耦合电感(双语)课件

Linear Circuit Analysis Magnetically Coupled Circuits and Transformers1.Introduction Experimental evidence shows that a change in the current i1 will generate a voltage v2,called the induced voltage,across the open circuit.Although physically isolated,each pair of coils in figures 18.1a and 18.1b is said to be magnetically coupled.How does one quantitatively account for magnetic coupling?The answer is to introduce a new circuit quantity,called mutual inductance,for coupled coils.Magnetically coupled circuits have many important engineering application.An extremely important magnetically coupled devices is the transformer,which is used to transform voltages and currents from one level to another.In addition,transformers have numerous uses in electronic systems,including(1)stepping ac voltages up or down,(2)isolating parts of a circuit from dc voltages,and(3)providing impedance level changes to achieve maximun power transfer between devices.2.Mutual Inductance and the Dot Convention Experimental evidence demonstrates that if the two coils in figure 18.1a are stationary,the induced voltage v2(t)is proportional to the rate of change of i1(t).Confining our attention initially to the magnitudes of voltage and current only,we can write the linear relationship as where M21 is a constant of proportionality linking v2 to the change in i1.if,in figure 18.1a,the voltage source is moved to coil 2 while coil 1 is open-circuited,one observes a similar relationship,viz.,(18.1)(18.2)As verified in a later section,M21 and M12 are equal and therefore can be designated by a single positive proportionality constant,Called the mutual inductance of the coupled inductors.Removing the absolute-value signs from equations 18.1 and 18.2,and using equation 18.3,implies that Equation 18.1 also suggests a possible experimental procedure for determining the value of M:apply a ramp current i1(t)=Kr(t)(k0)to coil 1,and measure the constant open-circuit voltage v2 induced in coil 2.Then M=v2/K.(18.3)(18.4)(18.5)To determine the sign,consider figure 18.2 and apply an increasing current i1(t)to coil 1.Since di1/dt0 for all t,one of the terminals of coil 2 must be at a higher potential relative to the other at all times.This terminal is easily identified by connecting a voltmeter to coil 2.Place a dot on the terminal at which enters.Let be increasing,.Place another dot on the terminal with the higher potential that is on the other open-circuited coil.Rule For Induced Voltage Drop Due to Mutual Inductance The voltage drop across one coil,from the dotted terminal to the undotted terminal,equals M times the derivative of the current through the other coil,from the dotted terminal to the undotted terminal.Or,equivalently,The voltage drop across one coil,from the undotted terminal to the dotted terminal,equals M times the derivative of the current through the other coil,from the undotted terminal to the dotted terminal.(18.6a)(18.6b)Example18.1 In figure 18.2,if i1(t)=2tu(t)A,a voltage vCD=-10u(t)mV is observed.Determine the(new)placement of the dots and the value of M.Since i1(t)is increasing and D is at a higher potential than C,the dots must be placed at(A,D)or(B,C).From equation 18.1,the value of the mutual inductance is SOLUTIONExample18.2 Figure 18.3b shows two waveforms i1(t)and v2(t)as displayed on an oscilloscope.Determine the placement of the dots and the value of the mutual inductance for the circuit of figure 18.3a.1-112+-Signalsource2-2(b)Figure 18.3SOLUTIONFor the time interval 0t0.5 ms,i1(t)is a ramp function and v2(t)is constant,The information is similar to that given in example 18.1.Therefore we can solve the problem in the same way.The current i1 is increasing and v2 is positive.According to figure 18.2,the dots must be placed at(A,C)or(B,D).Equation 18.5 becomesThe measured values during 0t0.5 s give v2=2 V and di1/dt=1/0.0005=2000 A/s.Thus M=2/2000=0.001H.Example18.3 In the circuit of figure 18.3a,if i1(t)=2(1-e-100t)u(t)A,find v2(t).SOLUTIONIn this case,di1/dt=200e-100tu(t).From example 18.2,M=0.001 H.From equation 18.5,It is worthwhile to remember the following basic facts:1.Let the numbers of turns of the two inductors in figure 18.1 be N1 and N2.Then the self and mutual inductances have approximately the ratio2.If two inductors are placed in a nonmagnetic medium,bringing the inductors closer together increases the value of M.3.If one inductor of a pair is rotated,then a larger value of M results when the axes of the inductors are parallel to each other.The smallest value of M occurs when the axes are perpendicular to each other.4.Changing the core on which two inductors are wound from a nonmagnetic material to a ferromagnetic material may increase the values of L1,L2,and M by several thousand times.3.Differential Equation,Laplace Transform,and Phasor Models of Coupled Inductors Figure 18.4 shows a pair of inductors with mutual inductance M.Besides the mutual inductance,each inductor in the pair also has a self-inductance,denoted by L1 and L2,respectively.The reference directions for voltages and currents may be chosen arbitrarily,with those shown in figure 18.4 being typical.Our objective here is to develop equations relating the currents and voltages for the dot placements of figures 18.4a and 18.4b.