10-F1127PA050SC-L169E09 Output Inverter Application flow 7PACK 1 1200 V / 50 A General conditions 3phase SPWM V GEon = 15 V V GEoff = -15 V R gon = 8 Ω R goff = 8 Ω Figure 1 IGBT Figure 2 Typical average static loss as a function of output current P loss = f(I out) 80 Ploss (W) 120 Ploss (W) FWD Typical average static loss as a function of output current P loss = f(I out) Mi*cosfi = 1 100 Mi*cosf i= -1 70 60 80 50 40 60 30 40 20 20 10 Mi*cosfi = 1 Mi*cosfi = -1 0 0 0 10 20 30 40 50 60 70 0 80 10 20 30 40 50 60 Iout (A) At Tj = 150 80 Iout (A) At Tj = °C Mi*cosφ from -1 to 1 in steps of 0,2 150 °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 P loss = f(I out) 100,0 fsw = 16kHz 90,0 FWD Typical average switching loss as a function of output current Ploss (W) Ploss (W) 70 80,0 P loss = f(I out) 45,0 fsw = 16kHz 40,0 35,0 70,0 30,0 60,0 25,0 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 10 20 30 40 50 60 70 0 80 10 20 30 40 50 60 Iout (A) At Tj = 150 80 Iout (A) At Tj = °C DC link = 600 V f sw from 2 kHz to 16 kHz in steps of factor 2 copyright Vincotech 70 150 °C DC link = 600 V f sw from 2 kHz to 16 kHz in steps of factor 2 1 27 Jul. 2015 / Revision 1 10-F1127PA050SC-L169E09 Output Inverter Application flow 7PACK 1 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) 80 Iout (A) 1200 V / 50 A Th = 60°C 70 60 80 Th = 60°C 70 60 50 50 Th = 100°C 40 40 30 30 20 Th = 100°C 20 10 10 0 -1,0 -0,8 -0,6 -0,4 -0,2 0,0 0,2 0,4 0,6 0,8 0 1,0 1 10 100 Mi*cos φ At Tj = 150 fsw (kHz) At Tj = °C DC link = 600 V f sw = 4 kHz T h from 60 °C to 100 °C in steps of 5 °C 150 °C DC link = 600 V Mi*cos φ =0,8 T h from 60 °C to 100 °C in steps of 5 °C Figure 7 Phase Figure 8 -1,00 -0,80 Iout (A) Phase Typical available 0Hz output current as a function of switching frequency I outpeak = f(f sw) Iout (Apeak) Typical available 50Hz output current as a function of Mi*cos φ and switching frequency I out = f(f sw, Mi*cos φ ) 80 70 Th = 60°C -0,60 60 -0,40 -0,20 56,0-64,0 0,00 48,0-56,0 0,20 40,0-48,0 50 Mi*cosfi 64,0-72,0 40 30 0,40 32,0-40,0 0,60 20 24,0-32,0 0,80 16,0-24,0 1,00 1 2 4 8 16 32 10 64 Th = 100°C 0 1 10 100 fsw At Tj = fsw (kHz) 150 °C At Tj = DC link = 600 Th = 80 V °C DC link = 600 V T h from 60 °C to 100 °C in steps of 5 °C Mi = copyright Vincotech 2 150 °C 0 27 Jul. 2015 / Revision 1 10-F1127PA050SC-L169E09 Output Inverter Application flow 7PACK 1 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 P out=f(T h) Pout (kW) 1200 V / 50 A 35,0 2kHz 30,0 100,0 2kHz 99,0 98,0 25,0 97,0 96,0 20,0 16kHz 95,0 16kHz 15,0 94,0 93,0 10,0 92,0 5,0 91,0 0,0 60 65 70 75 80 85 90 95 90,0 100 0,0 5,0 10,0 15,0 20,0 25,0 30,0 Th (oC) At Tj = 150 DC link = Mi = cos φ= f sw from 600 V 1 0,80 2 kHz to 16 kHz in steps of factor 2 40,0 45,0 Pout (kW) °C Figure 11 35,0 At Tj = 150 DC link = Mi = cos φ= f sw from 600 V 1 0,80 2 kHz to 16 kHz in steps of factor 2 °C Inverter Overload (%) Typical available overload factor as a function of motor power and switching frequency P peak / P nom=f(P nom,fsw) 400 350 300 250 200 150 Switching frequency (kHz) 100 Motor/ nominal power (HP/kW) 10,00 7,36 15,00 / 11,03 20,00 / 14,71 25,00 / 18,39 30,00 / 22,07 40,00 / 29,42 1 374 250 187 150 125 0 2 374 250 187 150 125 0 4 374 250 187 150 125 0 8 332 221 166 133 111 0 16 258 172 129 0 0 0 At Tj = 150 DC link = Mi = cos φ= f sw from Th = 600 V 1 0,8 1 kHz to 16kHz in steps of factor 2 80 °C °C Motor eff =0,85 copyright Vincotech 3 27 Jul. 2015 / Revision 1