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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

Emitter Follower

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3 Answers 3

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

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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.

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  • \$\begingroup\$ 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? \$\endgroup\$
    – Arsenal123
    Commented Apr 30, 2016 at 11:05
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$$I_b + I_c = I_e$$

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:

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

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