Question 7.5: A JFET has a drain current of 5 mA. If IDSS = 10 mA and VGS ...

I_{D}=I_{DSS}\left[1-\frac{V_{GS}}{V_{GS}(off) } \right] ^{2}

Substituting values,          5=10\left[1+\frac{V_{GS}}{6 } \right] ^{2}

\sqrt{\frac{5}{10}}=1+\frac{V_{GS}}{6}

or,                              1+\frac{V_{GS}}{6}=0.707

or,                              V_{GS}= 6(0.707 – 1)

or,                              V_{GS} = -1.76 V

V_{P} = -V_{GS} (\text{off}) = 6 V