Chapter 4 THE SERIES RESONANT CONVERTER T he objective of this chapter is to describe the operation of the series resonant converter in detail. The concepts developed in chapter 3 are used to derive closed-form solutions for the output characteristics and steady-state control characteristics, to determine operating mode boundaries, and to find peak component stresses. General results are presented, for every continuous and discontinuous conduction mode using frequency control. The origin of the discontinuous conduction modes is explained. These results are used to consider three design problems. First, the variation of peak component stresses with the choice of worst-case operating point is investigated, and some guidelines regarding the choice of transformer turns ratio and tank characteristic impedance are discussed. Second, the effects of variations in input line voltage and output load current are examined using the converter output characteristics. Finally, switching frequency variations are considered, and the tradeoff between transformer size and tank capacitor voltage is exposed. Vg Q3 D3 D1 Q4 D4 D2 + - Q1 iS + Q2 vS + vT L C Fig. 4.1. Series resonant converter schematic. D5 I D6 + vR D7 CF V - - + D8 R Principles of Resonant Power Conversion 4 . 1 . Subintervals and Modes The series resonant converter, Fig. 2.1, is reproduced in Fig. 4.1. It can be seen that the instantaneous voltage vT(t) applied across the tank circuit is equal to the difference between the switch voltage vS(t) and the rectifier voltage vR(t): vT(t) = vS(t) – vR(t) (4-1) These voltages, in turn, depend on the conducting state of the controlled switch network and uncontrolled rectifier network. A subinterval is defined as a length of time for which the conducting states of all of the semiconductor switches in the converter remain fixed; during each subinterval, vS(t), vR(t), and vT(t) are constant. For example, consider the case where transistors Q1 and Q4 conduct, and iL(t) is positive so that diodes D5 and D8 also conduct, as in Fig. 4.2a. In this case, we have vs = +Vg vR = +V (4-2) vT = Vg – V (a) The applied tank voltage is therefore constant and equal to Vg–V. In normalized form, one obtains MT = VT = 1 – M Vg (4-3) Hence, according to section 3.3, the normalized state plane trajectory for this subinterval is a circular arc centered at MT = 1-M, as shown in Fig. 4.2b. The radius depends on the initial conditions. Note that, since we have assumed that iL(t) is positive and diodes D5 and D8 conduct, this particular switch conducting state can occur only in the upper half-plane (jL > 0). For negative jL, diodes D6 and D7 would conduct instead, MT + vT - + vC Vg +- iL > 0 L direction of current flow C CF + V R - vT = Vg - V (b) jL > 0 . would be changed, and an arc centered at a MT = 1 - M 1 mC different location would be obtained. A subinterval utilizing the switch conduction state described Fig. 4.2 Q1 conduction subinterval, in which Q1 and Q4 conduct, and iL > 0 so above and in Fig. 4.2 is referred to in shorthand that D5 and D8 conduct: a) circuit; form as subinterval Q1. b) normalized state plane trajectory. 2 Chapter 4. (a) Vg +- + iL < 0 vT - L C direction of CF current flow + (a) + vC - V iL < 0 Vg +- - . L + C CF V - vT = -Vg + V jL (b) jL < 0 1 - direction of current flow vT = Vg + V (b) jL vT + vC - + R The Series Resonant Converter -1 MT = 1 - M . MT = -1 + M mC mC jL < 0 Fig. 4.3 D1 conduction subinterval, in which iL < 0 such that diodes D1, D4, D6, and D7 conduct: a) circuit; b) normalized state plane trajectory. + vT (a) + vC - Vg +- iL > 0 L C CF direction of current flow + V - VT = -Vg - V (b) Fig. 4.4 Q2 conduction subinterval, in which Q2 and Q3 conduct, and iL < 0 so that D6 and D7 conduct: a) circuit; b) normalized state plane trajectory. Many other switch conduction states can occur. Subinterval D1 is similar to subinterval Q 1, except that the tank current iL(t) is negative. The conducting devices are antiparallel diodes D1 and D 4, and output rectifier diodes D6 and D 7. The applied tank voltage is therefore VT = Vg+V, or in normalized form, . jL > 0 MT = 1 + M (4-4) The circuit and state plane trajectory for this mC subinterval are summarized in Fig. 4.3. Note that MT = -1 - M -1 this switch conduction state can only occur in the negative half-plane (jL < 0). Fig. 4.5 D2 conduction subinterval, in Symmetrical switch conduction states Q2 which iL > 0, such that diodes D2, D3, (Fig. 4.4) and D2 (Fig. 4.5) can also occur, in D5 and D8 conduct: a) circuit; b) normalized state plane trajectory. which iL, vS, vR, and vT have the opposite polarity from states Q1 and D1 respectively. These correspond to MT = -1+M (Q2) and MT = -1-M (D2). 3 Principles of Resonant Power Conversion (a) + vT + (a) - iL > 0 L C open circuit - o oo Vg +- vT iL > 0 + Vg +- V L C + V direction of current flow - - oo vT = -V (b) (b) jL . jL = 0 jL jL > 0 . mC mC does not change mC MT = -M Fig. 4.6 Subinterval X, in which all four rectifier diodes D5, D6, D7 and D8 are reverse-biased. The inductor current remains at zero, and the tank capacitor voltage does not change: a) one possible circuit topology; b) normalized state plane trajectory. Fig. 4.7 Subinterval P1, in which D2 and Q4 (or Q1 and D3) conduct, and iL > 0 so that D5 and D8 also conduct: a) circuit; b) normalized state plane trajectory. + vT (a) Under certain conditions, it is possible for all four uncontrolled rectifier diodes (D5, D 6, + D 7, D8) to become simultaneously reverse- Vg biased. When this occurs, the circuit topology is as given in Fig. 4.6. The tank inductor is then zero, and the tank capacitor voltage remains at its (b) initial value. This switch conduction state is denoted “X”. When phase control is used, two other subintervals can occur: P 1 , which occurs for i L > 0, is summarized in Fig. 4.7, and P 2 , which occurs for iL < 0, is summarized in Fig. - + vC - o iL < 0 L + C direction of current flow CF V - vT = V jL . MT = M mC jL < 0 4.8. An operating mode is defined by a Fig. 4.8 Subinterval P2, in which D1 and Q3 (or Q2 and D4) conduct, and iL < 0 so that sequence of subintervals which combine to form D6 and D7 conduct: a) circuit; b) normalized state plane trajectory. a complete switching period. Discontinuous 4 Chapter 4. The Series Resonant Converter conduction modes contain at least one X subinterval, while continuous conduction modes contain no X subintervals. As seen later in this chapter, the different modes cause the series resonant converter to exhibit widely varying terminal characteristics. 5 Principles of Resonant Power Conversion 4 . 2 . State Plane and Charge Arguments for the k=1 Continuous Conduction Mode i L(t) State plane trajectory Typical inductor current iL(t), capacitor voltage vC(t), and applied tank voltage vT(t) waveforms are diagrammed in Fig. 4.9 for the k=1 continuous conduction mode. This mode is defined by the subinterval sequence Q1-D1-Q2-D2. In this mode, the switching period begins when the control circuit switches transistors Q1 and Q4 on, with the inductor current iL(t) positive. The state plane trajectory for this subinterval is given in Fig. 4.10a; it begins at ω0t = 0 with some initial values of tank inductor current and capacitor voltage. The tank rings with a circular state plane trajectory centered at MT = 1–M until, at ω0t = β, the inductor current rings negative, and subinterval D1 begins. The normalized state plane trajectory then follows a circular arc centered at MT = 1+M. At time ω0t = β+α ≡ γ (one half switching period), the control circuit switches transistors Q1 and Q4 off, and Q2 and Q3 are switched on. Subinterval Q2 begins, and as shown in Fig. ω0 t v C (t) +VC1 -VC1 v T(t) Vg + V Vg - V -Vg + V -Vg - V 4.10c, the trajectory continues along a circular arc centered at MT = -1+M until the inductor current again reaches zero. The output bridge rectifiers then switch, and subinterval D2 begins. The trajectory follows an arc centered at MT = -1-M for the remainder of the switching period. The switching period ends when the control circuit switches Q2 and Q3 off, and Q1 and Q4 on. If the converter operates in equilibrium, then the trajectory begins and ends at the same point in the state plane, and the tank waveforms are 6 ωT γ= 0 S 2 γ γ β subinterval : Q1 α D1 β Q2 α D2 Fig. 4.9 Typical tank inductor current iL, tank capacitor voltage, vC, and applied tank voltage vT waveforms, for the k = 1 continuous conduction mode. Chapter 4. The Series Resonant Converter periodic. Otherwise, a transient occurs in which the trajectory for each switching period begins at a different point, and follows a different path in the state plane. If the circuit is stable, then the trajectory eventually converges to a single closed path, and the waveforms become periodic. To find the converter steady-state characteristics, we need to solve the geometry of this closed path, and to relate it to the load current using charge arguments. jL (a) ω0 t = 0 radius MA1 jL Q1 subinterval β . (c) . ω0 t = β 1 - M + MA1 mC 1-M . . . ω0 t = γ + β -1 + M mC radius MA1 β ω0 t = γ Q2 subinterval (b) jL jL (d) D2 subinterval . . 