BJT AC Analysis Problem

Discussion in 'Homework Help' started by Digit0001, Oct 13, 2011.

Joined:
Mar 28, 2010
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Hi

Can someone explain to me how they got part c) finding the small signal; voltage gain in the following document.
View attachment ques5_tut4.doc

This is the AC equivalent circuit i used.

P.S
2. GeoracerModeratorStaff Member

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Please don't upload .doc or .docx formats. If a simple .png image isn't enough, use the .pdf at least.

Many members are reluctant to use the .doc format.

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3. t_n_kSenior Member

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Personally I think the solution to part (c) is overcomplicated.

Since it is required to find Vo/Vs, I would simply assume a voltage source of value Vs is applied to the emitter.

From there one can observe that

$v_{be}=-\frac{r_\pi}{r_\pi+R1||R2}v_s=-\frac{6.97k}{6.97k+37.5k}v_s=-0.157v_s$

Given that, one can then determine the output as a function of vs since

$v_o=-g_mv_{be}R_c||R_L=-17.93[mA/V](-0.157v_s)2.4[k\Omega]v_s=6.744v_s$

or Av=6.744 [which seems a better estimate to me than 6.73]
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