(2) Analysis☆Assuming V, to be sinusoidal, at the line frequency w=2f,①ACsidePhasorequation（交流侧相量方程)：V=V+V=V+ joLI(18 -23)L1convlconvlslSSPhasordiagram(相量图)：Veonvi9=180°7Vt0=00vs161?1V.8LT(c)Veconvi(b)Figure 18-9 Rectification and inversion: (a) general phasor diagram; (b) rectificationat unitypowerfactor; (c)inversionatunitypowerfactor
(2) Analysis☆ (18 −23) s s1 + jL I Assuming vs to be sinusoidal, at the line frequency ω=2πf, ① AC side Phasor equation (交流侧相量方程): V =V +V =V s conv1 L1 conv1 ② Phasor diagram (相量图):
3Theimportant equationsV2(comlsin)sconvlcoso)PVsVoLsoV-Vconvl(18 -26)sljoL,>For a given Vsand the chosen Ls, P and Q can beobtainedbycontrollingthemagnitudeandthephaseof Vconv1
③ The important equations: (18 −26) s s conv1 s1 Vs V V 2 V V 2 P = jL Q s (1− conv1 cos ) Ls s ( conv1 sin ) Ls Vs = V −V I ➢For a given vs and the chosen Ls , P and Q can be obtained by controlling the magnitude and the phase of vconv1
Inversion modeRectifier modeVeonvi6=0°la1V.★9=180°18VLIVLI8T下v.VconvcosO= -1cos= 1
Rectifier mode Inversion mode cos = 1 cos = −1
(3) Switch-Mode Converter ControlThere are various ways to implement the current loopcontrol:>Indirectcurrentcontrol（间接电流控制)>Directcurrentcontrol（直接电流控制）
(3) Switch-Mode Converter Control There are various ways to implement the current loop control: ➢Indirect current control (间接电流控制) ➢Direct current control (直接电流控制)
@ Indirect Current Control-----Base on phasor equationRectifier mode6=0°V.lg11v, = V, sin Ot8VlTVLi= OL,Is1VconviK, tanOL,VVconvlconvlcossin(ot -) →v.Vconv Sconvlconvl
① Indirect Current Control-Base on phasor equation: Rectifier mode ˆ ˆ conv1 conv1 conv1 conv1 conv1 s s Vˆ v =V = →Vˆ sin(t − ) → v s cos Is1 = s → Ls Vˆ vL1 = Ls Is1 Vˆ tan v =V sint