1+M radius MA2 ω0 t = α + β = γ α D1 subinterval α ω0 t = β mC . . -1 - M ω0 t = γ + α + β = 2γ radius MA2 . mC ω0 t = γ + β Fig. 4.10 Construction of the state plane trajectory for one complete switching period, in the k=1 CCM: a) subinterval Q1; b) subinterval D1; c) subinterval Q2; d) subinterval D2. Capacitor charge arguments The inductor current waveform of Fig. 4.9 contains only one positive-going and one negative-going zero crossing per switching period; this is true throughout the k=1 continuous conduction mode. In consequence, the discussion of section 3.1, regarding tank capacitor charge variation, applies directly to this case and leads to a result nearly identical to Eq. (3-6). The tank 7 Principles of Resonant Power Conversion inductor current coincides with the tank capacitor current in the series tank, and hence the tank capacitor voltage vC(t) increases when the inductor current iL(t) is positive. During the half switching period where iL(t) is positive, the capacitor voltage increases from its negative peak -VC1 to its positive peak +VC1. The total change in vC is the peak-to-peak value 2VC1. This corresponds to a total increase in charge q on the capacitor, given by the integral of the positive portion of the iL as shown in Fig. 4.11. area = q Hence, we have: i L(t) q = C (2VC1) (4-5) area = -q This q is also directly related to the dc load current I. The load current is the dc component, or average value, of the rectified tank inductor current | iL |: I = < | iL | > = = 1 1 T 2 s ω0 t v C (t) 1T s 2 +VC1 | iL(τ) | dτ (4-6) 0 2q Ts since the integral in Eq. (4-6) is equal to the charge q. We can now eliminate q from Eqs. (4-5) and (4-6), and solve for VC1: VC1 = ITs 4C -VC1 area = q |i L(t)| I=<|iL|> (4-7) or, in normalized form: MC1 = Jγ 2 (4-8) where MC1 = VC1 / Vg. Fig. 4.11 Use of charge arguments to relate the peak capacitor voltage VC1, load current I, and charge quantity q in the series resonant converter for k=1 CCM. This is a useful result, because it allows us to relate the load current to the peak capacitor voltage. In the normalized state plane, we can now find the circle radii directly in terms of the normalized load voltage and current M and J. 8 Chapter 4. The Series Resonant Converter 4 . 3 . Solution of the k=1 Continuous Conduction Mode Characteristics The radii MA1 and MA2 of the state plane trajectory, re-drawn in Fig. 4.12, can now be found. At time ω0t = β, it can be seen that the radius MA1 of the Q1 subinterval is MA1 = (MC1) – (1–M) = Jγ –1+M 2 (4-9) Jγ –1–M 2 (4-10) and the radius MA2 of the D1 subinterval is MA2 = (MC1) – (1+M) = So we know the radii and centers of the circular arcs in terms of the normalized output voltage M, current J, and control input γ (or switching frequency fs = π f0 / γ). jL D2 subinterval ω0 t = 2γ . -MC1 ω0 t = γ + β α ω0 t = 0 . . . . -1 + M 1+M -1 - M β Q1 subinterval 1-M radius MA1 . MC1 = ω0 t = β Jγ 2 mC radius MA2 ω0 t = γ D1 subinterval Q2 subinterval Fig. 4.12 Complete normalized state-plane trajectory, for steady-state operation in the k=1 CCM. Finally, it is desired to find a closed-form expression that relates the steady-state output voltage, output current, and the control input; i.e., we want to directly relate M, J, and γ. In steady-state, the endpoint of the state plane plot after one switching interval ( at time ω 0 = 2γ = ω0Ts) coincides with the initial point (at time ω0t = 0), and the trajectory is closed. For a given M and J in the k=1 CCM, there is a unique set of values of γ, α, β, MA1, and MA2 which cause this 9 Principles of Resonant Power Conversion jL π − (π − α) − (π − β) = γ − π MA2 = _J_γ 2 -1-M Α M 1 A J_γ_ = 2 1 + M to happen, and which can be found by solving the geometry of the state plane. With these values, the triangle of Fig. 4.13 is formed, whose solution yields the converter steadystate characteristics. The lengths of the two radii are already known, and the length of the triangle base is the distance between the and D2 Q1 C α . B π−α π−β −1 − Μ . β 1−Μ mC 2 Fig. 4.13 Magnification of Fig. 4.12, for the solution of steady-state conditions. subinterval centers, or 2. Hence, the lengths of the three sides of the triangle are known, and are functions of only M, J, and γ. The included angles can be found using simple geometry. The two angles adjacent to the base are (π – α) and (π – β), which are functions of the unknowns α and β. The remaining angle Θ b c a2 = b2 + c2 - 2bc cosΘ a Fig. 4.14 The Law of Cosines can be found knowing that the three angles of the triangle must sum to π, and is given by π – (π – α) – (π – β) = γ – π (4-11) since γ = α + β. Note that this is a function of the control input γ alone, and does not depend separately on α or β. The Law of Cosines, Fig. 4.14, can now be used to relate the top angle and the three sides, and hence to find how M and J depend on γ. One obtains: (2)2 = (Jγ2 – 1 – M)2 + (Jγ2 – 1 + M)2 Jγ Jγ – 2( – 1 – M)( – 1 + M) cos(γ–π) 2 2 10 (4-12) Chapter 4. The Series Resonant Converter Jγ Jγ – 1)2 + 2M2 + 2[( – 1)2 – M2] cos(γ) 2 2 (4-13) Simplification yields: 4 = 2( which can be rearranged to obtain: 1 = (Jγ2 – 1)2 1 + cos γ 1 – cos γ + M2 2 2 (4-14) Trigonometric identities can now be used to obtain: M2 sin2 γ Jγ γ + ( – 1)2 cos2 = 1 2 2 2 (4-15) This is the desired closed-form solution for the series resonant converter operating in the k=1 continuous conduction mode, with frequency control. Output characteristics At a given switching frequency fs, corresponding to a given γ = π f0 / fs, Eq. (4-15) shows that the relation between M and J is an ellipse, centered at M = 0 and J = 2/γ. A typical ellipse is plotted in Fig. 4.15. no solutions The uncontrolled output J for M ≥ 1 rectifier diodes do not allow the (DCM instead) load current to be negative, so we must have J > 0. Also, with valid passive load requires a passive load, M must be solutions I and V to have same positive when J is positive. polarity 1 Hence, the portions of the M ellipse that lie in the second, bridge rectifier does not allow I < 0 third, and fourth quadrants are not valid physical solutions. Also, it is shown in section 4.4 Fig. 4.15 Elliptical output characteristic M vs. J (Eq. 4-15), that the solution is not valid for for a given γ in the k = 1 CCM. M≥1; instead, the k=1 discontinuous conduction mode occurs for M=1. Hence, the solution is valid only for 0 ≤ M < 1 and J > 0. Equation (4-15) is plotted in Fig. 4.16. It can be seen that as the load current (or J) is increased, the output voltage (or M) decreases. Hence, the output impedance of the open-loop converter is substantial. It is instructive to examine some limiting cases. 11 Principles of Resonant Power Conversion 5 F = 0.95 F = 1.0 F = 0.925 4 F = 0.9 J 3 2 F = 0.8 F = 0.7 1 2 π F = 0.6 F = 0.5 0 0 0.2 0.4 0.6 0.8 1 M Fig. 4.16 At Output characteristics of the series resonant converter, operating in the k = 1 continuous conduction mode. Solutions occur over the range 0 ≤ M <1, 2/π ≤ J < ∞; solutions not shown here for J > 6. F = 0.5 (half resonance): Then fs = 0.5 f0, and γ = π / 0.5 = 2π. The output characteristic, Eq. (4-15), becomes M 2 ⋅ 0 + ( π J – 1) 2 ⋅ 1 = 1 (4-16) or, J = 2/π (4-17) which is independent of M. The ellipse collapses to a horizontal line, and the converter operates as a current source. At F = 1.0 (resonance): Then fs = f0, and γ = π / 1 = π. The output characteristic, Eq. (4-15), becomes M 2 ⋅ 1 + ( J π – 1) 2 ⋅ 0 = 1 2 (4-18) or, M = 1 (4-19) which is independent of J. The ellipse collapses to a vertical line, and the converter operates as a voltage source. For 0.5 < F < 1, the converter operates as neither a voltage source nor a current source. 12 Chapter 4. The Series Resonant Converter Value of J when M = 1 The critical minimum value of J occurs when M = 1; for J less than this value, the converter does not operate in k = 1 CCM, and Eq. (4-15) is not valid. Plugging M = 1 into Eq. (415) yields γ 1 – sin2 2 2 = 1 (J2π – 1) = (γ ≠ π) (4-20) γ cos2 2 or, J = 4/γ (4−21) which varies between 2 / π and 4 / π for F between 0.5 and 1. Output short circuit current J S C (4-23) Equation (4-23) is plotted in Fig. 4.17. It can be seen that the converter short-circuit current is inherently limited, except at resonance. 10 8 6 Jsc When M = 0, Eq. (4-15) becomes (4-22) (JSC2 γ – 1)2 cos2 2γ = 1 Solve for JSC: γ JSC = 2 (1 + | sec |) = 2F (1 + | sec π |) 2 2F γ π 4 2 0 0.5 0.6 0.7 0.8 0.9 1 F The characteristics of a given load can be Fig. 4.17 Normalized short-circuit output current JSC vs. switching frequency, in superimposed on the converter output the k=1 CCM. characteristics, allowing graphical determination of output voltage vs. switching frequency. For example, with a linear resistive load, I = V/R (4-24) or, in normalized form, J = MQ (4-25) with Q = R0 / R. Equation (4-25) describes a line with slope Q, as shown in Fig. 4.18. The intersection of this load line with the converter elliptical output characteristic is the steady-state operating point for a given switching frequency. As shown in the example of Fig. 4.19, nonlinear load characteristics can also be superimposed, and the operating point determined graphically. 13 Principles of Resonant Power Conversion F = 1.0 5 F = 0.95 4 F = 0.925 J F = 0.9 3 ine l load 2 F = 0.8 F = 0.7 F = 0.6 1 2 π Q= 0 0 R0 R 0.2 F = 0.5 0.4 0.6 0.8 1 M Fig. 4.18 Resistive load line superimposed over the converter output characteristics. F = 1.0 5 F = 0.95 4 F = 0.925 J F = 0.9 3 e 2 ad F = 0.8 lin lo F = 0.7 1 2 π F = 0.6 F = 0.5 0 0 0.2 0.4 0.6 0.8 1 M Fig. 4.19 Nonlinear load characteristic superimposed over the converter output characteristics. 