Discussion
A junction device is self biased with \( R_{S} = 1 \ \mathrm{k\Omega} \), \( I_{DSS} = 10 \ \mathrm{mA} \) and \( V_{P} = -4 \ \mathrm{V} \). The drain current settles near
Answer: C
The bias line and the transfer curve must agree, so the current satisfies both:
\[ \begin{aligned} I_{D} \\ &= 10\left(1 - \tfrac{-I_{D} \times 1000}{-4}\right)^{2} \end{aligned} \]
whose solution is about 2.2 mA, giving a gate-source voltage near -2.2 volts. Solving a device curve against a circuit line is exactly the graphical method's algebraic equivalent, and iterating two or three times converges quickly.
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