Discussion
A stage has \( C_{\mu} = 4 \ \mathrm{pF} \) and a gain of \( -150 \). The Miller capacitance at its input is about
Answer: A
Multiplying by one plus the gain magnitude:
\[ C_{M} = (1+150) \times 4 = 604 \ \mathrm{pF} \]
giving about 604 picofarads, over a hundred times the device's own figure and certain to dominate the input capacitance. Dividing rather than multiplying gives 37 pF, and ignoring the gain leaves 4 pF. The multiplication is why a stage's measured input capacitance bears so little resemblance to its data sheet value.
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