14 Chapter 4. The Series Resonant Converter Control plane characteristics It is also instructive to plot the voltage conversion ratio M vs. normalized switching frequency F. Doing so requires knowledge of the load characteristics, so that J can be eliminated from Eq. (4-15). In the case of a resistive load satisfying V = I R, Eq. (4-25) can be substituted into Eq. (4-15), yielding: M2 sin2 γ MQγ γ +( – 1)2 cos2 = 1 2 2 2 (4-26) Now solve for M: M2 [sin2 γ Qγ 2 2 γ + cos 2 2 2 ] – M Qγ cos2 2γ γ + (cos2 – 1) = 0 2 (4-27) Use of the quadratic formula yields: Qγ 2 M = γ Qγ tan2 + 2 2 2 1+ 2 Qγ 1± 2 (tan2 2γ ) (tan2 2γ + Qγ 2 2 ) (4-28) 1 Q=1 0.8 Q=2 0.6 M Q=5 0.4 Q = 20 0.2 0 0.5 0.6 0.7 0.8 0.9 1 F Fig. 4.20 Steady-state control characteristics M vs. F for various values of Q = R0/R in the k=1 continuous conduction mode, Eq. (4-28). 15 Principles of Resonant Power Conversion To obtain the correct solution, in which M > 0, the plus sign should be used. Equation (4-28), together with the identity γ = π / F = π f0 / fs, is a closed-form representation of the control characteristics M vs. F for the k=1 CCM. It is plotted in Fig. 4.20 for various values of Q, or load resistance R. As R is decreased, corresponding to heavy loading, the Q is increased. Loading the converter causes the output voltage to decrease, and results in a peaked characteristic near resonance. Control of diode conduction angle α Another popular scheme for controlling the series resonant converter when it operates below resonance is known as diode conduction angle control, or “α control”. Rather than using a voltage controlled oscillator to cause the switching frequency and γ to be directly dependent on a control signal (known as frequency control, or “γ control”), the α controller causes the diode conduction time and angle α to be directly dependent on the control signal. The switching frequency varies indirectly, and depends on both α and the load current. This control scheme requires a current monitor circuit which senses the zero crossings of the tank current waveform. A timing circuit then causes the transistors to switch on after a delay which is proportional to a control voltage. The transistor off time, which coincides with the diode conduction time, is therefore proportional to the control voltage. To understand the converter characteristics under α control, we need to eliminate γ from the k=1 CCM solution, Eq. (4-15), in favor of α. This can be done by again referring to Fig. 4.13. Application of the Law of Cosines, using the included angle at vertex C, yields Jγ –1+M 2 2 = 22 + Jγ –1–M 2 2 – 2 (2) Jγ – 1 – M cos(π–α) 2 (4-29) This equation can be solved for γ: (1 + M) (1 – cos( α)) γ = 2 J (M – cos(α)) (4-30) This describes how the switching frequency varies for a given range of α and load. Equations (430) and (4-15) can now be used to eliminate γ, and to determine the α control characteristics. The result is J = where (1 + M) (1 – cos( α)) sin(α) (M – cos(α)) π – tan–1 M – cos(α) (4-31) 0 ≤ tan–1(•) ≤ π/2, and M > cos(α). Equation (4-31) describes the output characteristics under α control, and is plotted in Fig. 4.21. It can be seen that the output characteristics resemble hyperbolae, with vertical asymptotes 16 Chapter 4. The Series Resonant Converter M = cos(α). Also, comparison with Fig. 4.16 reveals that the switching frequency approaches resonance (fs →f0) as α→0, and fs→0.5f0 as α→π. Decreasing α causes M and/or J to increase. It can also be seen that, for α < π/2, the converter short-circuit current is not inherently limited. 6 5 4 α = .3π α = .2π α=0 α = .4π J 3 α = .5π 2 α = .6π α = .7π 2/π 0 α=π 0 0.25 0.5 0.75 1 M Fig. 4.21 k=1 CCM output plane characteristics, diode conduction angle control. Mode boundaries, k=1 CCM So far, we have studied only the k=1 continuous conduction mode, characterized by the subinterval sequence Q1–D1–Q2–D2. The state plane diagram of Fig. 4.12 and the succeeding analysis are both based on the assumption that the transistors and diodes conduct in the order given by this sequence. As stated previously, the diode conduction angle α (i.e., the lengths of the D1 and D2 subintervals) vanishes as the switching frequency approaches resonance. Increasing the switching frequency beyond resonance must therefore cause a different subinterval sequence to occur. Likewise, as fs approaches 0.5f0, the diode conduction angle α and transistor conduction angle β both approach π, or a complete resonant half-period. For the series resonant converter, no ringing subinterval can extend through an angle of more than π radians, because the state plane centers and output diode switching boundary both lie on the jL = 0 axis. Therefore, decreasing the switching 17 Principles of Resonant Power Conversion frequency below fs = 0.5f0 must cause new subintervals to occur. Hence, the k=1 CCM and Eq. (4-15) are restricted in validity to the range 0.5f0 ≤ fs ≤ f0. By examination of Fig. 4.16, it can be seen that the k=1 CCM solutions do not extend beyond the range J ≥ 2/π. If it is desired to operate the converter at light loads corresponding to J < 2/π, then a different mode must be used, most likely with a different range of switching frequencies. In addition, as shown in the next section, the k=1 CCM is restricted to the range 0 ≤ M < 1. 4 . 4 . Discontinuous Conduction Modes At light loads, all four bridge rectifier diodes can become reverse-biased during part of the switching period, causing the converter to operate in a discontinuous conduction mode. More that one discontinuous conduction mode is possible, depending on the load current. Each of these is characterized by a sequence of subintervals ending in subinterval X, Fig. 4.6. The k=1 discontinuous conduction mode This mode is defined by the subinterval sequence Q1-X-Q2-X. As shown in Fig. 4.22, transistor Q1 conducts for a complete tank half-period. The four bridge rectifier diodes then become reverse-biased, and subinterval X occurs for the remainder of the half switching period. As in the k=1 continuous conduction mode, the dc load current I is given by: 2q Ts where q, shown in Fig. 4.22, is: I = <| iL |> = (4-32) Ts/2 q = iL (t) dt (4-33) 0 The average input current is: Ts/2 < ig > = 1 1 2T s iL (t) dt (4-34) 0 since ig = iL when Q1 conducts. Substitution of Eqs. (4-32) and (4-33) into Eq. (4-34) yields: < ig > = 2q = I Ts (4-35) 18 Chapter 4. Q1 X Q2 The Series Resonant Converter So the converter dc input and output currents are equal. If the converter is lossless and operates in equilibrium, then the input and output powers must be equal; this implies that the voltages are also equal: X iL q Pin = Vg < ig > = Pout = V I ω 0t γ−π π Use of Eq. (4-35) yields: -q Vg = V vC or: VC1 M = 1 (4-37) Hence, in the k=1 DCM, the converter dc conversion ratio M is unity, and is independent of the values of load current and switching frequency. The usual tank capacitor charge arguments can be used to complete the solution and compute the peak tank capacitor voltage VC1. During the Q1 -VC1 vT : (4-36) Vg -V -Vg +V q q |iL| <|iL|> conduction subinterval, the charge on the tank capacitor changes by an amount q, corresponding to an increase in voltage of 2VC1 (see Fig. 4.23). Hence: q = C · (2VC1) ω 0t Elimination of q using Eq. (4-32) yields: I Ts VC1 = 4C Fig. 4.23 Tank inductor current and capacitor voltage waveforms for the k = 1 DCM. or, in normalized form: MC1 = (4-38) Jγ 2 (4-39) (4-40) Equation (4-40) happens to be identical to the result for the k=1 CCM, Eq. (4-8). Beware, this does not occur for all other operating modes. This mode cannot occur above resonance. The switching period must be long enough that the tank can ring through one complete Q1 subinterval of length π during each half switching period of length γ. Hence, a necessary condition for the occurrence of the k=1 DCM is: 19 Principles of Resonant Power Conversion γ>π (4-41) or, in terms of F: F>1 (4-42) An additional necessary condition for occurrence of the k=1 DCM is given in the next subsection. Reason for occurrence of the k=1 DCM Why does the tank stop ringing at the end of the Q1 subinterval? As suggested previously, the reason is that all four bridge rectifier diodes become reverse-biased at this instant. Physical arguments are used in this subsection to prove this assertion, and to derive the conditions on load current and frequency which lead to operation in this mode. These arguments also have a very simple state-plane interpretation. As seen in Fig. 4.1, the voltage applied to the tank inductor vL is: vL = L diL = vS – vC – vR dt (4-43) During subinterval Q 1, vS = V g, and vR = V. We have already shown that V = Vg in this mode (Eq. (4-37)), and so vL becomes vL = L diL = – vC dt (4-44) for subinterval Q1. vL(t) is plotted in Fig. 4.23. At ω0t = 0, this is a positive quantity since, as shown in Fig. 4.22, vC(0) = -VC1. So initially, diL/dt is positive and iL increases. At ω0t = π/2, vL(t) passes through zero, and iL begins to decrease. At ω0t = π, iL reaches zero. Can the inductor current iL continue to decrease for ω0t ≥ π? This is possible only if the applied inductor voltage vL continues to be negative. Note that, if iL rings negative, then the bridge rectifier will switch from vR = +V to vR = –V, and subinterval D1 will occur. The applied tank inductor voltage, Eq. (4-41), would then become: vL = L diL = Vg + V – vC = 2Vg – vC dt (4-45) for subinterval D1, with vC(π) = VC1. Note that it is possible for this voltage to be either positive or negative at ω0t = π, depending on whether or not VC1 is greater than 2Vg. In the k=1 discontinuous conduction mode, 2Vg–VC1 is a positive quantity. As a result, diodes D6 and D7 cannot turn on at ω0t = π: doing so would require that iL become negative, which cannot occur if vL is positive (since iL(π)=0). Instead, all four bridge rectifier diodes become reverse-biased. 