aug-29-2003 1 BCR158... pnp silicon digital transistor ? switching circuit, inverter, interface circuit, driver circuit ? built in bias resistor ( r 1 =2.2k ? , r 2 =47k ? ) BCR158/f/l3 BCR158t/w eha07183 3 2 1 c e b r 1 r 2 type marking pin configuration package BCR158 BCR158l3 BCR158f BCR158t BCR158w wis wi wis wis wis 1=b 1=b 1=b 1=b 1=b 2=e 2=e 2=e 2=e 2=e 3=c 3=c 3=c 3=c 3=c - - - - - - - - - - - - - - - sot23 tsfp-3 tsfp-3 sc75 sot323
aug-29-2003 2 BCR158... maximum ratings parameter symbol value unit collector-emitter voltage v ceo 50 v collector-base voltage v cbo 50 emitter-base voltage v ebo 5 input on voltage v i(on) 10 collector current i c 100 ma total power dissipation- BCR158, t s 102c BCR158f, t s 128c BCR158l3, t s 135c BCR158t, t s 109c BCR158w, t s 124c p tot 200 250 250 250 250 mw junction temperature t j 150 c storage temperature t stg -65 ... 150 thermal resistance parameter symbol value unit junction - soldering point 1) BCR158 BCR158f BCR158l3 BCR158t BCR158w r thjs 240 90 60 165 105 k/w 1 for calculation of r thja please refer to application note thermal resistance
aug-29-2003 3 BCR158... electrical characteristics at t a = 25c, unless otherwise specified parameter symbol values unit min. typ. max. dc characteristics collector-emitter breakdown voltage i c = 100 a, i b = 0 v (br)ceo 50 - - v collector-base breakdown voltage i c = 10 a, i e = 0 v (br)cbo 50 - - collector-base cutoff current v cb = 40 v, i e = 0 i cbo - - 100 na emitter-base cutoff current v eb = 5 v, i c = 0 i ebo - - 164 a dc current gain 1) i c = 5 ma, v ce = 5 v h fe 70 - - - collector-emitter saturation voltage 1) i c = 10 ma, i b = 0.5 ma v cesat - - 0.3 v input off voltage i c = 100 a, v ce = 5 v v i(off) 0.4 - 0.8 input on voltage i c = 2 ma, v ce = 0.3 v v i(on) 0.5 - 1.1 input resistor r 1 1.5 2.2 2.9 k ? resistor ratio r 1 / r 2 0.042 0.047 0.052 - ac characteristics transition frequency i c = 10 ma, v ce = 5 v, f = 100 mhz f t - 200 - mhz collector-base capacitance v cb = 10 v, f = 1 mhz c cb - 3 - pf 1 pulse test: t < 300s; d < 2%
aug-29-2003 4 BCR158... dc current gain h fe = ? ( i c ) v ce = 5v (common emitter configuration) 10 -1 10 0 10 1 10 2 ma i c 0 10 1 10 2 10 3 10 - h fe collector-emitter saturation voltage v cesat = ? ( i c ), h fe = 20 0 0.1 0.2 0.3 v 0.5 v cesat 0 10 1 10 2 10 ma i c input on voltage v i (on) = ? ( i c ) v ce = 0.3v (common emitter configuration) 10 -1 10 0 10 1 10 2 v v i(on) -1 10 0 10 1 10 2 10 ma i c input off voltage v i(off) = ? ( i c ) v ce = 5v (common emitter configuration) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 v 1 v i(off) -3 10 -2 10 -1 10 0 10 1 10 ma i c
aug-29-2003 5 BCR158... total power dissipation p tot = ? ( t s ) BCR158 0 20 40 60 80 100 120 c 150 t s 0 50 100 150 200 mw 300 p tot total power dissipation p tot = ? ( t s ) BCR158f 0 20 40 60 80 100 120 c 150 t s 0 50 100 150 200 mw 300 p tot total power dissipation p tot = ? ( t s ) BCR158l3 0 20 40 60 80 100 120 c 150 t s 0 50 100 150 200 mw 300 p tot total power dissipation p tot = ? ( t s ) BCR158t 0 20 40 60 80 100 120 c 150 t s 0 50 100 150 200 mw 300 p tot
aug-29-2003 6 BCR158... total power dissipation p tot = ? ( t s ) BCR158w 0 20 40 60 80 100 120 c 150 t s 0 50 100 150 200 mw 300 p tot permissible pulse load p totmax / p totdc = ? ( t p ) BCR158 10 -6 10 -5 10 -4 10 -3 10 -2 10 0 s t p 0 10 1 10 2 10 3 10 - p totmax / p totdc d = 0 0.005 0.01 0.02 0.05 0.1 0.2 0.5 permissible pulse load r thjs = ? ( t p ) BCR158 10 -6 10 -5 10 -4 10 -3 10 -2 10 0 s t p -1 10 0 10 1 10 2 10 3 10 k/w r thjs 0.5 0.2 0.1 0.05 0.02 0.01 0.005 d = 0 permissible puls load r thjs = ? ( t p ) BCR158f 10 -6 10 -5 10 -4 10 -3 10 -2 10 0 s t p -1 10 0 10 1 10 2 10 k/w r thjs d=0.5 0.2 0.1 0.05 0.02 0.01 0.005 0
aug-29-2003 7 BCR158... permissible pulse load p totmax / p totdc = ? ( t p ) BCR158f 10 -6 10 -5 10 -4 10 -3 10 -2 10 0 s t p 0 10 1 10 2 10 3 10 p totmax / p totdc d=0 0.005 0.01 0.02 0.05 0.1 0.2 0.5 permissible puls load r thjs = ? ( t p ) BCR158l3 10 -7 10 -6 10 -5 10 -4 10 -3 10 -2 10 0 s t p -1 10 0 10 1 10 2 10 r thjs 0.5 0.2 0.1 0.05 0.02 0.01 0.005 d = 0 permissible pulse load p totmax / p totdc = ? ( t p ) BCR158l3 10 -7 10 -6 10 -5 10 -4 10 -3 10 -2 10 0 s t p 0 10 1 10 2 10 3 10 p totmax / p totdc d = 0 0.005 0.01 0.02 0.05 0.1 0.2 0.5 permissible puls load r thjs = ? ( t p ) BCR158t 10 -6 10 -5 10 -4 10 -3 10 -2 10 0 s t p -1 10 0 10 1 10 2 10 3 10 k/w r thjs d=0.5 0.2 0.1 0.05 0.02 0.01 0.005 0
aug-29-2003 8 BCR158... permissible pulse load p totmax / p totdc = ? ( t p ) BCR158t 10 -6 10 -5 10 -4 10 -3 10 -2 10 0 s t p 0 10 1 10 2 10 3 10 p totmax / p totdc d=0 0.005 0.01 0.02 0.05 0.1 0.2 0.5 permissible puls load r thjs = ? ( t p ) BCR158w 10 -6 10 -5 10 -4 10 -3 10 -2 10 0 s t p -1 10 0 10 1 10 2 10 3 10 k/w r thjs 0.5 0.2 0.1 0.05 0.02 0.01 0.005 d = 0 permissible pulse load p totmax / p totdc = ? ( t p ) BCR158w 10 -6 10 -5 10 -4 10 -3 10 -2 10 0 s t p 0 10 1 10 2 10 3 10 - p totmax / p totdc d = 0 0.005 0.01 0.02 0.05 0.1 0.2 0.5
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