T SERIES RESONANT CONVERTER - University of Colorado …

1 Chapter 4 THE SERIES RESONANT CONVERTERhe objective of this chapter is to describe the operation of the SERIES RESONANT CONVERTER indetail. The concepts developed in chapter 3 are used to derive closed-form solutions forthe output characteristics and steady-state control characteristics, to determine operating modeboundaries, and to find peak component stresses. General results are presented, for everycontinuous and discontinuous conduction mode using frequency control. The origin of thediscontinuous conduction modes is results are used to consider three design problems. First, the variation of peakcomponent stresses with the choice of worst-case operating point is investigated, and someguidelines regarding the choice of transformer turns ratio and tank characteristic impedance arediscussed.

2 Second, the effects of variations in input line voltage and output load current areexamined using the CONVERTER output characteristics. Finally, switching frequency variations areconsidered, and the tradeoff between transformer size and tank capacitor voltage is exposed.+-D5D6D7D8D1D2D3D4Q4Q3Q2Q1 VgIR+-V+-vSiSLC+-vR + vT - CFFig. RESONANT CONVERTER of RESONANT Power and ModesThe SERIES RESONANT CONVERTER , Fig. , is reproduced in Fig. It can be seen that theinstantaneous voltage vT(t) applied across the tank circuit is equal to the difference between theswitch 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 anduncontrolled rectifier subinterval is defined as a length of time for which the conducting states of all of thesemiconductor switches in the CONVERTER remain fixed; during each subinterval, vS(t), vR(t), andvT(t) are constant.

3 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. In this case, we havevs = +VgvR = +V(4-2)vT = Vg VThe applied tank voltage is therefore constant andequal to Vg V. In normalized form, one obtainsMT = VTVg = 1 M(4-3)Hence, according to section , the normalizedstate plane trajectory for this subinterval is acircular arc centered at MT = 1-M, as shown in The radius depends on the initial that, since we have assumed that iL(t) ispositive and diodes D5 and D8 conduct, thisparticular switch conducting state can occur only inthe upper half-plane (jL > 0). For negative jL,diodes D6 and D7 would conduct instead, MTwould be changed, and an arc centered at adifferent location would be obtained.

4 A subintervalutilizing the switch conduction state describedabove and in Fig. is referred to in shorthandform as subinterval = 1 - M1mCjL > 0(b)+-VgR+-VLC CF+ vC -vT = Vg - ViL > 0+ vT -(a) direction of current flowFig. Q1 conduction subinterval, inwhich Q1 and Q4 conduct, and iL > 0 sothat D5 and D8 conduct: a) circuit;b) normalized state plane 4. The SERIES RESONANT Converter3 Many other switch conduction states canoccur. Subinterval D1 is similar to subinterval Q1,except that the tank current iL(t) is negative. Theconducting devices are antiparallel diodes D1 andD4, and output rectifier diodes D6 and D7. Theapplied tank voltage is therefore VT = Vg+V, or innormalized form,MT = 1 + M(4-4)The circuit and state plane trajectory for thissubinterval are summarized in Fig.

5 Note thatthis switch conduction state can only occur in thenegative half-plane (jL < 0).Symmetrical switch conduction states Q2(Fig. ) and D2 (Fig. ) can also occur, inwhich iL, vS, vR, and vT have the opposite polarityfrom states Q1 and D1 respectively. These correspond to MT = -1+M (Q2) and MT = -1-M (D2)..MT = 1 - M1mC(b)jL < 0jL+-VgR+-VLC CF+ vC -+ vT -(a)iL < 0 vT = Vg + Vdirection of current flowFig. conduction subinterval, inwhich iL < 0 such that diodes D1, D4,D6, and D7 conduct: a) circuit; b)normalized state plane (b)jL < 0jLMT = -1 + M+-Vg+-VLC+ vC -+ vT -(a)iL < 0direction of current flowvT = -Vg + VCFFig. Q2 conduction subinterval, inwhich Q2 and Q3 conduct, and iL < 0 sothat D6 and D7 conduct: a) circuit;b) normalized state plane of current flow+-Vg+-VLC CF+ vC -iL > 0+ vT -(a)VT = -Vg - > 0(b)MT = -1 - MFig.

