enter image description here

From Bartlett's bisection theorem. Common-node.

So I don't get, shouldn't \$ i_{b1} \$ be:

\$ \frac{V_{ic}}{r_\pi + 2 R_{ee} } \$


\$r_\pi\$ is in series with \$2 R_{ee}\$, so they have same current. No?

I also had shown in red what the current is across \$2 R_{ee}\$

  • 1
    \$\begingroup\$ As you already know the Iee is equal to Iee = Ib+Ib*beta = Ib*(beta + 1). Additional from KVL we have Vin = Ib * r_pi + Iee*2Ree = Ib * r_pi + Ib*(beta +1)*2Ree. And now we can solve for Ib = Vin/(r_pi + (beta+1)*2Ree) \$\endgroup\$
    – G36
    Nov 19, 2017 at 8:29

1 Answer 1


The output voltage of this circuit is equal to minus the collector current flowing in the collector resistance times the collector resistance: \$V_{out}=-i_cR_C\$ as shown in the below sketch: enter image description here

The base current is equal to the voltage across \$r_\pi\$ divided by \$r_\pi\$. That is \$i_b=\frac{V_{in}-V_E}{r_\pi}\$. The emitter voltage is the drop across the emitter resistance \$R_E\$ in which a current made of the sum of the base current \$i_b\$ with the collector current \$i_c\$ flows. Therefore, \$V_E=R_E(i_b+i_c)=R_E(i_b+\beta i_b)=R_E(\beta+1)i_b\$. Now extract \$i_b\$ from this definition and substitute it in the first expression we derived. You should have \$i_b=\frac{V_{in}}{(\beta+1)R_E+r_\pi}\$. You know that the output voltage \$V_{out}\$ is equal to \$-i_cR_C=-\beta i_bR_C\$. Substitute the last definition of \$i_b\$, rearrange and you should obtain \$\frac{V_{out}}{V_{in}}=-\frac{R_c\beta}{r_\pi+R_E(\beta+1)}\$. If the transistor gain \$\beta\$ is large enough, this expression simplifies to \$\frac{V_{out}}{V_{in}}\approx -\frac{R_C}{R_E}\$.

A quick dc bias point simulation with a Mathcad sheet shows the result is correct:

enter image description here enter image description here


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