Question 7.3: An n-channel JFET has IDSS = 12 mA and pinch-off voltage, Vp...
An n-channel JFET has I_{DSS} = 12 mA and pinch-off voltage, V_{p} = –4V. Calculate the drain current for V_{GS} = –2 V. If the tranconductance, g_{m,} of a JFET with the same I_{DSS} at V_{GS} = 0 is 4×10^{-3} mho, find the pinch-off voltage.
From Shockley’s equation we have,
I_{D}=I_{DSS}\left[1-\frac{V_{GS}}{V_{P}} \right] ^{2}
Given, I _{DSS} = 12 mA , V_{p} = -4 V, V_{GS} = -2 V
Substituting the given values,
I_{D}=12\left[1-\frac{(-2)}{(-4)} \right] ^{2} mA
=12\times \frac{1}{4} mA
= 3 mA.
g_{m}= \frac{∂}{∂ V_{GS}}\mid V_{DS}= Constant
= \frac{∂}{∂ V_{GS}}I_{DSS} \left[1-\frac{V_{GS}}{V_{p}} \right] ^{2}
= \frac{-2 I_{DSS}}{ V_{p}} \left[1-\frac{V_{GS}}{V_{p}} \right]
= g_{mo} \left[1-\frac{V_{GS}}{V_{p}} \right]
where g_{mo} =\frac{-2I _{DSS}}{V_{p}}
g_{mo} is the value of g_{m} when V_{GS} = 0.
Therefore, V_{p}=\frac{-2I_{DSS}}{g_{mo}}=-\frac{2\times 12 m A}{4 millimho} =- 6 volts.