Discussion
A transimpedance stage has \( R_{f} = 10 \ \mathrm{k\Omega} \) and \( C_{in} = 5 \ \mathrm{pF} \). Its bandwidth is about
Answer: D
The dominant pole is set by the feedback resistance and the input capacitance:
\[ \begin{aligned} f &= \frac{1}{2\pi (10^{4})(5 \times 10^{-12})} \\ &\approx 3.2 \ \mathrm{MHz} \end{aligned} \]
giving about 3.2 MHz. Reducing the feedback resistor to one kilohm would give ten times the bandwidth and a tenth of the transimpedance, which is the trade in its plainest form. The same relation appears wherever a resistance and a capacitance set a limit.
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