hw05sol

# hw05sol - 1 ECE 315 Homework 5 Solution Fall 2007 1(BJT...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1 ECE 315 Homework 5 Solution Fall 2007 1. (BJT large-signal and small-signal characteristics in realistic designs) For a npn transistor in the forward active mode with emitter area A E =100 μ 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 μ m, the emitter width W E =0.1 μ m, V bi =0.96V for the BE junction, the electron mobility μ n =1000cm 2 /Vs, the hole mobility μ p =400cm 2 /Vs, and the minority lifetime τ n = τ p =10-7 s, (a) If you need to evaluate the current at the BE junction (i.e., the emitter current I F0 ), what is the largest current component among electron/hole drift/diffusion in emitter/base (8 current components) and recombination in emitter/hole (two current components)? Briefly explain. (4 pts). The base width W B =0.5 μ m is short (in comparison with the electron diffusion length in base as L n = m μ τ D n n 1 . 16 = . Therefore the recombination current will be very small. The minority drift is smaller than the minority diffusion at the junction. The majority drift is assumed small until the series resistance dominates. The minority hole diffusion current at the emitter side is smaller than the minority electron current at the base side since N E >> N B . Therefore, the dominant current component should be the electron diffusion current on the base side. (b) Find I F0 in the Ebers-Moll model (notice that this only needs to include the largest current component above. 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. Notice that fA is a typical number for the ideal I F0 . We can express this in the normalized area, which will be around 1nA/cm 2 . This is a practical number to remember. (c) 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 β can be estimated by (D nB N E W E )/(D pE N B W B ). Calculate β and α F (4 pts). Briefly explain this approximation from definition of β =I C /I B (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 . β = (D nB N E W E )/(D pE N B W B ) = 500 , and α F =0.998....
View Full Document

{[ snackBarMessage ]}

### Page1 / 7

hw05sol - 1 ECE 315 Homework 5 Solution Fall 2007 1(BJT...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online