Discussion
A fixed bias stage has \( R_{B} = 470 \ \mathrm{k\Omega} \) from a \( 12 \ \mathrm{V} \) supply. The base current is about
Answer: B
Subtracting the junction drop and dividing:
\[ \begin{aligned} I_{B} &= \frac{12 - 0.7}{470 \ \mathrm{k}} \\ &\approx 24 \ \mathrm{\mu A} \end{aligned} \]
giving about 24 microamps. With a beta of 150 that is 3.6 mA of collector current, and with a beta of 300 it would be 7.2 mA - the very spread that makes fixed bias unusable in a linear stage. The calculation is easy; the problem is what it depends on. Working one numerical case through both schemes side by side fixes the distinction better than any rule of thumb.
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