20 Chapter 4. k = 1 DCM iL The Series Resonant Converter k = 1 CCM iL ω 0t Q1 vC X Q2 ω 0t X vC Q1 D1 Q2 D2 VC1 -VC1 vL vL Vg -V-vC(t) Vg+V-VC1 > 0 Vg -V-vC(t) Vg +V-vC(t) 2Vg 2V 2Vg 2V -Vg +V-vC(t) Fig. 4.23 -Vg +V-vC(t) Comparison of tank waveforms of the k=1 continuous and discontinuous conduction modes. The inductor voltage and current remain at zero for the rest of the half-switching-period. Inductor voltage and current waveforms for the k=1 DCM and k=1 CCM are compared in Fig. 4.23. Hence, the requirement for the k=1 DCM to occur is: Vg + V – vC(π) = 2Vg – vC(π) > 0 In normalized form, this can be written: 21 (4-46) Principles of Resonant Power Conversion 1 + M – MC1 > 0 Substitution of Eqs. (4-37) and (4-40) into this expression yields: (4-47) J < 4γ (4-48) This is the basic condition for operation in this mode. It can be seen that the k=1 DCM occurs at light load. A simple state plane interpretation The above arguments can be given a simple geometrical interpretation in the state plane. As shown in chapter 3, for the series tank circuit the state plane trajectories evolve in the clockwise direction about the applied tank voltage. Consider the hypothetical state plane trajectories of Fig. 4.24a. At ω0t = π, jL reaches zero and mC = MC1. The figure is drawn for the case MC1 < (1+M). Note that a D1 subinterval cannot occur for ω0t > jL (a) Q1 D1 A ? . .. Q1 D1 (1-M) (1+M) mC mC = MC1 at ω0 t = π D1 B ? (b) jL Q1 π, since such a subinterval would involve a trajectory centered at mC = (1+M), and given either X . .. by hypothetical trajectory A or B. Trajectory A is impossible, because subinterval D1 cannot occur except for negative jL. Trajectory B is also impossible, because it does not travel clockwise about the center mC = (1+M). Hence, there can be no D1 subinterval, and instead an X subinterval occurs (as described in Fig. 4.6) in which mC remains constant and equal to MC1, as shown in (1-M) (c) jL MC1 (1+M) mC Q1 . .. Q1 D1 mC Fig. 4.24b. (1-M) (1+M) MC1 In contrast, a CCM trajectory is shown in D1 Fig. 4.24c. In this case, MC1 > (1+M), so that for negative jL the trajectory is able to evolve in the Fig.4.24 Hypothetical state plane trajectories: (a) for MC1 < (1+M), clockwise direction about the center mC = (1+M). trajectories A and B are impossible; (b) Thus, the geometry of the state plane actual k=1 DCM trajectory, with MC1 < (1+M); (c) actual k=1 CCM trajectory, trajectory gives a simple interpretation of the with MC1 > (1+M). boundary condition between the continuous and discontinuous conduction modes. 22 Chapter 4. The Series Resonant Converter The k=2 discontinuous conduction mode In the k=2 discontinuous conduction mode, the tank rings for two complete half-cycles during each half-period of length Ts/2. After the two complete half-cycles, the output bridge rectifier diodes become reverse-biased. Waveforms for this mode are given in Fig. 4.25. For this mode to occur, the switching period must be at least as long as twice the tank natural period T0. Hence, γ ≥ 2π, or F ≤ 1 2 (4-49) This is a useful mode, because the output is easily controllable: the converter output behaves as a Q1 D1 π π X Q2 D2 π π X iL γ vL=LdiL dt γ Vg -V-vC Vg +V-vC 2V 2V Vg -V-vC(2π) 2Vg vC VC2 VC1 -VC1 Fig. 4.25 Tank waveforms for the k=2 discontinuous conduction mode. 23 ω0 t Principles of Resonant Power Conversion current source, of value controllable by the switching frequency. Also, the switch turn-on and turn-off transitions both occur at zero current, so switching losses are low. A disadvantage of this mode is its higher peak transistor currents than in the continuous conduction mode, and hence higher conduction losses. As is shown in this subsection, the load current and switching frequency are directly related for the k=2 DCM, and hence a wide load current specification implies that the switching frequency must vary over a wide range. This is less of a disadvantage than one might at first think, because the transformer can be sized to the maximum switching frequency (0.5 f0 ). Operating point variations that cause the switching frequency to vary below this value do not necessitate use of a larger, lower frequency transformer. Analysis First, capacitor charge arguments are used to relate the peak tank capacitor voltage to the dc load current. The tank inductor current iL(t), which coincides with the tank capacitor current, is re drawn in Fig. 4.26. The total charge contained in the negative portion of the iL(t) waveform is defined as –q. The peak-to-peak tank capacitor voltage is 2VC2, which represents a change in capacitor charge of q. iL total charge = -q ω0 t Fig. 4.26 Tank inductor current waveform, with emphasis on total charge flowing through the tank capacitor. Hence: 2 VC2 = q C (4-50) The average output current I is: I = <| iL |> = 2q Ts (4-51) Elimination of q from Eqs. (4-50) and (4-51) yields: 24 Chapter 4. The Series Resonant Converter I = 4CVC2 Ts (4-52) Normalization of Eq. (4-52) and solution for the normalized peak capacitor voltage MC2 yields: MC2 = Jγ 2 (4-53) This result is again similar to the k=1 DCM and k=1 CCM cases. It states that the peak tank capacitor voltage is proportional to the load current. jL Q1 ) M 1- D2 M . . . . -MC2 -( . . . . MC2 = MC1 -MC1 (-1+M) (-1-M) C2 (1-M) X X Jγ 2 (1+M) mC D1 MC2 - (1+M) Q2 Fig. 4.27 State plane diagram for the k=2 discontinuous conduction mode. The state plane diagram for the k=2 DCM is given in Fig. 4.27. The switching period begins with (mC, jL) = (–MC1,0). Subintervals Q1 and D1 are of length π, and are given by semicircles in the state plane. Their radii are MC2–(1-M) and MC2–(1+M), respectively, as indicated on the diagram. Subinterval X ends the half switching period, with (mC, jL) = (+MC1,0). The converter output characteristics for this mode can be found in a manner similar to that used for the k=1 CCM. If the converter operates in steady state, then the state plane diagram is closed, as indicated in Fig. 4.27. The ending point of the D2 subinterval must therefore coincide with the beginning of the Q1 subinterval. The portion of the state plane in the vicinity of this point is magnified in Fig. 4.28. It can be seen that, in steady state, the sum of the radii of the D2 and Q1 subintervals is equal to the distance between their centers, or 2 = (MC2 – 1 – M) + (MC2 – 1 + M) 25 (4-54) Principles of Resonant Power Conversion Simplification yields MC2 = 2 = jL Jγ 2 (4-55) radius = MC2 -1-M . Now solve for the normalized load current J: J = 4 = 4F γ π (4-56) Hence, the load current depends on switching frequency but not on output voltage, and the converter behaves as a current source in this mode. As sketched in Fig. 4.29, the k=2 DCM output characteristics are straight horizontal lines. With a resistive load, we have J = MQ = 4 F π (-1-M) radius = MC2 -1+M . -MC1 (1-M) 2 Fig. 4.28 Illustration of relations between MC1, MC2 and M. k=1 M=1 DCM J k=1 CCM (4-57) F = 1/2 2 π with Q = R/R0. Solution for M then yields the control plane characteristic: M 1 This result is used in the next section, for derivation of the mode boundaries. 1 3 1/π k=1 DCM; M=1 M (4-59) k=2 DCM boundaries 1 Fig. 4.29 Output plane characteristics, emphasizing k=2 DCM. π Q= the state plane diagram of Fig. 4.27, it can be seen that MC1 is equal to MC2, minus twice the radius of the D1 subinterval: 1 3 .5/π Thus, the control plane characteristics (M vs. F , for a linear resistive load) vary linearly with F. The solution can be completed by solving for the normalized voltage MC1. By inspection of MC1 = MC2 – 2(MC2 – 1 – M) = 2M F = 1/4 k=2 DCM J = 4F π F=0 (4-58) Q=0 M = 4 F πQ mC k=2 DCM M=4 F πQ Q = 2/ 3/π = Q 4/π Q= k=1 CCM other modes As noted previously, this mode is restricted to the frequency range γ ≥ 2π, or F ≤ 0.5. Since, according to Eq. (4-56), J and F are directly related, this restriction also places an upper limit on 26 1 4 1 2 Fig. 4.30 Control plane characteristics, emphasizing k=2 DCM. F Chapter 4. The Series Resonant Converter the load current J: J = 4F ≤ 2 π π (4-60) Hence, the k=2 DCM is restricted to the portion of the output plane below J = 2/π. In addition, for the k=2 DCM to occur, the output bridge rectifier (1) must continue to conduct at ω0t = π (between the Q1 and D1) subintervals, and (2) must become reverse-biased at ω0t = 2π (after the D1) subinterval. By use of the state plane, it can be seen that, requirement (2) leads to the constraint: MC1 > 1 – M (4-61) and requirement (1) leads to: MC2 > 1 + M (4-62) By substitution of Eqs. (4-55) and (4-59), and after a small amount of algebra, these two constraints become: 1 > M > 1 3 (4-63) To summarize, the k=2 DCM boundaries are: 1 > M > 1 3 2 > J > 0 π 1 2 (4-64) > F > 0 The k=2 DCM output plane boundaries are given in Fig. 4.29, and the control plane characteristics in Fig. 4.30. As M→1, the converter enters the k=1 DCM, while for M→1/3, the converter enters a higher-order (k≥2) continuous or discontinuous conduction mode. 4.5. A General Closed-Form Solution The