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  2001. 11. 30 1/2 semiconductor technical data KTC4077 epitaxial planar npn transistor revision no : 2 low noise amplifier application. features  high voltage : v ceo =120v.  excellent h fe linearity : h fe (0.1ma)/h fe (2ma)=0.95(typ.).  high h fe : h fe =200  700.  low noise : nf=1db(typ.), 10db(max.).  complementary to kta2017. maximum rating (ta=25 1 ) dim millimeters a b d e 1. emitter 2. base 3. collector usm 2.00 0.20 1.25 0.15 0.90 0.10 0.3+0.10/-0.05 2.10 0.20 0.65 0.15+0.1/-0.06 1.30 0.00-0.10 0.70 c g h j k l k 1 3 2 e b d a j g c l h mm n n m 0.42 0.10 n 0.10 min + _ + _ + _ + _ + _ electrical characteristics (ta=25 1 ) note : h fe classification gr(6):200  400 bl(8):350  700 h rank type name marking d fe characteristic symbol rating unit collector-base voltage v cbo 120 v collector-emitter voltage v ceo 120 v emitter-base voltage v ebo 5 v collector current i c 100 ma base current i b 20 ma collector power dissipation p c 100 mw junction temperature t j 150 1 storage temperature range t stg -55  150 1 characteristic symbol test condition min. typ. max. unit collector cut-off current i cbo v cb =120v, i e =0 - - 0.1  a emitter cut-off current i ebo v eb =5v, i c =0 - - 0.1  a dc current gain h fe (note) v ce =6v, i c =2ma 200 - 700 collector-emitter saturation voltage v ce(sat) i c =10ma, i b =1ma - - 0.3 v transition frequency f t v ce =6v, i c =1ma - 100 - mhz collector output capacitance c ob v cb =10v, i e =0, f=1mhz - 4.0 - pf noise figure nf v ce =6v, i c =0.1ma f=1khz, rg=10k u - 1.0 10 db
2001. 11. 30 2/2 KTC4077 revision no : 2 10 100 10k 1k 100 10 1k 10k 100k common emitter v =6v f=1khz ce 12 10 8 6 4 3 2 nf=1 db nf=1db 2 3 4 6 8 1 0 12 c g nf - r , i signal source resistance r ( ? ) g collector current i ( a) c ta=-25 c ta=25 c ta=100 c collector current i (ma) dc current gain h 0.3 10 50 30 100 1 fe 500 300 1k common emitter 100 v =6v 310 ce 30 c 400 h - i fe c h parameter 1 500 100 30 1 collector-emitter voltage v (v) ce h parameter - v ce 10 3 3 10 30 100 300 common emitter i =-1ma f=270hz ta=25 c e h fe ie h re h oe h ( x k ? ) ( x 10 ) ( x ) -5 ? c - v cb collector-base voltage v (v) 0 ob 1 collector output capacitance ob cb c (pf) 10 20 30 40 50 60 70 80 3 5 10 30 50 f=1mhz i =0 ta=25 c e


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