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