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An N-channel JFET has IDSS = 1 mA and VP = -8V. Its maximum transconductance is ________
1. 4S
2. 0.00025 S
3. 0.2 S
4. 0.001 S

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Correct Answer - Option 2 : 0.00025 S

Concept:

The drain current in saturation for a JFET is given by:

\({{I}_{D}}\left( sat \right)={{I}_{DSS}}{{\left( 1-\frac{{{V}_{GS}}}{{{V}_{p}}} \right)}^{2}}\)   ---(1)

IDSS = Saturation Drain Current

Vp = Pinch Off Voltage

The transconductance (gm) is defined as the change in ID with respect to a change in the gate to source voltage (VGS), i.e.

\(g_m=\frac{\partial {{I}_{D}}}{\partial {{V}_{gs}}}\)

Taking the differentiation of Equation (1), we get:

\(g_m=\frac{2{{I}_{DSS}}}{\left| {{V}_{p}} \right|}\left( 1-\frac{{{V}_{gs}}}{{{V}_{p}}} \right)\)

The maximum value of gm occurs when VGS = 0

\({{\left. {{g}_{m}} \right(max)}}=\frac{2{{I}_{DSS}}}{\left| {{V}_{p}} \right|}\)

Calculation:

Given:

IDSS = 1 mA, Vp = -8V

We can write:

Maximum transconductance

gm0 = \(\frac{2I_{DSS}}{|V_p|}\)

\(\frac{2\;\times\;1}{8}=0.25~mS\)

= 0.00025 S

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