Note for ECE 432 at OSU 02
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This 6 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at Ohio State University taught by a professor in Fall. Since its upload, it has received 18 views.
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Date Created: 02/06/15
i39 39 397Yngriw WiaWL 5 a 3M quotBk gBasei WMWWW WWWK wmwmmm gvarm Emi g M C Em 3 gig 3 WW XXV 5 I p3 M W 3 is 3quot E g New B wozksheeter 3 2 3 gal bag vegflee w 3 M Vie refore a wag 253 i M Ef gfn 3 L3 7 w W ESE Dif lfgfEi 1W g We H w gs L be The LP quotr and quot NAB pEO 2 4 rewrite 3 in terms of the doping density DE N w A 2 l v 3 A2 a s A l 439 M a w g 4 Ln g 39 72gt N e L quot 5 a IS a measure of how m gh of t e emitter current IS used i he g put collector The remainder goes to the base Since we want base current to be small we want a to be as close to l as possible 356 emll p N 5 45 NEE 5 LON Now using mm t w Wei3313 a e g M 3 W3 e M quotM a w fa 3 Ina 7 LE Now you can see that if a is close to one the denominator of this is going to be very small and so the 8 becomes large The BJT current gain equations are now solved You can easily find the intermediate quotef ciencyquot factors in terms of the device parameters 2 Emitter anCCIIOD efficrency y 45 3 a W W quot A 9 392 0lt Y aa a a 2 f Base transport factor 2 ar 2 Lei 115 l n g Ln J There39s one factor we didn39t discuss which is the collector efficiency factor we assumed this to be 1 It is a measure of the reverse current in the collectorjunction Collector efficiency M 153 1 53 Cu Cn Note that we assumed that hole component of the reverse biased collector base junction was zero and so M 1 2 New BJT worksheetnb w Q Next we find the hole current flowing into the emitter junction 15p This is just the given by the ideal diode equation we derived earlier mama Ti 56 633 W 2 v D F t i 96 quot39 m MW Ep i 9 i e Hwy 3 quot m3 7 ijfe W E10 m 3 The last part is to find the recombination current in the base 1389 To do this we assume that the base width is short enough so that the minority carrier distribution is a straight line We then find the stored charge in the base QB which is just the area of the triangle above Once we know the stored charge we can find the recombi nation current by dividing this charge by the minority carrier recombination time 3 g w i gqiagd ike ry Cfe bags WW gt meg5i QB M Aquot gt W 1 x Q 2 339 9 j a e W 3 33 R wmioi j f V 9 215 395 IBrec9 avg ga 6 Mi gt gfa 39 TN Now the main principle is that we want the input or base current to be small The base current supplies two components here 15 and Brec and so we need to design our BJT to make them small 4 Now we39re ready to find a First we write IC En Bree W ire 57quot W5 l tn fem MT 6 w 5 bmw G gi I gt i a NCXtIEIEPIE q j 4 Ev
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