the current constant in the time interval from t = 0 to t = 10 msec?Q3. Is the current constant in the time interval from t = 10 msec to the last timeshown? If current is proportional to the time derivative of voltage, there is only one time segment, from t = 0 to 1 millisecond, where the voltage derivative is non-zero, so thats the only time there is any currrent that is non-zero. After one millisecond has elapsed, the voltage derivative goes to zero, so there isnt any current then. If there isnt any current, then the voltage stays constant because no charge is flowing in or out. Remember, current is charge flow! The voltage derivative is constant from t = 0 to 1 millisecond. If thats true, then the current is constant in that period. Now, you should be able to compute the current.Energy In Capacitors Storing energy is very important. You count on the energy stored in your gastank if you drove a car to school or work today. Thats an obvious case of energystorage. There are lots of other places where energy is stored. Many of them arenot as obvious as the gas tank in a car. Here are a few. Youre reading this on a computer, and the computer keeps track of the date and time. It does that by keeping a small part of the computer running when you think that the computer is turned off. Theres a small battery that stores the energy to keep the clock running when everything else is turned off. If you own a stereo or television that you have to plug into the wall plug, then you should realize that the wall plug voltage becomes zero 120 times a second. When that happens, the system keeps running because there are
  • 8. capacitors inside the system that store energy to carry you through those periods when the line voltage isnt large enough to keep things going! Capacitors cant really be used to store a lot of energy, but there are manysituations in which a capacitors ability to store energy becomes important. In thislesson we will discuss how much energy a capacitor can store.Capacitors are often used to store energy. When relatively small amounts of energy are needed. Where batteries are not desired because they might deteriorate. For larger power/short duration applications - as in power supply filters, or to keep power up long enough for a computer to shut down gracefully when the line power fails. To calculate how much energy is stored in a capacitor, we start by looking atthe basic relationship between voltage and current in a capacitor. i(t) = C dv(t)/dt Once we have this relationship, we can calculate the power - therate of flow of energy into the capacitor - by multiplying the currentflowing through the capacitor by the voltage across the capacitor. P(t) = i(t)v(t) Given the expression for the power: P(t) = i(t)v(t) And given the expression for the current: i(t) = C dv(t)/dt We can use the expression for current in the power expression: P(t) = (C dv(t)/dt) v(t) We can recognize that power is simply rate of energy input. P(t) = dE/dt = (C dv(t)/dt) v(t)
  • 9. Now, the derivative of energy can be integrated to find the total energyinput. P(t) = dE/dt = (C dv(t)/dt) v(t)givesNow, assuming that the initial voltage is zero (there is no energy stored inthe capacitor initially, we find that the energy stored in a capacitor isproportional to the capacitance and to the square of the voltage across thecapacitor. Ec = (1/2)CV2The expression for the energy stored in a capacitor resembles other energystorage formulae.For kinetic energy, with a mass, M, and a velocity, v. EM = (1/2)MV2For potential energy, with a spring constant, K, and an elongation, x. ESpring = (1/2)Kx2Since the square of the voltage appears in the energy formula, the energystored is always positive. You cant have a negative amount of energy in thecapacitor. That means you can put energy into the capacitor, and you cantake it out, but you cant take out more than you put in.Power in to the capacitor can be negative. Voltage can be positive whilecurrent is negative. Imagine a capacitor that is charged. You could charge acapacitor by putting a battery across the capacitor, for example. Then, ifyou placed a resistor across the capacitor, charge would leave the capacitor- current would flow out of the capacitor - and the energy in the capacitor
  • 10. would leave the capacitor only to become heat energy in the resistor. When energy leaves the capacitor, power is negative. When you use capacitors in a circuit and you analyze the circuit you need to be careful about sign conventions. Here are the conventions we used, and these conventions were assumed in any results we got in this lesson.Frequency Dependent Behavior For A Capacitor We start with a capacitor with a sinusoidal voltage across it.where:  vC(t) = Voltage across the capacitor  iC(t) = Current through the capacitor  C = Capacitance (in farads) We will assume that the voltage across the capacitor is sinusoidal:vC(t) = Vmax sin( t) Knowing the voltage across the capacitor allows us to calculate the current:iC(t) = C dvC(t)/dt = C Vmax cos( t) = Imax cos( t)where Imax = C VmaxComparing the expressions for the voltage and current we note the following. The voltage and the current are both sinusoidal signals (a sine function or a cosine function) at the same frequency. The current leads the voltage. In other words, the peak of the current occurs earlier in time than the peak of the voltage signal. The current leads the voltage by exactly 90o. It will always be exactly 90o in a capacitor. The magnitude of the current and the magnitude of the voltage are related: Vmax/Imax = 1/ C
