# $i_{b1}$ in this simple bjt circuit

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 $2 R_{ee}$

• 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)
– G36
Nov 19, 2017 at 8:29

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:
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 $\frac{V_{out}}{V_{in}}\approx -\frac{R_C}{R_E}$.