# need help going crazy

Discussion in 'Homework Help' started by acelectr, Mar 26, 2011.

1. ### acelectr Thread Starter Member

Aug 28, 2010
73
0
Hi i need help about this equation, the eqation and the circuit related are attached. How did Ib become like that? Should not Ib be the source signal over the seen resistance across its terminals? If so wher did this Rb//Rsig came from? And what is with the Rb/(Rb+Rsig) term? I'm completely lost, still working on it. If i can get any feedback i'll appreciate. Thnx

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2. ### Georacer Moderator

Nov 25, 2009
5,142
1,266
It's an application of the Thevenin's theorem.

Since you know the input resistance, forget the transistor network and replace anything after the arrow with the mentioned resistance.

Suppose you have a network that includes the signal source, Rsig and Rb. Its output pin is at the base of the transistor. Its Thevenin equivalent is a voltage source of mangitute $V_{sig} \frac{R_B}{R_{sig}+R_B}$ in series with a resistance of $R_{sig} // R_B$Ω.

When you re-connect that network (in series) to the input resistance of the transistor, the current will be given by the formula
$I= \frac{V}{R} \\
=\frac{V_{sig} \frac{R_B}{R_{sig}+R_B}}{R_th + R_in}$

Substituting Rth and Rin will give you the given formula of your book.

3. ### acelectr Thread Starter Member

Aug 28, 2010
73
0
I still do not understand ... how it is possible to neglect the transistor? Is it possible to explain from another point of view?

4. ### Fraser_Integration Member

Nov 28, 2009
142
5
You don't neglect the transistor, rather you use Thevenin's theorem as described to get an equivalent voltage and resistance for everything BEFORE the transistor, which you will then put back into the old circuit, and you are left with a V(th), and an R(th) and R(in) - given by the expression with the arrow in the diagram - in series, meaning current is easy to solve.

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5. ### acelectr Thread Starter Member

Aug 28, 2010
73
0
thnx alot I think I'm getting somthng yyyyyayyyy