2002. 2. 20 1/2 semiconductor technical data mpsa27 epitaxial planar npn transistor revision no : 1 darlington transistor features h complementary to mpsa77. maximum ratings (ta=25 ? ) to-92 dim millimeters a b c d f g h j k l 4.70 max 4.80 max 3.70 max 0.45 1.00 1.27 0.85 0.45 14.00 0.50 0.55 max 2.30 d 1 2 3 b a j k g h f f l e c e c m n 0.45 max m 1.00 n 1. emitter 3. collector 2. base + _ electrical characteristics (ta=25 ? ) characteristic symbol test condition min. typ. max. unit collector cut-off current i cbo v ce =50v, i e =0 - - 100 na emitter cut-off current i ebo v eb =10v, i b =0 - - 100 na collector-emitter breakdown voltage v (br)ces i c =100 a, i e =0 60 - - v collector-base breakdown voltage v (br)cbo i c =100 a, i e =0 60 - - v dc current gain h fe (1) * v ce =5v, i c =10ma 10k - - h fe (2) * v ce =5v, i c =100ma 10k - - collector-emitter saturation voltage v ce(sat) * i c =100ma, i b =0.1ma - - 1.5 v base-emitter voltage v be * v ce =5v, i c =100ma - - 2 v characteristic symbol rating unit collector-base voltage v cbo 60 v collector-emitter voltage v ces 60 v emitter-base voltage v ebo 10 v collector current i c 500 ma collector power dissipation p c 625 mw junction temperature t j 150 ? storage temperature range t stg -55 q 150 ? * pulse test : pw ? 300 s, duty cycle ? 2%.
2002. 2. 20 2/2 mpsa27 revision no : 1 collector current i (ma) c collector-emitter saturation voltage v (v) base-emitter saturation voltage v (v) be(sat) v , v - i ce(sat) c base-emitter voltage v (v) collector current i (ma) v =5v 0 30 50 1 3 5 10 c 100 ce 200 i - v be be 0.4 0.8 1.2 1.6 2.0 2.4 c ce(sat) be(sat) dc current gain h fe 1 collector current i (ma) c h - i fe c 3 10 30 100 300 1k 1k 3k 10k 30k 100k 300k 1000k v =5v ce i /i =1000 c b collector current i (ma) c collector-emitter voltage v (v) ce safe operating area 100 30 3 110 550 single nonrepe- titive pulse tc=25 c curves must be derated linearly with increase in temperature * i max(pulsed) c * c i max(continuous) 100 s * * 1ms * 1s dc operation tc=25 c v max. ceo ta= 25 c 10 30 50 100 300 500 1k 2k 4 10 30 100 300 500 v be(sat) v ce(sat) 0.1 0.3 1 0.5 3 10 5
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