An NPN amplifier and bandwidth issues

Exercise 4.7 An NPN amplifier and bandwidth issues

a)

An answer:

pict

\begin{array}{rcl} a_{c} & = \frac{v_{out}}{v_{in}} \\ v_{out} = - g_{m} v_{be} R_{2} \\ v_{be} = \frac{(\frac{βfe}{g_{m}} ∕ ∕ R_{1}) v_{in}}{\frac{1}{jω C_{in}} + R_{3} + (\frac{βfe}{g_{m}} ∕ ∕ R_{1})} \\ \Rightarrow \\ a_{v} & = - \frac{g_{m} R_{2} (\frac{βfe}{g_{m}} ∕ ∕ R_{1})}{\frac{1}{jω C_{in}} + R_{3} + (\frac{βfe}{g_{m}} ∕ ∕ R_{1})} \\ = - g_{m} R_{2} \frac{jω (\frac{βfe}{g_{m}} ∕ ∕ R_{1}) C_{in}}{1 + jω (R_{3} + (\frac{βfe}{g_{m}} ∕ ∕ R_{1})) C_{in}} \end{array}

Where the shorthand notation for parallel impedances is used: $A ∕ ∕ B$ is the impedance of $A$ and $B$ in parallel.

b)

An answer:
The big chunk of the work was done in the previous answer. From that is can readily be derived that the transfer function is first order high-pass. The high-frequency voltage gain is:

\begin{array}{rcl} a_{v} (\infty) & = - g_{m} R_{2} \frac{\frac{βfe}{g_{m}} ∕ ∕ R_{1}}{R_{3} + (\frac{βfe}{g_{m}} ∕ ∕ R_{1})} \end{array}

The pole (and zero) is:

\begin{array}{rcl} ω_{0} & = \frac{1}{(R_{3} + (\frac{βfe}{g_{m}} ∕ ∕ R_{1})) C_{in}} \end{array}

c)

An answer:

pict

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