Discussion
A tail current of \( 20 \ \mathrm{\mu A} \) charges a \( 30 \ \mathrm{pF} \) compensation capacitor. The slew rate is about
Answer: A
Dividing the available current by the capacitance:
\[ \begin{aligned} \frac{20 \times 10^{-6}}{30 \times 10^{-12}} \\ &\approx 0.67 \ \mathrm{V/\mu s} \end{aligned} \]
giving about 0.67 volts per microsecond, close to the figure a general-purpose operational amplifier quotes. That such a familiar specification falls out of two internal quantities is worth noticing, since it explains why faster devices run more tail current or use less compensation.
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