  • 11. Now, with these observations in hand, it is possible to see that there may bean algebraic way to express all of these facts and relationships. The methodreduces to the following. If we have a circuit with sinusoidally varying voltages and currents (as we would have in a circuit with resistors, capacitors and inductors and sinusoidal voltage and current sources) we associate a complex variable with every voltage and current in the circuit. The complex variable for a voltage or current encodes the amplitude and phase for that voltage or current. The voltage and current variables can be used (using complex algebra) to predict circuit behavior just as though the circuit were a resistive circuit. We need to do two things here. First, we can illustrate what we mean with anexample. Secondly, we need to justify the claim above. We will look at an examplefirst, and we will do two examples. The first example is jsut the capacitor - all byitself. The second example will be one that you have considered earlier, a simpleRC low-pass filter.Example 1 - The Capacitor In a capacitor with sinusoidal voltage and currents, we have:where: vC(t) = Voltage across the inductor vC(t) = Vmax sin( t) iC(t) = Current through the inductor iC(t) = C Vmax cos( t) = Imax cos( t) C = Capacitance (in farads) We represent the voltage with a complex variable, V. Considering this as acomplex variable, it has a magnitude of Vmaxand and angle of 0o. We would write:V = Vmax/0o Similarly, we can get a representation for the current. However, first note:iC(t) = C Vmax cos( t) = Imax cos( t) = Imax sin( t + 90o)
  • 12. (Here you must excuse the mixing of radians and degrees in the argument of thesine. The only excuse is that everyone does it!) Anyhow, we have:I = Imax/90o = j Imax = j C VmaxWhere j is the square root of -1. Then we would write:V/I = Vmax/j C Vmax = 1/j Cand the quantity 1/j C is called the impedance of the capacitor. In the nextsection we will apply that concept to a small circuit - one you should have seenbefore. Before moving to the next section, a little reflection is in order. Here aresome points to think about. A phasor summarizes information about a sinusoidal signal. Magnitude and phase information are encoded into the phasor. Frequency information is not encoded, and there is a tacit assumption that all signals are of the same frequency, which would be the case in a linear circuit with sinusoidal voltage and current sources. We looked at a case where we encoded a signal Vmax sin( t) into a phasor of Vmax/0o. That was completely arbitrary, and many others would have encoded Vmax cos( t) into a phasor of Vmax/0o. Phasors are intended only to show relative phase information, and it doesnt matter which way you go.Using Impedance In the last section we began to talk about the concept of impedance. Let usdo that a little more formally. We begin by defining terms. A sinusoidally varying signal (vC(t) = Vmax sin( t) for example) will berepresented by a phasor, V, that incorporates the magnitude and phase angle ofthe signal as a magnitude and angle in a complex number. Examples include thesetaken from the last section. (Note that these phasors have nothing to do with anyTV program about outer space.) vC(t) = Vmax sin( t)
  • 13. is represented by a phasor V = Vmax/0o iC(t) = Imax sin( t + 90o)is represented by a phasor I = Imax/90o va(t) = VA sin( t + )is represented by a phasor Va = VA/ Next, we can use the relationships for voltage and current phasors to analyzea circuit. Here is the circuit. Now, this circuit is really a frequencydependent voltage divider, and it is analyzeddifferently in another lesson. However,here we will use phasors. At the end of thisanalysis, you should compare how difficult itis using phasors to the method in the otherlesson. We start by noting that the current inthe circuit - and there is only one current -has a phasor representation: I = Imax/0oWe will use the current phase as a reference, and measure all other phases fromthe currents phase. Thats an arbitrary decision, but thats the way we will start. Next we note that we can compute the voltage across the capacitor. VC = I/j CThis expression relates the current phasor to the phasor that represents thevoltage across the capacitor. The quantity 1/j C is the impedance of thecapacitor. In the last section we justified this relationship. We can also compute the phasor for the voltage across the resistor.VR = IR
  • 14. This looks amazingly like Ohms law, and it is, in fact, Ohms law, but it is in phasorform. For that matter, the relationship between voltage and current phasors in acapacitor - just above - may be considered a generalized form of Ohms law! Now, we can also apply Kirchhoffs Voltage Law (KVL) to compute the phasorfor the input voltage. VIN = VR + VC = IR + I/j C = I(R + 1/j C) You should note the similarities in what happens here and what happens whenyou have two resistors in series. If you have a resistor, R, and a capacitor, C, in series, the current phasor can be computed by dividing the input voltage phasor by the sum of R and 1/j C. If you have two resistors in series (call them R1and R2), the current can be computed by dividing the input voltage by the sum of R1and R2.Example Consider a series circuit of a resistor and capacitor. The series impedance is: Z = R + 1/j CThats the same as we showed just above. The impedance can be used to predictrelationships between voltage and current. Assume that the voltage across theseries connection is given by: vSeries(t) = Vmax cos( t)That corresponds to having a voltage phasor of: V = Vmax/0oWe also know that the impedance establishes a relationship between the voltageand current phasors in the series circuit. In particular, the voltage phasor is theproduct of the current phasor and the impedance. V=IZ
  • 15. For our particular impedance, we have: V = I*(R + 1/j C)So, we can solve for the current phasor: I = V / (R + 1/j C)Now, we know the voltage phasor and we know the impedance so we can computethe current phasor. Let us look at some particular values.Assume: R = 1.0 k C = .1 f = 10-7 f f = 1 kHz, so = 2 103 Vmax = 20 vThen: ZR = 1.0 k ZC = 1/(j C) = 1/(j2 103 10-7 ) = j 1.59 kAnd, the total impedance is: Z = ZR + ZC = (1.0 + j 1.59) kThis impedance value can also be expressed in polar notation: Z = 1.878 /62oNow, compute the current phasor: I = V / (R + 1/j C)Substituting values, we find: I = V / Z = Vmax/0o / 1.878 /62o =20/0o / 1.878 /62o I = V / Z = (20 / 1.878) /-62o = 10.65 /-62oamps
  • 16. And, we need to examine exactly what this means for the current as a function oftime. But that isnt very difficult. We can write out the expression for thecurrent from what we have above. iC(t) = 10.65 cos( t - 62o) amps

×