steady-state solutions for all frequency-controlled continuous and discontinuous conduction modes are stated here. Type k CCM The type k continuous conduction mode occurs over the frequency range f 0 < f < f0 S k+1 k (4-65) 27 Principles of Resonant Power Conversion CCM WAVEFORMS type k CCM; k odd iL γ Q1 Q1 π ∫ ∫ π D1 Q1 π D2 π D1 ω0 t symmetrical Q2 (k-1) complete half-cycles type k CCM; k even γ iL Q1 D1 π π D1 Q1 π ∫ ∫ Q1 π D1 D2 Q2 symmetrical k complete half-cycles Fig. 4.31 Tank inductor current waveforms for the general type k continuous conduction mode. The output plane characteristics are elliptical, and are described by the equation γ Jγ γ + 1 ( + (-1)k)2 cos2( ) = 1 ) 2 2 2 2 ξ where ξ is the subharmonic index, 2 M2 ξ sin2( 1 + (-1)k 2 The voltage conversion ratio M is restricted to the range (4-66) ξ = k+ (4-67) 0 ≤ M ≤ 1 ξ (4-68) At light load, the converter may enter a discontinuous conduction mode. Typical tank current waveforms are shown in Fig. 4.31. For k even, diode D1 conducts first, for a fraction of a half resonant cycle. If k is odd, then transistor Q1 conducts first, for a time 28 Chapter 4. The Series Resonant Converter less than one complete half-cycle. In either case, this is followed by (ξ - 1) complete half-cycles of ringing. The half-switching-period is then concluded by a subinterval shorter than one complete resonant half-cycle, in which the device that did not initially conduct is on. The next half switching period then begins, and is symmetrical. The steady-state control plane characteristic can be found for a resistive load R by substituting J = M Q into Eq. (4-66), where Q = R0 / R. Use of the quadratic formula and some algebraic manipulations yields: Qγ 2 M = γ Qγ 4 ξ tan2 + 2 2 2 (-1)k+1 + 1+ (ξ2 – cos2 2γ ) (ξ4tan2 2γ Qγ 2 γ cos2 2 2 + Qγ 2 2 ) (4-69) This is the closed-form relationship between the switching frequency and the voltage conversion ratio, for a resistive load. It is valid for any continuous conduction mode k. Type k DCM, k odd The type k discontinuous conduction modes, for k odd, occur over the frequency range f s < f0 k (4-70) In these modes, the output voltage is independent of both load current and switching frequency, and is described by (4-71) M = 1 k This mode occurs for the range of load currents 2(k + 1) 2(k – 1) > J > γ γ (4-72) In the odd discontinuous conduction modes, the tank current rings for k complete resonant half-cycles. All four output bridge rectifier diodes then become reverse-biased, and the tank current remains at zero until the next switching half-period begins, as illustrated in Fig. 4.32. A dc equivalent circuit for the SRC operating in an odd discontinuous conduction mode is given in Fig. 4.33; it is a “dc transformer” with effective turns ratio 1:1/k. Converters are not usually designed to operate in these modes, because the output voltage cannot be controlled by variation of the switching frequency. Nonetheless, all converters designed to perform below resonance can operate in one or more odd discontinuous conduction modes if the load current is sufficiently small, and hence the converter designer should be aware of their existence. 29 Principles of Resonant Power Conversion (a) iL γ Q1 π π X Q1 ∫ ∫ D2 π D1 X ω0 t Q2 symmetrical k complete half-cycles (b) 4 π J 1.0 8 3π 12 5π k=1 DCM k=3 k=5 DCM DCM etc. 1 5 Fig. 4.32 1 3 1 General type k discontinuous conduction mode, k odd: a) tank inductor current waveform; (b) output characteristics. Ig 1:1 k I 0 0 Fig. 4.33 M 0 + + Vg V - - 0 Steady-state equivalent circuit model for an odd discontinuous conduction mode: an effective dc transformer. 30 Chapter 4. The Series Resonant Converter Type k DCM, k even The type k discontinuous conduction modes, for k even, occur over the frequency range f fs < 0 k (4-73) These modes have current source characteristics, in which the load current is a function of switching frequency and input voltage, but not of the load voltage. The output relation is: J = 2k γ (4-74) Operation in this mode occurs for: (4-75) 1 > M > 1 k–1 k+1 In the even discontinuous conduction modes, the tank current rings for k complete resonant half-cycles. All four output bridge rectifier diodes then become reverse-biased, and the tank current remains at zero until the next switching half-period begins, as illustrated in Fig. 4.34a. The series resonant converter possesses some unusual properties when operated in an even discontinuous conduction mode. A dc equivalent circuit is given in Fig. 4.35; it is a gyrator with gyration conductance g = 2k/γR0. The gyrator has the property of transforming circuits into their dual networks; in this case, the gyrator characteristic effectively turns the input voltage source Vg into its dual, an output current source of value g V g [9]. Some very large converters have been designed to operate purposely in the k = 2 DCM at light load [ref Hughes] and even at full load [10]. 31 Principles of Resonant Power Conversion (a) iL γ Q1 Q1 π π ∫∫ π D1 X π D2 D1 Q2 ω0 t symmetrical k complete half-cycles (b) J 1.0 2 π F = .2 k=4 DCM F = 0.4 etc. k=2 F = 0.25 DCM F = 0.1 F=.05 1 5 Fig. 4.34 1 3 M 1 General type k discontinuous conduction mode, k even: (a) tank inductor current waveform; (b) output characteristics. Ig = gV I = gVg 0 0 + g Vg 0 Fig. 4.35 - + V g = 2k γR0 - 0 Steady-state equivalent circuit model for an even discontinuous conduction mode: an effective gyrator. The converter exhibits current source characteristics. 32 Chapter 4. 1.0 The Series Resonant Converter Q=0.2 0.9 Q=0.2 0.8 Q=0.35 0.7 Q=0.5 0.6 Q=0.35 Q=0.75 M 0.5 0.3 Q=1 Q=0.5 0.4 Q=0.75 Q=1 Q=1.5 Q=1.5 Q=2 Q=2 Q=3.5 Q=5 0.2 0.1 Q=3.5 Q=5 Q=10 Q=20 Q=10 Q=20 0.0 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 F Fig. 4.36. Complete control plane characteristics of the series resonant converter, for the range 0.2 ≤ F ≤ 2. M k=1 DCM 1 k=2 DCM 1 5 k=3 DCM k=4 DCM k=5 DCM etc. 1 5 1 4 k=2 CCM 1 3 k=0 CCM k=1 CCM k=3 CCM 1 2 1 3 1 2 1 Fig. 4.37. Complete control plane characteristics for continuous and discontinuous conduction mode boundaries. 33 F Principles of Resonant Power Conversion Composite characteristics The complete control plane characteristics can now be plotted using Eqs. (4-65) - (4-75). The result is shown in Fig. 4.36, and the mode boundaries are explicitly diagrammed in Fig. 4.37. It can be seen that, for operation above resonance, the only possible operating mode is the k=0 CCM, and that the output voltage decreases monotonically with increasing switching frequency. Reduction in load current (or increase in load resistance, which decreases the Q) causes the output voltage to increase. A number of successful designs that operate above resonance and utilize zerovoltage switching have been documented in the literature. Operation below resonance is complicated by the presence of subharmonic and discontinuous conduction modes. The k=1 CCM and k=2 DCM are well behaved, in that the output voltage increases monotonically with increasing switching frequency. Increase in load current again causes the output voltage to decrease. Successful designs which operate in these modes and employ zero-current switching are numerous. However, operation in the higher-order modes (k=2 CCM, k=4 DCM, etc.) is normally avoided. Given F and Q, the operating mode can be evaluated directly, using the following algorithm. First, the continuous conduction mode k corresponding to operation at frequency F with heavy loading is found: k = INT 1 F (4-76) where INT(x) denotes the integer part of x. Next, the quantity k1 is determined: k1 = INT 1 + 2 1 + Qπ 4 2F (4-77) The converter operates in type k CCM provided that: k1 > k (4-78) Otherwise, the converter operates in type k1 DCM. A computer listing is given in Appendix 1, in which the conversion ratio M is computed for a given F and Q. First, the above algorithm is used to determine the operating mode. Then, the appropriate equation (4-69), (4-71), or (4-74) is evaluated to find M. The function works correctly for every frequency-controlled mode. Output plane characteristics for the k=0 CCM, plotted using Eq. (4-66), are shown in Fig. 4.38. The constant-frequency curves are elliptical, and all pass through the point M=1, J=0. Output plane characteristics which combine the k=1 CCM, k=1 DCM, and k=2 DCM are shown in Fig. 4.39. These were plotted using Eqs. (4-66), (4-71), and (4-74). It can be seen that the constant-frequency curves are elliptical in the continuous conduction mode, vertical (voltage source characteristic) in the k=1 DCM, and horizontal (current source characteristic) in the k=2 DCM. 34 Chapter 4. The Series Resonant Converter 6 F = 1.05 F = 1.07 5 F = 1.10 4 J F = 1.01 3 F = 1.15 2 F = 1.30 1 0 0 0.2 0.4 0.6 0.8 1 M Fig. 4.38 Output characteristics in the k=0 continuous conduction mode (above resonance). F = 1.0 J F = .93 3 F = .96 F = .90 2.5 F = .85 2 k=1 CCM 1.5 F = .75 4 π k=1 DCM 1 F = .5 2 π F = .25 F = .1 k=2 DCM 0 0 0.2 0.4 0.6 0.8 1 M Fig. 4.39 Composite output characteristics for the k=1 CCM, k=1 DCM and k=2 DCM modes. Mode boundaries are indicated by heavy lines. 35 Principles of Resonant Power Conversion 4 . 