In JFET, the transconductance in the saturation region is given by:

A. \(\rm \sqrt{\frac{V_G+V_{bi}}{V_p}}\)

B. \(\rm g_{max}\left(1-\sqrt{\frac{V_G+V_{bi}}{V_p}}\right)\)

C. \(\rm 2qN_Da\mu \frac{Z}{L}\left(1-\sqrt{\frac{V_G+V_{bi}}{V_p}}\right)\)

D. \(\rm \left(g_{max}-\sqrt{\frac{V_G+V_{bi}}{V_p}}\right)\)

Choose the correct answer from the options given below:

Concept: 

JFET Drain Current:

 \(I_D = G_0\left[V_{DS}-\frac{2}{3}V_p\left(\left(\frac{V_{bi}-V_{DS}-V_{GS}}{V_p}\right)^{3/2}-\left(\frac{V_{bi}-V_{GS}}{V_p}\right)^{3/2}\right)\right]\)

JFET Transconductance: The transconductance, g_m of a JFET is given by \(g_m = \frac{\partial I_D}{\partial V_{GS}}\). Hence, we get

\(g_m = G_{0}\left[\left({V_{bi}-V_{GS}+V_{DS}}\over V_p\right)^{1/2}-\left({V_{bi}-V_{GS}}\over V_p\right)^{1/2}\right]\)

where, G₀ = 2qN_Dμa\(\frac{Z}{L}\) represents the conductance of the channel at the equilibrium.

Calculation:

In the saturation region, substituting V_DS=V_p-(V_i-V_G), we get

\(g_m = G_{0}\left[1-\left({V_{bi}-V_{GS}}\over V_p\right)^{1/2}\right]\)

From the above expression, g_max = G₀

Since V_GS is negative for n-channel JFET, hence substituting G₀ = g_max and V_GS = -V_G, we get

\(g_m = g_{max}\left[1-\left({V_{bi}+V_{G}}\over V_p\right)^{1/2}\right]\)

Substituting the value of g_max, we get

\(g_m = 2qN_D\mu a\frac{Z}{L}\left[1-\left({V_{bi}+V_{G}}\over V_p\right)^{1/2}\right]\)

Additional Information In the linear region, V_D << (V_i - V_G) and the transconductance g_m can be approximated as

\(g_m =\frac{G_{0}V_D}{2\sqrt{V_p(V_i-V_G)}}\)