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IRK.105 データシート(PDF) 2 Page - International Rectifier |
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IRK.105 データシート(HTML) 2 Page - International Rectifier |
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2 / 8 page ![]() IRK.105 Series 2 Bulletin I27133 rev. I 09/04 www.irf.com I T(AV) Max. average on-state current (Thyristors) 180o conduction, half sine wave, I F(AV) Max. average forward T C = 85 o C current (Diodes) I O(RMS ) Max. continuous RMS on-state current. As AC switch I TSM Max. peak, one cycle 1785 t =10ms No voltage or non-repetitive on-state 1870 t =8.3ms reapplied I FSM or forward current 1500 t =10ms 100% V RRM 1570 t =8.3ms reapplied 2000 t =10ms T J = 25 o C, 2100 t =8.3ms no voltage reapplied I2t Max. I2t for fusing 15.91 t =10ms No voltage 14.52 t =8.3ms reapplied 11.25 t =10ms 100% V RRM 10.27 t =8.3ms reapplied 20.00 t =10ms T J = 25 o C, 18.30 t =8.3ms no voltage reapplied I2√t Max. I2√t for fusing (1) 159.1 KA2√s t =0.1to10ms,no voltage reappl. TJ=TJ max V T(TO) Max. value of threshold 0.80 Low level (3) voltage (2) 0.85 High level (4) r t Max. value of on-state 2.37 Low level (3) slope resistance (2) 2.25 High level (4) V TM Max. peak on-state or I TM = π x IT(AV) V FM forward voltage I FM = π x IF(AV) di/dt Max. non-repetitive rate T J = 25 o C, from 0.67 V DRM, of rise of turned on 150 A/µs I TM =π x IT(AV), I g = 500mA, current tr < 0.5 µs, tp > 6 µs I H Max. holding current 250 T J = 25 o C, anode supply = 6V, mA resistive load, gate open circuit IL Max. latching current 400 TJ=25oC,anode supply=6V,resistive load Parameters IRK.105 Units Conditions 235 On-state Conduction Initial T J = TJ max. A KA2s V mΩ 1.64 V or I (RMS) I (RMS) ELECTRICAL SPECIFICATIONS Voltage Ratings Type number VRRM , maximum VRSM, maximum VDRM , max. repetitive IRRM Voltage repetitive non-repetitive peak off-state voltage, IDRM Code peak reverse voltage peak reverse voltage gate open circuit 130°C -V V V mA 04 400 500 400 06 600 700 600 08 800 900 800 IRK.105 10 1000 1100 1000 20 12 1200 1300 1200 14 1400 1500 1400 16 1600 1700 1600 Sinusoidal half wave, Initial T J = TJ max. 105 T J = TJ max TJ = TJ max TJ = 25°C (1) I2t for time t x = I 2√t x √t x (2) Average power = V T(TO) x IT(AV) + rt x (IT(RMS)) 2 (3) 16.7% x p x I AV < I < p x IAV (4) I > p x I AV |
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