6 . Converter Losses If the transistors and diodes can be modeled as constant voltage drops while they conduct, then there is a relatively easy way to adapt the ideal analysis of the previous sections to include these losses. The transistor and diode forward voltage drops cause a different voltage to be applied to the tank circuit during each subinterval, which can be accounted for by changing the effective Vg and V seen by the tank circuit. Let us consider the half-bridge example of Fig. 4.40. The semiconductor Vg device forward voltage drops are defined as follows: Vg VT transistor on-state voltage VDin antiparallel diode forward voltage VDout output diode forward voltage VDin + VT - VDout + VT + V - The applied tank voltage waveform vT is shown in Fig. 4.41 for the k=1 CCM. Since the applied tank voltage is (by Fig. 4.40 Half-bridge circuit with nonideal semiconductor devices. assumption) constant during each subinterval, the normalized state plane trajectories are again a series of circular arcs, and the preceding analytical method can be used to solve this converter. In fact, it is not necessary to rederive the entire analysis, because the converter circuit of Fig. 4.42 (in which the input and output voltages are modified but the converter is otherwise ideal) exhibits exactly the same tank voltage waveform, vT of Fig. 4.41. In consequence, the ideal k=1 CCM solution previously derived, Eqs. (4-15) and (4-28), also apply to this converter, but with the input and output voltages replaced by the effective values: Vg eff = 1 2 VD in – 1 2 VT + actual Vg (4-79) VT Vg +VDin +V+2VDout Vg -VT -V-2VDout Q1 D1 Q2 D2 -Vg +VT +V+2VDout -Vg -VDin -V-2VDout Fig. 4.41 Actual tank voltage waveforms for the half-bridge circuit of Fig. 4.40. 36 ω0 t Chapter 4. 1 VDin 2 1 VT 2 actual Vg + 1 VDin 2 Fig. 4.42 ideal switches - - + ideal diodes (no forward drop) + actual Vg 1 VT 2 The Series Resonant Converter VT 1 VDout + 2 1 VDin + 2 + 1 VT 2 - + + - + actual V - Half-bridge circuit with explicit voltage drops due to non-ideal diodes and transistors. Veff = 2 VD out + 1 2 VD in + 1 2 VT + actual V (4-80) Note that these effective values must also be substituted into the normalization base quantities. The result is given below: Meff = Veff Vg eff (4-81) Jeff = I R0 Vg eff (4-82) 2 M2eff ξ sin2( γ J γ γ + 1 ( eff + (-1)k)2 cos2( ) = 1 ) 2 2 2 2 ξ (4-83) The effect of the semiconductor forward voltage drops is primarily to shift the output plane characteristics somewhat to the left, and hence to lower the actual output voltage. It is also possible, but much more complicated, to account for resistive losses such as tank inductor and capacitor equivalent series resistance (esr), shown in Fig. 4.43. These losses effectively damp the tank circuit, and cause the normalized state plane trajectories to become spirals of decreasing radius. Closed-form solutions are not known; instead, computer iteration can be used to evaluate the highly transcendental equations that are obtained. The results show that the converter is very sensitive to small resistive loss elements when operated at large values of normalized current J. This coincides with operation near resonance with a large Q = J/M. The 37 Principles of Resonant Power Conversion same conclusions can be obtained more simply using the approximate sinusoidal methods of chapter 2. RS RS Re L C esr Vg RS RDout RDout RDout RDout + RS V _ Fig. 4.43 Half-bridge circuit with explicit resistances due to non-ideal diodes and switches. 4 . 7 . Design Considerations It may not be initially apparent how to best design a series resonant converter to meet given design objectives, yet a sub-optimal design may exhibit significantly higher tank current and voltage stresses than necessary, and/or may cause the switching frequency to vary over an unacceptably wide range. In this section, component stresses are analyzed, and the results are overlaid on the output plane plots derived in the previous sections. Also, specifications on the input voltage and output power ranges are translated into a region in the same output plane, so that it can be seen how component stresses and switching frequency vary with operating point. One can then choose the transformer turns ratio and tank characteristic impedance, as well as tank inductance and capacitance, to obtain a good converter design. A 600W full bridge example is given. Approximate peak component stresses In the continuous conduction mode, the tank inductor current and tank capacitor voltage peak magnitudes are both approximately proportional to the load current, and nearly independent of the converter input and output voltages. This is true because the dc load current I is equal to the average rectified tank inductor current <| iL |>. With a 1:n transformer turns ratio, we have 38 Chapter 4. (a) L iL I = 1n <| iL |> C + vC - I = 1n < | iL | > (4-84) To the extent that the waveshape of iL(t) does not 1:n 1 |iL| n I Cf + V – (b) The Series Resonant Converter vary with converter operating point, there is a direct relationship between the peak value of iL and its average rectified value <|iL |> = nI. Indeed this is the case near resonance, where iL is nearly sinusoidal. Equation (4-84) then becomes: Ts I = 1 n Ts 0 ILP | sin ωt | dt = 1n 2ILP π (4-85) Hence: |iL(t)| ILP = π I 2 ILP ≅ nπ I 2 ILP I = <|iL|> (c) J (4-86) Or, in normalized form, one obtains: JLP ≅ π J 2 (4-87) Thus, we expect the peak tank current ILP to be directly proportional to the load current I, and 3 essentially independent of the load voltage. We JLP = 4 can plot contours of constant peak tank current in J = 3 LP 2 the output plane, to see how component stresses JLP = 2 vary with operating point; Eq. (4-87) predicts that such contours should appear as in Fig. 4.44(c). 1 The approximation used above coincides with the sinusoidal approximation used in chapter 1 M 2. It is accurate for operation in the continuous Fig. 4.44 Peak stress approximation where conduction mode, near resonance. However, in (a) series resonant converter has a 1:n the discontinuous conduction mode, or in the transformer turns ratio; (b) dc load continuous conduction mode near the DCM current equals average rectified tank inductor current; (c) result shows that boundaries, the tank current is highly peak tank current is directly proportional nonsinusoidal, and Eqs. (4-84) - (4-87) become to load current. poor approximations. The tank capacitor voltage waveform is directly related to the tank inductor current, since these components are connected in series. Hence, the peak tank capacitor voltage is also directly proportional to the load current, and is essentially independent of the load voltage. In the case of 39 Principles of Resonant Power Conversion sinusoidal tank waveforms, the peak tank capacitor voltage VCP is related to ILP through the characteristic impedance R0: VCP ≅ ILP R0 ≅ nπ I R0 2 (4-88) Discussion Suppose that we have constructed a converter which is capable of producing the given rated output power Pmax at the normalized operating point M = 0.5, J = 5. This converter could also produce twice the rated power, or 2Pmax, by operating at the point M = 1.0, J = 5; i.e., by doubling the output voltage without changing the load current. This is true because the peak component stresses are independent of M and depend only on J. The component currents will be nearly the same in both cases, and hence the losses will be almost the same also. Furthermore, a converter with lower peak currents could be constructed by halving the transformer turns ratio n, so that the converter operates at rated output voltage and power with M = 1. Equation (4-84) predicts that this would cause the peak tank current to be reduced by a factor of two. Component stresses and losses would be reduced accordingly. The conclusion is that the converter should be designed to operate with M as close to unity as other considerations will allow. This implies that the transformer turns ratio should be minimized, and it leads to low peak tank and transistor currents. Converters designed to operate with lower-than-necessary values of M do not fully utilize the power components. Exact peak component stresses Exact expressions for the peak tank current and voltage can be derived using the state plane diagram. For example, the state plane diagram for the k=1 CCM is reproduced in Fig. 4.45, with the peak values of jL and mC identified. It can be seen that the peak normalized tank inductor current is given by the circle radius during the Q1 conduction subinterval: JLP = Jγ –1+M 2 (4-89) The peak normalized tank capacitor voltage MCP, shown in Fig. 4.45, was previously found. It is: MCP = MC1 = Jγ 2 (4-90) We wish to plot these component stresses in the output plane to determine how stresses vary with operating point. To do so, we need to express JLP and MCP as functions of J and M, with γ eliminated. The easiest way to do this is to solve for γ/2 in Eqs. (4-89) and (4-90): 40 Chapter 4. jL The Series Resonant Converter peak value of jL: Jγ jLP = - 1 + M 2 (circle radius during Q1 conduction interval) jLP Q1 Jγ 2 -1 D2 +M . . . . . -1 + M . Jγ 2 mC MC1 = 1+M -1 - M 1-M D1 peak value of mC: Jγ mCP = MC1 = 2 Q2 Fig. 4.45 State plane diagram for k=1 CCM. γ J +1–M = LP (4-91) 2 J γ = MCP (4-92) 2 J These expressions are then inserted into the converter output characteristic, Eq. (4-15), to eliminate γ. Solving for J, one then obtains JLP + 1 – M J = (4-93) 2–1 (J – M) LP π – tan–1 1 – M2 J = π – tan–1 MCP (MCP – 1)2 – 1 1 – M2 (4-94) Here, it is necessary to use care to select the correct branch of the arc tangent function. When the denominator is written as shown, then the correct answer will be obtained when the arc tangent function is defined to lie in the domain – π/2 ≤ tan–1(•) ≤ + π/2. Equations (4-93) and (4-94) can now be used to overlay contours of constant peak tank stress on the output plane characteristics. The result is given in Fig. 4.46. It can be seen that the contours are nearly horizontal lines, and do not differ much from Fig. 4.44. 