me6209 v0 2 page 1 of 7 lo w power consumption ldo me6209 series general descript ion features selection guide t he me 62 09 series are a group of positive voltage ou tput,three ?pin regulator,that provide a high current even when the input/output voltage differerntial is small.low power consumption and high accuracy is achieved through cmos technology.they allow input voltages as high as 18v. t ypical application l cameras, video recorders l voltage regulator for microprocessor l voltage regulator for lan cards l wireless communication equipment l audio/video equipment t ypical application circuit l ultra low quienscent current: 3.0ua(typ) l high input voltage (up to 18v) l low dropout voltage :80mv@iout=40ma ?v out=3.3v ? l output voltage accuracy ? 2 ? l maximum output current ? 250ma ? within max.power dissipation,vout=3.3v ? l low temperature coefficient l package ? sot23-3 0 to-92 0 sot89-3 v?mw`r?y?b? w w w . g o f o t e c h . c o m
me6209 v02 page 2 of 7 pin configuration sot23-3 sot89-3 to-92 pin assignment me6209axx pin number sot89-3/to-92 sot23-3 pin name functions 1 1 v ss ground 2 3 v in input 3 2 v out output absolute maximu m ratings parameter symbol ratings units input voltage v in 18 v output voltage v out vss-0.3 ?^v in +0.3 v output current iout 500 ma operating temper ature range t opr -40 ?^?85 ! storage temperature range t stg ?
40 ?^? 125 ! sot89-3 500 to-92 500 power dissipation sot23-3 p d 300 mw n ~n?t?m?w3^v?mw`r?y?b?g ?pqls? tel: 0755-8398 3377 / 135 9011 2223 http://www.gofotech.com v?mw`r?y?b? w w w . g o f o t e c h . c o m
me6209 v02 page 3 of 7 block diagram electrical characteristics me6209a33 (v in = v out +1.0v ?c in= c l =10uf ? ta=25 o c, unless otherwise noted ) parameter symbol conditions min. typ. max. units output voltage v out (e) (note 2) i out =40ma, v in =vout+1v x 0.98 v out (t) (note 1) x 1.02 v input voltage v in 18 v maximum output voltage i out _max v in =vout+1v 250 ma load regulation ? v out v in =vout+1v, 1ma i out 60ma 15 40 mv dropout voltage (note 3) v dif i out =40ma 80 mv supply current i ss v in =vout+1v 3 4 ?a line regulations ? v out ? v in v out i out =40ma vout+1v v in 18v 0.1 0.2 %/v %3 vout/ %3 ta tempera ture coefficient v in =vout+1v, i out =40ma -40 ! |