A transimpedance amplifier

An answer:

\begin{array}{rcl} H (jω) & = \frac{v_{out}}{i_{in}} \\ v_{out} = A (v_{+} - v_{-}) = - A v_{-} \\ v_{-} = (\frac{v_{out} - v_{-}}{R_{2}} - i_{in}) \frac{1}{jω C_{1}} \\ = \frac{v_{out} - i_{in} R_{2}}{1 + jω R_{2} C_{1}} \\ v_{out} = - A \frac{v_{out} - i_{in} R_{2}}{1 + jω R_{2} C_{1}} \\ v_{out} = \frac{\frac{A i_{in} R_{2}}{1 + jω R_{2} C_{1}}}{1 + \frac{A}{1 + jω R_{2} C_{1}}} \\ H (jω) & = \frac{A R_{2}}{1 + jω R_{2} C_{1} + A} \\ = \frac{A}{1 + A} \cdot \frac{R_{2}}{1 + jω (\frac{R_{2} C_{1}}{1 + A})} \end{array}

From which is follows that

\begin{array}{rcl} H (0) = \frac{A}{1 + A} \cdot R_{2} \\ ω_{0} = \frac{1 + A}{R_{2} C_{1}} \end{array}

The magnitude part of a Bode plot is:

pict