2005. 2. 28 1/4 semiconductor technical data KTX215U revision no : 0 switching application. interface circuit and driver circuit application. features including two devices in us6. (ultra super mini type with 6 leads.) with built-in bias resistors. simplify circuit design. reduce a quantity of parts and manufacturing process. dim millimeters a b d g us6 2.00 0.20 1.25 0.1 2.1 0.1 0.2+0.10/-0.05 0-0.1 0.9 0.1 0.65 0.15+0.1/-0.05 b1 h c t g 1 3 2 b b1 d a h t 6 4 c c a1 1.3 0.1 a1 5 + _ + _ + _ + _ + _ 1. q (emitter) 4. q common (emitter) 3. q out (collector) 5. q in (base) 6. q (collector) 1 2 2. q (base) 1 1 2 2 equivalent circuit q1 maximum rating (ta=25 ) e c q 1 b r1 r2 common out q 2 2 in q r1=10k ? r2=10k ? 1 q1 23 65 4 q2 equivalent circuit (top view) * total raing. characteristic symbol rating unit collector-base voltage v cbo -15 v collector-emitter voltage v ceo -12 v emitter-base voltage v ebo -6 v collector current i c -500 i cp * -1 characteristic symbol rating unit output voltage v o 50 v input voltage v i 30, -10 v output current i o 100 characteristic symbol rating unit power dissipation p d * 200 junction temperature t j 150 storage temperature range t stg -55 150 bj type name 123 4 6 5 q2 maximum rating (ta=25 ) q1, q2 maximum rating (ta=25 ) marking epitaxial planar pnp/npn transistor * single pulse pw=1ms.
2005. 2. 28 2/4 KTX215U revision no : 0 q1 electrical characteristics (ta=25 ) characteristic symbol test condition min. typ. max. unit. output cut-off current i o(off) v o =50v, v i =0 - - 500 dc current gain g i v o =5v, i o =10 50 80 - output voltage v o(on) i o =10 , i i =0.5 - 0.1 0.3 v input voltage (on) v i(on) v o =0.2v, i o =5 - 1.8 2.4 v input voltage (off) v i(off) v o =5v, i o =0.1 1.0 1.2 - v transition frequency f t * v o =10v, i o =5 - 200 - input current i i v i =5v - - 0.88 note : * characteristic of transistor only. q2 electrical characteristics (ta=25 ) characteristic symbol test condition min. typ. max. unit collector cut-off current i cbo v cb =-15v, i e =0 - - -100 na collector-base breakdown voltage v (br)cbo i e =-10 a -15 - - v collector-emitter breakdown voltage v (br)ceo i c =-1ma -12 - - v emitter-base breakdown voltage v (br)ebo i e =-10 a -6 - - v dc current gain h fe v ce =-2v, i c =-10ma 270 - 680 - collector-emitter saturation voltage v ce(sat) i c =-200ma, i b =-10ma - -100 -250 mv transition frequency f t v ce =-2v, i c =-10ma, f t =100mhz - 260 - mhz collector output capacitance c ob v cb =-10v, i e =0, f=1mhz - 6.5 - pf
2005. 2. 28 3/4 KTX215U revision no : 0 c collector current i (ma) base-emitter voltage v (v) be i - v cbe ta=125 c v =-2v i /i =20 ce v =-2v ce v =-2v ce cb i /i =20 cb i /i =50 c b i /i =20 cb i /i =10 cb ta=25 c ta=-4 0 c ta=125 c ta=25 c ta=-40 c ta=25 c ta=125 c ta=25 c ta=-40 c ta=25 c ta=125 c ta=25 c ta=-40 c 10 1k 30 50 100 300 500 v - i c collector current i (ma) ce(sat) c ce(sat) collector-emitter saturation voltage v (mv) -1 -3 -1 -10 -30 -100 -300 -1k -3 -5 -10 -30 -50 -100 -300 -500 -1k v - i c collector current i (ma) ce(sat) c ce(sat) collector-emitter saturation voltage v (mv) -1 -3 -1 -10 -30 -100 -300 -1k -3 -5 -10 -30 -50 -100 -300 -500 -1k 0 -1 -0.5 -1.0 -1.5 -3 -5 -10 -30 -50 -100 -300 -500 -1k f - i c collector current i (ma) tc t transition frequency f (mhz) -1 -3 -10 -30 -100 -300 -1 k -100 -10k -300 -500 -1k -3k -5k v - i c collector current i (ma) be(sat) c be(sat) base-emitter saturation voltage v (mv) -1 -3 -10 -30 -100 -300 -1k 10 1k 30 50 100 300 500 -1 -3 -10 -30 -100 -300 -1k dc current gain h fe collector current i (ma) q (pnp transistor) c h - i fe 1 c
2005. 2. 28 4/4 KTX215U revision no : 0 c - v , c - v cb collector-base voltage v (v) eb emitter-base voltage v (v) q (pnp transistor) ob 1 c ob cb ib collector input capacitance c (pf) ob collector output capacitance c (pf) -0.1 -0.3 1 -1 -3 -10 -30 -100 3 5 10 30 50 100 300 500 1k ib c ib eb i =0a f=1mhz ta=25 c e collector lpower dissipation p (mw) 0 c 0 ambient temperature ta ( c) pc - ta 25 50 75 100 125 50 100 150 200 250 150 0.5 100 50 30 10 5 3 1 30 10 3 1 0.3 i(on) o output current i (ma) 0.3 o 100 0.1 i - v q2 v =0.2v o ta=100 c ta=25 c ta=-25 c input on voltage v (v) i(on) i(off) input off voltage v (v) o output current i ( a) i - v o i(off) o v =5v q2 10k 5k 3k 1k 500 300 100 50 2.0 1.8 1.6 1.4 1.2 1.0 0.8 30 0.6 ta=100 c ta= 25 c ta=-25 c o output current i (ma) i dc current gain g g - i io o v =5v 300 100 50 30 10 100 30 5 10 3 1 0.5 ta=100 c ta=25 c ta=-25 c q2
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