# Voltage gain of an Emitter Follower

Please see attached image. Can you please help me understand how they went from the first to the second? Not too sure how they got (B+1) in there.

Thanks

With simple algebra, from $(5.317)$ you get: $$v_\pi=\frac{v_{out}}{R_E}\frac{r_\pi}{g_mr_\pi+1}$$ Now: $$\beta\triangleq\frac{i_c}{i_b}\\ i_b=\frac{v_\pi}{r_\pi}\\ i_c=g_mv_\pi\\ \implies \beta=g_mv_\pi\frac{r_\pi}{v_\pi}=g_mr_\pi$$ And plugging this result in the first equation you get $(5.318)$: $$v_\pi=\frac{v_{out}}{R_E}\frac{r_\pi}{\beta+1}$$

B=R(pi)xGm On first line, make a common denominator on left side. You will get: V(pi)(1+(GmxR(pi))/R(pi) Replace R(pi)xGm with B and solve for V(pi) and you get the second equation.

• Sorry I'm still not seeing it. How does B=R(pi)xGm? Also how does R(pi)/(1+B) = 1/Gm? Also how does (1+Gm.R(pi)) =B+1? Commented Apr 30, 2016 at 11:05

$$I_b + I_c = I_e$$

Since: $$\frac{I_c}{I_b} = \beta$$

Then:

\begin{align} I_b + \beta(I_b) &= I_e \\\\ \Rightarrow (\beta+1)Ib &= Ie \tag 1 \end{align}

From the circuit diagram:

\begin{align} I_b = \frac{v_\pi}{r_\pi} &&\text{and} && I_e = \frac{v_{out}}{R_E} \end{align}

Substituting in (1)

$$\frac{v_{out}}{R_E} = \frac{(\beta+1)V_\pi}{r_\pi}$$

Then by re arranging the terms:

$$V_\pi = \frac{r_\pi}{(\beta+1)}\cdot\frac{v_{out}}{R_e}$$