41 Principles of Resonant Power Conversion 5 F=0.975 JLP =7 M CP=7 4 F=0.95 JLP =5 3 J M CP=5 F=0.9 JLP =3 2 F=0.8 M CP=3 F=0.7 1 F=0.6 JLP =1.5 MCP=2 F=0.5 0 0 0.25 0.5 0.75 1 M Fig. 4.46 Superposition of peak tank current and voltage stress curves on the normalized output characteristics, for the k=1 continuous conduction mode (below resonance). Solid lines: curves of constant switching frequency; dashed lines: contours of constant peak tank capacitor voltage; shaded lines: contours of constant peak tank inductor current. Similar analysis can be used to derive the component stresses for above resonance operation. The results for the k=0 CCM are MCP (MCP + 1)2 – 1 1 – M2 J = tan–1 JLP – 1 + M J = tan–1 –1 + J = tan–1 (JLP + M)2 – 1 1 – M2 J2 1 + LP 1 – M2 JLP 1 – M2 (4-95) , 1 – M2 < JLP M (4-96) , 42 1 – M2 > JLP M Chapter 4. The Series Resonant Converter 5 M CP =7 JLP =7 F=1.05 4 F=1.07 M CP =5 JLP =5 3 J F=1.1 M CP =3 F=1.15 2 JLP =3 F=1.3 M CP =1 1 F=1.5 JLP =1 0 0 0.25 0.5 0.75 1 M Fig. 4.47. Superposition of peak stress curves on the normalized output plane, above resonance (k=0 CCM). Solid curves are contours of constant switching frequency, dashed curves are contours of constant peak tank capacitor voltage, and shaded curves are contours of constant peak tank inductor current. Two cases occur in Eq. (4-96), depending on whether the peak occurs at the beginning of the switching period or at some time in the middle of the period. Equations (4-95) and (4-96) can now be used to generate the plot of Fig. 4.47, where peak tank stresses are overlaid on the converter output characteristics for the k=0 CCM. Use of the output plane In a typical voltage regulator design, the output voltage is regulated to a given constant value V. The input voltage Vg and output power P, as well as the output current I = P/V, vary over some specified range: Vgmax ≥ Vg ≥ Vgmin (4-97) Pmax ≥ P ≥ Pmin (4-98) 43 Principles of Resonant Power Conversion where Pmax = V Imax and Pmin = V Imin To regulate the output voltage, the controller will vary the switching frequency fs over some range, and consequently the operating point will vary. It is desired to choose the transformer turns ratio and the values of tank inductance and capacitance such that a good design is obtained, in which the tank capacitor voltage and inductor current are low, the range of switching frequency variations is small, and the transformer and tank elements are small in size. Note that the specifications (4-97) and (4-98) do not, by themselves, determine the operating region of the normalized operating plane. By changing the tank characteristic impedance R0 and transformer turns ratio n, the operating region can be moved to any arbitrary range of M and J. This follows from the definitions of M and J: M = V n Vg (4-99) Here, V and the range of Vg are specified, but the transformer turns ratio n is not. So we can choose n, and hence also scale the range of M, arbitrarily, subject to M ≤ 1. Also: J = n R0 I Vg (4-100) The range of I and Vg are specified, and n is given by the choice of M from Eq. (4-99) above. But we can still choose the tank characteristic impedance R0 arbitrarily, and hence we can also scale the range of J arbitrarily. So selection of an operating region, or M and J, is equivalent to choosing n and R0. Let us, therefore, map Eqs. (4-97) and (4-98) into the normalized output plane, for some given values of n and R0. Equations (4-97) and (4-99) imply that, as the input voltage varies from its maximum to minimum values, the voltage conversion ratio will also vary from Mmin to Mmax, where: Mmin = V n Vgmax Mmax = V n Vgmin (4-101) The output power can be written: P = IV = (4-102) J Vg V 2 = J V n R0 M R0 The expression on the right of Eq. (4-102) is the most useful for design because Vg has been eliminated in favor of the constant specified value of V. Solution for J yields: 2 J = M n R0 P V2 (4-103) 44 Chapter 4. The Series Resonant Converter For a given value of power P, J is proportional to M: if the input voltage varies (causing M to vary as given by Eq. (4-99) then J will also vary as given by Eq. (4-103). The peak J is given at maximum power and minimum input voltage: n Pmax R0 Jmax = Vgmin V (4-104) while the minimum J occurs at minimum power and maximum input voltage: (4-105) n Pmin R0 Jmin = Vgmax V J Jmax P Jmin = Pm Vg = Vgmin ax Vg = Vgmax The region defined by the specifications can be plotted using Eqs. (4-101) and (4-103) - (4-105), as in Fig. 4.48. This region can be overlaid on the converter output characteristics, to see how switching frequency and component stresses must vary as the operating point changes. For the k=0 CCM example shown, the design (Mmax, Jmax) = (0.8, 4) has been selected. Given values for Pmax , P min , Vgmax, and V gmin, the values of Mmin and Jmin can be evaluated, and the operating region is in P = Pm constructed. It can be seen that the maximum Mmax M Mmin switching frequency will occur at the point (Mmin, J min ), i.e., at maximum input voltage Vgmax and Fig. 4.48 Voltage regulator operating region, plotted in the output plane. minimum power P min . The component stresses are highest at maximum power Pmax. 600W design example The preceding arguments are next applied to the example of a 500W off-line full-bridge dcdc converter. The converter specifications are input voltage range 255 ≤ Vg ≤ 373 output power range 60 ≤ Pout ≤ 600 maximum switching frequency fsmax = 1 MHz regulated output voltage V = 24 Designs operating below resonance, for various choices of Mmax and Jmax , are compared in Table 4.1. Given the specifications and the choice of Mmax and Jmax , values of Mmin and Jmin can be computed from Eqs. (4-101), (4-104), and (4-105). The converter operating region can then be superimposed on the converter characteristics, as in Fig. 4.48, and the normalized peak stresses 45 Principles of Resonant Power Conversion and switching frequency variations can be determined graphically. Alternatively, the exact equations can be evaluated to find these quantities. The tank resonant frequency f0 is then chosen such that the maximum switching frequency, which occurs at the point (Mmax, Jmax) for below resonance operation, coincides with 1 MHz. The required transformer turns ratio n is found by evaluation of Eq. (4-101), and the required tank characteristic impedance R0 can then be found using Eq. (4-104). Values of L, C, ILP, and VCP can then be evaluated. A Table 4.1. Comparison Mmax Jmax Mmin Jmin fsmin kHz 0.9 5.0 0.62 0.34 280 of designs, L, f0, MHz µH 1.04 75 below resonance C, ILP, VCP, mode nF A V 0.3 4.2 2080 k=1 CCM, B 0.9 1.5 0.62 0.10 95 1.2 20 0.9 4.7 710 C 0.9 0.5 0.62 0.068 68 2.5 3.0 1.2 9.9 510 D 0.45 1.5 0.31 0.068 104 1.3 9.0 1.6 8.7 780 point k=2 DCM k=1 CCM, k=2 DCM k=2 DCM k=1 CCM, k=2 DCM All of the designs of Table 4.1 operate in the k=2 discontinuous conduction mode at light load; design C operates in DCM even at full load. It is assumed that the transistors are allowed to conduct only once per half-switching-period, and hence the converter operates in k=2 DCM for all switching frequencies below 0.5 f 0 , and for all values of J below 2/π. To ensure that the rated output voltage can be obtained when semiconductor forward voltage drops and other losses are accounted for, the maximum allowable value of M is taken to be 0.9 rather than 1.0. These designs all operate with zero current switching. Point A, (Mmax, Jmax) = (0.9, 5.0), exhibits the lowest peak current, 4.2A referred to the transformer primary. However, the peak tank capacitor voltage is very high, 2080V referred to the primary side. This voltage can be reduced by reducing the choice of Jmax , for example, to 1.5 as in point B. This has little effect on the peak current, but the peak capacitor voltage is reduced to 710V. Further reductions in Jmax further reduce the peak capacitor voltage to some extent; however, the peak tank current is increased as the discontinuous conduction mode boundary is approached. At point C, where the converter operates in DCM for all load currents, the peak tank current is approximately twice that of points A and B. Point D illustrates the effect of a suboptimal choice of Mmax, 0.45 rather than 0.9; this also approximately doubles the peak current. Several designs for the same specifications are illustrated in Table 4.2 for the above resonance case. No discontinuous conduction mode occurs in this case. The design procedure is similar to the below resonance case, except that the maximum switching frequency now occurs at minimum output power and maximum input voltage, i.e., at the point (Mmin, Jmin). These designs operate with zero voltage switching. 