hw05sol - ECE 315 Homework 5 Solution Spring 2009 1. (BJT...

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Unformatted text preview: ECE 315 Homework 5 Solution Spring 2009 1. (BJT large-signal and small-signal characteristics one more time) For an npn transistor in the forward active mode with emitter area A E =100 m 2 , N E = 10 20 cm-3 , N B = 10 17 cm-3 , N C =10 15 cm-3 , the base width W B =0.5 m, the emitter width W E =0.1 m, V bi =0.96V for the BE junction, the electron mobility n =1000cm 2 /Vs, the hole mobility p =400cm 2 /Vs, and the minority lifetime n = p =10-7 s, (a) Find I F0 in the Ebers-Moll model. Please notice that this is a short-base diode case). (4 pts) This is an ideal forward-biased short-base diode, and therefore, the saturation current for the EB junction is: + = D i n p A i p n N n W D N n W D qA I 2 2 . We can see directly that the first term dominates since N D in the emitter is much larger than N A in the base if the neutral emitter region is about the same order of magnitude of the neutral region of base ( W B = W p here). Hence, B i B n F N n W D qA I 2 = = 1.87 10-16 A=0.187fA. We can express I F0 in the normalized area, which will be around 1nA/cm 2 . This is a practical number to remember for silicon technology. (b) The base current contains two components: the recombination current and the hole back injection into emitter. If we assume that back injection dominates and both emitter and base are short-based, the current gain can be estimated by (D nB N E W E )/(D pE N B W B ). Calculate and F (4 pts). We use D nB instead of D n above to remind you that D n is the minority diffusivity in the base region, which is p-type doped in npn . = (D nB N E W E )/(D pE N B W B ) = 500 , and F =0.998. If we ignore recombination in the base region, the collector current is basically all of the electron injection current in base being swept over at the BC junction, which is ( 29 1 / 2- = kT qV B i B n C BE e N n W D qA I . If the base current is dominated by the back injection into emitter, ( 29 1 / 2- = = kT qV E i E p Ep B BE e N n W D qA I I . Therefore, in the ideal diode case, = I C /I B = (D nB N E W E )/ (D pE N B W B ). For advertisement to the interested readers only: Notice that the important relation here is that is inversely proportional to W B , which will be used in the base-width modulation calculation below. There are two major assumptions in our conclusion above. Recombination is small in the base, and n i is the same in the emitter and base regions. Recombination is not that small actually, but it is still inversely proportional to W B for the region that recombination can happen. Often the device design will come up with the recombination current roughly equal to the back injection, and therefore will be halved. n i in the emitter region will be larger due to an effect called bandgap narrowing in 1 the heavily doped semiconductors. This typically will further reduce by 2 or 3. Therefore, practical design of BJT often comes up with between 50 and 300. between 50 and 300....
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This note was uploaded on 03/11/2009 for the course ECE 3150 taught by Professor Spencer during the Spring '07 term at Cornell University (Engineering School).

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hw05sol - ECE 315 Homework 5 Solution Spring 2009 1. (BJT...

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