10-FZ062PA100SA-P994F08 10-PZ062PA100SA-P994F08Y datasheet Output Inverter Application flow PHASE0 600 V / 100 A General conditions 3phase SPWM VGEon = 15 V VGEoff = -15 V Rgon = 4 Ω Rgoff = 4 Ω Figure 1 IGBT Figure 2 FWD Typical average static loss as a function of output current Typical average static loss as a function of output current Ploss = f(Iout) Ploss = f(Iout) 140 Ploss (W) Ploss (W) 180 160 140 120 Mi*cosf i= -1 Mi*cosfi = 1 100 120 80 100 80 60 60 40 40 20 20 Mi*cosfi = -1 Mi*cosfi = 1 0 0 0 At Tj = 20 40 150 60 80 100 120 0 140 160 Iout (A) At Tj = °C Mi*cosφ from -1 to 1 in steps of 0,2 40 150 60 80 100 120 140 Iout (A) 160 °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 Ploss = f(Iout) Ploss (W) Ploss (W) 20 60,0 Ploss = f(Iout) 25,0 fsw = 16kHz fsw = 16kHz 50,0 20,0 40,0 15,0 30,0 10,0 20,0 5,0 10,0 fsw = 2kHz fsw = 2kHz 0,0 0,0 0 At Tj = 20 150 40 60 80 100 120 0 140 Iout (A) 160 At Tj = °C DC link = 320 V fsw from 2 kHz to 16 kHz in steps of factor 2 copyright Vincotech 20 150 40 60 80 100 120 140 160 Iout (A) °C DC link = 320 V fsw from 2 kHz to 16 kHz in steps of factor 2 1 02 Dec. 2015 / Revision 2 10-FZ062PA100SA-P994F08 10-PZ062PA100SA-P994F08Y datasheet Output Inverter Application flow PHASE0 Figure 5 Phase Figure 6 Typical available 50Hz output current as a function Mi*cosφ Iout = f(Mi*cos φ) Phase Typical available 50Hz output current as a function of switching frequency Iout = f(fsw) Iout (A) 160 Iout (A) 600 V / 100 A Th = 60°C 160 140 140 120 120 Th = 60°C 100 100 Th = 100°C 80 80 60 60 40 40 20 20 0 -1,0 At Tj = 0 -0,8 -0,6 -0,4 150 DC link = 320 fsw = 4 Th from Th = 100°C -0,2 0,0 0,2 0,4 0,6 0,8 1,0 Mi*cos φ 1 10 °C At Tj = V kHz DC link = 320 Mi*cos φ =0,8 60 °C to 100 °C in steps of 5 °C 150 Th from Figure 7 Phase V 60 °C to 100 °C in steps of 5 °C Phase Typical available 0Hz output current as a function of switching frequency Ioutpeak = f(fsw) Iout (Apeak) -1,00 -0,80 120 100 -0,60 Th = 60°C -0,40 80 120,0-135,0 105,0-120,0 100 °C Figure 8 Typical available 50Hz output current as a function of Mi*cos φ and switching frequency Iout = f(fsw, Mi*cos φ) Iout (A) fsw (kHz) 90,0-105,0 75,0-90,0 0,00 60,0-75,0 45,0-60,0 Mi*cosfi -0,20 60 Th = 100°C 0,20 30,0-45,0 40 0,40 0,60 20 0,80 0 1,00 1 2 4 8 16 32 64 1 fsw (kHz) At Tj = 10 fsw (kHz) 150 °C At Tj = DC link = 320 Th = 80 V °C DC link = 320 V Th from 60 °C to 100 °C in steps of 5 °C Mi = copyright Vincotech 2 150 100 °C 0 02 Dec. 2015 / Revision 2 10-FZ062PA100SA-P994F08 10-PZ062PA100SA-P994F08Y datasheet Output Inverter Application flow PHASE0 Figure 9 Inverter Figure 10 Inverter Typical efficiency as a function of output power efficiency=f(Pout) efficiency (%) Typical available peak output power as a function of heatsink temperature Pout=f(Th) Pout (kW) 600 V / 100 A 40,0 35,0 100,0 99,0 2kHz 98,0 2kHz 30,0 97,0 16kHz 25,0 96,0 20,0 95,0 16kHz 94,0 15,0 93,0 10,0 92,0 5,0 91,0 0,0 90,0 60 65 70 75 80 85 90 At Tj = 150 DC link = Mi = cos φ= fsw from 320 V 1 0,80 2 kHz to 16 kHz in steps of factor 2 95 100 Th ( o C) 0,0 °C Figure 11 5,0 10,0 15,0 20,0 25,0 30,0 At Tj = 150 DC link = Mi = cos φ= fsw from 320 V 1 0,80 2 kHz to 16 kHz in steps of factor 2 35,0 40,0 45,0 Pout (kW) °C Inverter Overload (%) Typical available overload factor as a function of motor power and switching frequency Ppeak / Pnom=f(Pnom,fsw) 450 400 350 300 250 200 Switching frequency (kHz) 150 Motor nominal power (HP/kW) 100 7,50 / 5,52 10,00 / 7,36 15,00 / 11,03 20,00 / 14,71 25,00 / 18,39 30,00 / 22,07 1 484 363 242 181 145 121 2 476 357 238 178 143 119 4 461 345 230 173 138 115 8 431 323 216 162 129 0 16 378 284 189 142 114 0 At Tj = 150 DC link = 320 Mi = 1 cos φ= fsw from Th = °C V 0,8 1 kHz to 16kHz in steps of factor 2 80 °C Motor eff =0,85 copyright Vincotech 3 02 Dec. 2015 / Revision 2