Discussion
For a JFET \(I_{DSS}\) = 8 mA and peak voltage \(V_p\) = -8 V, what will be the drain current for gate to source voltage of -2 V ?
Answer: A
The drain current is 4.5 mA (option A), from the JFET square-law relation in the saturation region.
\[ I_D = I_{DSS}\left(1 - \frac{V_{GS}}{V_p}\right)^2 \]
\[ I_D = 8\left(1 - \frac{-2}{-8}\right)^2 \]
\[ I_D = 8(0.75)^2 = 4.5\ \text{mA} \]
The value 8 mA (B) is the current at \(V_{GS} = 0\), where the bracket is 1. The values 16 mA and 12 mA are larger than \(I_{DSS}\), which a JFET cannot exceed at a gate voltage that reverse biases the channel.
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