10-FZ062PA075SA-P993F08 10-F0062PA075SA-P993F09 datasheet Output Inverter Application flow PHASE0 600 V / 75 A General conditions 3phase SPWM V GEon = 15 V V GEoff = -15 V R gon = 2 Ω R goff = 2 Ω Figure 1 IGBT Figure 2 Typical average static loss as a function of output current P loss = f(I out) 120 Ploss (W) 160 Ploss (W) FWD Typical average static loss as a function of output current P loss = f(I out) 140 100 Mi*cosfi = 1 120 Mi*cosf i= -1 80 100 60 80 60 40 40 20 20 Mi*cosfi = -1 0 At Tj = 20 150 40 60 80 100 0 120 140 Iout (A) At Tj = °C Mi*cosφ from -1 to 1 in steps of 0,2 20 150 40 60 80 100 120 Iout (A) 140 °C Mi*cosφ from -1 to 1 in steps of 0,2 Figure 3 IGBT Typical average switching loss as a function of output current Figure 4 FWD Typical average switching loss as a function of output current P loss = f(I out) Ploss (W) Ploss (W) Mi*cosfi = 1 0 0 40,0 35,0 fsw = 16kHz P loss = f(I out) 20,0 18,0 16,0 fsw = 16kHz 30,0 14,0 25,0 12,0 10,0 20,0 8,0 15,0 6,0 10,0 4,0 5,0 2,0 fsw = 2kHz fsw = 2kHz 0,0 0,0 0 At Tj = 20 150 40 60 80 100 0 120Iout (A) 140 At Tj = °C DC link = 320 V f sw from 2 kHz to 16 kHz in steps of factor 2 copyright Vincotech 20 150 40 60 80 100 120 Iout (A) 140 °C DC link = 320 V f sw from 2 kHz to 16 kHz in steps of factor 2 1 04 Jun. 2016 / Revision 2 10-FZ062PA075SA-P993F08 10-F0062PA075SA-P993F09 datasheet Output Inverter Application flow PHASE0 Figure 5 Phase Figure 6 Typical available 50Hz output current as a function Mi*cosφ I out = f(Mi*cos φ ) Phase Typical available 50Hz output current as a function of switching frequency I out = f(f sw) Iout (A) 120 Iout (A) 600 V / 75 A Ts = 60°C 100 120 Ts = 60°C 100 80 80 Ts = 100°C 60 60 40 40 20 20 0 -1,0 At Tj = Ts = 100°C 0 -0,8 -0,6 -0,4 150 -0,2 0,0 0,2 0,4 0,6 0,8 1,0 Mi*cos φ 1 At Tj = °C DC link = 320 V f sw = 4 kHz T s from 60 °C to 100 °C in steps of 5 °C 10 150 DC link = 320 Mi*cos φ =0,8 T s from Figure 7 Phase 100 °C V 60 °C to 100 °C in steps of 5 °C Figure 8 Typical available 50Hz output current as a function of Mi*cos φ and switching frequency I out = f(f sw, Mi*cos φ ) Phase Typical available 0Hz output current as a function of switching frequency I outpeak = f(f sw) Iout (Apeak) -1,00 -0,80 Iout (A) fsw (kHz) 100 90 -0,60 80 -0,40 70 -0,20 60 Ts = 60°C 100,0-110,0 80,0-90,0 70,0-80,0 0,00 60,0-70,0 50,0-60,0 40,0-50,0 Mi*cosfi 90,0-100,0 50 0,20 40 0,40 30 0,60 20 0,80 10 1,00 0 Ts = 100°C 30,0-40,0 1 2 4 8 16 32 64 1 fsw (kHz) At Tj = 10 fsw (kHz) 150 °C At Tj = DC link = 320 Ts = 80 V °C DC link = 320 V T s from 60 °C to 100 °C in steps of 5 °C Mi = copyright Vincotech 2 150 100 °C 0 04 Jun. 2016 / Revision 2 10-FZ062PA075SA-P993F08 10-F0062PA075SA-P993F09 datasheet Output Inverter Application flow PHASE0 Figure 9 Inverter Figure 10 Inverter Typical efficiency as a function of output power efficiency=f(P out) efficiency (%) Typical available peak output power as a function of heatsink temperature P out=f(T s) Pout (kW) 600 V / 75 A 30,0 25,0 100,0 99,0 2kHz 98,0 2kHz 97,0 20,0 16kHz 96,0 15,0 95,0 16kHz 94,0 10,0 93,0 92,0 5,0 91,0 0,0 90,0 60 65 At Tj = DC link = Mi = cos φ = f sw from 70 75 150 °C 320 1 V 80 85 90 95 100 Ts ( o C) 0,0 At Tj = DC link = Mi = cos φ = f sw from 0,80 2 kHz to 16 kHz in steps of factor 2 Figure 11 5,0 10,0 150 °C 320 1 V 15,0 20,0 25,0 30,0 35,0 Pout (kW) 0,80 2 kHz to 16 kHz in steps of factor 2 Inverter Overload (%) Typical available overload factor as a function of motor power and switching frequency P peak / P nom = f( P nom , f sw) 500 450 400 350 300 250 200 Motor nominal power (HP/kW) Switching frequency (kHz) 150 100 5,00 / 3,68 7,50 / 5,52 10,00 / 7,36 15,00 / 11,03 20,00 / 14,71 25,00 / 18,39 1 538 359 269 179 135 0 2 531 354 265 177 133 0 4 517 344 258 172 129 0 8 489 326 245 163 122 0 16 439 293 220 146 0 0 At Tj = 150 DC link = 320 °C V Mi = 1 cos φ = f sw from Ts = 0,8 1 kHz to 16kHz in steps of factor 2 80 °C Motor eff =0,85 copyright Vincotech 3 04 Jun. 2016 / Revision 2