(1)Differential Equation of coupled inductors The voltage developed across each inductor is the sum of the voltage due to the self-inductance Lk(dik/dt)and the voltage due to the mutual inductance M(dij/dt),with the signs duly considered.In figure 18.4a,currents are entering at the dotted terminals,and voltage drops are from the dotted terminals to the undotted terminals;hence,the mutual terms have plus signs,according to equation 18.6a.Applying superposition leads to the following set of equations governing the circuit of figure 18.4a,(18.7a)(18.7b)On the other hand,the voltage v2(t)in figure 18.4b is from the undotted terminal to the dotted terminal of coil 2.The current through coil 1,from the undotted terminal to the dotted terminal,is i1,Therefore,according to equation 18.6b,(18.8b)In figure 18.4b,v1(t)is the voltage drop from the dotted terminal to the undotted terminal of coil 1.The current through coil 2,from the dotted terminal to the undotted terminal,is i2.Therefore,according to equation 18.6a,(18.8a)Again,the development based on figure 18.1 presupposes linearity.If the two inductors are placed in a nonmagnetic medium,this is true.If the inductors are coupled through a ferromagnetic medium,then the linear relationships hold only if both currents are sufficiently small so that the magnetic medium has not reached saturation,a phenomenon discussed in other courses or more advanced texts.Our investigation considers only the linear case.Example18.4 A pair of coupled inductors is connected in two different ways,as shown in figure 18.5.In each case,find the differential equation relating v and i,and then find the equivalent inductance“seen”at the two terminals of each box.SOLUTIONFor figure 18.5a,we can apply equations 18.7 directly to obtain+-+-Figure 18.5 Equivalent inductance of two series-connected inductors.Comparing this relationship with We obtainSimilarly,for figure 18.5b,we apply equations 18.8 to obtainTherefore,The preceding example implies that(18.9)Applying the Laplace transform to equations 18.7 and 18.8,while invoking the assumption that i1(0-)=i2(0-)=0,yields Comparing equation 18.10 with equations 18.7 and 18.8 yields only two differences:1.V(s)and I(s)replace v(t)and i(t).2.The Laplace transform variable s replaces the operator d/dt.(18.10a)(18.10b)(2)Laplace Transform of coupled inductorsExample18.6 In the circuit of figure 18.7,R1=4,R2=2,L1=8H,L2=6H,and M=4H.If E=36V,i2(0-)=0,and S is closed at t=0:(a)Find i1(t)and i2(t).(b)Show that V2(t)and di1/dt are positive for all t 0.SOLUTION(a)Writing two loop equations in the s-domain producesFigure 18.7EIn matrix notation,Solving for I1 and I2 by Cramers rule,we obtainInverse Laplace transforming I1(s)and I2(s)yields(b)Using the expressions found in part(a)impliesandFrom equation 18.11 and 18.12,V2(t)0 and di1/dt 0 for all t 0,for t0 The Laplace transform method is very useful for finding the complete solution.In practice,however,there are many occasions where the excitation to a stable circuit is sinusoidal and only the steady-state solution is of interest.The phasor method of analysis proves easier than the Laplace transform method in such cases.Phasor equations resemble Laplace transform equations under the assumption of zero initial conditions.The only difference is that V and I are now phasors(complex numbers),With s replaced by j.In phasor form,equations 18.10 becameThe following example illustrates the use of the phasor method.(18.13a)(18.13b)(3)Phasor models of coupled inductorsExample18.7 Consider the circuit of figure 18.8.Find the steady-state components of V1(t)and V2(t)at the frequency of 1 rad/s for the following two cases:(a)The 2-F capacitor is disconnected.(b)The 2-F capacitor is connected.(a)With the capacitor disconnected,=0.Hence induces no component in ,and the equation for isFigure 18.8SOLUTIONFirst we note that=1rad/s and that the phasor for the voltage source is 10 V.and(b)With the 2-F capacitor connected,the two loop equations areFrom which it follows thatAnd,since ,The solutions of these equations are:so,in which case Clearly,are out of phase!Observe that are in phase.Figure18.84.Applications:Automobile Ignition and RF AmplifierExample 18.8Figure 18.9 shows an automobile ignition system and a simplified circuit model thereof.This ignition system is typical of older model cars.The ignition coil is a pair of inductors wound on the same iron core.The coil connected to a power source is called the primary,whereas the coil connected to a load is called the secondary.The primary has a few hundred turns of heavy wire,the secondary about 20000 turns of very fine wire.When the ignition point(or contact)opens by cam action,a voltage exceeding 