Discussion
A follower of \( 60 \ \Omega \) output impedance replaces a \( 10 \ \mathrm{k\Omega} \) source ahead of a stage with \( 600 \ \mathrm{pF} \) of Miller capacitance. The break moves from
Answer: A
Computing the two break frequencies:
\[ \begin{aligned} f_{1} &= \tfrac{1}{2\pi(10^{4})(600\mathrm{p})} \approx 27 \ \mathrm{kHz} \\ f_{2} &= \tfrac{1}{2\pi(60)(600\mathrm{p})} \approx 4.4 \ \mathrm{MHz} \end{aligned} \]
so 27 kHz to 4.4 MHz, a factor of over one hundred and sixty for the price of one transistor. A buffer that takes no voltage gain can still transform a stage's usefulness, which is the clearest possible argument for impedance transformation.
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