6 D2 conduction subinterval, inwhich iL > 0, such that diodes D2, D3,D5 and D8 conduct: a) circuit; b)normalized state plane of RESONANT Power Conversion4 Under certain conditions, it is possiblefor all four uncontrolled rectifier diodes (D5, D6,D7, D8) to become simultaneously reverse-biased. When this occurs, the circuit topology isas given in Fig. The tank inductor is thenzero, and the tank capacitor voltage remains at itsinitial value. This switch conduction state isdenoted X .When phase control is used, two othersubintervals can occur: P1, which occurs foriL > 0, is summarized in Fig. , and P2,which occurs for iL < 0, is summarized in operating mode is defined by asequence of subintervals which combine to forma complete switching period.

7 Discontinuous+-VgLCiL > 0+ vT -(a)oo+-Voo = 0mCmC does not change(b)Fig. X, in which all fourrectifier diodes D5, D6, D7 and D8 arereverse-biased. The inductor currentremains at zero, and the tank capacitorvoltage does not change: a) one possiblecircuit topology; b) normalized stateplane trajectory.+-VgLCiL > 0+ vT -(a)o+-Vdirection of current flowvT = (b)jL > 0MT = -M Fig. P1, in which D2 andQ4 (or Q1 and D3) conduct, and iL > 0 sothat D5 and D8 also conduct: a) circuit;b) normalized state plane (b)jL < 0jLMT = M +-Vg+-VLCCF+ vC -+ vT -(a)iL < 0direction of current flowovT = V Fig. P2, in which D1 and Q3(or Q2 and D4) conduct, and iL < 0 so thatD6 and D7 conduct: a) circuit; b)normalized state plane 4. The SERIES RESONANT Converter5conduction modes contain at least one X subinterval, while continuous conduction modes containno X subintervals.

8 As seen later in this chapter, the different modes cause the SERIES resonantconverter to exhibit widely varying terminal of RESONANT Power Plane and Charge Arguments for the k=1 Continuous ConductionModeState plane trajectoryTypical inductor current iL(t), capacitorvoltage vC(t), and applied tank voltage vT(t)waveforms are diagrammed in Fig. for the k=1continuous conduction mode. This mode isdefined by the subinterval sequence this mode, the switching period begins when thecontrol circuit switches transistors Q1 and Q4 on,with the inductor current iL(t) positive. The stateplane trajectory for this subinterval is given in ; it begins at 0t = 0 with some initial valuesof tank inductor current and capacitor voltage. Thetank rings with a circular state plane trajectorycentered at MT = 1 M until, at 0t = , theinductor current rings negative, and subinterval D1begins.

9 The normalized state plane trajectory thenfollows a circular arc centered at MT = 1+ time 0t = + (one half switchingperiod), the control circuit switches transistors Q1and Q4 off, and Q2 and Q3 are switched Q2 begins, and as shown in , the trajectory continues along a circular arccentered at MT = -1+M until the inductor currentagain reaches zero. The output bridge rectifiersthen switch, and subinterval D2 begins. Thetrajectory follows an arc centered at MT = -1-M forthe remainder of the switching period. Theswitching period ends when the control circuitswitches Q2 and Q3 off, and Q1 and Q4 the CONVERTER operates in equilibrium,then the trajectory begins and ends at the samepoint in the state plane, and the tank waveforms areVg - VVg + V-Vg + V-Vg - Vsubinterval : Q1 D1 Q2 D2vT(t)+VC1-VC1vC(t)iL(t) 0t = 0TS2 Fig.

10 Tank inductor current iL,tank capacitor voltage, vC, and appliedtank voltage vT waveforms, for thek = 1 continuous conduction 4. The SERIES RESONANT Converter7periodic. Otherwise, a transient occurs in which the trajectory for each switching period begins ata different point, and follows a different path in the state plane. If the circuit is stable, then thetrajectory eventually converges to a single closed path, and the waveforms become periodic. Tofind 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 - M 1 - M + 0t = radius MA1(a) 0t = 0Q1 0t = 1 + Mradius MA2 0t = + = (b)D1 0t = + 0t = + + = 2 -1 - M(d)radius MA2D2 + M. radius MA1. 0t = 0t = + (c)Q2 subintervalFig.