46 Chapter 4. The Series Resonant Converter Table 4.2. Comparison of designs, above resonance L, C, ILP, VCP, mode point Mmax Jmax Mmin Jmin fsmin f0, µH nF kHz A V kHz A 0.9 5.0 0.62 0.34 511 495 157 0.6 4.0 1940 k=0 B 0.9 1.5 0.62 0.10 220 200 117 5.4 3.9 545 C 0.9 1.0 0.62 0.068 157 137 113 12 3.8 350 D 0.9 0.2 0.62 0.014 45 28 111 290 3.9 50 E 0.45 1.0 0.31 0.068 133 95 81 34 8.8 287 CCM k=0 CCM k=0 CCM k=0 CCM k=0 CCM Point A is again at Mmax = 0.9, Jmax = 5.0. The design is not significantly different from the corresponding design below resonance. The peak current is again approximately 4A, and the peak tank capacitor voltage is again very high (approximately 2kV). Reducing Jmax has the favorable effect of reducing this voltage. In fact, since there is no discontinuous conduction mode, VCP can be made arbitrarily small by reducing Jmax a sufficient amount. The peak tank inductor current is slightly smaller than in the below-resonance case. The problem with doing so is the large range of switching frequency variations. For example, point D illustrates a design with a peak capacitor voltage of 50 volts and peak current of 3.9A; the problem with this design is the minimum switching frequency of 45kHz. This design will require quite large magnetics. A compromise is point C, for which the peak capacitor voltage is 350V, with low peak inductor current of 3.9A and a minimum switching frequency of 157kHz. Point E illustrates again the effect of a suboptimal choice of Mmax; this leads to increased peak tank current. It is apparent that, for operation above resonance, large switching frequency variations can occur. This is undesirable because it requires that the transformer and filter components be sized to the minimum switching frequency, which can be quite low. The benefits of high frequency operation are then lost. Several control schemes have been described in the literature which circumvent this problem. Constant frequency operation can be obtained by duty cycle [7] or phase control [13-15] of the transistor bridge. Large switching frequency variations can also occur below resonance, but these do not lead to a large transformer. The reason for this is the occurence of the k=2 DCM at all switching frequencies below 0.5 f 0 . In this mode, no voltage is applied to the transformer during the discontinuous (X) subintervals. The transformer can therefore be designed as if its minimum frequency is 0.5 f0, and the low values of fsmin in Table 4.1 are not a problem. Filter size is also not adversely affected by the wide range of switching frequency variations in this case, because the low frequency operating points coincide with the low output current points, where less filtering is needed. The output capacitor value is therefore determined by the maximum power point, which occurs at high frequency. 47 Principles of Resonant Power Conversion REFERENCES [1] F.C. Schwarz, “An Improved Method of Resonant Current Pulse Modulation for Power Converters,” IEEE Power Electronics Specialists Conference, 1975 Record, pp. 194-204, June 1975. [2] R. King and T. Stuart, “A Normalized Model for the Half Bridge Series Resonant Converter,” IEEE Transactions on Aerospace and Electronic Systems, March 1981, pp. 180-193. [3] V. Vorperian and S. Cuk, “A Complete DC Analysis of the Series Resonant Converter,” IEEE Power Electronics Specialists Conference, 1982 Record, pp. 85-100, June 1982. [4] R. King and T.A. Stuart, “Inherent Overload Protection for the Series Resonant Converter,” IEEE Transactions on Aerospace and Electronic Systems, vol. AES-19, no. 6, pp. 820-830, Nov. 1983. [5] R. Oruganti and F.C. Lee, “Resonant Power Processors, Part 1: State Plane Analysis,” IEEE Transactions on Industry Application, vol. IA-21, Nov/Dec 1985, pp. 1453-1460. [6] A. Witulski and R. Erickson, “Steady-State Analysis of the Series Resonant Converter,” IEEE Transactions on Aerospace and Electronic Systems, vol. AES-21, no. 6, pp. 791-799, Nov. 1985. [7] Steven G. Trabert and Robert W. Erickson, "Steady-State Analysis of the Duty Cycle Controlled Series Resonant Converter,” IEEE Power Electronics Specialists Conference, 1987 Record, pp. 545-556, (IEEE Publication 87CH2459-6). [8] R.L. Steigerwald, “High Frequency Resonant Transistor Dc-Dc Converters,” IEEE Transactions on Industrial Electronics, vol. IE-31, no. 2, pp. 181-191, May 1984. [9] S. Singer, “Loss-Free Gyrator Realization,” IEEE Transactions on Circuits and Systems, no. 2, pp. 26-34, January 1988. [10] Chambers, 15kHz 35kW high voltage converter using $15 SCR’s. [11] Y. Cheron, H. Foch, and J. Salesses, “Study of a Resonant Converter Using Power Transistors in a 25kW X-ray Tube Power Supply,” IEEE Power Electronics Specialists Conference, Proceedings ESA Sessions, pp. 295-306, June 1985. [12] A. Witulski and R. Erickson, “Design of the Series Resonant Converter for Minimum Component Stress,” IEEE Transactions on Aerospace and Electronic Systems, July 1986. [13] SRC phase control reference Vandelac PESC 87 [14] F.S. Tsai, P. Materu, and F.C. Lee, “Constant Frequency, Clamped Mode Resonant Converters,” IEEE Power Electronics Specialists Conference, 1987 Record, pp. 557-566, June 1987. [15] SRC phase control ref GE scheme (high voltage converter) PESC 90 [16] K.D.T. Ngo, “Analysis of a Series Resonant Converter Pulsewidth-Modulated of Current-Controlled for Low Switching Loss,” IEEE Power Electronics Specialists Conference, 1987 Record, pp. 527-536, June 1987. 48 vol. CAS-28, Chapter 4. The Series Resonant Converter P ROBLEMS 1. Using the phase plane, derive the k=0 continuous conduction mode characteristics. Sketch your results in the output plane (M vs. J) and control plane (M vs F for resistive load). 2. Analyze the k=3 discontinuous mode: 3. a) Draw the phase plane diagram; b) Draw waveforms for the inductor current, capacitor voltage, and inductor voltage; c) Use tank capacitor charge arguments to relate the normalized load current to normalized capacitor voltage boundary values; d) Solve for the output characteristics; e) Determine the complete set of conditions on normalized switching period and load current which guarantee operation in this mode. It is desired to obtain a converter with current source characteristics. Hence, a series resonant converter is designed for operation in the k=2 discontinuous mode. The switching frequency is chosen to be fs = 0.225 fo, where fo is the tank resonant frequency (consider only open-loop operation). The load R is a linear resistance whose value can change to any positive value. a) Plot the output characteristics (M vs J), for all values of R in the range [0,∞]. Label mode boundaries, evaluate the short-circuit current, and give analytical expressions for the output characteristics. b) Over what range of R (referred to the tank characteristic impedance Ro) does the converter operate as intended, in the k=2 discontinuous mode? 4. Derive the equations for the peak tank stresses in the k=0 continuous conduction mode, Eqs. (4-95) and (4-96). 5. Design of a high density series resonant converter L C I v in CF V o ut Design a half-bridge series resonant converter, as shown above, to meet the following specifications: Input voltage Vin: Output voltage Vout: Output current I: Maximum switching frequency: Output voltage ripple: 134–176 volts 48 volts 1–10 amperes 750 kHz no greater than 1 volt peak-to-peak 49 Principles of Resonant Power Conversion Design the “best” converter that you can, which combines high efficiency with small volume. Use your engineering judgement to select the operating mode, tank elements L and C, and transformer turns ratio to attain what you consider to be the best combination of small transformer size, small C F , high minimum switching frequency, low peak tank capacitor voltage, and low peak transistor current. The volume of a 50V, 1µF, X7R ceramic chip capacitor is approximately 50 mm 3 . Capacitor volume scales as the product of capacitance and voltage rating, so that the volume of a 100V 5µF capacitor of the same dielectric material is 500 mm 3 . Capacitors are available with voltage ratings of 50V, 75V, 100V, 200V, 300V, 400V, and 500V, and essentially any capacitance value. You must choose a capacitor with a voltage rating at least 25% greater than the actual peak voltage applied by your design. Use a ferrite EE core for the transformer. Estimate the transformer size using the Kg method, allowing a fill factor Ku of 0.5, total copper loss Pcu no greater than 0.5W, and peak flux density Bmax of 0.1T. The core geometrical constant Kg of the center-tapped transformer is defined as follows: 2 2 Kg = ρ λ 1 i1 108 Pcu Ku B2max cm5 where ρ is the resistivity of copper (1.724·10-6Ω·cm), λ1 is the applied primary volt-seconds, and i 1 is the applied primary rms current. For this problem, you may neglect core loss. This is not valid in general, especially for 750kHz transformers, but is a simplifying assumption for this problem. The estimated volumes of transformers constructed in this manner are as follows: Core type vol., mm3 Kg , cm5 EE12 675 0.73·10-3 EE16 1125 2.0·10-3 EE19 1550 4.1·10-3 EE22 EE30 EE40 2425 8575 14700 8.3·10-3 86·10-3 0.21 EE50 EE60 31500 42250 0.91 1.4 Hence, attempt to minimize the total volume V tot of the transformer, output filter capacitor, and tank capacitor, while keeping the peak currents reasonably low. You may neglect the size of all other components. Specify: (1) your choices for L, C, and transformer turns ratio; (2) the range of M, J, and fs over which your design will operate; (3) the transformer core size required (don’t bother to compute number of turns or wire size); (4) the value of CF required; and (5) the total volume Vtot as defined above. 50

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