Discussion
A device has \( I_{DSS} = 8 \ \mathrm{mA} \) and \( V_{P} = -4 \ \mathrm{V} \). Its transconductance at zero gate voltage is
Answer: B
The maximum transconductance follows from the square law's slope:
\[ \begin{aligned} g_{m0} &= \frac{2 I_{DSS}}{|V_{P}|} \\ &= \frac{16}{4} \\ &= 4 \ \mathrm{mS} \end{aligned} \]
giving 4 millisiemens, against the forty a bipolar device would give at eight milliamps. The factor of two in the numerator comes from differentiating the square, and omitting it is the usual slip here.
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