20000V is induced across the secondary,causing the spark plug to fire.Todays ignition systems do not typically have points and a condenser;switching in done electronically.The box in dashed lines in figure 18.9a is replaced by an electronic ignition module.Nevertheless,the basic idea of generating a high voltage to cause a spark to occur at the plug is accomplished by a basic RLC circuit containing a switch that represents the point of the ignition system.SolutionAs mentioned,the following figure is a simplified circuit model for the ignition system.Suppose that the switch has been closed for a long time.Since the secondary is open-circuited,it has no effect on the solution for the primary current.Using the model for an initialized inductor results in the s-domain equivalent circuit of figure 18.10.Accordingly,at ,we have Figure 18.9b Its simplified circuit model+_The primary current is simply the net driving voltage divided by the total impedance in the series circuit,Substituting the given component values into the equation yieldsTaking the inverse Laplace transform of yieldsHaving obtained ,we calculate from the basic relationship,using the plus sign in the present case of dot markings:From this expression,the voltage reaches a magnitude of 36000V in about (one-fourth of a cycle of the oscillations).This voltage is high enough to cause the spark plug to fire.In practice,when the engine is running,a resistance wire or an actual resistor is placed in the primary circuit to limit the amount of current flow through the breaker point.The capacitor(condenser)serves a similar purposethat of protecting the breaker point by suppressing the arc that results when the point opens.5.Coefficient of Coupling and Energy Calculation We begin this section with a justification of our assumption that M12=M21=M.The justification stems from a physical property that a pair of stationary coupled coils cannot generate average power.One can justify that M12=M21=M by the principles of magnetic circuits,but this is beyond the scope of this text.As a consequence,we will show that the mutual inductance M has upper bound ,i.e,the mutual inductance can never exceed the geometric mean of the self-inductances.PASSIVITY PRINCIPLE FOR INDUCTORSA pair of stationary coupled inductors is a passive system;i.e.,they cannot generate energy and,hence,cannot deliver average power to any external network.Justification that M12=M21=M This property is a consequence of the principles of electromagnetic field theory and cannot be proven by a pure mathematical argument.The principles of field theory,however,are not commonly known by students studying introductory circuit analysis.To make our approach more accessible,we will build our justification on the passivity principle for inductors.Suppose that M12M21.Then,instead of having equations 18.7,the differential equations for the coupled inductors of figure 18.4a must account for this difference and take the form Let us apply i1=sin(t)A and i2=cos(t)A to the inductors.From equations 18.20,the terminal voltages are (18.20a)(18.20b)The total instantaneous power delivered to the inductors is,therefore,(18.21)To calculate ,the average power delivered to the coupled inductors,we use the identities .It follows immediately that the first and last terms in equation 18.21 make no contribution to ,whereas the terms involving and lead to(18.22)This result shows very clearly that with ,the average power of coupled inductor is always zero for arbitrary sinusoidal excitations.Hence coupled inductors are said to be lossless.Calculation of Stored energy(18.23a)(18.23b)Consider the coupled inductors shown in figure 18.4.The voltage-current relationships at the terminals are given byHaving proved that ,we shall now show that there is a limit to the value of that is attainable once and are specified.This is again done by the use of the passivity principle.Where d(i1,i2)is the total derivative of the product i1i2 and is equal to i1di2+i2di1.Next we apply driving sources to the inductors to bring the currents up to i1=I1 and i2=I2 at t=T.The energy delivered to the inductors during the time interval(0,T)is(18.24)For this reason,the energy given by equation 18.24 is called the stored energy.The physics of the situation shows that the energy is stored in the magnetic field produced by the currents in the inductors.A couple of things about this result are worth noticing:1.The final integral in equation 18.24 depends only on the final values I1 and I2.The exact waveforms of i1(t)and i2(t)during 0t T are immaterial.2.The energy W(T)delivered by the sources during 0t T is not lost,but merely stored in the system.18.24Upper Bound for M and the Coefficient of Coupling The energy W(T)must be nonnegative for arbitrary values of I1 and I2.Otherwise the inductors will be generating energy during the time interval 0t T and thus violate the passivity principle.To ensure a nonnegative W(T)for all I1 and I2,the values of L1,L2,and M must