2003. 12. 12 1/4 semiconductor technical data 2N3906E epitaxial planar pnp transistor revision no : 0 general purpose application. switching application. features low leakage current : i cex =-50na(max.), i bl =-50na(max.) @v ce =-30v, v eb =-3v. excellent dc current gain linearity. low saturation voltage : v ce(sat) =-0.4v(max.) @i c =-50ma, i b =-5ma. low collector output capacitance : c ob =4.5pf(max.) @v cb =-5v. complementary to 2n3904e. maximum rating (ta=25 ) dim millimeters a b d e esm 1.60 0.10 0.85 0.10 0.70 0.10 0.27+0.10/-0.05 1.60 0.10 1.00 0.10 0.50 0.13 0.05 c g h j 1 3 2 e b d a g h c j 1. emitter 2. base 3. collector + _ + _ + _ + _ + _ + _ characteristic symbol rating unit collector-base voltage v cbo -40 v collector-emitter voltage v ceo -40 v emitter-base voltage v ebo -5 v collector current i c -200 ma base current i b -50 ma collector power dissipation p c 100 mw junction temperature t j 150 storage temperature range t stg -55 150 type name marking z a
2003. 12. 12 2/4 revision no : 0 electrical characteristics (ta=25 ) characteristic symbol test condition min. typ. max. unit collector cut-off current i cex v ce =-30v, v eb =-3v - - -50 na base cut-off current i bl v ce =-30v, v eb =-3v - - -50 na collector-base breakdown voltage v (br)cbo i c =-10 a, i e =0 -40 - - v collector-emitter breakdown voltage * v (br)ceo i c =-1ma, i b =0 -40 - - v emitter-base breakdown voltage v (br)ebo i e =-10 a, i c =0 -5.0 - - v dc current gain * h fe (1) v ce =-1v, i c =-0.1ma 60 - - h fe (2) v ce =-1v, i c =-1ma 80 - - h fe (3) v ce =-1v, i c =-10ma 100 - 300 h fe (4) v ce =-1v, i c =-50ma 60 - - h fe (5) v ce =-1v, i c =-100ma 30 - - collector-emitter saturation voltage * v ce(sat) 1 i c =-10ma, i b =-1ma - - -0.25 v v ce(sat) 2 i c =-50ma, i b =-5ma - - -0.4 base-emitter saturation voltage * v be(sat) 1 i c =-10ma, i b =-1ma -0.65 - -0.85 v v be(sat) 2 i c =-50ma, i b =-5ma - - -0.95 transition frequency f t v ce =-20v, i c =-10ma, f=100mhz 250 - - mhz collector output capacitance c ob v cb =-5v, i e =0, f=1mhz - - 4.5 pf input capacitance c ib v be =-0.5v, i c =0, f=1mhz - - 10 pf input impedance h ie v ce =-10v, i c =-1ma, f=1khz 2.0 - 12 k voltage feedback ratio h re 1.0 - 10 x10 -4 small-signal current gain h fe 100 - 400 collector output admittance h oe 3.0 - 60 noise figure nf v ce =-5v, i c =-0.1ma, rg=1k , f=10hz 15.7khz - - 4.0 db switching time delay time t d v out total 4pf c 10k ? 275 ? v =-3.0v cc 300ns -10.6v 0.5v 0 t ,t < 1ns, du=2% r in v f - - 35 ns rise time t r - - 35 storage time t stg 20 s 1n916 or equiv. -10.9v 9.1v v out total 4pf c v =-3.0v cc 275 ? 10k ? v in 0 t ,t < 1ns, du=2% rf - - 225 fall time t f - - 75 2N3906E * pulse test : pulse width 300 s, duty cycle 2%.
2003. 12. 12 3/4 2N3906E revision no : 0 c 0 collector current i (ma) collector current i (ma) 0 c 0 base-emitter voltage v (v) be be c i - v 10 dc current gain h fe -3 -1 -0.3 -0.1 collector current i (ma) c 0 collector-emitter voltage v (v) ce ce c i - v h - i c collector current i (ma) collector-emitter saturation ce(sat) v - i base-emitter saturation collector current i (ma) c be(sat) v - i -1 -2 -3 -4 -20 -40 -60 -80 -100 common emitter ta=25 c i =-0.1ma b -0.4 -0.8 -1.2 -1.6 -40 -80 -120 -160 -200 ta=125 c ta =25 c ta=-55 c common emitter v =-1v ce fe c -10 -30 -100 -300 30 50 100 300 500 1k common emitter v =-1v ce ta=125 c ta=25 c ta=-55 c ce(sat) c voltage v (v) common emitter -0.01 -0.1 -0.3 -1 -0.1 -0.03 -0.05 -0.3 -0.5 -1 -3 -10 -30 -100 i /i =10 c -300 b ta=125 c ta=25 c ta=-55 c be(sat) c voltage v (v) -3 ta=125 c -0.3 -0.1 -0.1 -1 -0.5 -0.3 -1 -3 -10 ta=25 c ta=-55 c -30 -100 -300 common emitter i /i =10 -5 -10 c e -0.2 -0.3 -0. 4 -0.5 -0.6 -0.7 -0.8 -0.9 -1
2003. 12. 12 4/4 2N3906E revision no : 0 capacitance c (pf) ob -3 -1 -0.3 -0.1 reverse voltage v (v) cb c - v , c - v v - i b base current i (ma) -0.001 -0.1 -1 -10 ce 0 collector-emitter voltage v (v) ce b -0.2 -0.4 -0.6 -0.8 -1.0 -0.01 common emitter ta=25 c i =1ma c i =10ma c c i =30ma c i =100ma ob cb ib eb c (pf) ib v (v) eb -10 -30 0.5 1 3 5 10 30 50 f=1mhz ta=25 c c c ib ob
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