satisfy the inequalityorTo see this,we rewrite equation 18.24 in the form(18.26a)(18.26b)Equation 18.27 shows that is negative whenever is negative.Now is a second-degree polynomial in with a positive coefficient for the term.From analytic geometry,depending on the sign of the discriminant ,the curve may or may not intersect the axis,as illustrated in figure 18.12.18.27Figure 18.12 Plot of stored energy versus current ratio xFrom the figure it is obvious that if ,there will be some current ratio that yields a negative and hence a negative ,again violating the passivity principle.Therefore,which yields equaton 18.26From equations 18.27 and 18.28,(18.28)(18.29)The degree to which M approaches its upper bound is expressed by a positive number,called the coefficient of coupling,defined asWhen ,is also zero,and the inductors are uncoupled.When ,the inductors have unity coupling,an idealized situation impossible to realize in practice.Example18.10 In both circuits of figure 18.4,suppose that L1=5H,L2=20H,M=8H,I1=2A,and I2=4A.Find:(a)The coupling coefficient k.(b)The stored energy.SOLUTION(a)For both circuits,(b)For figure 18.4a,For figure 18.4b,Example18.11 In the circuit of figure 18.13,I1=6A.Find the minimum value of the stored energy and the corresponding value of I2.SOLUTIONFrom equation 18.24,Figure 18.13 Coupled inductors for calculating stored energy in example 18.11 This yields I2=2A,and the corresponding minimum stored energy Wmin=18-36+72=54J.Following the standard method in calculus for finding the maximum and minimum,we set to zero and solve for :6.Ideal Transformer as A Circuit Element And ApplicationsIdealization 1.The coupled inductors have unity coupling.i.e.,M2=L1L2,or the coupling coefficient k=1.Effect of Idealization 1.Under this idealized condition of unity coupling,the pair of coils has the voltage transformation propertyTwo coupled differential equations containing three parameters L1,L2,and M characterize the coupled coils of figure 18.4.By imposing two idealized conditions on these parameters,some very useful circuit properties result.Where a is a constant and both v1 and v2 are the voltage drops from dotted to undotted terminals of the coils,as the figure.To derive the condition of equation 18.30,note that the constraint M2=L1L2 implies that L1/M=M/L2.Denoting the ratio L1/M by a leads to L1=aM and M=aL2.Substituting these relationships into equations 18.7 and dividing equation 18.7a by equation 18.7b yields A unity coupling coefficient is an idealization that is not achievable in practice.However,coupling coefficients near unity are achievable by winding the turns of two inductors very closely together so that nearly all the flux that links one coil also links the other coil.With unity coupled coils,equation 18.30 indicates that the voltages v1(t)and v2(t),both from dotted to undotted terminals,always have the same polarity.With coupling less than unity,it is possible for v1 and v2 to have opposite polarities at some time instants,as shown in example 18.7 The constant a=L1/M=M/L2=in equation 18.30 is then simply the ratio of the numbers of turns,denoted by N1 and N2,of the two coils and is usually referred to as the turns ratio,i.e.,a=N1/N2.It is possible to show thatIdealization 2.In addition to unity coupling,the coupled coils have infinite mutual and self-inductances.Effect of Idealization 1 and 2.Under these two idealized conditions,the pair of coils has the current transformation property.Where both i1 and i2 are the currents entering the dotted terminals of the coils,as per figure 18.4a.Letting L1,L2,and M,L1/M=M/L2=a Ideal TransformerTwo coupled coils satisfying the relationshipsare said to be an ideal transformer,shown in figure 18.14a.In the figure,the two vertical bars serve as a reminder of the presence of a ferromagnetic core in the physical device.The term“ideal”may or may not appear in the schematic diagram.Again,the mathematical model of an ideal transformer depends only on the turns ratio a:1 and the relative dot positions.To avoid the negative sign in the current relationship,the alternative labeling of voltages and currents shown if figure 18.14b may be used.The subscript p stands for the primary coils,connected to a power source,and s for the secondary coil,connected to a load.Note that ip is entering at the dotted terminal and is leaving the dotted terminal.The notation of figure 18.14b is more commonly used in the study of electric power flow.For our present purposes in circuit analysis,we have defined an ideal transformer strictly from its terminal voltage-current relationships.In section 8 we will show how a practical transformer can be constructed to achieve approximately the idealized conditions k=1 and(L1,L2,M).One important simplification resulting from the idealizations is that an ideal transformer is characterized by two algebraic equations containing a single parameter a,the turns ratio.This is to be contrasted with a pair of coupled coils,characterized by two d